EX hey guys good evening and welcome back welcome back again to your unacademy n English Channel I hope all of you doing doing amazing all of you doing great so people quickly let me know in the chats if all of you can hear me if I am perfectly Audible and visible to everyone of you quickly let me know in the chats with some fire emojis yes with some fire emojis quickly in the chats let me know quickly in the chats if all of you can hear me people with some fire emojis yeah so what's up how are you all doing what's happening people what's happening what is happening I believe everyone of you are doing wonderful are doing great so my dear students as you all must be knowing today it's going to be our day 27 right it is going to be our day 27 of of the Series game of neat so till now we are done with 26 chapters of your physics chemistry biology right and my dear students today it's going to be the 27th chapter which we are going to do yeah well it does not feel like that we have almost complete we have complete 26 chapters yet yeah right and I'm amazed like now students are following this particular series with 100% dedication with 100% honesty that's amazing I would want you guys to have the same J throughout this particular neat 2024 Journey right some four months are left let's utilize these four months properly so that we can score really good when it comes to the neat 2024 examination right so the chapter which I'll be doing today that is going to be solution one of the important chapters I would say in your physical chemistry part you can expect some two to three questions right in this year neat 2024 examination from this particular chapter as well and as you all must be knowing this chapter again we are going to start from the basics from the scratch right if you have studied this chapter before or if you have not studied this chapter before no worries at all everything will be done and dusted in detail that to from the basics from the scratch whatever theory is involved whatever problem patterns are involved in this particular chapter I'll be doing every single thing in this particular session and people are asking can J students watch this session as well absolutely you guys are most welcome you guys are most welcome too yeah perfect and people what is going to be the duration of The Da session let me tell you this chapter might take 7 hours 8 hours I have no idea how much how much time it's going to take but till the chapter gets completed I'll be with you right I'll be teaching you and I would want you guys to be with me till the end of the session so are you guys ready are you guys ready people are you guys ready yeah are you guys ready are you guys ready with your pen and paper yes quickly in the chats in the chats people in the chats in the chats I want the chat rate to be super high I want the chat rate to be super high are you all ready with your pen and paper quickly everyone let me see the Joe of the session first let me see the Jo of the session first wonderful and the ones who have not smashed that that like button yet I would want you guys to smash that like right now yes yes yes smash that like button right now and Mark your attendance every one of you and you know everyone means everyone yeah everyone means everyone people are copying my dialogue everyone means everyone yeah perfect just give me a second and we shall be starting just a second people just a second just a second all right so guys let me quickly show you the topics which we shall be covering in the today's session have a look exactly on the session flow right from where exactly we are going to start until what topic we are going to cover we are going to cover the whole chapter we shall be starting with the first topic that is your concentration terms we shall be discussing the terminologies which are related to liquid solutions which involves the vapor pressure boiling point freezing point etc etc right apart from that we shall be discussing the RS law and its applications after doing the RS La we shall be doing the ideal non- ideal Solutions their examples their problem patterns then we shall be moving on to the cative properties the first cative property which we shall be discussing that's going to be the relative lowering in Vaper pressure then comes your elevation in bing point then comes your depression in freezing point and then it's going to be the osmotic pressure and my dear students at the end of the session at the end of the session it is going to be the most important topic of this particular chapter which we shall be covering that is going to be the van hops Factor so this is going to be the session flow these are the topics which we shall be be covering in the today's chapter that is solution so everyone in the chats are you all ready are you all ready with me are you all ready with me again I'm repeating the same thing duration of the session I have no idea I have no idea it might take 7 hours it might take 8 hours it might take 9 hours it might take 6 hours also who knows yeah it is just you have to be with me till the end perfect so let me have some water and let's get going all right peoples so let's get going with the first topic of the session that is going to be your concentration terms okay well first of all before starting the concentration terms let me quickly give you certain basic things which are required understand the concentration terms let me exactly give you some basic things which are required to understand this particular topic that is concentration terms my dear people every one of you must be knowing what exactly a solution is right as you all must be familiar with the fact solution it exactly has got two components solution consists of two components exactly you must be knowing this this is complete Basics one is called as solute and one is called as solvent one is called as solute and one is called as solvent what is a solute what is a solvent let's get to know about that my dear students solute is basically that component of the solution which is present in lesser amount that component of the solution which is present in lesser amount we call that component of the solution as solute similarly what is the solvent solvent is basically that component of the solution which is present in larger amount which is present in larger amount perfect so solution it conss of two components one is called as solute one is called as solvent generally solute is present in lesser ount and your solvent is present in larger amount Point number one point number two related to solvent Point number two related to solvent do remember do remember the state of solution the state of solution is same is same as that of is same as that of the state of solvent state of solution is same as that of state of solvent what exactly is meant by this particular term let's have a look over this one example my dear students let's say I'm taking sugar and you know what is the state of sugar it is solid over here I'm going to introduce the sugar into water let's say this is H2O liquid let's say this is H2O liquid right first thing imagine that I have taken some 5 g of sugar right I'm going to introduce these 5 g of sugar in 100 ml of water what I'll be getting exactly when I'll be introducing the sugar into water I'll be getting a solution I'll be getting a solution now if you ask me what is going to be the state what is going to be the physical state of this particular solution as I told you the state of solution will be same as that of state of solvent so since the the physical state of solvent is liquid over here so I would say the physical state of solution it's again going to be liquid only it is again going to be liquid only so do remember whatever is going to be the state of solvent same is going to be the state of solution this is one very important point which I would want you guys to remember from now on okay perfect Point number one point number two in this particular chapter solvent I'll be representing by number one and Ute I'll be representing by number two this is one more important thing and basic thing right in order to reduce the calculations afterwards we'll be using this particular Point solvent will be representing with number one or you can represent solvent by a solute by number two or B the choice is all yours again I'm telling you solvent is represent when number one or letter a salute is represented by number two or letter B correct my dear students in this particular chapter whenever you see a term W written anywhere w w w in this particular chapter it stands for what it stands for the mass of substance W stands for the mass of substance in grams Point number one whenever you see a term e e stands for equivalent mass of the substance equivalent mass of the substance similarly my dear students whenever in this particular chapter you see a term n n stands for the number of moles n stands for the number of moles now if this particular point is clear if this particular point is clear then there is one more thing that is MM which stands for what which stands for the molar mass of the substance okay now you tell me in the chat for example if I'm writing something like this if I'm writing W2 can you let me know the meaning of W2 here what is going to be the meaning of W2 you know W stands for mass and 2 stands for solute so W2 what is its significant W2 simply stands for the mass of solute it stands for the mass of solute in Gs correct for example for example I'm writing N1 what is meant by the term N1 first of all n stands for moles and one stands for solvent so this N1 will be representing the number of moles of solvent the number of moles of solvent similarly for example if I write if I write E2 what is E2 e stands for equivalent Mass 2 stands for solute so E2 stands for the equivalent mass of the solute correct I believe these basic things are are absolutely clear to you I believe these basic things are absolutely clear to you right and people at the same time let me add up one more thing since I told you solution it consists of solute and solvent solution consist of solute and solvent do remember mass of the solution is always equal mass of the solution is always equal the mass of salute plus the mass of solvent mass of solution it is always equal to the mass of solute plus the mass of solvent this is one more thing which I would want to share with you I believe this is clear apart from this there are some basic more terminologies which you must know which you must know my de students somewhere in this particular chapter you'll come across a term binary solution you'll come across a term binary solution or somewhere it'll be mentioned that we have got a tary solution or somewhere it will be mentioned that we have got a quary solution right somewhere in this particular chapter you'll find a term written as aquous solution what is meant by these particular Solutions what is a binary solution what is a tary solution what is a coary solution what is an aqua solution you should know their meanings as well my dear students binary solution is the one which consists of two components exactly binary solution it consists of two components and in these two components one component will be solute and one more component will be solvent one component will be solute and one more component will be solvent if I talk about tary solution in case of tary solution how many components exactly do we have we have exactly got three components in the tary solution we have got three components among which there will be one solvent and there'll be two components of solute two solute components and one solvent component right similarly if you come across the term quary solution quary solution is the one which consists of exactly four components which consist of how many components four components and among these four components there will be three components of salute and there will be one component of the solvent there will be one component of the solvent right there will be one component of the solvent and in the similar way if you look at this particular terminology the aquous solution what is an aquous solution my students you will find such kind of the solution in which the solvent used in which the solvent used is water whenever you see such kind of the solution in which we use water as the solvent we shall be calling that particular solution as the aquous solution so I believe these four terminologies are exactly clear to you if yes let me know in the chats with some fire emojis quickly let me know in the chats with some fire emojis quickly quickly people quickly everyone yes these are some basic things which you must be knowing before starting this particular chapter right I'll be using this particular term frequently aquous solution aquous solution is the one in which the solvent used is water which the solvent used is water perfectly fine all right if this is perfectly fine then let's get going let's get started with the first topic what is that going to be that is going to be the concent ation terms that's going to be the concentration terms now my dear students in the concentration terms what exactly do we have to study we will have to study all the types of concentration terms number one we shall be discussing marity of dilution we shall be discussing marity of mixing of Ideal Solutions of non- Ideal Solutions right all those things I shall be discussing in detail so starting with the first terminology what is that that's going to be your weight by weight percentage weight by weight percentage of salute I hope you are writing all the things with me I want you guys to make the running notes with me weight by weight percentage of salute or you can call it as the mass percentage of the solute you can call it as the mass percentage of the solute now how do you exactly Define this particular term how do you exactly Define this particular ter there's a simple definition which you need to remember weight by weight percentage of solute it is defined as the mass of solute the mass of solute in grams the mass of solute in grams which is present in which is present in 100 G of solution the mass of solute in Gams which is present in 100 G of solution the mass of solute which is present in 100 G of solution right you shall be calling that as the weight by weight percentage of solute what is meant by that imagine you have taken 100 G of any solution imagine you have taken 100 G of any solution people right in 100 G of solution if you find 10 G of salute you can say weight by weight percentage of salute is 10 if by chance in 100 G of solution you find 20 G of solute you can say weight by weight percentage of solute is 20 in general if in 100 G of solution you find x g of solute you can say weight by weight percentage of solute is X so what is weight by weight percentage of solute it is basically that mass of solute which is present in 100 G of solution I believe this is clear now try to understand try to understand people what I'm going to talk about imagine this this is your container which I have imagine I taken the container and in this particular container imagine I have got a solution let's assume that in this container we have got 500 G of solution imagine that in this container we have got 500 G of solution now if there is solution in the container what does it mean that means in the container there will be solute as well as solvent evident since there is solute solution in the container that means in the container automatically there will be solute as well as solvent imagine my dear students in 500 G of solution imagine there are 50 g of solute imagine that 50 g of solute are present in 500 G of solution this is a scenario which I have created over here now I'm going to write the same scenario in the form of a statement try to understand why am I doing it I'll be writing something like this 500 G of solution 500 G of solution contains 50 g of solute 500 G of solution contains 50 g of solute if I use the unitary method can you let me know in 1 G of solution in 1 G of solution how many gram of solute will be there absolutely you should be able to let me know that right 500 G of solution contains 50 g of solute so 1 G of solution contains 50 ID 500 G of solute right if I ask you 100 G of solution contains how much you'll say 100 G of solution contains 50 divid 500 multiplied by 100 G of solute now if I ask you what exactly did we calculate here if I ask you you what exactly did we calculate here you should be able to understand this particular statement what is the statement try to understand I'll say this is the mass of solute this particular term it is the mass of solute which is present in 100 G of solution and as per definition that mass of solute which is present in 100 G of solution that mass of solute that mass of solute which is present in 100 gam of solution what do you call that as you call that as the weight by weight percentage of salute so indirectly I would say that I have calculated the weight by weight percentage of the solute in the solution right why did I do this what is the need to do all this because I want to generalize a formula I do not want you guys to do this procedure every time no I just want to derive one formula what has to be the formula try to understand try to understand what has to be the formula if you want to calculate Weight by weight percentage of solute one formula you can make try to understand over here in this box how many terms we have we have got three terms numerator 50 denominator 500 multipli 100 what was this 50 exactly this 50 was the mass of salute so if you make the result in the numerator you will have to write the mass of solute in G divided by in the denominator what do we have 500 what was this 500 it was the mass of solution so in the denominator you will be exactly writing the mass of solution in grams and eventually at the end you'll be multiplying it with the number 100 right so this is one direct result which I would want all of you to remember from now onwards in order to calculate the weight by weight percentage of solute so weight by weight percentage of solute in any solution is calculated like this mass of solute in Gams divided by mass of solution in Gams multiplied 100 that's all that's all is it clear people is it clear people is it clear to everyone yes now similarly similarly similarly if I ask you how do we calculate Weight by weight percentage of solvent you should be in a position to answer me this one how do we calculate Weight by weight percentage of solvent it is going to be mass of solvent in grams divided by mass of solution in grams multiplied by what multiplied 100 so this particular term is giving you the weight by weight percentage of solvent as simple as that right perfect perfect guys I want you guys to understand how these results exactly come how do we make these results that's very important okay that is very important now people apart from this let me tell you one more simple thing I'm writing one statement over here for example I'm writing 20% weight by weight of NaOH Aqua Solution can you try to decode this particular statement for me you should be you should be in a position to decode this kind of the statement have a look I wrote a statement 20% weight by weight of Na aquous solution aquous means the solvent which we have taken that is water so in water we have introduced NaOH now 20% weigh by weight NaOH what does it mean it simply means that 20 G of NaOH are present in are present in 100 G of solution be careful with the terminologies I'm not writing 100 G of water I'm writing 100 G of solution right similarly similarly for example I'm writing a statement you need to tell me the answer in the charts let's say I'm writing 30% weight by weight of Na Aqua Solution what does it mean try to decode it quickly quickly people what does it mean you should be knowing this it means that 30 G of solute which is your NAC 30 G of your solute is present in 100 G of solution I believe every single thing till here is absolutely clear to everyone yes and my dear students take a note of one simple thing take a note of one simple thing and do remember this from now onwards what is the note note is pretty much simple if you have got a binary solution if you have got a binary solution for the binary solution do remember weight by weight percentage of salute plus weight by weight percentage of solvent it is always equal to 100 it is always equal to 100 so once you get weight by weight percentage obsolute you can easily calculate Weight by weight percentage of solvent as well right yes perfect guys is it clear to you I believe this is clear and these are the things which you are supposed to remember related to we by weight nothing else these are the things which you have to remember that's all let me know once in the chats with the fire OT everyone then only I can move on to the next terminology first of all I'll be giving you a lot of terminologies and then once we are done with the terminologies then we shall be doing a lot of questions as well yeah so tell me quickly I'm mov on to the second terminology second concentration term what is that that is weight by volume percentage weight by volume percentage of salute now if I ask you how to define this particular statement how to define this particular statement you should be able to do it you should be able to do it weight by volume percentage of solute how do we Define it weight by volume percentage of solute is defined as the mass of solute the mass of solute in grams present in present in now here instead of 100 G I'll be writing present in 100 ml of solution 100 ml of solution so so for example you have got 100 ml of solution imagine in 100 ml of solution you have got 20 G of solute just imagine in 100 ml of solution you have got 20 G of solute you will say weight by volume percentage of solute is 20 imagine in 100 ml of solution there are 40 G of solute you'll say weight by volume percentage of solute is 40 in general if I say in 100 ml of solution there are x g of solute you'll say weight by volume percentage of salute is X as simple as that right now guys I'm not going to derive its result I'm going to give you the result I'm going to give you the result K is saying board is not clearly visible K on my side connectivity is all okay everything is okay I can see the board is absolutely clear from my side it is just you need to change your settings I think you are watching it at very low resolution right just change the settings perfect so people what is going to be the result by means of which you can calculate Weight by volume percentage of solute let me give you the result directly weight by volume percentage of solute is always equal to mass of solute in grams divided by the volume of solution in milliliters multiplied by 100 this is how exactly you can calculate weight by volume percentage of solute this is how you can calculate Weight by volume percentage of solute number one number two if you want to calculate Weight by volume percentage of solvent if you want to calculate Weight by volume percentage of solvent you can easily do that as well you can easily do that as well it is going to be mass of solvent in G divided by volume of solution in milliliters multiplied by 100 multiplied by 100 this is one more thing which I would want you guys to remember one more thing now dear students now dear students just try to decode this statement I'm writing something like this I'm writing 20% weight by volume of NaOH aquous solution NaOH aquous solution can you let me know what is the exact meaning of this particular statement quickly let me know the meaning of this particular statement everyone of you everyone means every everyone in the chats everyone what does it mean look at this particular terminology it is weight by volume and here I have mentioned aquous solution that means the solvent which is used that is water right so I can say 20 G of solute 20 G of solute are present in here I'm not going to use 100 G here I'll be using 100 ml of solution 100 ml of solution right simple and basic stuff simple and basic stuff I'm teaching you right now simple and basic stuff I'm teaching you right now tell me the meaning of this particular thing 30% weight by volume of NAC aquous solution quickly let me know in the chats what is the meaning of this particular terminology 30% weight by volume of NAC aquous solution what is meant by it I want you guys to say it I want you guys to write it in the chats it simply means that 30 G of your solute which is NAC is present in 100 ml of solution so I'm pretty much sure from now onwards I'm pretty much sure from now onwards you should be able to decode these sort of simple simple statements whenever they are in your questions right and people one direct result one direct result which is going to ease out a lot of things afterwards which is going to ease out a lot of things afterwards when it comes to calculations do remember weight by volume percentage of solute weight by volume percentage of solute is always equal weight by weight percentage of solute weight by weight percentage of solute multiplied by density of solution but this density of solution has to be in G per ml I'm not deriving this result right just remember this particular result directly because I'm going to use this result many of times afterwards do remember this particular thing okay weight by volume percentage of salute is always equal weight by weight percentage of solute multiplied by what multiplied by density of solution in grams per ml well I can see we have got one amazing person in the chats shiage Kix sir guys do you know him do you know him do you know this guy who is in the chats you have no idea who is right and sooner or later you are going to get one amazing surprise sooner or later you are going to get one amazing surprise be ready for that be ready for that be ready for that be ready for that yeah okay uh no Ashu I do not know which Nam ma'am you talking about can you write the full name maybe I would be knowing yeah soon soon you'll get the surprise just wait for like 20 to 25 days okay perfect guys perfect perfect I think you have guessed it correctly right so let's keep it a surprise for now yeah let's keep it a surprise for now and eventually you'll get to AED eventually you're are going to get to aard all right I'm moving ahead so I hope you have copied this particular statement as well I shall be using this statement in the questions afterwards weight by volume percentage of salute is always equal to weight by weight percentage of salute multiplied by what multiplied by density of solution and the density of solution has to be taken in G per ml yeah Perfect all right moving on to the third terminology which you must be knowing what is that let's have a look I dear students the third terminology which I'm writing over here that is weight by sorry that is volume by volume percentage that is volume by volume percentage of salute now you should be in a position you should be in a position to Define this particular terminology as well you should be in a position to Define this particular ter ology as well how do we exactly Define it let me know in the chats is going to be it is defined as the volume of solute in milliliters the volume of solute in milliliters present in present in 100 ml of solution it is defined as the volume of solute which is present in 100 ml of solution so imagine that that you have got 100 ml of solution imagine that you have got 100 ml of solution my dear students if in 100 ml of solution imagine there are 20 ml of solute in 100 ml of solution imagine you have got 20 ml of solute so you'll directly say volume by volume percentage of solute is 20 imagine in 100 ml of solution there are 30 ml of solute you'll directly say volume by volume percentage of solute is 30 and in general in general if in 100 ml of solution there are X mL of solute you'll directly say volume by volume percentage of salute is X that's all that's all people that's all right so I believe you can easily make the result over here well I'm not deriving the result I'm giving you the result and it's application part I'll be letting you know in some time so I'm writing volume by volume percentage of solute volume by volume percentage of salute is equal is equal volume of solute in milliliters divided by volume of solution in milliliters multiplied 100 multiplied by 100 this is the result by means of which we can calculate volume a volume percentage of solute Point number one point number two if by chance you want to calculate volume by volume percentage of solvent you can do that as well and how exactly you will be wrting volume of solvent in milliliters divided by volume of solution in milliliters and you'll be multiplying this particular terminology by the number 100 eventually you'll be getting the volume by volume percentage of solvent volume by volume percentage of solvent now you tell me now you tell me I'm writing something like this 20% volume by volume of NaOH aquous solution of n AOH aquous solution Tell me the meaning of this particular terminology Tell me the meaning of this particular terminology quickly quickly guys everyone Tell me the meaning of this particular terminology what does it mean quickly in the chats it means that it means that 20 ml of solute is exactly present in 100 ml of solution it means that 20 of solute is present in 100 ml of solution yes pretty much simple pretty much simple right for example I'm writing something like this try to decode this I'm writing 30% 30% volume volume of NaOH what is meant by this quickly quickly what does it what does it mean it means that 30 ml of NaOH is present in 100 ml of Solution that's all nothing else it's pretty much simple okay till now we have covered three terminologies I'm not going to give you the questions yet we have to cover one more terminology then I'll be giving you the questions okay one terminology which you would be knowing from your class 11th what is that that is mole fraction that is mole fraction which is represented by Kai my dear students tell me how do you exactly Define the term MO fraction how do you define the term MO fraction quickly hello siia how are you quickly mole fraction how do you exactly Define the term mole fraction so it is defined as the ratio of the ratio of moles of a component the ratio of moles of a component to that of to that of total moles present in the solution to that of total moles present in the solution now what is meant by it have a look this is very much simple and easy to understand imagine that imagine that we have got a solution in the container if we have got a solution in the container that automatically tells you in the container you have got solute and at the same time you have got solvent as well now my dear students let's assume that let's assume that there are N1 there are N1 moles of solvent in the solution and there are N2 moles of solute in the solution just imagine it imagine there are N1 moles of solvent in the solution N2 moles of solute in the solution now if I want to define the mole fraction of solute how would you define the mole fraction of solute quickly mole fraction of solute will be simply equal number of moles of solute in the solution divided by the total moles present in the solution divided by total moles present in the solution now number of moles of solute I'm representing with N2 divided by total moles present in the solution is just going to be N1 plus N2 N1 plus N2 right so this particular result is going to give you the mole fraction of solute similarly my dear students if you want to calculate the mole fraction of solvent as well you can do that mole fraction of solvent is equal is equal is equal it is going to be number of moles of solvent which is N1 divided by total moles present in the solution which is N1 plus N2 right so this particular terminology is going to give you the mole fraction of solvent now at the same time at the same time whenever you see a binary solution whenever you see a binary solution for the binary solution do remember mole fraction of solvent plus mole fraction of solute it is always equal to one for the binary solution mole fraction of solute plus small fraction of solvent is always equal to one right perfect this is something which you need to know now guys in the questions they might confuse you by some fancy terminologies they might confuse you by some fancy terminologies they might give you the data they might use the term mole percentage they might use the term mole percentage now once you see mole percentage you know you'll you'll think a lot about it right mole percentage is nothing what is it it is mole fraction multiplied by 100 mole percentage is nothing it is mole fraction multiplied 100 for example for example I'm defining mole percentage of solvent it means nothing it is just mole fraction of solvent multiply that term with 100 if I want to define the mole percentage of solute you can do that as well it is mole fraction of solute multiplied by 100 right whenever mole fraction is Multiplied with 100 whenever mole fraction is Multiplied with 100 you call it as mole percentage you call it as mole percentage right right people the way I told you the sum of mole fractions the sum of mole fractions of all the components of the solution the sum of mole fractions of all the components in the solution is equal to one you know that in the similar way do remember in the similar way do remember the sum of the sum of mole percentage of all the components the sum of mole percentage of all the components of a solution solution is equal to 100 is equal to 100 this is one more statement which I would want everyone of you to take a note of all of this is clear all of this is clear say yes or no in the chats all these terminologies are clear if all these terminologies are clear let's try to solve some basic questions question number one on your screen one basic question this is one basic question this is look at the question carefully calculate the mole fraction of solute calculate the mole fraction of solute in 10% weight by weight of NaOH aquous solution so people can you tell me what is the meaning of this particular statement can you tell me what is the meaning of this particular statement what does it mean it means simply that 10 G of solute 10 G of solute are present present in 100 G of Solution that's all right that's all 10 G of solute are present in 100 G of solution 10 G of solute that means mass of solute which we represent with W2 you got to know as 10 G right if mass of solute is 10 G mass of solution is 100 G what do you think about the mass of solvent you know mass of solvent plus mass of solute that's 100 right mass of solute plus mass of solvent that's 100 right so mass of solvent is going to be how much 100 minus 10 which comes out to be 90 gam this comes out be 90 G what am I supposed to calculate look at it carefully I'm supposed to calculate mole fraction of solute this particular term I'm supposed to calculate mole fraction of solute and as for the result mole fraction of solute is nothing it is moles of solute divided by total moles present in the solution as for the formula you know it now my dear students in order to calculate mole fraction of solute what do I need I need N2 and I need the N1 N2 and N1 are to be calculated now first thing what is meant by N2 N2 means number of moles of solute number of moles of solute will be mass of solute in grams divid by mol mass of solute N1 moles of solvent mass of solvent in grams divide by m mass of solvent simple formula I'm using nothing else right now what is the mass of solute mass of solute is 10 G divided by what is the molar mass of solute molar mass of NaOH that's 40 so the value comes out be 1X 4 right these are the moles of solute in the solution similarly moles of solvent will be equal mass of solvent that is 90 G molar mass of solvent what is my solvent solvent is water and molar mass of water you all must be knowing that's 18 the value comes out be five so you got N1 as well as N2 if you got N1 as well as N2 you can put it here in this expression N2 value you got to know that as 1X 4 N1 value is 5+ 1x 4 the value comes out to be how much 1 iD 21 so 1 iD 21 this is going to be the mole fraction of solute which I was supposed to calculate yes right people this is giving you the mole fraction of solute now for example in the same question if they ask you if they ask you to calculate mole fraction of solvent as well how you going to do that you know in case of the binary solution kai1 + K2 value is equal to 1 therefore kai1 is going to be 1 - K2 it's going to be 1 - 1 upon 21 the value comes out be 20 / 21 so this is the mole fraction of solvent in the same solution right is it perfectly clear to everyone is it perfectly clear to everyone is it perfectly clear to everyone quickly quickly my dear students yeah I hope these sort of questions you can easily solve from now on this was our first type of question based on mole fraction second type of question look at the question carefully and try to solve this look at the equation carefully and try to solve this look at the question carefully and try to solve this people everyone everyone give it a try everyone give it a try the question is saying that what mass of water what mass of water in grams should be added to 16 G of methanol to make its mole fraction as 0.25 make its mole fraction as 0.25 what is meant by this have a look have a look guys imagine this is your container imagine this is your container in the container what do we have we have got 16 G of methanol we have got 16 G of methanol in the container now as per the question what mass of water should be added what mass of water should be added such that the mole fraction of methanol becomes 0.25 so basically we have to make the mole fraction of methanol as how much we have to make the mole fraction of methanol as 0.25 as for the question as for the question we have to make mole fraction of methanol at 0.25 now they asking us how many GRS of water are required to make the mole fraction of methanol as 0.25 let's assume let's assume x g of water are required let's assume that x g of water are required to make the mole fraction of methanol as 0.25 let's do this assumption let's use this assumption right let's use this assumption assumption is I am adding x g of water I'm adding x g of water to 16 G of methanol then only its mole fraction is becoming 0.25 this is my assumption right so people the question is done and dusted if I ask you now if if I ask you now if I ask you now how many G of methanol are there in the container you'll say there are 16 G of methanol in the container if I ask you how many G of water are there in the container you will say x g of water are there in the container we have added x g of water then only mole fraction of methanol is coming out be 0.25 now people tell me one thing mole fraction of methanol means number of moles of methanol if divided by total moles divided by total moles in the container as for the formula this has to be equal to 0.25 right moles of methanol do we know the moles of methanol no let's exactly get to know how many moles of methanol are there in the container number of moles of methanol will be equal mass of methanol in Gs divided by molar mass of methanol the value comes out be5 right similarly how many moles of water are there in the container mass of water in the container that's xide M mass of water that's 18 right right people perfect now you got number of moles of methanol you got number of moles of water put these two terms in this expression so what will I get from here moles of methanol in the container .5 divided by moles of methanol that's .5 moles of water xid 18 the value is coming out be 0.25 do the cross multiplication part and get the value of x and when you solve this the value of X comes out to be 27 G so can you let me know the meaning of this 27 G can you let me know the meaning of this 27 G can you let me know the meaning of this particular 27 G quickly quickly in the chats quickly everyone is it clear people so I would say we are supposed to add 27 G of water we are supposed to add 27 G of water in the container then only mole fraction of methanol will become 0.25 yeah perfect is it clear people is it clear to everyone is it clear to everyone quickly quickly guys let's talk about something called as marity how do you exactly Define the term marity how do you define the term marity uh you can consult these sort of sessions for J Mains as well no worries okay all right people marity how do we Define it you should be in a position to Define it marity is defined as the number of moles of solute the number of moles of solute present in present in 1 lit of solution this is the definition of marity number of moles of solute present in 1 ler of solution imagine in 1 lit of solution there are 2 moles of solute you'll directly say marity of the solution is two imagine in 1 liter of solution there are 5 moles of solute you'll directly say marity of the solution is five imagine in 1 lit of solution there are X moles of solute you'll directly say marity of the solution is X so what is it number of moles of solute which will be present in 1 liter of solution which will be present in one liter of Solution that's all that's it right and people how do you how do you get its results I'm not deriving the results I'm giving you the results and I'll show you its application part have a look people marity the first result to calculate marity is equal number of moles of solute divided by volume of solution in liters this is the first and the basic result by means of which you can calculate marity of the solution number two number of moles of solute number of moles of solute can be written as mass of solute in grams divided by molar mass of solute in the denominator you had volume of solution in liters instead I'll use volume of solution in ml and in the numerator I'll be multiplying with th this is one more result which you have to remember directly which you have to remember directly right and my dear students is that it no that's not it there are few more results okay marity is equal to 10 multiplied by weight by weight percentage of solute 10 multiplied by weight by weight percentage of solute multiplied by density of solution in G per ml divided by molar mass of solute divided by m mass of solute this is one more result which you have to remember to calculate marity of the solution there will be lot of questions in which weight by weight percentage of solute will be given in which density of the solution will be given and we will be directly supposed to calculate marity if you will remember this result it will not take you more than 10 seconds to solve the question right do we have one more result to calculate marity I would say yes see guys if you remember in the beginning only I told you in the beginning only I've told you weight by weight percentage of solute multiplied by density of solution weight by weight percentage of solute multiplied by density of solution what do you call that as you call that as weight by volume percentage of solute so this is 10 * by weight by volume percentage of solute divided by molar mass of solute this is one more result which I would want every one of you to remember remember so that we can solve the questions in very lesser time yes is it clear people is it clear people quickly someone is saying sir please Dance you should have been in front of me I would have made you do MRA not dance I hope you know what mu is nonsense okay so these are 1 2 3 four four results by means of which you will be calculating marity of the solution right right people right Perfect all right look at few more things guys look at few more things look at few more things if I write a statement like this I'm writing two m NAC Aqua Solution can you let me know what is meant by this particular statement 2 m NAC Aqua Solution 2 m NAC Aqua Solution how do you decode this statement it is very very much simple it means that 2 moles of your solute are present in are present in 1 liter of solution 2 moles of solute are present in 1 liter of solution correct now if you ask me what are the units of marity if you ask me what are the units of marity moles in the numerator volume in the denominate denominator so units of marity will be moles per liter units will be moles per liter or you can call it directly as molar you can call it directly as molar now guys few more things related to m ity which you must remember from now onwards sometimes sometimes in the questions they will give a statement like this semi solution sometimes they'll write a statement like semar solution sometimes they can write decolar solution decolar solution sometimes they can write ctim solution ctim solution and sometimes they can write Millar solution you should exactly know what is the meaning of all these particular statements which I'm mentioning over here you should know it you should know it you should know it my dear students semi solution do remember it is that solution whose marity is equal to 0.5 what is meant by marity is equal to 0.5 it means that 0.5 moles of of solute will be present in 1 lit of solution right decim solution decim solution that solution whose molarity will be 0.1 right ctim solution that solution whose marity will be 0.01 Millar solution that solution whose marity will be 0.01 do remember these terminologies as well do remember these terminologies as well and my dear students do remember one more thing if you ask me whether marity is temperature dependent or not I would say marity marity is temperature dependent marity is temperature dependent do remember any concentration term if you look at any concentration term in the formula of any concentration term in the formula of any concentration term if you find find the term volume right you have got different concentration terms you have got their different results if any result if any result of any concentration term you find a term volume do remember that concentration term is going to be volume dependent if you look at the expression of marity there's a term volume here there's a term volume here if there's a term volume here that means it's going to be temperature dependent that means it's going to be temperature dependent do you remember when we increase the temperature when we increase the temperature I will directly say volume of the solution increases when you increase the temperature of the solution volume of the solution increases if volume of the solution increases that tells you that the molarity of the solution decreases do remember this particular statement as well right upon increasing the temperature volume of the solution increases which eventually is going to tell you that marity of the solution decreases I hope this is clear I hope this clear to everyone people yes I hope this is clear to everyone okay perfect okay one last thing tell me what is the meaning of this particular thing 5 capital M NaOH Aquis can you quickly tell me what is the meaning of this particular statement in the chats everyone everyone in the chats quickly quickly quickly quickly everyone in the chats quickly what is meant by it I want everyone to write it it means 5 moles of NaOH are present in one liter of solution yeah perfectly done perfectly done perfectly done now guys similarly the way we defined m marity in the similar way you are going to define the term mity as well mity which is represented by small M now how do we Define exactly this particular term again look at the definition and eventually do everything related to the definition mity what is mity it is defined as the number of moles of solute number of moles of solute present in present in 1 kg of solvent it is defined as the number of moles of salute which are present in 1 kg of solvent imagine you have taken 1 kg of solvent imagine you have taken 1 kg of solvent in the container and you are introducing 5 moles of solute into that I can say in 1 kg of solvent there are 5 moles of solute right and and number of moles of solute present in 1 kg of solvent that defines mity as simple as that right that defines mity number of moles of solute which are present in 1 kg of solvent if if in 1 kg of solvent you are introducing 5 moles of solute the mity of the solution will be five in 1 kg of solvent if you are introducing 10 moles of solute mity 10 in 1 kg of solvent you are introducing X moles of solute mity X that's it okay be careful with the definition that's it now what is required here formulas are required formulas are required mity which is represented by small M the first Formula number of moles of solute divided by mass of solvent in kilogram this is the first Formula by means of which you can calculate mity of the solution Point number one point number two instead of moles of solute you can write mass of solute divided by mol mass of solute instead of W1 in kg you can write W1 in Gs and you can multiply it with, here this is the second formula this is the second formula are there some other formulas yes there are some other formulas yes there are some other formulas do remember people mity can be calculated like this as well mole fraction of solute multiplied th000 divided by 1 minus mole fraction of solute multiplied by mol mass of solvent this is one direct result this is one direct result by means of which you can calculate mity there will be lot of questions in which mole fraction will be given and you will be supposed to calculate mity directly use this result solve the equations in 10 seconds no need to use your brain just apply the formula that's it just apply the formula that's it there will be a lot of questions in which mity will be given and you will be supposed to calculate mole fraction use the formula and kill it that's all yes is there one more result I would say yes there is one more result by means of which mity can be calculated do remember that what is that that is going to be 1,000 multiplied by marity of the solution divided by, multiplied density of solution in G per ml minus marity of the solution multiplied by marol mass of solute this is one more result you will find a particular set of questions in concentration terms wherein wherein molarity of the solution will be given and you will be supposed to calculate mity or indirectly you will be given mity and you will be supposed to calculate marity do remember this result kill the question in question in 10 seconds not more than that not more than that now if you are thinking what are going to be the units of mity just look at it in the numerator moles in the denominator Mass so it's moles per kg moles per kg or you can write it directly as mol m o l a l mol right you can write it like this and people for example I'm writing a statement like this 2 mol NAC Aqua Solution can you quickly let me know what is the meaning of this terminology two M NAC aquous solution quickly it means that we have got a solution whose mity is two now what does that mean it simply means that two moles of solute are present in as for the definition as present in 1 kg of solvent that's it is that it yes that's it now if I ask you is molarity going to be temperature dependent or independent look at all the formulas in any formula do you see the term volume there's no volume anywhere mentioned if there is no volume mentioned of course it's going to be temperature independent so this is temperature independent term mity of the solution is temperature independent now now now if I ask you a question if I ask you a question among molarity and mity which one is preferred which one always gives you the correct results quickly what do you think marity or mity do you remember mity is preferred over mity why why because mity is temperature independent mity of the solution does not change when you change the temperature of the solution but marity of the solution can change when you change the temperature of the solution right so I would want you guys to remember this mity is preferred over what over marity yes or no yes or no someone is preparing tomato rice while listening to the class that's nice good prepare prepare prepare good for you I believe people whatever I've said till now every single thing is clear to you yeah now before solving the questions before solving the questions there are two more things which I would want you guys to remember what are those things let's see one by one let's see one by one let's see one 1 by one there is something called as parts per million parts per million of solute this is one more terminology PPM of solute PPM of solute how do you define PPM of solute it's simple guys how do you define PPM solute it is defined as it is defined as the mass of solute in gr present in 10 ra^ 6 G of solution whatever mass of solute in G will be present in 10 ra^ 6 G of solution whatever mass of solute and Gams will be present in 10 rais G of solution you'll be called that particular term as a PPM of solute PPM of solute now here you need to remember one result directly result is parts per million of solute is equal mass of solute in G divided by mass of solution in G multiplied 10^ 6 multiplied 10^ 6 and in the similar way if you anytime want to calculate parts per million of solvent it is going to be mass of solvent in Gs divided by mass of solution in Gs multiplied by what multiplied by 10^ 6 this is the result to calculate PPM of solvent result to calculate p PM of solvent yes now people now tell me one thing tell me one thing can I write this PPM of solute like this PPM of solute is equal can I write it like this W2 in Gs divided by W solution instead of 10 ra power 6 can I write 100 mtip 10^ 4 I can do that I can do that right now this particular term we have already discussed what do you call this particular term as I'll be writing it finally PPM of solute is equal this term is called as if you remember this is called as weight by weight percentage of salute multiplied by 10 ra^ 4 sometimes you'll find some questions in which weight by weight percentage of solute will be given and you will be supposed to calculate PPM of solute use this result solve it that's it similarly similarly PPM of solvent PPM of solvent will be equal this term I can write it like this weight by weight percentage of solvent multipli by 10^ 4 correct do remember these direct expressions and in sometimes in sometime people you'll understand in sometime you'll understand you'll understand exactly where do I use these direct results are these clear are these clear are these terms clear let let me know quickly guys there is one last concentration term and then there are going to be a lot of questions which we shall be doing then there are going to be a lot of questions which we shall be doing what is that last conent okay by the way a lot of students must be thinking why he did not teach normality yeah I hope you guys will be thinking that why I did not discuss normality because normality is the topic which we have discussed in detail in the chapter redox reaction which was done few days back I hope you remember normality was a concept which I have discussed in detail in the chapter reduxx reactions okay perfect now the last concentration term that is parts per billion parts per billion of solute how do you define it how do you define it instead of 10 ra^ 6 you'll be writing 10 ra^ 9 that's it that's it people so make the result directly it is mass of solute in Gams divided by mass of solution in Gs multiplied by instead of 10^ 6 right 10^ 9 perfect for example I'm asking you ppb of solvent parts per billion of solvent how we are going to do it mass of solvent in G divided by mass of solution in G now multiply it with 10 ra^ 9 10 ra^ 9 can you convert these results in weight by weight yes you can do that you can write it as weight by weight percentage of solute multiplied 10^ 7 you can write it as weight by weight percentage of solvent multiplied by 10 ra^ 7 these are further more results which you need to remember right so these were all the results people which you have to remember related to concentration terms okay perfect the first result was weight by weight percentage of solute W2 divide by W solution into 100 weight by weight percentage of solvent W1 divide W solution into 100 right second weight by volume percentage of solute W2 divided by volume of solution in ml multiped 100 weight by volume of solvent W1 in G divide volume of solution in ml multip 100 there was one direct result weight by volume percentage of solute is equal weight by weight percentage of solute multiplied by density of solution right this was one more result volume by volume percentage of solute was discussed V2 divide by V solution multipli 100 volume by volume percentage of solvent V1 divide by V solution multi 100 mole fraction of solute N2 divid N1 plus N2 mole fraction of solvent N1 N1 plus N2 right similarly marity marity first result N2 divide volume of solution in liters W2 multi th000 divided M2 multip volume of solution in ml right similarly there were other results too 10 multiplied by density of solution multiplied by weight by weight percentage of solute divide by m mass of solute one more result 10 multip weight by volume percentage of solute divided by m mass of solute then comes mity N2 / W1 kg W2 * 1000 divide m 2 multiplied W1 in G right similarly there was one more result if you remember K2 multiped th000 divid 1- K2 into M1 one more result th000 multi capital M divid th000 MTI density minus capital M MTI M2 I hope you know I hope you remember I hope you remember yeah I hope you remember people Perfect all right people so let's try to do a few questions let's try to do few questions now you should be easily killing these questions guys you should be easily killing these questions look at the first question calculate the weight by weight percentage 1.5 m ch3 co aquous solution whose density is given what are we supposed to calculate we are supposed to calculate Weight by weight percentage of solute what is given we are given with the marity of solution I'm just writing the data which is given marity of solution is given density of solution is also given 0.8 G per ml 0.8 G per ml look at this particular solution right look at this particular solution ch3 Co is your solute so molar mass of ch3 Co mol mass of solute when you solve it it will come out to be 60 G per mole right my dear students do you remember any result which connects all these terms have we discussed any result which connects all these terms yes there was one result marity of the solution is equal 10 multiplied by density of solution in grams per ml multiplied by weight by weight percentage of solute divided by molar mass of solute this was one result now use the data M value is 1.5 is equal to 10 multipli density of solution8 weight by weight percentage of solute needs to be calculated divided by mol mass of solute is 60 one equation one unknown solve it and get the value of weight by weight percentage of solute which was to be calculated that's it and as far as I remember this value will come out to be 11.25 right so what is meant by this 11.25 what is meant by this 11.25 this is the weight by weight percentage of solute it means that 11.25 g of solute are present in 100 G of solution yes perfectly done moving on to one more question look at this question guys look at this question quickly look at this question quickly calculate the mity of a two molar NaOH aquous solution whose density is given look at the question remember the results and kill it marity is given mity needs to be calculated how do you solve it there is one direct result which connects mity and marity if you remember mity is equal th000 multip marity of the solution divide by th000 multipli by density of solution minus marity multipli M mass of salute this is one result which we discussed few minutes back only yeah now marity of the solution is two correct density of the solution is given as 1.2 G per ml correct this is your solute so molar mass of solute M mass of n is 40 G per mole I believe everything is given this is given this is given this is given this is this you know solve it and get the value of mity nothing to do here as well nothing to do here as well right okay guys will you be able to solve these quickly quickly people yeah perfectly done try to solve this question try to solve this question it's a nice it's a nice one it's a nice one people the question is dissolving 120 g of UA dissolving 120 g of UA in 1000 G of water right dissolving 120 g of Uria in 1,000 G of water so as for the question 120 g of Ura 120 g of Uria are to be dissolved in 1,000 G of water and when 120 g of Uria is dissolved in 1,000 G of water what do we get we get a solution we get a solution and density of this particular solution is given to us density of this particular solution is given to us 1.15 G per ml what are we supposed to calculate we are supposed to calculate marity of this particular solution we are supposed to calculate marity of this particular solution how exactly you going to do it try to understand what exactly I'm going to do marity marity which is capital M there is one of the result to calculate marity I had given you already mass of solute in Gams multiplied by 1,000 divided molar mass of solute and this is volume of solution in in ml do you remember this result I've given you long back now people when you are mixing 120 g of UA with th000 G of water among these two which one is lesser in amount Ura is lesser in amount so Ura is my salute UA is my solute right UA is my salute so what is the mass of solute it's 120 g mass of solute is 120 g multiplied by th000 divided by mol mass of solute M mass of Uria when you calculate it it will be 60 now this volume of solution is not given to me volume of solution is not given to me now how will I calculate volume of solution that is the point how will I calculate volume of solution look at this term density of solution is given instead of density of solution I can write mass of solution divide by volume of solution right this is given to me as 1.15 G per M correct so I'm using the parameter density density of solution is given so density of solution is mass of solution divide volume of solution now tell me what is the mass of solution mass of solution means mass of solute plus mass of solvent mass of solute plus mass of solvent this plus this so that is 11 120 1120 G divided by volume of solution needs to be calculated so take volume here take this here so this is 1.15 G per ml I'm taking in the denominator and here I'm getting volume of solution in milliliters right people yes when you solve this particular term 1120 divid 1.15 it will come out to be 973 ml so what did we calculate from here I calculated the volume of solution in milliliters so put it here it is 120 * 1,000 ID 60 and this is 973 when you solve this the value comes out to be 2.05 so what is this 2.05 this is the marity of the solution correct you can solve this question by one more approach as well first calculate Weight by weight percentage of solute then use the formula marity is equal 10 multip density of solution multip weight by weight percentage of solution div by m mass of salute you can do that as well no issues at all no issues at all you can do that okay you can do that give this question a try quickly give this question a try quickly calculate the mity calculate the mity of a KCl Aqua Solution which is prepared by dissolving 7.45 G of KCl in 500 mL of solution what is it what is meant by it try to understand try to understand you have made a solution you have made a solution right in which water is solvent KCl is solute as per the question 7 .45 G of KCl are present in 500 mL of solution this is mentioned 7.45 G of KCl are present in 500 mL of solution this is mentioned right what do I need to calculate I need to calculate mity there is one formula for mity if you remember W2 in G multiplied by th000 divided by m 2 and this was W1 in G I hope you remember this result I have given you long back I've given you long back now the solute is KCl and what is the mass of solute mass of solute W2 is 7.45 G multiplied by th000 divided by mol mass of solute M mass of KCl when you solve it it is 74.5 G per mole W1 W1 is the mass of solvent in G but I do not not know what is the mass of solvent I do not know what is the mass of solvent how do I get the mass of solvent try to understand I'll be using this parameter density of solution right density of solution instead of density of solution I'll write mass of solution in GRS divided by volume of solution in milliliters density is equal mass by volume right is equal 1.2 correct is equal 1.2 G per ml now as for the question volume of solution is given volume of solution is given 500 so this is 500 so mass of solution will be 500 into 1.2 I'll say the mass of solution will be this multiplied by this which comes out be 600 G but in the formula do I need the mass of solution no I do not need the mass of solution I need the mass of solvent what did I get I got the mass of solution you know mass of solution is basically mass of solute plus mass of solvent so this has to be 600 G corre what is the mass of salute 7.45 G Plus W1 is equal 600 so you can calculate W1 from here which will come out to be 59255 G so this is the mass of solvent so put it here instead of W1 what exactly you going to write you are going to write 59255 solve this and get the mity that's all nothing to do nothing to do in this particular question as well I hope this is clear I hope this is clear and people in the concentration terms this is your homework question because all these types we have we are done we are given with a solution whose marity is given density of solution is given first calculate mity you can do that use a direct formula right once you calculate mity then calculate Weight by weight percentage of solute you can do that too right weight by volume percentage of solute you can do that too all the direct results which I gave you mole fraction of solute once you calculate mity there is one result which connects mity with mole fraction right perfect guys so this is going to be your homework question this is going to be your homework question which you shall be doing and do let me know its answer in the comment section at the end okay is everything going perfect till now is everything going perfect till now is everything going perfect till now yeah now we have got one more amazing topic guys and this is frequently used many of times someone is saying fourth part confusion fourth part fourth part was the mole fraction of solute right so first of all I believe you can easily calculate marity from here right mity you can use with direct result once you calculate marity you can use this result mole fraction of solute multip th000 1 minus mole fraction of solute molar mass of solvent right this you would have already calculated right M mass of solvent you know water is solvent 18 so one equation one unknown from here you can calculate K to that's it nothing to worry yes is it perfectly fine people everyone everyone everyone in the chats in the chats quickly everyone in the chats quickly everyone in the chats quickly sir I have been following you consistently since last 6 months and there has been massive progress in my approach to chemistry glad to know that glad to know that okay let me move ahead people marity of dilution before discussing dilution before discussing dilution there is before discussing dilution there is one thing which I want you to remember what is that have a look have a look my dear students few minutes back I give you a result to calculate marity of the solution that was number of moles of solute divided by volume of solution in liters divide by volume of solution in liters this I told you FES back right now people can I write the same result like this can I write N2 is equal N2 is equal to this multiplied by this marity of the solution multiplied by volume of solution in liters I can do that no issues at all right so what is meant by this particular statement this is important this is important why see whenever you will be having a solution whenever you'll be having a solution whose marity will be given to us whose volume in liters we know multiply marity of that solution with its volume in liters you will be getting the number of moles of solute in that particular solution for example you got a solution whose marity is two let's say volume of that solution is 5 l so this multiplied by this is equal 10 so that means there are 10 moles of solute in that solution that means there will be 10 moles of solute in that solution right agreed perfect number one number two in the same equation in the same equation instead of volume of solution in liters if I use volume of solution in milliliters instead of volume of solution in liters if I write volume of solution in milliliters now I'm not going to call this as the number of moles of solute I will be calling this as the number of Mill moles of solute I will be calling this as the number of Mill moles of solute this is one more thing which I want you guys to know so one is moles of solute in the solution moles of solute in the solution is always equal marity of that solution multiplied by volume of that solution in liters now similarly Mill moles of solute in the solution will be equal marity of that solution multiplied by volume of the same solution but in milliliters but in milliliters what what is what is the difference between the two there's no difference these are just the fancy terminologies again I'm saying Mill moles are nothing when you multiply moles by th you get millimoles when you multiply moles by th you get Mill moles so if I write milles of solute milles of solute will be moles of solute multiplied with nothing else nothing else people right just to confuse you nothing else I believe this is clear now if this is clear then it is a high time to discuss the marity of dilution understand what exactly I'm going to do and try to remember the concept do not mug up the formula here try to remember the concept for example this is the container which I have right imagine in this container I've got a solution I've got a solution that means in this particular container we have got solute as well as solvent we have got solute as well as solvent in this container we have got a solution in the container that means there is solute as well as Sol right imagine that imagine that imagine that imagine that marity of this particular solution which is there in the container marity of the solution which is there in the container is for example M1 imagine volume of the same solution in the container is for example is for example B1 L let's assume that I've taken the solution whose marity is M1 volume of the same solution is V1 L okay my dear students if I ask you how many moles of solute are there in the solution how many moles of solute are there in the solution few minutes back I told you marity multiplied by volume in liters that gives you the moles of solute so if I ask you how many moles of solute are there in the solution right now marity of the solution is M1 multiply it with volume of solution in liters so this multiplied by this it's going to give me the number of moles of solute in the solution agreed agreed now people imagine I doing I'm adding imagine I'm adding extra solvent in the same container imagine I'm adding extra solvent in the same container the process of addition of extra solvent into the solution the process of addition of extra solvent into the solution is something which you call as dilution so basically what am I doing I'm adding extra solvent in the solution that means I'm doing the dilution that means I'm doing the dilution now tell me when I will be adding extra solvent into the solution tell me will the volume of solution remain same is it going to be V1 no it will change it will increase let's say the final volume of the solution is V2 L let's assume after the dilution the final volume of the solution is V2 L now you know marity is inversely proportional to volume so if I'm adding extra solvent in the solution due to which volume of the solution is increasing and if volume of the solution is increasing that means marity of the solution would have decreased right if the marity of the solution initially was M1 let's say finally the marity of the solution would be M2 now if I want you guys to compare between M1 and M2 can I say can I say M2 will be less than M1 because marity is inversely propal to volume we are adding extra solvent due to which volume of solution is increasing if volume of solution is increasing marity of the solution will be decreasing right yes now now now tell me one thing tell me one thing after the dilution is done what is the new marity of the solution M2 what is the new volume of solution V2 If I multiply these two terms If I multiply these two terms what do I get what do I get If I multiply these two terms can I say I again got the moles of solute present in the solution can I say this was the case of before dilution and this is the case of after dilution this was the case these were the moles of solute present in the container before dilution and these are the moles of solute present in the container after dilution yes if you look properly what did we do what did did we do exactly we added extra solvent we added extra solvent did we do anything with salute did I add extra salute no did I took did I take salute out no I did nothing with salute I just added what I just added the solvent I did nothing with the salute I did nothing with the salute I just added extra solvent agreed agreed so can I use one simple statement can I say number of moles of solute before dilution has to be equal number of moles of solute after after dilution because I did not do anything with the solute whatever solute was there in the container before same amount of solute will be in the container afterwards as well because I just added solvent I did nothing with solute right so people moles of solute before dilution moles of solute before dilution is equal m M1 V1 moles of solute up dilution is equal M2 V2 right this is one equation which I got over here and you call this particular formula as the dilution formula you call this particular formula as dilution formula right now now now if I ask you one simple thing if I ask you one simple thing tell me the volume of what is the volume of extra solvent added what is the volume of extra solvent which we added how do we calculate that how do we calculate the volume of extra solvent added how do we calculate that see initially the volume of solution was V1 then you added some extra solvent finally the volume of the solution is V2 so final volume of the solution minus initial volume of the solution that is always going to give me what volume of extra solvent added that's always going to give me the volume of extra solvent added right V2 minus V1 correct if this concept is clear to you you can easily kill this question you can easily kill this question easily kill this question how much water how much water should be added to 200 mL of two M solution so basically we have got a solution whose marity whose initial marity is 2 whose initial volume is 200 mL right we have got a solution whose initial marity is 2 whose initial volume is 200 mL now in this solution what we are doing we are adding extra solvent and the solvent which we are adding here that is water so we are adding extra water now due to the addition of extra water what's happening what's happening final marity of the solution is becoming 0.2 right let's assume the final volume of the solution is to can I say this is the simplest case of dilution this is the simplest case of dilution in which I shall be using the dilution formula M1 V1 has to be equal M2 V2 so what is M1 2 what is V1 200 what is M2 0.2 from here you can calculate V2 and when you solve this V2 it will come out to be 2,000 ml but this is the final volume of the solution was I supposed to calculate final volume of the solution look at the question I was not supposed to calculate final volume of the solution I was supposed to calculate the extra solvent which I had added so I'll say volume of the extra solvent which I have added is equal to vs2 minus V1 V2 we got to know as 2,000 and V1 is 200 the value comes out to be 1,800 ml so what is meant by this 1800 ml so can I say I had to add I had to add 1,800 mL of extra water then only the marity of the solution changed from 2 to 0.2 is this clear to everyone quickly my dear students absolutely all these are your pyqs they have been asked in different competitive examination all of these all of these let me know once in the chats with the fire emojis if every single thing is clear to them quickly quickly people quickly quickly quickly everyone everyone everyone you know guys do I need to repeat it all the time everyone means everyone do I need to repeat it everyone means everyone and what am I seeing all the students have not liked the session yet what does it what does it mean should I end the session yeah you're forcing me to end the session so smash that like right now yeah smash that like button right now right now means right now right now means right now right now means right now everyone means everyone ion will be done too just wait for it let's complete this chapter first guys everything will be done in detail relax relax right there were a lot of students who came in the beginning sir you will stop this series after 10 lectures and all right there were a lot of students who came up with this thought yeah so I would I would I would tell those students to just have a chill pill and watch the sessions those are all the paid users by the way a users means you know right what I'm talking about first we have to finish all these concentration terms right then I'll be giving you a break first let's finish it off okay marity of mixing marity of mixing marity of mixing marity of mixing see I'm going to break this marity of mixing into two cases case one which is the simplest case case one imagine these are the two containers which I have imagine in this container we have got HCL Aqua Solution imagine in this container again we have got HCL Aqua Solution so how many solutions do we have we have two solutions we have two solutions imagine marity of this solution is M1 volume of this solution is for example V1 L imagine marity of this solution is M2 volume of this solution is V2 L for example now my dear students if you mix them if you mix them in the final container this is the final container this is the container in which you're mixing these two solutions now this is HCl Aqua Solution this HCL Aqua Solution are these reacting Solutions do they react this HCL this HCL have you seen HCL HCL reacting no so these are non-reacting solutions so when I mix them what do I get I get a result in solution I get a final solution this is my resulting solution this is my resulting HCL aquous solution this is my resulting HCL aquous solution now people what is the final volume of this resulting aquous solution which I'm representing with VR volume of the resulting solution volume of the final solution that will be V1 plus V2 if the resulting solution was ideal if the resulting solution was ideal assume that marity of the resulting solution is Mr Mar it of the resulting final solution is Mr now people what concept should I use here can you tell me before mixing before mixing how many moles of solute were here in the container it was M1 multiplied by V1 right these were the moles of solute in this container how many moles of solute were in this container it was M2 multiplied V2 in lit if I add them if I add them should I be calling this particular term as the total moles of solute total moles of solute present in both these containers before mixing yes these are my total moles of solute present in both the containers before mixing now after mixing is done what is the marity of the final solution Mr what is its volume VR so if I multiply Mr with VR what should I be calling this particular term marity multiplied volume in liters what what do that that give us this is something which will give us the number of moles of salute in the final container is it before mixing or after mixing this is after mixing this is after mixing so I'll say before mixing some salute was here in this container some solute was here in this container when you mix them up all the solute which was present in both these containers all that salute came here in this container so tell me the concept what concept should I be using I should be saying number of moles of solute total moles of solute before mixing has to be equal total moles of solute after mixing correct as simple as that as simple as that as simple as that tell be moles of solute before mixing that is M1 V1 plus m2v2 is equal this is going to be Mr multip VR VR is nothing that V1 + V2 now from this particular equation you can easily calculate Mr marity of the final resulting solution which is M1 V1 + M2 V2 / V1 + vs2 if the resulting solution is ideal this is from this equation you get one type of question from this equation you get one type of question this was my case one now what about my case two what about my case two what about my case two people look at this case look at this case and after this case we'll be solving certain questions over here I'm taking three containers not two I'm taking three containers these are the three containers which I have imagine in this container we have got HCL Aqua Solution imagine in this container we have got HCL Aqua Solution imagine in this container there is only water pure water nothing else pure water nothing else okay marity of this solution is M1 volume of this solution is V1 this is M2 this is vs2 volume of this water is foral V volume of this water is foral V now in the chats tell me how many solutions do we have remember carefully what I'm saying listen listen to me very carefully I'm asking you how many solutions do we have listen carefully how many solutions do we have three or two this is a solution this is a solution but this is a pure solvent we have got two solutions we have got two solutions and in the third container we have got a pure solvent right now people now people tell me one thing did I mix them yet I did not mix them yet I did not mix them yet so this is the case of before mixing this is the case of before mixing tell me if I ask you how many moles of salute were there in this container before mixing you will say M1 V1 M1 V1 moles of solute present in the first container before mixing in the second container moles of solute before mixing is m2v2 in this container there is no solute there's no solute right so if I add them up what this term is going to give me it's going to give me number of moles of solute it's going to give me total moles of solute present with us before mixing before mixing right before mixing now people if I mix all these three containers in the final container this is my final container I will be getting a resulting solution I'll be getting a final solution of HCL assume that assume that the volume of the final solution when you mix them up volume of the final solution is equal to VR now this VR has to be equal V1 plus V2 plus v V1 + vs2 plus v imagine imagine marity of this final solution is Mr Right Now people if I just multiply these two terms what do I get if I multiply Mr with VR If I multiply Mr with VR If I multiply Mr with VR I got the total moles of solute present in this particular container is it before mixing or after mixing this is after mixing right this is after mixing this is after mixing now tell me the concept before mixing there was some salute ha there was some solute here right and now you're mixing them up can I say all the solute which was there in these two containers it finally reached in this container right yeah so what is the concept which I'll be using total moles of solute which we had before mixing has to be equal total moles of solute after mixing right so I'll be using this particular statement here total moles of solute before mixing has to be equal total moles of solute after mixing perfect now tell me how many moles of solute are there before mixing it is M1 V1 + M2 V2 it has to be equal total moles of salute after mixing look at that it is M multiplied by what multip VR and you know VR is nothing that is V1 plus V2 plus v what was v v was the volume of water added my dear students if you got to know how to make this particular equation you're sorted if you got to know how to make this particular equation you sorted because from this equation what they'll ask you they'll ask you in the exam to calculate the volume of water added how much volume of water we have added you'll be getting equ question question from this particular equation okay you'll be getting a equation from this particular equation now now now I believe case 1 case 2 is over let's see what kind of questions have been asked from these two cases one simplest among all kill it kill it quickly kill it quickly 200 mL of 3 m HCL Aquis is mixed with 300 mL of 2 m HCL AAS and the final volume is made up of made up to 1500 ml calculate the marity of the resulting solution is it even equation no we have got the first solution whose marity is how much whose marity is is three volume of the first solution is 200 mL marity of the second solution is equal to 2 volume of the second HCL solution is equal to 300 ml now what we are doing we are mixing them up and what we'll be getting we'll be getting a resulting solution volume of that resulting solution is given as 1500 ml marity of that resulting solution needs to be calculated which equation case one or case two case one or case two case one or case two I'll be using case two these are non-reacting Mr R will be equal M1 V1 + M2 V2 divided by V final V final is how much V final is how much, 1500 ml right right people right so use all the parameters here and kill this question nothing else you have to do nothing else you have to do 2.4 will be the answer yes absolutely right wonderful good job done good job done good job done okay if this is clear just read the question and tell me only one thing read the question and tell me only one thing is it based on case one or case two is it based on case one or case two quickly till then I can have some water is it based on case one or case two you need to tell me is it based on case one or case two yeah Perfecto guys wonderful Sun ising breakfast yes it's morning right now yes yes you need breakfast see guys look at this particular question carefully as for this question we have got three containers 1 2 and three okay in the first container you have got HCL Aqua Solution in the second container you have got HCL Aqua Solution in the third container you only have water you only have water marity of the first solution which is M1 is equal 0.6 and its volume is 250 so 0.6 its volume is 250 ml m ity of the second solution is nothing but 0.2 and its volume is how much its volume is 750 ml imagine in the third container you have got some V mL of water imagine you have got some V mL of water here now what you are doing you mixing them when you mix them what do we get we get a resulting solution Final Solution you get a resulting solution a final solution right perfect marity of this resulting solution is given to me as for the question 0. 25 and volume of this resulting solution will be V1 + V2 + V so 750 + 250 plus v which comes out be 1,000 + V right 1,000 + V now you know it is case to correct so make the equation quickly the equation has to be M1 V1 + M2 V2 it has to be equal Mr multiplied VR which is V1 + V2 plus v that comes out to be directly how much 1,000 plus volume of water which was there in third container right now this is something which we know 0.6 * 250 plus M2 0.2 * V2 750 is equal Mr 0.25 multiplied by what, plus volume of water added one equation one unknown from here you can calculate the volume of water which was added how much it's coming after solving quickly quickly quickly everyone everyone you have to be very quick so 200 mL so 200 mL of water was added while mixing that's something which we were supposed to calculate I believe it is done and dusted solve this question as well quickly quickly quickly quickly my dear students everyone 29.2% weight by weight weight by weight solution of HCL wow whose density is given whose density is given occupies 8 ml volume calculate the final volume of the solution whose concentration becomes 0.4 M on dilution look at the question and see what is mentioned in the question and what we are supposed to calculate break after 15 to 20 minutes be there do not go anywhere be there look at this particular question we have got HCL aquous solution correct density of of the solution is given okay weight by weight percentage of solute is given right perfect volume of this solution is given now what exactly we are doing the solution which we had taken we are doing its dilution we are adding extra solvent we are adding extra solvent when we are adding extra solvent what do we get we get the final HCL Aqua Solution right marity of this final HCL Aqua Solution that is M2 is 0.4 is 0.4 and let's assume its final volume is V2 and that is something which we have to calculate so tell me just one thing is it the case of dilution or this is the case of mixing this is the case of dilution now I should be using the dilution formula which is M1 V1 has to be equal M2 V2 now M1 do we know we do not know the initial marity of the solution so my first point is to calculate the initial marity of the solution initial mity of the solution is equal to 10 multiplied by density of solution multiplied by weight by weight percentage of solute divided by m mass of solute right so M1 will be equal 10 multipli by density of solute is 1.25 weight by weight of solute is 29.2 and molar mass of solute molar mass of HCL 36.5 when you solve it I believe it comes out be 10 I believe it comes out be 10 right so M1 is 10 V1 is 8 M2 is 0.4 V2 you are supposed to calculate you are sorted so 80 / 0.4 the value comes out be 200 m that is going to be the final volume of the solution if you got it let me know in the chats with some fire emojis that to quickly people that too quickly quickly quickly the Jo has to be high the J has to be high we have to complete this chapter right in detail all the theory all the problem patterns we have to do correct from this question two to three questions from this chapter two to three questions you'll be getting two to three question so 12 marks you'll be securing by this chapter and it is very easy chapter trust me on that trust me on that it's a it is very easy chapter when it comes to neat and it is the most difficult chapter when it comes when it comes to J Advanced when it comes to J Advanced it is the most difficult chapter when it comes to neat it the easiest chapter yeah J Main's neat level is now same all right guys now now there is one one more type of the question there is one more type of the question mixing of non- Ideal Solutions mixing of non- Ideal Solutions so first of all this particular term the nonideal solutions we have to discuss in detail afterwards but right now I just need its one point afterwards I'll be discussing non ideal Solutions in detail but right now I just need its one point what is that one point have a look imagine imagine I got two containers okay this container contains liquid one this container contains liquid 2 now you are mixing these liquids you are mixing these liquids in the final container this is the final container so what did I get I got a resulting solution I got a resulting solution over here okay I got a resulting solution over here if the volume of the first liquid is V1 volume of the second liquid is V2 volume of the resulting solution which I got over here it is VR it is VR now remember one thing remember one thing remember one thing if VR comes out to be equal V1 + V2 if VR comes out to be equal to V1 + V2 then at that point of time I shall be calling this resulting solution as the ideal solution as the ideal solution number one number two if VR is not equal to V1 + V2 I shall be calling this resulting solution as the then it's going to be my nonideal solution just one important point I used I did not discuss ideal not non ideal with you yet something which we have to discuss in some time I just need one point that's it if VR is equal to V1 plus V2 ideal if VR is not equal to V1 plus V2 non ideal that's all that's all okay now people now remember one more thing no matter no matter whether the solution no matter whether the resulting solution is ideal or non ideal mass of resulting solution will be always equal W1 + W2 mass of resulting solution will be always equal W1 + W2 if the resulting solution is ideal or non ideal nothing to do with that mass of resulting solution will be always equal mass of liquid 1 plus mass of liquid 2 this is the concept which I'll be using here this is the concept which I'll be using here tell me one thing tell me one thing if I analyze this particular Point properly tell me if VR if volume of resulting solution comes out to be greater than V1 + V2 if volume of resulting solution comes out to be greater than V1 plus V2 tell me what would have happened at that point of time can I say expansion has taken place volume is increasing I was expecting when I was mixing these two liquids I was expecting the volume of res solution to be V1 plus V2 but it's coming out to be greater than that mean expansion has taken place right and people if VR comes out to be less than V1 plus V2 should I be calling this as expansion no be calling this as contraction I'll be calling this as contraction a particular type of question is asked wherein they will ask you how much was the volume contraction how much was the volume expansion when you mixed to liquids when you mix two liquids how much was the volume contraction how much was the volume expansion how to deal with that particular question look here how to deal with that particular question look at the question carefully look at the question carefully look at the question carefully people on mixing 15 mL of ethy alcohol with 15 mL of pure water at 4° Cen the resulting non-ideal solution is found to have density of 0.924 G per ml calculate the contraction in volume calculate the contraction in volume understand what it says so you have got a container you have got a container this is your first container this is your second container and this is your third container try to understand what exactly I'm going to say in the first container you have got ethy alcohol in the second container you have got pure water at 4° Cen okay right in the first container we have got 15 mL of ethy alcohol 100l of ethy alcohol right density of ethy alcohol is given to us as 0.792 G per ml correct now this water this water it is kept at 4° Cen it is kept at 4° Cen and volume of this water is also given 15 mL so volume of the water in the second container is 15 ml now the water is present at 4° Cen density of water at 4° Cen you must be knowing it is 1 G per ml correct now people the two liquids which we have we are mixing them and we are getting a final resulting solution we are getting a final resulting solution now now now let's assume that let's assume that the final volume of this resulting solution is VR density of the resulting solution is is given density of the resulting solution is given 0.924 G per ml I have to check the contraction in volume first of all understand how and when volume would have contracted I'll say contraction in volume is only possible if VR will be less than V1 plus V2 if VR will be less than V1 plus V2 then only I can say contraction has happened contraction of volume has happened they are asking me how much is this volume contraction how much is this volume contraction how to deal with this particular question look here what is D1 density of this liquid density is equal to mass by volume so instead of D1 can I write W1 by V1 that is 0.792 do we know the V1 that is 15 15 multiplied by 0 something can you let me know the exact value in the chats quickly 0.792 multiplied 15 can you let me know the exact value which you'll be getting from here quickly quickly everyone everyone I think it'll be 10 or 11 something like that but I want the exact value from your side quickly quickly quickly solve it yeah quickly be first be fast how much this value is coming out to be what is saying 10.5 is it coming out to be 10 10.5 are you sure yeah someone is saying 19.5 wow it comes out to be 11.8 11.88 G I calculated the mass of first liquid now D2 what is D2 density of this water which is mass of water divide by volume of water which is given to me as 1 G per M so can I say W2 is equal 1 MTI V2 1 multi V2 it will be 15 G correct correct people I calculate the mass of first liquid I calculate the mass of second liquid tell me tell me tell me no matter if the resulting solution is ideal or non- ideal no matter if the resulting solution is ideal or nonideal mass of resulting solution as I told you it's always equal to W1 + W2 now what is W1 11.88 what is W2 15 the value comes out be 26.8 G this is the mass of the resulting solution right now people use this parameter density of resulting solution means mass of resulting solution divide volume of resulting solution which is 0.94 right perfect so volume of resulting solution will be equal um mass of resulting solution that is 2688 divided by 0.924 and this value when you solve it this value will come out to be 29 ml so what is this 29 ml what is this 29 ml I use the density parameter mass of resulting solution divide volume of resulting solution right mass of resulting solution we know correct so I'll be getting volume of resulting solution 29 ml volume of resulting solution should have been 15 + 15 30 ml but it's only 29 ml that means volume contraction is happening so how much is the contraction in volume how much is the contraction in volume which I'm representing with Delta V final volume minus initial volume final volume is 29 initial volume is 15 + 15 30 so contraction in volume is 1 ml this is something which we were supposed to calculate so volume contraction of 1 ml has happened now you must be thinking why this contraction in volume happened that is something for which you have to wait for one or two hours yeah if you want to know that you have to wait for it for like 1 to two hours I believe all the things still here are absolutely clear so our concentration terms are done and dusted now comes the liquid solution now comes the topic vapor pressure but I believe a lot of students are asking about the break so let's take a quick break I'm not giving you a long break right now long break the dinner break will be taking afterwards one short break of 15 minutes I'm giving you but you have to promise that you'll be back you have to promise that you'll be back the time is right now 1922 so 722 it'll be 732 737 7 okay right guys be back now I'm going to start something amazing and you are going to understand all these things the way you have never studied trust me on that but be back be back okay be back on time is everyone back uh just give me a second people yeah back with full energy is is that is everybody back is everybody back so people whatever we have discussed till now let me know in the chats if every single thing is clear let me know in the chats if every single thing is clear till here let me know in the chats if every single thing is clear till here quickly quickly people are joining in let's wait for one more minute let's wait for one more minute so till then you can let me know if every single thing is clear uh Vijay is asking sir is Avengers batch completed physics is completed chemistry and bio are getting complete on 10th of January in Avengers badge 10th of January their cabus is getting completed and then they are starting the revision PDF will be available on my telegram no no yakob no I'm not so should we start I want you to light up the chats now now again I want you to light up the chats again should we get going now next chapter I'll let you know next chapter I'll let you know perfect guys so let's start with a called as vapor pressure this is something important it forms the bases of the liquid solutions this particular topic which I'm going to teach you now okay vapor pressure so first of all how do we exactly Define the term vapor pressure guys be careful with what I say be careful with what I say take a note of every single thing now vapor pressure is defined as the pressure exerted the pressure exerted by the vapors the pressure exerted by the vapors of a volatile liquid when when the liquid and its Vapors when the liquid and its Vapors are in equilibrium with each other when the liquid and its Vapors are in equilibrium at a particular temperature at a particular temperature now you must be thinking what is meant by this try to understand people carefully try to understand imagine I'm taking a container over here this is the container for example in this container what exactly I'm going to keep I'm going to keep a volatile liquid till here I have kept a volatile liquid in this particular container and I believe this is the top surface of the volatile liquid so over here I have taken a volatile liquid in this particular container right imagine this particular liquid is it it is present at some constant temperature right temperature over here is kept constant perfect now first of all you must be thinking what is meant by volatile liquid let me tell you volatile liquid is the one which has got tendency to form Vapors that liquid which has got the tendency to get converted into its Vapors that is what you call as volatile liquid any liquid any liquid which can form its Vapors which can form its Vapors is what you call as is what you call as volatile liquid is what you call as volatile liquid for example you have got water water can get converted into its Vapor so water is the volatile liquid so basically I have taken a volatile liquid over here now guys as soon as I kept the volatile liquid in this container what will happen some of the water molecules will start getting converted into its Vapors some of the some of the volatile liquid for example which was water in the container once I introduced water in this container some of the water molecules they escaped they got converted into its vapers right so that is my first step once I introduced water in the container what happened I'll say evaporation started I'll say evaporation started after some time you'll observe one more phenomenon after some time you'll observe one more phenomenon you'll observe some of the vapors will start getting converted into liquid right so initially evaporation started after some time condensation started initially some of the liquid molecules started getting converted into Vapors after some time the vapor molecules they started getting convert into liquid as well perfect perfect now people try to understand initially evaporation started then condensation started there will be a scenario when when one of the liquid molecule gets converted into vapor at the same time one Vapor molecule gets converted into liquid so basically there will be a time when equilibrium will be established between the liquid and its Vapors there will be a time when equilibrium will be established between the liquid and its Vapors for example I had the Liquid X initially what happened it started getting converted into its Vapors right so what happened in the beginning evaporation started after some time these Vapors again started getting converted into liquid right so after some time I'll say condensation started right then a time will reach when rate of evaporation and condensation becomes equal at that point of time I would say equilibrium equilibrium is established between the liquid and its Vapors now people try to understand if equilibrium is established between liquid and its Vapors what does that mean that means if one liquid molecule is getting converted into vapor at the same time one vapor is getting converted into liquid so can I say once equilibrium is established can I say the amount of vapors generated over here they become fixed once equilibrium is established I'll say amount of vapors generated ha will become fixed because if one liquid molecule is getting converted into vapor at the same time one Vapor molecule is getting converted into liquid right so at equilibrium at equilibrium I must say at equilibrium I must say amount of vapors generated are fixed amount of vapors generated are fixed now people since equilibrium is established now we have definite amount of vapors over here right now tell me will these Vapors be in rest or they'll be in motion they'll be in motion can I say these Vapors will be colliding with each other yes they'll be colliding with the walls of container as well so can I say these Vapors are exerting pressure these V vapers are exerting pressure because they're colliding with each other at the same time they are colliding with the walls of the container they are colliding with the topmost surface of the liquid as well right so can you see these Vapors are exerting pressure over here they're generating some pressure that pressure which is exerted by the vapors of the volatile liquid when liquid and Vapors are in equilibrium with each other that pressure is something which you call as vapor pressure let me know once if you got to know what this wapor pressure is first of all in order to define the vapor pressure your liquid and its Vapors they must be in equilibrium they must be in equilibrium right the pressure exerted by the vapors the pressure exerted by the vapors of the volatile liquid when the volatile liquid and its Vapors are at equilibrium with each other at a particular temperature that pressure exerted is something which you call as vapor pressure I hope you got the meaning of the term vapor pressure right now people what are the factors on which this vapor pressure depends on what are the factors before talking about the factors on which vapor pressure depends let me tell you the factors on which vapor pressure does not depend let me tell you vapor pressure do remember it does not depend on the amount of liquid it does not depend on the amount of liquid it does not depend on the amount of liquid for example this was water in the container imagine there were 500 mL of water in the container now we have got one more container which contains only 100 mL of water if both the containers are kept at same temperature their vapor pressure will be equal so vapor pressure it does not depend on the amount of liquid vapor pressure it does not depend on the shape of the container in which liquid is kept do remember these two points vapor pressure first of all it does not depend on the amount of liquid number one it does not depend on the shape of the container in which liquid is kept okay now what are the factors then on which this vapor pressure depends let's have a look on them factors affecting vapor pressure it's pretty much simple the first Factor on which vapor pressure depends upon is something which you call as force of inter interaction force of interaction between the liquid molecules between the liquid molecules how we are going to justify it vapor pressure it depends on the force of interaction between the liquid molecules now it is it can be easily understood see below this layer we have got liquid on the top of this layer we have got Vapors right now if I talk about this particular bulk if I talk about this particular liquid can I say there will be liquid liquid molecule interactions here right let's say we have got one liquid molecule here one liquid molecule they'll be interacting with each other right there'll be force of attraction between the liquid molecules absolutely right right people see this is the liquid in the container if I take two molecules from the bulk for example from the bulk they'll be attracting each other basically there'll be force of interaction between them which will hold them together now people can I say more the attraction between the liquid molecules more the attraction between the liquid molecules can I say lesser will be the escaping tendency more the attraction between the liquid molecules more the attraction between the liquid molecules lesser will be escaping tendency if lesser is the escaping tendency less Vapors will get generated if less Vapors are generating that means less will be the vapor pressure right that's all that's all so do remember more the force of interaction between the liquid molecules lesser is going to be the escaping tendency lesser is going to be the escaping tendency and and eventually lesser is going to be the vapor pressure lesser is going to be the vapor pressure point number one Point number one point number two your vapor pressure it depends on temperature as well it depends on temperature as well these are the two main factors on which your vapor pressure depends now you must be thinking how VOR pressure depends on temperature this is something which you can easily understand tell me one thing increase the temperature when you increase the temperature of this liquid can I say there'll be more evaporation more evaporation means more Vapors are getting generated if more Vapors are getting generated more will the vapor pressure right so so on increasing the temperature on increasing the temperature what happens to the vapor pressure of the liquid vapor pressure it increases so can I say vapor pressure is directly proportional to temperature when you increase the temperature vapor pressure increases right and can I say Vaper pressure is inversely proportional to the force of interaction between the liquid molecules more the interaction between the liquid molecules lesser than V pressure now people there is one equation now people there is one equation which quantitatively there is one equation which quantitatively gives you the idea how much the vapor pressure of liquid increases upon increasing the temperature there is one equation there is one equation which quantitatively gives you the idea which quantitatively gives you the idea of how much the vapor pressure of liquid increases upon increase the temperature that equation is something which you call as that equation is what you call as clausus claperon equation I'm not going to derive this equation because that's of no use correct I will be giving you the equation and remember that and remember that so the equation is something like this log of P2 / P1 is equal Delta H vaporization ided R sorry divide by 2.303 ride by 2.303 r this is 1 upon T1 - 1 upon T2 now what these terms are what these terms are have a look have a look imagine I got a liquid in the container and that liquid that liquid is in equilibrium with its Vapors that liquid is in equilibrium with its Vapors correct that liquid is in equilibrium with its Vapors imagine at temperature T1 the vapor pressure was P1 now I'm changing the temperature from T1 to T2 and I'm assuming vapor pressure changed from P1 to P2 I'm assuming I'm changing the temperature from P1 to T2 and assuming that vapor pressure changed from P1 to P2 imagine people you have increased the temperature imagine you have increased the temperature if you have increased the temperature of this liquid if you have increased the temperature of this liquid tell me what will be the value of T2 minus T1 imagine you have the the liquid which was there in the container right the liquid which was there in the container imagine you are increasing its temperature if you are increasing its temperature so final temperature minus initial temperature will come out to be positive so this term will come out to be positive if you see it is T2 minus T1 so T2 minus T1 I'll say this term this term in the brackets will come out to be positive enthalpy of vaporization enthalpy of vaporization right it is vaporization is endothermic so it's always positive only so this term is positive this term is positive so can I say log of P2 by P1 will be positive can I say log of P2 / P1 will be positive when is that possible when is log X positive log of x is positive only if x is greater than one so I would say p2/ P1 here is greater than one which is clearly telling you that P2 is greater than P1 it is clearly telling you that P2 is greater than P1 what was P2 P2 was the vapor pressure of the liquid at temperature T2 P1 was the vapor pressure of the liquid at temperature T1 since you had increased the temperature you had increased the temperature and upon increase the temperature what happened to the vapor pressure VAP pressure increased that's it that's it I believe it's clear yeah is it clear my dear students quickly quickly quickly quickly everyone everyone so do remember when you in increase the temperature vapor pressure increases and this equation exactly lets you know how the vapor pressure increases upon increasing the temperature perfectly done now from this equation can they ask any question from this equation can they ask any question they might they might who knows they might who knows but before solving that question before solving that question you should know what boiling point is what is a balling Point how do you define the term balling Point how do you define the term balling point my dear students it's pretty much simple if I talk about the boing Point understand it is defined as the temperature the temperature at which the temperature at which vapor pressure of the liquid becomes equal the temperature at which vapor pressure of the liquid becomes equal to the external pressure to the external pressure which is generally your atmospheric pressure what is meant by this which is generally our atmospheric pressure what is meant by this let's get to know this let's get to know those people try to understand what exactly I'm going to say try to understand what exactly I'm going to say because these are some basic points which you should know then only this chapter can get clear to you otherwise not otherwise not okay so this is your container for example this is your container in this container imagine you have kept a volatile liquid the volatile liquid is present till this surface this is the top surface of the volatile liquid imagine this is the top surface of the volatile liquid imagine this liquid right now is at temperature 25° C just to make you understand this liquid for example this liquid is water imagine this volatile liquid is your water right so this liquid is right now at 25° Centigrade so can you say at 25° Centigrade this liquid will be in equilibrium with its Vapors absolutely this liquid will be in equilibrium with its Vapors this liquid will be in equilibrium with this Vapors at a particular temperature imagine the vapor pressure of the liquid at 25° Centigrade imagine the vapor pressure of the liquid at 25° Centigrade for example that is 0.1 ATM this is the vapor pressure of the liquid at 25° Centigrade for example right outside this container what do we have atmosphere so can you say outside the external pressure which we have that is basically the atmospheric pressure imagine that is 1 atm outside the atmospheric pressure is 1 atm now tell me one thing at 25° Centigrade vapor pressure of the liquid is less than that of atmospheric pressure yes now people tell me one thing if you want to increase the vapor pressure of this particular liquid what do we have to do if you want to increase the vapor pressure of this particular liquid can I say I have to supply heat I have to raise the temperature I have to raise the temperature in order to Inc increase the vapor pressure of the liquid if I increase the temperature if I keep on increasing the temperature I'll say vapor pressure of liquid will be increasing can I say there will be a particular temperature at which vapor pressure of the liquid becomes equal to external pressure can I say upon increasing the temperature continuously vapor pressure of the liquid will keep on increasing it will keep on increasing it will keep on increasing at some particular temperature vapor pressure of liquid becomes equal to ATM Eric pressure and that particular temperature that particular temperature at which vapor pressure of the liquid becomes equal to the external pressure becomes equal to the atmospheric pressure that particular temperature is what you call as boiling point right and you know in case of water that boiling point is how much that boiling point is 100° Cen normally right so initially I kept the water at 25° Cen at that point of time it's vapor pressure was 0.1 ATM now I slowly increased the temperature of water and when the temperature of water reached 100° when the temperature of water reached 100° at that point of time the vapor pressure of water became equal to the atmospheric pressure and that particular temperature at which vapor pressure becomes equal to your external pressure vapor pressure becomes equal to your external pressure that particular temperature is what you call us that's what you call us balling Point yes now guys try to understand one more thing I hope this particular scenario is completely clear I hope this particular scenario is completely clear just a second I hope this particular scenario is completely clear if you want to show it graphically if you want to show it graphically how do you show it graphically see imagine I'm I'm plotting a graph between vapor pressure versus temperature now you know when you increase the temperature of the liquid its vapor pressure increases when you increase the temperature of the liquid its vapor pressure it will increase when you increase the temperature of the liquid when you increase the temperature of the liquid vapor pressure of the liquid increases right at this point vapor pressure is zero if you go up vapor pressure is increasing increasing increasing let's say at this particular Point your vapor pressure is 18 at this particular point for example the vapor pressure is 1 atm right which is basically the atmospheric pressure which is basically the atmospheric pressure my dear students the temperature the temperature the temperature the temperature at which vapor pressure of the liquid becomes equal to the atmospheric pressure the temperature at which vapor pressure of the liquid becomes equal to its atmospheric pressure that is something which which you call as balling point so this particular point will be representing your balling Point yes and this was the curve for what for a volatile liquid I believe this particular curve is clear if this is clear let me know once in the chats quickly if this is clear let me know once in the chats quickly quickly my dear students everyone and you know everyone means everyone yeah right people shape of the graph is curved it's cured it's curved it's not a straight line it's cured okay now guys there are two things which you have to remember directly one is called as standard balling point one is called as normal balling point when we talk about standard balling point when we Define standard balling point at that point of time external pressure is kept as one bar when we Define normal boiling point external pressure is kept as 1 atm that is the difference that's the difference if external pressure is one bar that means we are talking about standard balling point if external pressure is 1 atm we are talking about we are talking about the normal balling point we are talking about the normal balling point I hope the term balling point is clear okay now guys I just want to see your understanding of the concept I just want to see the understanding of your concept tell me the answer of this question on increasing on increasing the external pressure on increasing the external pressure on increasing the external pressure boiling point of the liquid boiling point of the volatile liquid does it increase or decrease what do you think use your brain use your brain when I increase external pressure when I increase the external pressure will the boiling point of liquid increase or decrease quickly use your brain hair use your brain hair it's pretty much simple guys here the external pressure was 18 18m now in the second case for example external pressure is 5 ATM in the second case for example exteral pressure is 5 ATM so in the first case you are increasing the vapor pressure from 0.1 to 1 but here in the second case you have to increase the vapor pressure from 0.1 to 5 so here you have to increase the temperature by larger amount as compared to first so what does that mean that means vapor pressure increases sorry that means the balling Point increases right more the external pressure more the boiling point more the external pressure more should be the raise in temperature such that vapor pressure of liquid becomes equal to external pressure and if more is the rise in temperature that means more is the boiling point as simple as that yeah let me know once in the charts if this particular point is clear let me know once in the charts if this particular point is clear on increasing the external pressure the boiling point of the volatile liquid that increases yes or no in the chats yes or no in the chats yes or no in the chats these were two basic terminologies which you were supposed to know before starting the r law yes J students can also watch it absolutely now guys this is something very important from which short short question you guys are going to get short short question you guys are going to get okay RS law so what this RS law exactly is all about let me write its definition first then I'll make you understand what it means okay this is important all right if I talk about the RS law RS law says that the partial vapor pressure rods law says that the partial vapor pressure what am I writing here the partial vapor pressure of the component of a component over the over the liquid solution over the liquid solution is directly proportional to the mole fraction is directly proportional to the mole fraction mole fraction of the same component of the same component in liquid phase or you can say in the solution phase now what is meant by this particular statement try to understand very carefully this is my first container this is my second container in the first container till here I'm keeping a volatile liquid a this is a volatile liquid a in the first container and in the second container what exactly am I doing I'm keeping the volatile liquid B in the second container okay temperature is fixed temperature is fixed temper tempure is fixed now guys if this is a volatile liquid what does that mean what does that mean that means there will be the vapors of a and assume the vapors of liquid a are in equilibrium with the liquid imagine these Vapors are in equilibrium with the liquid imagine the vapors of B are also in equilibrium with the liquid right so we are in a position to Define their Vapor pressures let's assume the vapor pressure here is p a let's assume the vapor pressure here is PB now first of all what is p a and what is p b p a is the vapor pressure of liquid a when present in pure State when present in pure State you can see the first container in the first container we only have a there's nothing added to it right p b what is p b p b I'm calling as vapor pressure of B vapor pressure of B when present in pure State vapor pressure of B when present in pure State okay I've taken two volatile liquids which are in equilibrium with their Vapors so vapor pressure of a right here is p not a vapor pressure of B right here is PB okay now guys imagine the number number of moles of liquid a in the container the number of moles of liquid a in the container is na number of moles of liquid B in the container is NB I'm talking about the moles of liquid a in the container and moles of liquid B in the container right now people just understand if you mix these liquids if you mix these liquids if you mix these liquids what do you get what do you get if if you mix these liquids I'll be getting a solution of A+ B this is something which I'm calling as solution of A and B absolutely this is the solution of A and B this is the solution of A and B now a as well as B both were volatile so over here in the solution there are a particles as well as B particles right there are a particles as well as B particles so tell me one thing will head only be the vapors of a or there will be Vapors of b as well what do you think since both are volatile if you're mixing them in this particular space you'll find Vapors of a as well as Vapors of B right you'll find Vapors of a as well as Vapors of B so over here what is this particular thing these are vapors of a as well as B these are vapors of a as well as B okay now people the total VAP vapor pressure of the solution the total vapor pressure of the solution I'm representing by PS PS or PT choice is yours the total vapor pressure of the solution will the total vapor pressure of solution be due to Only The Vapors of a only the vapors of b or both will have contributed towards the total vapor pressure what do you think what do you think quickly I'll say both the vapors of a as well as B would have contributed towards the total vapor pressure as per Dalton's law Dalton's law the total vapor pressure will be will be equal the sum of contributions made by each the sum of contributions made by each this is as for which law this is not as for ral's law this is as for Dalton's law this is as for Dalton's law you should know this the total vapor pressure of the solution will be the sum of will be the sum of contributions made by each of the A and B Vapors and you call these contributions as the partial Vapor pressures this is something which I call as partial vapor pressure of a similarly this is something which I call as partial vapor pressure of B partial vapor pressure of B now what dton what what R slw exactly is all about now what R slw exactly is all about R SL say that the partial vapor pressure for example the partial vapor pressure of a r law says that the partial vapor pressure of a component for example the partial vapor pressure of a I hope you know what is the partial vapor pressure of a this is the partial vapor pressure of a rod says that the partial vapor pressure of a is directly proportional to mole fraction of a in the solution phase mole fraction of a in the solution phase more than the mole fraction of a in the solution phase more the mole fraction of a in the solution phase more will be the moles of a in the solution and if more moles of a are than the solution more Vapors of a would have got generated that means a would have contributed more towards total vapor pressure right right people right so the sky a what is the sky a this is this is the mole fraction of a in which phase in the solution phase in the liquid phase right correct again I'm telling you the same thing more the mole fraction of a in the liquid what does that mean that means more are the moles of a in the solution if more moles of a are there in the solution that means more Vapors of a would have got generated if more vapor of a would have got generated I'll say a would have contributed more towards the vapor pressure towards the total V pressure what does that mean that means more will be the partial V pressure of a more will be the partial V pressure of a so if I remove this proportionality if I remove this proportionality you'll get a constant and the constant here is p a multiplied what multiplied by K right so the partial vapor pressure the partial vapor pressure of a is equal vapor pressure of a in pure State multiplied by mole fraction of a in the solution phase in the similar way in the similar way people if I write PB partial VOR pressure of B partial VOR pressure of B that will be directly proportional to mole fraction of B in the solution phase and if you remove the proportionality sign you'll be getting p b multiplied by K B this is again one more equation which I got right and what is the sky B here skyb is something which I call as mole fraction of B in the solution phase mole fraction of B in the solution phase and I believe this p a p b already you know what is p a p a is the vapor pressure of a in which state in pure state in pure State and what is p b p b is the vapor pressure of B in the pure state right what is Pa Pa a is the partial vapor pressure it is the par partial vapor pressure of a right is the partial vapor pressure of a in the solution right what is PB PB is the partial VOR pressure of b or you call them as the contribution of A and B towards the total vapor pressure of the solution correct right people this is something which you call a switch law this is your RS law basically this is your simplest statement of the RS law which you have to remember now now now there are some other versions as well there are some other things which you need need to know about R FL as well what are those things the first thing which you need to know the first thing which you need to know if I ask you to plot a graph if I ask you to plot a graph between PA and Kai a what do you think will be the nature of the graph first one second one if I ask you to plot a graph between PB versus KY B what will be the nature of graph if I ask you what will be the nature of the graph between PA a versus KY B nature of the graph if I ask you what will be the nature of the graph between PB versus Sky a you should be in a position to make these graphs you should be in a position to make these graphs have a look you know as for your RS Law PA a is equal p a MTI K correct PA you are plotting along Y axis so this your y k a you are plotting along xaxis so this has to be m so Y is equal MX Y is equal MX yal MX straight line what about the slope of it it slope will be simply equal to p a right similarly we already know PB is nothing it is p not B multiplied by KB PB you are plotting along Y axis KYB you are plotting along x-axis this is your M so Y is equal MX again a straight line what about its slope its slope will be simply equal to p b right right now have a look here you know you know your PA you know it is p a multipli k what is Kai a Kai a is the mole fraction of a in the liquid phase in the solution phase mole fraction of a in the solution phase in the solution phase can you write it like this p a is equal p a understand understand k a is the mole fraction of a in the solution phase k b is the mole fraction of B in the solution phase can I say here can I say here K A Plus k b it has to be one K A Plus k b it has to be one if K A Plus K B has to be one so instead of K I can write 1us k b absolutely I can do that so PA a has to be equal p a minus p a k b correct PA a your plotting along y- axis i b you are plotting along x-axis so this has to be your M the sign is minus so this is C Y is = - mx + C Y is = - mx + C - mx + C slope slope is equal to minus p a intercept intercept is equal p a right intercept is equal p a same goes for the another curve right it'll be like this and what about the slope slope here will be guys I hope you got it I hope you got it similar minus P not B minus p b what about the intercept intercept will be PB right let me know once in the chats if it's clear quickly let me know in the chats if it's clear quickly let me know in the chats if it is clear okay if this is clear if this is clear then then let's move on to some special cases of the rod law special case of the RS law but before talking about this tell me once in the chats if all the things are clear and I want everyone to speak quickly Guys these basic things are very important if they are getting clear it will not take us more than 3 to 4 hours to cover the complete chapter trust me on that if these things are getting clear all right all right so if this is clear R slw for a solution containing volatile liquids the same scenario I mean the same scenario you know this was volatile liquid a volatile liquid B now you mix them up right let now you mix them up perfect this is a solution this is a solution containing A and B and these are the vapors of A and B here perfect the total vapor pressure of the solution is the sum of the contributions made by the vapors of A and B here correct right now guys understand what exactly I will have to do here let me keep this slide over here have a look we already know know total vapor pressure of the solution is equal to PA plus PB PA a plus PB now people PS is equal instead of p a i can write p a k instead of PB you know you can write P not b k b this is one of the equations which you have to remember from which you can get the questions first equation from which question is going to come right first equation from which question is going to come and simple question second second since we already know since we already know since we already know kai a plus k b is equal to 1 mole fraction of A and B in the solution phase is equal to one so what I'll do over here I'll write p a k a as such let me replace Sky B with what with 1us Sky a let me replace the sky B with 1us Sky a so I can write PS is equal p a k a plus this will be p b minus p b multiplied by K now people what we can do at the end this is K this is K take K common right but first I'll write p b let me write p b first after that I'm taking gu a common it'll be P A minus p b it'll be P A minus p b and what have I taken common I have taken k a common this is one more equation from which question is going to come one more equation from which question is going to come one more equation from which question is going to come what about the next equation from which the question is going to come let's have a look let's have a look since in this equation over here I replaced k b with 1us Kai a I could have done one more thing I could have replaced Kai a as well instead of k a i could have written 1us k b let's do that too so PS will be equal this is p not a instead of K I'm writing 1us KY B right and over here what do we have we have got p b multiplied by KB now what you need to do just make this equation in the proper format nothing else make this equation in the proper format nothing else so I'll write PS is equal p a I'm writing as such here I'm taking k b common so it will be p b minus p a multiplied by K B so this is one more equation for the RS law which we are supposed to remember well I would suggest you to remember this equation and these two eventually you can generate eventually you can generate them eventually you can generate them right eventually you can generate them now guys there is one very very very important graph which we need to understand I know majority of the students would have remembered that graph for now till now but you need to understand that graph that is very important okay so let me make that particular graph which is related to R law let me make the particular graph which is related to R law first of all before plotting the graph before plotting the graph I'm not showing you any questions yet questions will be done in some time right before showing you the graph let me make the scenario again this is a solution of A and B this is a solution of A and B and A and B both were volatile right so over here you have got Vapors of a as well as B Vapors of a as well as B correct right total vapor pressure of the solution was contribution made by each PA plus PB and your PA as per rods law is equal to p a k your PB is equal to what PB is equal to p b k B right where Kai A and K B these are the mole fractions of A and B in the liquid state and already there's one more thing Kai a plus k b already we know that is one these are the things which we already have discussed these are the things which already we have discussed these are the things which already we have discussed now guys it's time to plot one very important graph right which you need to understand basically which you need to understand which you need to understand which you need to understand have a look on this particular side I'm plotting vapor pressure on this particular side I'm plotting mole fraction Kai mole fraction so what is Kai KY is the mole fraction of the component in the solution phase in liquid phase guy is the mole fraction of the component in the liquid phase in the solution phase okay now my dear students try to understand what exactly I'll be doing try to understand what exactly I'll be doing imagine at this particular Point imagine at this particular Point imagine your k a is equal to zero for example if Kai a becomes zero that means Kai B will be equal to one so at this particular Point either you say k b is zero or either you say Kai is equal 0 or you can say k b is equal 1 now try to understand this if Kai a is equal to0 what does that mean that means mole fraction of a in the solution phase is zero that means a is not there in the container what is there in the container there is only B in the container there is only B in the container so I will say B is present in its pure state so whatever will be the vapor pressure right now that will be due to Pure B and vapor pressure due to Pure B is represented for example by P not B so this is for example your P not B I hope you got this particular Point what I said right I hope you got this particular point now people if you go in the forward Direction in the forward Direction what you'll observe Kai a will increase and Kai B will decrease see at this point Kai a was Zero k b was one at this particular Point Kai b or let me write it as Kai a Kai a will be one and Kai B will be zero 1 + 0 makes it one K A Plus KY B has to be one here it's one here it's one everywhere on this particular line K A Plus k b will be one only okay look at this particular point at this particular Point Kai a is one k b is zero what does that mean Kai a is one k b is z k b is zero means mole fraction of B mole fraction of B in the solution is zero that means there is no B in the container if there is no B in the container that means there's only a in the container if there is only a in the container what does that mean that means the vapor pressure right now will be due to Pure a and vapor pressure due to Pure a is is represent by P A imagine this is your p a this point is your p a I hope you got this I hope you got this now try to understand very carefully if you move in the forward Direction hair Kai is zero hair Kai is one so on moov in the forward Direction Kai a is increasing on mov in the forward Direction Kai is increasing if Kai a is increasing that means PA is increasing on moving in the forward Direction k a is increasing k a increasing means mole fraction of a in the solution phase is increasing if mole fraction of a in the solution phase increases can I say more Vapors of a will be there can I say more will be the contribution of a towards the total V pressure right do you get this do you get this point from here to here Kai a is increasing if Kai is increasing mole fraction of a in the solution phase is increasing right if mole fraction of a in the solution phase is increasing that means moles of a in the solution phase moles of a in the solution phase are increasing that means at that time you'll find more Vapors of a if you find more Vapors of a if Vapors of a gener if Vapors of a are getting increased what does that mean if Vapors of a generated gets increased that means contribution made by a towards the total vapor pressure also increases yes so what should be the nature of the graph the nature of the graph has to be like this on M from here to here right the contribution made by a towards the total W pressure the contribution made by a towards the total Vaper pressure that increases that increases from here to here correct and what will be the equation of this straight line it will be Pa a is equal to p a k this is the equation of the straight line right this is the equation of this particular straight line perfect you can check it out too at this point K is one if K is one that means it is p a perfect now try to understand it like this at this particular Point k b is one k b is one means in the container there is only B imagine there's only B here imagine there's only B here right so whatever Vapors will be here those are due to Pure B and vapor pressure due to Pure B is p b which are mentioned over here now tell me if you move forward if you move forward is the k b increasing or decreasing here it is one here it is zero if you move forward k b is k b is decreasing if k b is decreasing that means mole fraction of B in the solution phase is decreasing if mole fraction of B in the solution phase is decreasing that means that means Vapors of B over here will be decreasing as well if this term is decreasing that means this term will be decreasing Vapor of B will be decreasing here if Vapor of B will be decreasing what does that mean that means contribution made by B towards the total vapor pressure will be decreasing right so what will be the nature of the curve on mov from here to here contribution made by B towards the total V pressure that that decreases like this like this if I ask you what will be the equation for this line it will be PB is equal p b multip KB right and one thing which I've assumed from the graph this is your p b this is your p a i had already assumed p a is greater than p b you can assume the reverse p b is greater than p a the choice is all yours okay now my dear students if I join this particular point from here to here if I join if I join this particular point from here to here right this particular line Line This represents the total vapor pressure of the solution which is equal PA a plus PB or you can write P not a Kai a plus p b k b right tell me if this graph is absolutely clear to everyone or not quickly people quickly everyone in the chats I want everyone to let me know in the chats if this is absolutely clear to you or not quickly quickly people yeah guys I want everyone to say it yeah don't sleep don't sleep yeah just say it quickly just say it quickly all the things are clear clear perfect now there is one more question guys you have to remember all these Concepts which I'm giving you okay now there is one more concept then there'll be a lot of questions there'll be a lot of questions then dinner break will do at 9:30 dinner break will happen at 9:30 okay calculation of mole fraction of components in Vapor phase guys from this a particular type of question is asked from this a particular type of question is asked imagine this is your container and in this container what do we have for example we have got a solution of a plus b this is a solution of two volatile liquids A and B right this is the solution of two volatile liquids A and B correct imagine the moles of a in the liquid phase is na moles of B in the liquid phase is NB now if you understand it carefully on the top there'll be Vapors of a as well as B on the top there'll be Vapors of a as well as B correct let's say let's say the number of moles of a in the vapor phase is n a dash number of moles of B in the vapor phase is NB Dash is NB Das be careful with the terms this a This na represents Mo fraction of a in the liquid phase n a dash represents mole fraction of the vapors of a over here right if you look at this particular chamber this part of the container this part of the container it contains basically two Vapors Vapors of a as well as Vapors of B so I would say this part of the chamber contains a gaseous mixture this part of the chamber contains a gaseous mixture and that gaseous mixture contains two gases A and B correct it contains two gases A and B and as per Dalton's law the total vapor pressure will be the contributions made by each gas PA plus PB correct as per Dalton's law pa pa which is the contribution made by the vapors of a towards the total vapor pressure as per Dalton's law as per Dalton's law that has to be equal total pressure total pressure of this mixture which is basically total vapor pressure of the solution multiplied by mole fraction of a in the vapor phase so imagine guys a Das represents mole fraction of a in this mixture and here k a it represents mole fraction of a in this solution so these are two different things one is mole fraction of a in this mixture that is basically the mole fraction of gas a in the mixture this Sky a is the mole fraction of a in this solution these are two different things similarly k b Dash this is the mole fraction of B in this particular mixture and K B it is the mole fraction of B in this solution phase I hope these two things are clear if these things are clear pa pa is the contribution made by a towards the total V pressure as per Dalton's law it has to be equal total total pressure multiplied by mole fraction of a in this mixture right this is your Dalton's law this is your Dalton's law now as per ral's law what what is PA as per rods law your PA is p a MTI k a this K is mole fraction of a in the solution phase this is as for your RS law now see this term is giving you PA this term is giving you PA if you compare them when you compare them you get something like this PS multipli Kai a Das is equal p a multiplied by Kai and from this particular equation you can calculate Kai a Das which will be p p a * k a / PS so what is this particular term what is this particular amazing term guys what is this particular amazing term there will be some questions in which you will be asked to calculate the mole fraction of a in the wer phase right similarly you will be asked to calculate mole fraction of B in the V per phase k-b which will be equal p b * K b/ PS it is just you have to be careful with the terminologies you have to be careful with the terminologies let me write the terminologies again so that it won't be a confusion for you k a dash is equal it is the mole fraction of a in Vapor phase in Vapor phase k a is the mole fraction of a in which phase in solution phase or you can say liquid phase same goes for Kai B and K B Dash Okay so these two questions are asked after a few minutes you'll get to know about the questions as well but before doing the questions there is one more thing which you must know it is again very important which is again very important what is that that is vapor pressure of the solution vapor pressure of the solution containing a nonvolatile salute what is meant by this what is meant by this try try to understand people try to understand very carefully what it means because this forms this particular topic it forms the basis of your first ctive property that is related to lowering in V pressure this particular topic it forms the basis of your first topic that is relate to lowering in V pressure so you have to understand it very carefully I'm again taking two containers these are two containers imagine imagine or just a second just a second people see instead of two containers let me take only one container let's say this is your container and in this particular container imagine you have got a volatile liquid a this is a volatile liquid a in the container here's a concept here's a concept first thing if you look at the top surface can you say on the top surface there will be only a particles absolutely because a is right now in pure State there'll be only a particles on the top surface because a is in pure state of course a is in pure State okay so so I would say there'll be only the vapors of a here right so vapor pressure when a is present in its pure state is represented by P a right this is p a now guys imagine in the same container imagine in the same container you are introducing a nonvolatile salute imagine in the same container you are introducing a nonvolatile solute nonvolatile solute means that salute which cannot form its Vapors you're introducing one more liquid for example that liquid is nonvolatile Right introducing nonvolatile salute in the same container that that solute you are introducing which cannot form its Vapors now if you are introducing the nonvolatile salute what will happen what will happen in the container in the container what do I get I got a solution of let's say the name of this nonvolatile solu is B I got a solution of A and B again absolutely I got the solution of A and B but this solution has two components A and B since a is volatile but B is nonvolatile B is nonvolatile B is nonvolatile tell me one thing tell me one thing on the top surface on the top surface ha will there be only will there be only a particles or there will be B particles as well on the surface if I just need to make you understand if I just need to make you understand on this surface for example there will be a particles as well as B particles similarly in the bulk there will be a particles as well as B particles right let's say these are the B particles these are the B particles correct these are the B particles now over here as you see there are a as well as B here a as well as B here only a can get converted into Vapors only a can form its Vapors right only a can form its Vapors B cannot form its Vapors some from bulk will go top will go at the top right only a can form its Vapors B cannot form its Vapors now tell me one thing over here only a particles were there and all the a particles were volatile and all a particles right had got the tendency to get converted into V but here you have got a as well as b b is nonvolatile so can you think over it and let me know among these two cases where will be the probability of vapor formation less where will be the amount of vapor generated less quickly tell me that where will be the amount of vapor generated Less in this particular case there is a as well as B only a is volatile B is nonvolatile right okay so over here you can only find the vapors of a that to in minimal amount that to in minimal amount just just to make you understand it just to make you remember it over here at the surface there are b as well over here only there were all volatile particles here there are some nonvolatile particles as well which cannot form its Vapors right perfect so the vapor pressure of the solution the vapor pressure of this particular solution it was supposed supposed to be Pa a plus PB it was supposed to be Pa plus PB but B is nonvolatile B has no contribution B has no contribution towards the total vapor pressure right so can I say PS is nothing it is just PA I'll say PS will be equal PA a is p a multiplied by K as per RS law as per RS law as per RS law correct perfect so here one thing has to be addressed there one thing has to be addressed when you had only pure a in the container its VOR pressure was p a now you added a nonvolatile salute what happened to vapor pressure has vapor pressure increased or decreased the probability of vapor formation here was more the probability of vapor formation here is less because it contains some nonvolatile particles as well which cannot form its Vapors so can I say can I say on adding on adding nonvolatile solute on adding nonvolatile solute into a volatile liquid into a volatile liquid vapor pressure decreases vapor pressure decreases and and that decrease in the vapor pressure that decrease in the vapor pressure is called as lowering in vapor pressure is called as lowering in vapor pressure and that lowering in vapor pressure is represented by Delta P Delta P so if I ask you how much vapor pressure has decreased you will say initially it was P finally it is P K so how much vapor pressure has decreased initially it was p a and finally it is p a k a so P A minus p a k this will give me the decrease in the Press by the way this is the logic which I'm giving you so that you can remember the actual reason is the are the thermodynamic reasons I can explain the same scenario correctly on the base of entropy right but you need not to remember that you need not remember that okay just remember it like this in short perfect this is how you can remember it the actual reasons are entropy oriented the actual reasons are entropy oriented which you need need not to know okay perfect now people one more thing one more thing one more thing decrease in the VOR pressure initially it was p a now it's p a k this is something which I call as decrease in the VOR pressure this is something which you call as lowering in pressure if you take P not a common it'll be 1- Sky 1- SK is Sky B so you can write it like this as well P not a I take in common 1us sky is nothing that's Sky B perfect so this particular statement I believe is to everyone so when you add a nonvolatile solute into a volatile liquid when you add nonvolatile solute into volatile liquid when you when you put nonvolatile salute into volatile liquid what happens to the W pressure it decreases that decrease in the W pressure is what you call as lowering in V pressure right perfect now with this your roll SLO is over let's try to do some questions the first question on your screen the first question on your screen the normal boiling point of water is 373 Calin if you remember I gave you two terms one is normal balling point one is standard balling point when we talk about the normal balling point when we talk about the normal balling point at that point of time external pressure is kept at 1M if you remember and we are talking about the boiling point of water we are talking about the boiling point of water right which is how much 373 kin that means at 373 Kelvin the vapor pressure of water the vapor pressure of water the vapor pressure of water would have got equal to the atmospheric pressure external pressure which is 180 so this was the thing which was supposed to be concluded from here right this was the thing here which you had to conclude perfect the boiling point of water is given us 373 kin that means at 373 kin vapor pressure of water would be equal to what would be equal to external pressure which in normal boiling point case is 1 atm perfect now as for the question is concerned vapor pressure of water at temperature T Calin is 19 mm of HG so when the temperature is T Kelvin vapor pressure of water is how much it is equal to 19 mm of HG 19 M of HG so 1M you can write as 760 M of HG you can converted here only corre so if you look if you look at it carefully initially the vapor pressure was 19 M ofg then you have increased the temperature basically you have increased the temperature that's why vapor pressure has increased perfect Delta H VAP is given Delta H VOR is given let me write it over here let me mention it over here Delta vaporation is equal 40.67 KJ I'll be converting into jewles 10^ 3 Jew per mole then the temperature T would be we have to calculate this T this Ste we have to calculate right so if you look at this particular case carefully initially the vapor pressure was 19 then finally at the balling Point vapor pressure is 760 so boing vapor pressure is increasing how vapor pressure is increasing when you have increase the temperature so initial temperature is T and the final temperature is 373 kin initial vapor pressure is how much it is 19 and final VOR pressure is 760 now you have one equation that is clausus claperon equation log of p2/ P1 is equal Delta vaporization divided 2.303 this is R this is 1 upon T1 minus 1 upon T2 1 upon T2 right so you have all the parameters people you have P2 you have P1 value correct you have Delta VAP res correct R value 8.314 T1 is to be calculated P2 is given one equation one unknown that's it that's it when you solve this question you'll be getting the final result as 290 something okay correct cor I guys is it clear is it clear I hope this is clear to everyone moving on to one more simple question the vapor pressure of two pure liquids A and B that means they are giving the P A and P B value p a is 100 Tor and as per the question PB is how much it is 80 Tor it is at T calculate the total vapor pressure of the solution obtained by mixing two moles of a and three moles of B so moles of a is 2 moles of B are three what do I have to calculate I have to calculate the the vapor pressure of the solution you know it vapor pressure of the solution is equal p a k a plus what plus p b k b so PS is equal p a I'm writing as it what is k k means moles of a divided by total moles present in the container yes similarly this is going to be P not B moles of B divided by total moles present in the container correct so this is PS is equal p a what is p not a that's 100 number of moles of a in the solution phase that's two divide by total moles 2 + 3 5 plus P not B that's 80 moles of B 3 total moles 3 + 3 25 right solve it this will give you the total vapor pressure of the solution correct people are saying it's coming out to be 88 T of course it can come as 88 T absolutely it's coming out to be 88 T right perfect guys I believe this is clear let's move on to one more question let's move on to one more question look at this particular question carefully see what the question says two liquids A and B they form an ideal solution so we have got two liquids they forming that ideal solution at 300 Kelvin the vapor pressure of the solution the vapor pressure of the solution containing 1 mole of a the vapor pressure of the solution containing 1 mole of a and 3 mol of B is how much vapor pressure of the solution containing this much is equal 500 mm of HG at the same temperature if one more mole of B is further added so first of all you had a container which contained one mole of a and two moles of B in the solution phase three moles of B in the solution phase now you are introducing one more mole of B when you are introducing one mole of one more mole of B nothing will happens to the moles of a moles of a will remain the same but moles of B now will be 3 + 1 4 correct if one more mole of B is added the vapor pressure of the solution increased by 10 m of a so initially the vapor pressure of the solution was 500 now it will be 510 M of HG I hope you're understanding these things I I hope you're understanding these things guys is there any lag is there any lag is there any lag people no right because on my screen the cursor is continuously rotating I thought there is a lag or something all right so initially you had a container basically right initially you had a container initially you had a container in which the solution which we had in the solution there was one mole of a and three moles of B vapor pressure of that solution was given now you introduced one more mole of B so moles of B in the solution are now four moles of a are same due to which vapor pressure of the solution has increased by 10 m of HG so total W pressure of the solution now is going to be 510 M of HG we have to calculate what P not a p b so basically I have got two conditions if I use first condition PS has to be equal p a k here plus P not b k b here so PS is given to me as how much 500 is equal p not a as such K moles of a divide by total moles right now so it is 1x 4 it is 1x 4 plus this will be P not b as such mole fraction of B moles of B divide by total moles right now 3 + 1 4 so I got one equation I got one equation now look at the second case look at the second case total vapor pressure of the solution you know it has to be equal to p not a * K A Plus p b MTI by K B so total V of the solution is 510 is equal p not a right as such what is K a right now what is mole fraction of a right now it will be moles of a divide by total moles here 4 plus one 5 plus this is p not b and it's going to be 4id 5 now so if you look at these two equations you got two equations and two unknowns two equations two unknowns you can solve them easily by either substitution method or elimination method or cross multiplication method whatever methods you know by means of that you can easily get p a and you can get p b as well those were the two things which I was supposed to calculate nothing else I hope this sort of equation is clear to you you can solve this sort of equation look at one more question the vapor pressure of pure liquid a the vapor pressure of pure liquid a so basically they talking about p a the vapor pressure of pure liquid a is 70 t Okay the VOR PR of pure liquid a is 70 T it forms an ideal solution with another liquid B okay the mole fraction of B in the solution mole fraction of B in the solution is 0.2 that means mole fraction of a in the solution will be 0.8 total vapor pressure of the solution is also given 84 t total V pressure of the solution is also given 84 to what do we have to calculate we we have to calculate p b we have to calculate p b it's a simple question guys it's a simple question have a look I'll say total vapor pressure of the solution is nothing that is p not a Kai a plus p b k b correct what is PS PS is given as 84 T is equal p not a is 70 mole fraction of a is 0 uh 0.8 plus P not B you have to calculate k b is 0.2 so again one equation one unknown from here you can easily calculate PB that is something which you are supposed to calculate right correct so nothing to think in these questions guys you can easily kill them now you can easily kill them you can easily kill them for example there is one more question this can be asked too this can be asked too A solution has one is 4 molar ratio of pentane and hexane a solution has 1 is4 mol ratio of pentane and hexane if the P not values are this and this calculate the mole fraction of pentane in the vapor phase so let me make you understand the question first of all this is the container in this container you have got a solution this is a solution for example this is a solution of pentane and hexane there's a solution of pentane and hexane right molar ratio of pentane and hexane let me call this as a let me call this as B okay their mol ratio in the solution phase is given so Kai aide by K B Kai aide by K B is given to me as 1X 4 Kai a divided by K B is given to me as 1x4 right it is given to me as 1x4 P A and P B values are also given to me p a is equal how much 440 mm of HG and p b is given to me as 120 M of HG calculate the mole fraction of pentane in the vapor phase so you have to calculate k a Das basically right so few minutes back I told you how to calculate mole fraction of the component in the vapor phase it will be equal p a * K aide PS correct divide by PS right guys so can you solve it further can you solve it further so first of all p a is given to me right p a is given to me so let me do one thing let me first of all calculate PS total vapor pressure of the solution total vapor pressure of the solution it has to be equal p a k a plus p b k b correct so PS has to be equal what is p a p is 440 K A K by K B is equal 1X 4 what is does that mean that means if Kai is x k b will be 4X right correct so Kai a I'm calling as X Plus P not b k b will be how much 4X and p b is given to me as 120 so it's 120 into 4X so this value will come out to be 440x plus this will be 480x so PS comes out to be 0 842 448 9 92x this is your total pressure of the solution now put it here in this expression what do I get I'll get mole fraction of a in the vapor phase is equal p a p a is nothing that's 440 what is K A K A is X divide by what is PS that is 920x so XX canceled mole fraction of a in the wer phase is coming out be 440 divid 920 right this is the vapor this is the mole fraction of a in the vapor phase if you got the mole fraction of a in the vapor phase can you calculate mole fraction of B in the vapor phase absolutely Kai a Das plus k b Das will be 1 so k b Das will be 1 minus 440 / 920 solve this this will give you Kai B Dash is it clear I hope this is clear so with this our R is clear and now it is a time for ideal non- ideal Solutions so do you want to take a break here or continue do you want to take a break here or continue tell me that tell me that thing quickly do you want to take a break here or you want to continue everyone but are you promising that you guys will be back on time is it a promise with your teacher is it a promise with your teacher first everybody should write promise in the chats then only I can leave you and the chat should flow like anything chat should flow like anything chat should flow like anything don't you understand my angrezi quickly quickly quickly quickly quickly okay let me give a break for some time it's a dinner break go have a dinner but what time you'll be back just tell me that what time will you be back it's 9:10 now so I'll be resuming the session at at 9:45 sha everyone be back okay this chapter will still take some four hours this chapter will still take some four hours not less than that because I've taught everything in detail now okay perfect see you after the break then see you guys what's up people is everyone back yeah is everything everyone back all right so what's up how are you all doing how are you all doing sleepy yes I'm done with my dinner what about you guys are you done are you guys done perfect just give me a a second all right so tell me the topics which we covered till now are all the topics clear till here tell me the topics which we have covered till now are all the topics clear till now are all the topics clear till now let me know once in the chats everyone everyone yeah perfect guys so let's get going then without wasting a lot of a time let's get going let's have a look on the topic which is ideal Solutions so first of all how do you exactly Define the term ideal solution let me quickly write its definition first then I'll make you understand what these ideal Solutions are all about and what kind of questions are framed from the ideal nonideal Solutions have a look people how do we exactly Define ideal Solutions let me tell you these are the solutions that obey that obey rals law under all compositions the solutions which obey the the rals law under all compositions those are what we call as ideal Solutions now what is meant by it what is meant by it have a look people try to understand exactly what I'll be talking about see first of all if I talk about the vapor pressure if I talk about the vapor pressure of the solution which we represent by P vapor pressure of the solution okay dear students there are basically two ways there are two ways by means of which we can calculate the vapor pressure of solution there are two ways by means of which we can calculate vapor pressure of the solution one is experimentally experimentally with the help of manometer vapor pressure of the solution can be calculated by two ways one experimentally with the help of manometer and theoretically and theoretically it can be calculated with the help of with the help of R slw okay so in order to calculate the vapor pressure of the solution we have got two ways we can calculate vapor pressure of the solution experimentally with the help of manometer or we can calculate vapor pressure of the solution heretically with the help of RS law whenever we calculate vapor pressure of the solution experimentally that is what you call as the actual vapor pressure of the solution that's what you call as the actual vapor pressure of the solution or you can call it as The observed vapor pressure of the solution okay whenever you calculate the vapor pressure of the solution with the help of Ralts law you call that vapor pressure of the solution as the theoretical vapor pressure you call that vapor pressure of the solution as the theoretical vapor pressure and as per your RS law your theoretical vapor pressure is nothing thetical vapor pressure of the solution is equal p a k a plus p b KB correct so first of all I want you guys to remember these two things I want you guys to remember these two things there are two ways by means of which we can calculate the vapor pressure of the solution one experimentally one experimentally right with the help of manometer another one with the help of rals law and that vapor pressure of the solution which is calculated with the help of rals law you call that as the theoretical vapor pressure you call that as the theoretical vapor pressure perfect now my dear student students there are few Solutions there are few Solutions in which in which observed vapor pressure that means the actual vapor pressure which is calculated experimentally comes out to be equal to the theoretical vapor pressure which is calculated with the help of RS law and there are some types of the solutions in which observed vapor pressure and theoretical vapor pressure won't be equal they come out to be different they come out to be different let me tell you those Solutions those Solutions those particular solutions for which observed Vaper pressure and theoretical V pressure comes out to be equal you call those Solutions as the ideal solution okay and those particular solutions for which The observed vapor pressure which is calculated experimentally comes out to be different as calculated from the rals law you call those Solutions as the non ideal solution you call them as the non- ideal solution this is the broad classification this is the broad classification of the ideal and non- ideal solution okay this is the broader Way by means of which we can classify ideal and non- ideal solution right yes people is it clear to you till here is it clear to you till here again I'm repeating the same thing this is something important whenever we need to calculate the vapor pressure of the solution yes vij you your batch will be extended do do not spam here your batch excess will be extended don't worry about that okay vapor pressure of the solution we can calculate either with the help of manometer experimentally or we can calculate it theoretically with the help of RS law right that particular solution for which the actual vapor pressure which is what you call as observed vapor pressure which is calculated experimentally comes out to be equal to theoretical vapor pressure that's called as ideal solution and those solutions for which observed and thetical comes out be different those are called as non- ideal Solutions so let's talk about the ideal Solutions first of all let's talk about the ideal Solutions first of all let's see what are the parameters what are the points which are related to the ideal Solutions which we have to remember see guys I'm marking the heading as ideal Solutions I'm marking the heading as ideal Solutions okay all right try to understand very carefully what exactly I'm talking about for example I'm taking these two containers over here this particular container it carries it contains volatile liquid a this particular container it contains volatile liquid B and over here you have got Pure Vapors of a and vapor pressure of a here will be p a and here vapor pressure of B will be absolutely p b correct now my dear students if you mix them what you'll be getting if you mix them what you'll be exactly getting see when you mix these two solutions I mean when you mix these two liquids you get a solution of A and B this is your solution of A and B this is your solution of A and B right and over here you'll find the vapors of a as well as B correct and vapor pressure of the solution over here is represent by PS and as per roll slw it has to be equal to what it has to be equal to p a Kai a plus p b k b plus p b k b correct or you can say it has to be equal p a plus PB where p a is p not a k a PB is p b k b right this is something which you all must be knowing my dear students this particular solution which we got over here this solution of A and B as I told told you earlier I have got two ways to calculate the total vapor pressure of this particular solution I have got two ways to calculate the total vapor pressure of this particular solution one experimentally which is I'll be calling as observed vapor pressure and one theoretically which is calculated with the help of RS law which is calculated with the help of RS law so with the help of RS law the total vapor pressure should be equal p a k a plus p b k b plus p b k b with the help of R law the total vapor pressure of the solution should be this much right my dear students if you have such kind of the solution if you have such kind of the solution for which observed vapor pressure and theortical vapor pressure comes out to be equal you'll be calling this particular solution as the ideal solution you'll be calling it as the ideal solution that means observed vapor pressure when you calculate the vapor pressure experimentally it will also come out to be the same it it will also come out to be the same P not a Kai plus p b k b right whenever you see this sort of a solution for which observed and theoretical vapor pressure comes out to be equal I'll be calling that particular solution as the ideal solution I hope this particular point is clear I hope this particular point is Clear Point number one point number one right Point number one point number one point number two point number two in case of Ideal Solutions what all features do we have have a look people first of all you have got a liquid a over here right so in this particular liquid there will be AA interactions here in this particular liquid there will be BB interactions and here in this particular solution what kind of interactions you'll have here you'll have AB interactions and when we talk about the ideal Solutions in case of Ideal Solutions do remember a a interactions are same the magnitude of a interactions is equal to the magnitude of BB interactions is equal to the magnitude of ab interactions this is one very important point when you talk about the ideal Solutions when you talk about the ideal Solutions okay whatever a interactions will be there right BB interaction magnitude will be the same and ab interaction magnitude over here will be again the same point number one point number one okay imagine volume of this particular liquid over here is V 1 imagine volume of this particular liquid over here is V2 and volume of this resulting solution for example it is VR we say in case of Ideal Solutions Delta V mixing Delta V mixing Delta V mixing Delta V mixing is equal to zero how can you justify this statement how can you justify this particular statement Delta V mixing is zero what does that mean that means V final minus V initial V final minus V initial which will be equal V final is basically volume of the resulting solution minus V initial is V1 + V2 V initial is V1 + vs2 now if I ask you if all these interactions are same if all these interactions are same can I say V1 + V2 will be equal to VR and if V1 + V2 is equal to VR you'll categorically say Delta V mixing over here comes out to be zero in case of what in case of Ideal Solutions right I hope this particular this particular statement is clear in case of Ideal solution all the interactions are same if all the interactions are same if all the interactions are same what does that mean that means the final volume of the solution will be same as the sum of the initial volumes correct now imagine it like this if by chance these interactions were different imagine if AB interaction was weak imagine if if a interaction was weak then a and BB if AB interaction was weak then A and B particles would have been far right so final volume at that point of time would have been more than that of initial volume correct similarly if AB interactions would have been more if AB interactions would have been more than a BB then AB particles would have been closer so final volum would have been lesser at that time right but I'm telling you all the interactions here are same that clearly tells you that clearly tells you final volume will be equal to initial volume due to which Delta V mixing change in volume during mixing change in volume during mixing will be will be zero is this particular Point Clear is this particular Point Clear is this particular Point Clear to everyone there is one more thing related to Ideal Solutions what is that we say Delta H mixing is zero in case of Ideal Solutions what is many Me by that Delta is mixing is zero first of all over here in this liquid we have a interactions here we have BB interactions here AB interaction is getting formed so for this AB interaction to get formed can I say this interaction should break similarly this interaction should break and at the end this interaction should get formed right for the ab interaction to get form I'll say this bond has to break this bond has to break and this bond has to get formed right you know Bond breakage is endothermic requires energy Bond breakage that is endothermic requires energy Bond formation is exothermic energy is released now if all the interactions here are same that means the net energy absorbed here will be equal to the net energy released if net energy absorbed is equal to net energy released I would say Delta H mixing that has to be what that has to be zero that has to be zero right Perfect Since interactions are same everywhere this bond has to be broken down this interaction has to be broken down this interaction will get formed Bond breakage is endothermic requires energy Bond formation is exothermic energy is released since interactions are same so net energy absorbed in this case will be equal to what net energy released eventually Delta H mixing comes out to be what it comes out to be zero okay this was one more important point this was one more important Point similarly in case of Ideal Solutions do remember Delta s mixing Delta s mixing it is positive it is greater than zero entropy change is positive that means entropy changes positive what does it mean that means entropy is increasing during mixing which is evident Here My Dear students you only have a particles here you have got only B particles now when you are mixing them you have got a as well as B particles so can I say disorder over here will be more if disorder over here is more final disorder is absolutely greater than that of initial disorders right perfect so what does that mean if disorder is more that means entropy is more so during mixing entropy increases if entropy increases during mixing final entropy will be more than initial entropy therefore change in entropy will have to be positive change in entropy will have to be positive correct is this point clear is this point clear to everyone these are the questions which are directly asked Delta s mixing Delta H mixing Delta V mixing interactions okay perfect here you had only a particles only B particles here there are a as well as B particles correct so that means disorder is more final disorder is more if final disorder is more Delta s comes out to be positive similarly if you talk about Delta G if you talk about Delta G Delta G mixing what is Delta G as per thermodynamics Delta hus T Delta s Delta hus T Delta s in case of Ideal Solutions Delta H already zero Delta H already zero right Delta s is positive positive and negative makes it negative So eventually I would say Delta G mixing here is negative as well Delta G mixing here is negative as well so these are some very very very important points which you have to remember related to Ideal Solutions let me quickly summarize Point number one in case of Ideal Solutions observed vapor pressure with the help of experiments comes out to be same as that of calculated with the help of RS law right which is only possible if all the interactions are same number one all the interactions are same number one number two number two Delta V mixing is zero Delta H mixing is zero Delta s mixing is positive Delta G mixing that comes out be negative right perfect these are the points which you have to remember related to what related to Ideal Solutions yeah these are the points which you have to remember related to Ideal Solutions is this clear is this clear people yes okay if this is clear if this is clear then I would want you guys to remember few examples of these ideal Solutions right these are some of the examples of the ideal Solutions which you can remember directly right which you can remember directly examples of the ideal Solutions which you have to remember directly okay I'm not going into the details of it I am going to talk about the non- ideal Solutions now which is very important I'm going to talk about the non-ideal solutions now which is very important okay so first of all as I told you in case of non- Ideal Solutions as I told you in case of non ideal Solutions in case of nonideal solution solution as I told you in case of non- Ideal solution The observed vapor pressure The observed vapor pressure of the final solution it comes out to be different than that of theoretical than that of theoretical right theoretical means that vapor pressure of the solution which is calculated with the help of R law with the help of R law try to understand what exactly I'm going to talk about my dear students that particular solution whose observed vapor pressure comes out to be different as what we had calculated with the help of R's law we call that particular solution as non- adal solution right now the first case which arises here the first case which arises here if P observed is not equal to P theoretical that means P observed it can be greater than P theoretical or there can be one more case in which P observed will come out to be less than P theoretical right what is p theoretical P theoretical is nothing it is p not a K plus p b k b correct P theoretical is nothing it is P A Plus p b k so those Solutions in which those Solutions in which observed vapor pressure which is calculated experimentally comes out to be greater than theoretical vapor pressure comes out to be greater than theoretical paper pressure you call you say those non ideal Solutions show positive deviation you say those non idial solutions they show positive deviation from the RS law similarly over here these are the non ideal Solutions which show negative deviation from the RS law now let's talk about these non ideal solution showing positive and negative deviation in detail okay let's talk about them in detail let's talk about them in detail try to understand what exactly I'll be talking about this is something which is super important guys okay so my first topic here [Music] is non-ideal Solutions showing which deviation showing positive deviation non ideal Solutions showing positive deviation have a look exactly what I'm going to talk about try to understand carefully my dear students this is one container this is one more container here and this is over here one more container correct in the first container you know what do we have we have got pure a pure volatile liquid a in the container here you have got Pure volatile liquid B in the container and now what we are doing we are mixing them and we are getting a solution of A and B we are getting a solution of A and B over here right so first of all a is present in the pure State vapor pressure of a when present in pure form is basically your p a right similarly vapor pressure of B in pure state is basically your P not b and vapor pressure of the solution which is due to Vapors of a as well as B that is represented by PS This is the vapor pressure of solution and as as per your RTS law vapor pressure of the solution which you call as theoretical vapor pressure of the solution it has to be equal p not a K plus p b k b plus p b k b but as I told you in order to calculate the vapor pressure of the solution how many ways we have two ways one experimental through which you get observed vapor pressure of the solution one is theoretical with the help of RS law with the help of RS law right perfect now people try to understand understand what exactly I'm trying to convey here vapor pressure of the solution as I already mentioned we have got two ways one experimentally by means of which you get observed vapor pressure of the solution one with the help of rals law by which you get theoretical vapor pressure of the solution as per RS law vapor pressure of the solution should be what it should be p a k a plus P not B KB as per RS law as per RS law but this is the solution for which observed and thetical Vapor pressures come out to be different come out to be different this is the solution for which observed and theoretical vapor pressure comes out to be different that means your P observed is not equal to p a k plus p b k b right so first point your P observed what you get from the manometer right it is not equal p a k a plus p b k b right and over here I'm talking about so it is first of all the nonideal solution and right now I'm talking about that non- ideal solution right now I'm talking about that nonideal solution for which observed vapor pressure observed vapor pressure comes out to be greater than theoretical comes out be greater than this comes out be greater than this and we and whenever we get such kind of the solution for which observed vapor pressure comes out to be greater than comes out to be greater than this much right then we say that particular solution that particular non- ideal solution is showing positive deviation is showing positive deviation now people tell me one thing when is this particular equation possible when is this particular statement possible when observed Vaper pressure will come out to be greater than this when when my question to you when my question to you when when see if this was ideal if this was ideal at that point of time P observed would have been equal to this would have been equal to this but this is non ideal so P observed is not equal to this here in this particular case I'm telling you that P observed is for example greater than this which is only possible which is only possible if which is only possible if PA will be greater than P not a i and PB will be greater than P not be K right perfect perfect guys right I believe this particular point is clear this particular statement is only possible if the partial vapor pressure if the partial vapor pressure of a if the partial vapor pressure of a here this is your partial W pressure of a this is your partial W pressure of B right if this particular solution was supposed to be ideal then partial V pressure of a would have been equal to p a k and partial V pressure of B should have been equal to p b k b but but in this particular case P observed is coming out to be greater than this term which is only possible a partial vapor pressure of a is more than expected what was expected this was the partial VOR pressure of a which was expected with the help of rods law this was the partial vapor pressure of B which was expected with the help of ral's law but but the total vapor pressure of the solution is coming out to be greater than this particular term which is Possible only if the if the partial vapor pressure of the of a would come out to be more than expected partial vapor pressure of B comes out to be more than expected that is the only case right right people is this particular point clear now you tell me one thing you tell me one thing tell me one simple thing tell me one simple thing I'm telling you over here in this particular solution which showing which is showing positive deviation I'm telling you its total vapor pressure its actual total vapor pressure comes out to be greater than that of what greater than that of RS law greater than that of what that of calculator from RS law when is that possible that is possible only if a interactions will be lesser magnitude than that of a and BB interactions that is how that is how that is how more Vapors would have got generated over here than expected correct that's how more Vapors would have got generated over here than expected right people I believe you are getting this particular thing so do remember AB interactions AB interactions are less than that of a A or BB interactions if if a interactions would have been same if a interactions would have been same as of a BB then I should have been calling this as ideal if this was ideal then its observed vapor pressure should have been equal to theoretical vapor pressure correct but observed is coming out to be greater that means whatever interactions I was expecting interactions won't be that much stronger it'll be comparatively weaker due to which more Vapors would have got generated du which more Vapors would have got generated that is the reason that is the reason why the partial vapor pressure of a will be more than the expected that's why the partial vapor pressure of B will be more than the expected right expected means as per alls law and that is the reason why the total observed vapor pressure of the solution comes out to be more than that of theoretical vapor pressure which was expected as per alls law I hope this particular point is clear right now you tell me one thing since over here you have got a interactions over here you have got BB interactions over here you have got AB interactions right so in order to form these AB interactions this has to be broken this has to be broken and this has to get formed right and you know and you already know Bond breakage is endothermic energy is required right Bond formation is exothermic energy is released but here interactions are weak interactions are comparatively weak interactions are weaker than expected if interactions are weaker than expected can I say the magnitude of net energy released the magnitude of net energy released will be less than that of the magnitude of net energy absorbed the magnitude of net energy released will be less than that of the magnitude of net energy absorbed due to which the overall process is endothermic if the overall process is endothermic if the overall process endothermic you can categorically say Delta H mixing over here will be positive correct now if I ask you about Delta V mixing change in volume Delta V mixing interactions are weak if interactions are weak AB particles will be far if AB particles are far I'll say final volume of the solution will be greater I'll say final volume of the solution which is the volume of resulting solution it will be greater than V1 + V2 if VR is greater than V1 plus V2 then change in volume final minus initial it will come out to be what it will come out to be positive so Delta V mixing is positive here Delta V mixing is positive here right Delta V mixing is positive here and do remember what about Delta s mixing Delta s mixing Whenever there is mixing entropy increases entropy increase means delt s mixing has to be positive and do remember here Delta G mixing is negative Delta G mixing in case of that particular solution in case of that non- ideal solution which shows positive deviation for that Delta G mixing will be negative now you must be thinking how Delta G mixing is negative here Delta G is equal Delta H minus t Delta s right I told you Delta G mixing here is negative perfect you know this Delta H here is positive this is positive even delt s is positive then how come the overall sign will come out to be negative it is only possible if magnitude of T delt s here will be greater than that of magnitude of Delta so do remember in this particular case magnitude of T delt s will be greater than that of magnitude of Delta that is the reason why your Delta G mixing over here comes out to be negative in case of non ideal Solutions showing the positive deviation perfect showing the positive deviation Perfect People perfect is this clear let me know once in the chats let me know once in the chats quickly yeah clear is it absolutely clear so if the solution was supposed to be ideal then the interactions between A and B should have been same as that of a should have been same as of BB at that point of time P observed should have been equal to P theoretical but it's non ideal so P observed won't be equal to p thetical as per RS law p observed here is coming out to be more which is only possible if the partial vapor pressure of a would be more than expected if the partial pressure partial V pressure of B would be more than expected what was expected what was the expected partial V pressure of a that was p a k a what was the expected partial vure of B that was p b k b right perfect that is the reason why I wrote over here PA a will be greater than p a PB will be greater than p b KB right I believe this is clear to everyone okay now guys before showing you its graphical form before showing you its graphical form let me show you those non- ideal Solutions which show negative deviation after that I'll show you the graphical form of both the ones combinedly okay so now I'm going to talk about those non-ideal Solutions which show negative deviation which show negative deviation now you should be intelligent enough to understand this particular case on your own I believe right so first of all let me make it a little more clear to you again this is your container number one this is your container number two and this is your container number three right over here as you all must be knowing what do we keep we keep liquid a volatile liquid a here what do we have we have got volatile liquid B right and over here we are adding the two and what do we get we are getting a solution of a plus b a solution of a A plus b perfect and here you'll again find Vapors of a only so vapor pressure here is p a here the vapor pressure is what here the vapor pressure is p b here the vapor pressure of the solution which is represented with PS it as per RS law it should be as per RS law it should be p a k a plus p b k b right P not b k now again I'm telling you the same thing we have got two ways to calculate the vapor pressure of solution right to calculate vapor pressure of solution we have got two ways what all ways are there one is experimentally calculated which is called as observed total V pressure one is with the help of RS law which is called as theoretical vapor pressure and as for rals law the theoretical vapor pressure of solution should have been p a k plus p b k b as per alls law okay but which solution this is this is the solution which is showing negative deviation and in case of the solution showing negative deviation observed vapor pressure observed VOR pressure comes out to be less than that of thetical vapor pressure what does that mean that means observed V pressure comes out to be less than that of p a k a plus p not B K right when is this possible this is only possible if the expected if the if the actual partial VOR pressure of a comes out to be less than that of less than that of expected right similarly if the actual partial vapor pressure of B comes out to be less than that of expected and what is expected expected was P not B KB right opposite of of whatever we studied till now clear now guys understand it like this if the solution was ideal then P observed and P theoretical should have been equal if it was ideal correct that means in that case all the interactions would have been same but it is the non- ideal solution showing negative deviation observed vapor pressure is less than that of expected this is expected this observed observed is less than expected what can we conclude from there I can conclude in reality the actual amount of vapors generated hair would be less when is that possible that is only possible in reality if a interactions are more than that of a A or BB interactions right if the interactions are more less escaping tendency less Vapor formation correct now tell me one thing what about Delta V mixing is it going to be positive or negative Delta V mixing if the interactions are more here that means a particles will be Clos final volume will be less right final volume will be less right final volume will be less so VR will be less than V1 + vs2 which tells you that if final volume is less than that of initial that means change in volume final minus initial change in volume final minus initial change in volume final minus initial it comes out to be what it will come out to be negative it will come out be negative now similarly Delta H mixing Delta H mixing what will happen to this particular thing Delta mixing Delta mixing see for ab interaction to get formed over here first of all you have got a interactions here there are BB interactions here you have got a interactions for these interactions to get formed this has to be broken down this has to be broken down and this will get formed right breakage is endothermic right and over here it is exothermic Bond formation is exothermic right interactions here are stronger what does that mean that means energy released magnitude of energy released will be greater than that of magnitude of energy absorbed if magnitude of energy released will be greater than that of magnitude of energy absorbed that means the overall process will be exothermic and Delta H mixing will be negative similarly what about delt s mixing delt s mixing Whenever there is mixing Delta s is always positive and Delta G mixing will be negative a will be negative and if Delta G mixing is negative right you can make the condition you can make the condition as well for the Delta G mixing to be negative what does that mean that means Delta H minus t Delta mixing has to be negative now already we know this term is already negative this term is already negative this term is positive this term is positive positive negative makes it negative right positive negative makes it negative so it's it's already coming out to be negative only right Delta H is negative sorry Delta H is negative Delta s is positive Delta s is positive so positive negative makes it negative so there's a sign here negative here negative so overall sign will come out to be negative only perfect no condition you have to make over here I hope this particular point is all these points are clear to you I hope all these points are clear to you so let me quickly summarize all these things in just one slide in just one slide first of all I'm writing ideal solutions from this particular slide basically you'll begin getting the equations non ideal Solutions showing positive deviation non- ideal Solutions showing negative deviation let me quickly summarize all of them in one particular slide so that you can easily remember this so that you can easily remember this first of all in in case of Ideal Solutions observed vapor pressure is equal to theoretical VOR pressure right perfect or I can say observed vapor pressure is equal p a Kai a plus p b k b perfect which is Possible only if PA a is equal to P not a k a right and PB is equal p not b k b perfect all the intera are same all the interactions are same a interactions are same as that of BB as of ab right if I talk about Delta V mixing Delta V mixing in case of Ideal is zero right Delta H mixing is zero perfect Delta s mixing here is positive sorry yeah positive sorry and Delta G mixing there is negative these are the possible questions which can be asked in ideal Solutions anything out of these similarly non ideal solution showing positive deviation first point observed vapor pressure which is calculated with the help of experiments it is greater than that of theoretical that means the observed vapor pressure will be greater than that of p a k a plus p b k b whatever we had calculated with the help of RS FLW that is expected W pressure this is I mean this is expected W pressure which is also called as theoretical W pressure this particular scenario is only possible if PA is greater than p a k and PB is greater than what PB is greater than p b KB correct now AB interactions AB interactions are weaker than that of AA or BB interactions right similarly Delta V mixing is positive here Delta H mixing is positive here Delta s mixing is positive ha Delta G mixing is negative ha right possible questions which can be asked what do you think in the last case negative deviation in case of negative deviation will the actual W pressure be equal to the expected W pressure which we calculated from RS law no they'll be different how come I mean what difference exactly say observed W pressure there will be less than that of theoretical W pressure which we calculate from R law correct so I would say observed vapor pressure will be less than that of p a k a plus p b k b this is Possible only if PA if PA will be coming out to be less than that of p a k a and PB will be coming out to be less than that of p b KB correct uh interactions interactions talking about interactions a interactions are more than that of a A or BB interactions Delta V mixing here is negative Delta H mixing here is negative Delta s mixing a is positive Delta G mixing a will be negative correct one question which can be asked don't go into the details of that question one question those non-ideal Solutions which show the positive deviation those non- ideal solution which show the postive deviation they form minimum boiling eot trops this point do remember directly this is the question which can be asked which of the which of the following Solutions can form minimum boiling AER drops non- ideal Solutions showing positive deviation they form minimum boiling Gaz trops and they form maximum boiling gazi trops do remember them directly now guys I have to ask you certain questions here that's how I understand whether you got all these things properly or not okay and you will have to answer me in the chats first question first question heretical V pressure is calculated with the help of tell me in the chats heretical vapor pressure is calculated with the help of heretical vapor pressure is calculated with the help of quickly r slw yes observed vapor pressure is calculated with the help of observed vapor pressure is calculated with the help of experimentally with the help of manometer right tell me in case of Ideal Solutions in case of Ideal solution whatever amount of vapors we expected with the help of rolls law do we get the same amount of vapors experimentally as well say yes or no perfect tell me one thing tell me one thing tell me one thing in case of non ideal Solutions whatever amount of we Vapors we expected with the help of RS law experimentally do we get same amount of vapors more than expected or less than expected in case of positive deviation in case of positive deviation what ever Vapors were expected as per RS law do we get the same Vapor same amount of vapors experimentally no experimentally there will be more Vapor formation than expected which is Possible only if interactions between A and B are comparatively weaker right so more than expected in case of non ideal solution showing negative deviation negative deviation expected amount of Vapors and actual amount of vapors which one will be less less which one will be less expected or experimental the actual Vapor will be less than expected right that is the reason why actual vapor pressure is less than that of what thetical V pressure which is which was calculated as for alls law I hope all these points are clear I hope all these points are clear I hope all these points are absolutely clear to you now guys now now now see examples I would want you guys to remember directly these are the examples of Ideal Solutions right and you'll be getting this in the PDF format in the telegram perfect these are the examples of non-ideal solution showing positive deviation right and these are the examples of non- Ideal Solutions showing negative deviation you have to remember that well I'm not going into the examples here I just want to show you one simple thing what is that c I want to make three graphs here I want to make three graphs here three graphs I would want to make here and you will be telling me the difference among the among them first graph I'm making for ideal Solutions second graph is for those non ideal Solutions which show positive deviation third graph is for those nonideal Solutions which show negative deviation which show negative deviation just see how the graphs exactly will look like the same graphs which we discussed earlier as well try to understand what exactly they are I'm plotting vapor pressure on this side I'm plotting mole fraction of the component in the solution on this side right let's assume this at this point Kai a is equal zero and Kai B will be one at this particular Point Kai a will be one and Kai B will be zero we know it right at this particular point B is present in the pure form so let's call it as P not B right at this particular point a is present in the pure form right so let's call this as p a perfect I'll be making few curves I'll be making one graph I'll be making one graph if we move from left to right Kai is increasing if K is increasing that means VAP partial V pressure of a will be increasing perfect what will be the equation of this what will be the equation of this can I say in case of Ideal Solutions we observed let me write it over like this in case of Ideal Solutions PA will be exactly equal to how much it'll be exactly equal to p a k it'll be exactly equal to p a k right and this particular line this particular line if the solution is ideal then PB will be equal to P not B KB if the resulting solution is ideal this particular line will show that PS will be exactly equal to what it will be exactly equal to P not a K plus p b k b tell me one thing in case of non- Ideal Solutions will the graphs be same or different just tell me that will the graphs be same or different in case of non ideal Solutions will the graph be same or different in case of non ideal Solutions will the graphs be same or different this is vapor pressure this is mole fraction right first of all I'm making for ideal again I'm making for ideal again so Kai a is equal zero here k b is equal to 1 here Kai a is equal to 1 here and Kai B is equal to zero here for example right for example this particular Point represents your p b let's say this particular Point represents your p a perfect now tell me one simple thing this would have been this would have been the graph if this would have been the graph if PA would have been equal to p a k but in case of non- Ideal solution showing positive deviation is the actual partial vapor pressure of a equal to this or more than this it'll be more than this so if I talk about the actual graph will the actual graph be a straight line here or it'll go upwards like this it'll go upwards like this so this particular graph it is showing you that the actual the actual partial vapor pressure of a is more than that of expected which you calculate with the help of FRS law correct right people if the solution was ideal then this particular graph then this should have been the graph right and its equation should have been what its equation should have been its equation should have been PB is equal P not b k but in case of non ideal solution showing positive deviation partial Vaper pressure of B the actual partial V pressure of B will it be more than expected absolutely be more than expected more than what you expected through RS law so the actual graph will be like this right and this particular curve is telling you that PB will be actually greater than P not B KB perfect similarly this should have been in case of Ideal but we have got non ideal showing positive so the graph will be its graph will be like this correct so PS the actual total vapor pressure of the solution will be greater than that of p a Kai a plus p b k b perfect can you make for negative deviation on your own you should be in a position to make for negative deviation on your own it is just these graphs this graph will go downwards this graph will go downwards and this graph will go downwards correct correct so one will be like this I'm talking about this particular one one will be like this correct and and one would be like this correct and one more one more would be like this right all downwards all downwards lashika what you have to do with my salary yeah just keep it confidential it is a lot yeah okay I told you it's a lot it is it'll be your dream package the dream dream package okay I hope all these ideal non- ideal positive negative it's clear yeah taking 8h hour marathon has nothing to do with my salary if my salary would have been very less also I would be still taking these classes if it'll be like very very very high still it'll be the same there's nothing to do with this there's nothing to do with this yes saish I'm I'm seeing your reply okay I hope all these things are absolutely clear till here so ideal non ideal Solutions are clear to each and every one of you okay let me ask you questions once let me ask you a few questions once so that I'll get that confidence that you learned tell me one thing uh tell me one thing in case of non- Ideal Solutions showing negative deviation actual amount of vapors will that be less than expected more than expected or equal to expected quickly quickly quickly less than expected when is that possible when is that possible if AB interactions are more than expected or less than expected if a interactions are more than expected or less than expected more than expected wow nice good good guys Wonderful by the way all these things were they clear like this or they got clear now I'm pretty much sure you would have studied this chapter before were these points clear before or now they are getting clear you can tell me honestly that's great okay perfect now we are going to move into the last phase of the chapter that is cative properties right it's not that difficult or something okay so what are exactly the cative properties and what exactly we have to discuss in them let me first of all tell you these are these are the properties these are the properties of Ideal Solutions these are the properties of Ideal Solutions whose value whose value chains or let me write it like this these are the properties of the solution which depends on which depends only on which depends only on the number of particles of solute number of particles of solute and does not depend on and does not depend on the nature of solute does not depend on the nature of the solute so basically those properties of the solution those properties of an ideal solution whose value depends on the number of particles of solute so basically if you change the number of particles of solute in a solution that are are some properties there are some properties whose value chains right okay so if you talk about the ideal Solutions in case of Ideal Solutions if you take if you change the number of particles of solute if you increase or decrease the number of particles of solute there are some properties of the ideal Solutions whose value Changs whose value only depends on whose value only depends on the number of particles of solute does not depend on the nature of the solute you call those properties exactly as the cative properties there are four cative properties which we have to discuss one by one I'm just giving you the overview of the things the first one is the first one is the relative lowering in VOR pressure relative lowering in vapor pressure this is the first ctive property which we have to discuss second one is going to be elevation in bing Point second ctive property third one is going to be depression in phrasing point and the last one is something which you call as osmotic pressure these are those properties of the ideal solution these are basically those properties of the ideal Solutions whose value only depends on the amount of solute only depends on the number of particles of the solute only depends on the number of particles of the solute they do not depend on the nature of the solute and all those properties whose value only depends on the number of particles of solute and does not depend on the nature of the solute what do you call those properties as you call them as the colligative property now one by one we shall be discussing these colligative properties the first cative property which we have that is relative lowering in vapor pressure that is relative lowering in vapor pressure let's try to decode this relative lowering in vapor pressure but my dear students if you remember if you remember if you remember I told you one statement already on adding on adding a nonvolatile solute in a volatile liquid in a volatile liquid what happens to vapor pressure does vapor pressure increase or decrease we have discussed this before vapor pressure if you remember vapor pressure decreases right and that decrease in the vapor pressure if you remember and that decrease in the vapor pressure that decrease in the vapor pressure is called as what is called as lowering in vapor pressure which is represented by lowering in VOR pressure is represent by Delta P which used to be equal to P A minus PS do you remember this do you remember this we have discussed this anyways I'm just going to give you the quick overview of this particular thing again so that you can understand it properly for example this is a volatile liquid a in the container volatile liquid a in the container and over here vapor pressure of a is p a now you are introducing a nonvolatile salute B in the same container you introducing a nonvolatile solute B in the same container so what do I get here I got a solution of A and B I got the solution in B and vapor pressure of this particular solution will be equal to PS right which should have been equal to P plus PB which should have been equal to PA plus PB but B is nonvolatile it won't contribute towards V pressure so PS is nothing that is just equal to PA and Pa already you know that is p not a so when we add a nonvolatile solute in inter volatile liquid vapor pressure decreases basic and that decrease in the VOR pressure is what you call us that decrease in the VOR pressure is what you call as lowering in VOR pressure so this is what you will be calling as lowering in W pressure now guys if I divide this lowering in vapor pressure by the initial vapor pressure of the pure solvent right this particular term this particular this whole term over here right let me tell you this particular term lowering in vapor pressure with respect to initial VOR pressure of pure solent this particular term is what you call as relative lowering in vapor pressure this particular term is what you call as a relative lowering in pressure which is equal which is equal this is P A minus PS divided by P A now you know it can be written as P aus us PS is again p a k a as you can see there divided by P A so p a p a p not everywhere cancelled right so this is 1 minus Kai a and 1 minus k a comes out to be k b so I got one very very very important result that is relative lowering in vapor pressure is equal to basically P A minus PS / p a which is equal to mole fraction of non volatile solute which is equal to mole fraction of nonvolatile solute which can be further written as mole fraction of nonvolatile solute in the solution mole fraction of nonvolatile solute in the solution which can be written as number of moles of nonvolatile solute in the solution divide by total moles present in the solution divide by total moles present in the solution I hope this is clear I hope this particular point is clear now guys hair hair do remember if this non if this solution if the solution is very dilute if the solution is very dilute if the solution is very dilute if the solution is very dilute at that point of time you can say number of moles of number of moles of solute is far less than that of number of moles of solvent at that point of time the relative lowering in wer pressure which is basically P A minus p is ided by P A it can be written as I'll write NB divided by instead of Na plus NB I can just write na so this particular equation is not valid everywhere no it is only valid for dilute Solutions it is valid for dilute Solutions wherein the amount of solute will be far less than that of the solvent okay this is one of the equations which we'll be using in the questions number one number two there are few more equations which we are going to use in the numericals for numerical purposes I'm going to make certain equations for numerical purposes I'm going to make certain equations and what are those equations see first of all P A minus PS This is called as lowering in pressure correct earlier if you remember I divided it with p a but now I'm not going to divide it with p a I'll be dividing it with PS let's see what expression we get I'll be getting something like this P A minus PS is basically p a k and here PS is equal p a k correct perfect so p a p a everywhere canceled 1 minus K is KB divided by Kai a what is K a mole fraction of nonvolatile solute in the solution mole fraction of volatile solvent in the solution correct so this will be NB / na Plus n b divid by this will be na / na plus NB so na plus NB na plus NB everywhere cancelled so I got one equation equation is simple P A minus PS / PS it will be equal NB / n a this particular equation I won't be calling it as relative lowering in V pressure because related to vapor pressure related to lowering in V pressure it is only defined when you divide with what when you divide with p a but here I did not divide with p a i divide with PS this is not relate to lowering but this equation this equation it can be used everywhere as I told you here as I told you here this particular equation is only valid for dilute Solutions now no matter if the solution is dilute or concentrated this particular equation is used everywhere right everywhere people everywhere number one number one I'm going to give you one more equation which can ease out a lot of things which can help you out in solving the questions pretty much easily look at the same equation I got to know P A minus PS / PS is equal to NB what is NB number of moles of nonvolatile salute in the solution divided by na na na means number of moles of solvent which can be written as mass of solvent divide by m mass of solvent M mass of solent I'm writing Here If I multiply with th in the numerator and divide with th000 nothing will happen I divided and multiplied with th000 now why did I do so because if you look at this particular term now if you look at this particular term now if you look at this particular term now you can write it like this P A minus PS / PS is equal this particular term will be mity of the solution multiplied by molar mass of solvent divided by what divided by th000 there will be some questions in which they'll ask you to calculate mity you can directly use this particular equation and get mity I hope this is clear let me know once if it is clear so that I can solve few questions here tell me all these clear all these are clear so how many Expressions did we get three Expressions first P A minus PS / p a is equal NB / na plus NB if the solution is dilute then in the denominator instead of Na plus NB you will be writing na second expression valid everywhere where whether solution is concentrate or dilute whatever that is P A minus PS / PS is equal NB / na right but that is not the cative property cative property basically that is not your related to low ring related low ring is when you divide with P not a right the third expression is this one p Aus PS PS isal mity of the solution multip what multip Ma th000 okay now let's try to utilize them in the questions one very simple and basic question I'm just giving you the idea of solving it look at it calculate the relative lowering in VOR pressure calculate the relative lowering in vapor pressure relative lowering in vapor pressure If 100 G of a non volatile salute so what is the mass of nonvolatile solute mass of nonvolatile salute is 100 G right molar mass of this nonvolatile salute is equal to 100 G per mole dissolved in 432 G of water so mass of solvent mass of volatile solvent 432 G right and molar mass of solvent molar mass of water that is 18 G per mole this is something which all of you must be knowing already right what do I have to calculate I have to calculate the relative lowering in VAP pressure that means I need to calculate this value P A minus PSID p a this is something to be calculated which is basically equal NB / na plus NB so what do I need to calculate I need to calculate the rela lowering of vapor pressure that means this particular term its value has to be calculated and its value is equal NB ID na plus NB now do you know the value of NB and na NB number of moles of nonvolatile solute will be mass of solute divided by molar mass of solute number of moles of solvent will be equal mass of solvent divid by molar mass of solvent right so WB MB given so 100 divided 100 that's one right 100 divid 100 that's one na a w divid by Ma so 432 divide 18 432 divide 18 so you got NB you got na put it here and get the value of related to lowering in Vaper pressure I hope you can easily solve this sort of equation yeah people are saying it's 1 upon 25 whatever I mean you can have the answer with you look at this particular equation look at this particular equation the vapor pressure of an aquous solution of glucose so you have made an aquous solution of glucose right you made an aquous solution of glucose at 373 Calin okay is found to have 750 M HG W pressure the mity of the solution at the same temperature will be see guys what is the scenario this is the container which had water in it this is pure water correct this is pure water and temperature is kept as 373 kin so 373 Kelvin is basically the boiling point of water and at the boiling point vapor pressure becomes equal to atmospheric pressure so I would say vapor pressure of water right now V pressure of water right now will be equal to atmospheric pressure that is 760 mm of HG this is clear I believe correct because this something which I got from the first statement itself from the temperature right now as for the question in this water you have added glucose which is a nonvolatile salute you have added glucose which is a nonvolatile salute correct so what you have got you have got an Aqua Solution here this is the Aqua Solution of glucose Aqua Solution of glucose whose vapor pressure of course will be less right and that vapor pressure of the Aqua Solution is given to me as 750 m ofg what do I need to calculate I need to calculate mity of this particular solution few minutes back I gave you one result directly what was that that was P A minus PS / PS is equal it was mity multiplied m mass of solvent divid th000 correct right people so P not a is given PS is given molar mass of solvent molar mass of water 18,000 so from here you can easily get the mity of the solution correct this can be solved it is a simple question it is a very very very simple question very simple question just a second uh just people there is a solution here okay look at this particular question look at at this particular question you should be able to solve this one as well at a given temperature vapor pressure of pure Benzene as per the question you have a pure Benzene this is pure Benzene right vapor pressure of pure Benzene which I'm representing with p a that is given to me as 200 mm of EDG at the same temperature vapor pressure of the solution containing 2 G of nonvolatile solute so you are introducing a nonvolatile solute right how many Gs of nonvolatile solute 2 G of nonvolatile solute right in 78 G of benzene so initially you had how many grams of benzene 78 G of benzene you had initially in the container now in this Benzene you are introducing a nonvolatile solute after that what is happening you are getting a solution you are getting a solution this is the solution which you are getting right perfect and vapor pressure of this particular solution solution comes out to be 195 mm of HG as for the question because you know when a nonvolatile solute is added what happens to vapor pressure it decreases initially it was 200 now it's 195 calculate the mol mass of solute we have to calculate the M mass of salute again a simple question which equation I'll be using p a minus PS / PS is equal NB / na this equation is used everywhere as I told you you need not to check whether it is dilute or concentrated or whatever so p a is 200 PS is 195 PS is 195 NB number of moles of solute will be equal mass of solute divide by m mass of solute divided by number of moles of solvent means mass of solvent divide by m mass of solvent M mass of solvent molar mass of benzene solvent is your Benzene right it's m mass of 78 when you solve the this you'll be getting the M mass of nonvolatile salute which I was supposed to calculate correct you can easily solve it easily means easily okay one more question look at this carefully look at this carefully guys you need to give it a try as well what weight of nonvolatile solute Uria needs to be dissolved in 100 G of water in order to decrease it vapor pressure by 20% by 20% can you give it a try can you give it a try see guys what the scenario is let me make you familiar with the scenario first of all so as per the question you have got 100 G of water you have got 100 G of volatile liquid that's water correct I'm assuming its vapor pressure right now was p a I'm assuming its vapor pressure was p a now as per the question you are adding a nonvolatile solute Uria into it you adding a nonvolatile solute Uria into it what's going to happen after adding nonvolatile solute UA will the vapor pressure increase or decrease it will decrease for sure how much is decreasing it is decreasing by 20% that means vapor pressure of the solution will be initial vapor pressure and it has decreased by how much of amount 20% of p a has decreased I hope you got this particular statement I hope you got this statement initially it was P now we are adding a non volatile salute what has happened V pressure has decreased how much 20% so what will be the final vapor pressure of the solution initial minus 20% of this so it will be vapor pressure of the solution will be P A minus 0.2 it will be 0.8 p a correct this something which I got till here now the question says what weight of nonvolatile solute let's say we have added some WB G of this nonvolatile solute you have to calculate how much nonvolatile solute was added to decrease the vapor pressure by 20% that was the equation so I'll be solving it like this P A minus PS / PS is equal NB / na P A minus PS p a minus PS that will come out to be P A minus 0.8 p a which will be 0 point this ter will be 0.2 p a and here it'll be 0.8 p a so p a p a cancel 0.2 divid 0.8 comes out be 1X 4 is equal NB moles of salute will be equal to mass of solute which is to be calculated divide by molar mass of solute molar mass of Ura 60 number of moles of solvent mass of solvent solvent is water and we have taken it 100 G divide by m mass of solvent M mass of water so one equation one unkn known you can easily calculate WB from here you can calculate WB from here and the second question is calculate the mity for that you can use this expression which I gave you a few minutes back P A minus PSID PS is equal m multiplied mol mass of solvent divide by th000 all the things are given you just have to calculate this M over here is it clear guys there are some other questions which we have to do related to cative properties but those questions we cannot do now once we are done with with bof Factor then we can do those questions some good questions correct how long the session will be the session will be for two to three more hours two to three more hours are you ready to be with me till the end are you ready to be with me till the end are you sure are you sure people yeah we have to complete the chapter you have to 9% more battery so keep your phone on charge keep your phone on charge keep your phone on charge I cannot sleep for long because in the morning at 11: I have one more class Avengers batch okay let's not waste time let's complete this particular elevation and balling Point as well see guys again I'm telling you all these ctive prop properties to explain them basically there are thermodynamic reasons all these are explained with the help of entropy right but I'm not going there right now I'm giving you only that much amount of theory which is required to solve the questions okay yeah so what is elevation and balling Point all about how it happens for that I believe you know what is balling point I hope you remember what is B you remember what balling point exactly is first of if I want to define the boiling point the temperature at which vapor pressure of the liquid the temperature at which vapor pressure of the liquid becomes equal becomes equal to the external pressure which is generally atmospheric pressure right I hope you remember this is something which we have discussed correct this is something which we have discussed now people here comes few important things understand them carefully because I don't know whether you have I don't know whether you have I mean understood this particular scenario before or not see I'm not giving the thermodynamic Reason by the way I'll be giving you the classical reason to understand this elevation and boiling point imagine I'm taking two containers imagine these are two containers which I've have taken in this first container we have got a pure volatile liquid it is pure volatile liquid pure volatile liquid a okay pure volatile liquid a so first of all let's assume that let's assume that this volatile liquid is kept at temperature 25° Cen I'm assuming this volatile liquid is your water this water is at 25° Centigrade right this water is at 25° C I'm assuming the outside pressure is 760 mm of HG the external pressure is 760 mm of HG I'm assuming at 25° Cen the vapor pressure of this water for example it is 100 of NG just to make you understand the things just to make you understand the things this is water at 25° cenra at 25° Centigrade the vapor pressure of water for example is 100 and external pressure that is the atmospheric pressure for example that is 760 now people boiling point is that temperature at which vapor pressure of liquid becomes equal to atmospheric pressure right so first of all if I increase the temperature of this water what will happen it's vapor pressure will keep on increasing and there will be a temperature at which the vapor pressure becomes equal to atmospheric pressure that's called as balling point so imagine at 100° Centigrade imagine when the temperature of water came out to be 100° C at that point of time the vapor pressure of water became equal to the atmospheric pressure which is 760 so that particular temperature at which vapor pressure becomes equal to atmospheric pressure that is something which you call as balling point this is your balling point of water correct now people try to understand in the same container if I add a nonvolatile solute in the same container I have for example added a nonvolatile solute if I added a nonvolatile solute it becomes a solution now A plus b I've added a nonvolatile solute B in this container now it is becoming a solution imagine this solution was initially at 25° Cen 10 tell me at 25° Centigrade will its vapor pressure will the vapor pressure of solution be more than 100 or less than 100 you have to tell me vapor pressure of this particular solution would have been at 25° Cen it would be less than 100 right so vapor pressure of this solution at 25° Cen let's assume it is 50 m of HG outside pressure is same how much is that 760 M of FG now people tell me one thing if you have to boil the solution what you have to do you have to increase temperature when you increase this temperature vapor pressure of this solution will keep on increasing but people you have to increase the vapor pressure of solution from 50 to 760 where you were supposed to increase the V pressure from 100 to 760 where you have to increase the temperature more you have to increase the temperature more because you have to increase the vapor pressure from 50 to 760 the difference is 710 here the difference was 660 here you have to raise the temperature more so I'll say this solution this solution this solution for the vapor pressure of the solution to get equal atmospheric pressure I'll say temperature has to be higher so Bing point of the solution will be definitely more than that of 100 let's keep it as 105° let's keep it as 105 just to make you understand just to make you understand so basically initially when you had pure volatile liquid in the container you are supposed to increase this W pressure from 100 to 760 so accordingly you were raising temperature and one temperature arised one temperature arose at which the vapor pressure became equal to atmospheric pressure that was 100° the boiling point now this particular solution in order to boil this you have to increase this temperature from you have to increase this vapor pressure from 50 to 760 so the Gap here is more so you have to raise the temperature more so it's boiling point will be more right so tell me one thing on adding a nonvolatile solute into a pure solvent is the boiling point increasing or decreasing this was the boiling point of pure solvent this is the boiling point of solution tell me what happened tell me what happened can I write a simple statement like this can I write a simple statement like this on adding a nonvolatile salute into a pure volatile liquid into a pure volatile solvent first of all you know vapor pressure that decreases and at the same time balling Point increases and that increase in the balling point and that increase in the balling point is called as what that's called as elevation in balling point that is called as elevation in balling point so elevation in balling point is represented by Delta TB elevation in balling point is represented by Delta TB how much balling Point has elevated I'll say which one is at which one has got higher balling Point solution lower balling Point higher minus slower so how much balling Point has elevated I'll say balling Point has elevated here by 5° right so I can write it like this elevation in balling point will be equal high that is balling point of solution minus the Bing point of solvent boing point of solvent I hope this particular statement is again clear to you right yes now guys understand the same thing graphically a bit understand the same thing graphically a bit see if this is your vapor pressure and this is temperature vapor pressure versus temperature okay let's say this point represents atmospheric pressure for example which is 180 atmospheric pressure which is 1 atm for example right if I make a graph between vapor pressure and temperature for a pure solvent we know with the increase in temperature vapor pressure increases with the increase in temperature vapor pressure increases right so this is the graph for a solvent for example pure solvent perfect now from the solvent imagine I have made a solution from the solvent I have made a solution vapor pressure of solution will be less than that a pure solvent you know so graph for solvent will lie below this I mean graph for solution will lie below this perfect right graph for solution will lie below this because vapor pressure of the solution containing nonvolatile solute that is less than that of pure solvent right now tell me one thing look at this particular scenario can I say this is the temperature at which this is the temperature at which vapor pressure of solvent vapor pressure of pure volatile solvent becomes equal to atmospheric pressure so this point I'll be representing as boiling point of solvent boiling point of solvent right this point I'll be calling as balling point of solvent this point I'll be calling as boiling point of solent now if I extend the same line it touchs here and accordingly go little down right this is the point this is the temperature at which vapor pressure of solution becomes equal to atmospheric pressure and that particular temperature at which vapor pressure becomes equal to atmospheric pressure that is your Bing point so this is the vapor this is the Bing point of your solution so look from here to here you can calculate boiling point of solvent from here to here it is Bing point of solution as you can categorically say from the see from the graph boiling point of solution is greater than that of boiling point of pure solvent right if I ask you how much the Bing point is increased initially it was TB now it's TB solution so this Gap this Gap over here what does it represent it represent elevation in boiling point and this elevation in boiling point will be PB solution minus TB solvent TB solution minus TB solvent that's what I've written over here correct that's what I've written over here okay that's what I've written over here now guys mathematically elevation in bing point right I'm not deriving the results I'll be just giving you the results and let's apply them elevation in bing point is directly proportional to mity of the the solution more the mity of the solution more the mity of the solution right that means more the nonvolatile salute into it more will be the elevation in bing Point more will be the elevation in bing point and if you remove this proportionality sign you'll be getting a constant KB multiplied by m right this is one result which you have to remember if I would want to elaborate this result a bit if I would want to elaborate this result bit elevation and boiling point will be boiling point of solution minus What minus the boiling point of solvent right is equal to KB I'm writing as such which is a constant mity of the solution is mass of solute in gram multip th000 divided by molar mass of solute multiplied by mass of solvent in Gs okay right you can remember it in this format as well no issues at all correct guys now this KB what this KB is called as this KB over here is called as mol elevation constant this is called as mol elevation constant or you can call it as iosc opic constant and this KB this K KB this KB this KB mol elevation constant it is solvent dependent every solvent has got a particular value of KB every solvent has got a particular value of KB right what are going to be the units of this KB it can be written as de Centigrade per M you can write it as degree Centigrade per M or you can write it as Kelvin kg per mole perfect these are the units of this KB which is particularly solvent dependent right and these are going to be its units now if I ask you how do you define this KB how do you define this constant how do you define this constant KB see if mity of the solution is one at that point of time KB will be equal Delta TB right I'll say if mity of the solution is one at that point of time your KB is is equal Delta TB so how can I define it from here you can easily Define it I'll say mol elevation constant mol elevation constant is defined is defined as the elevation in balling point the elevation and balling point of a solution it is defined as the elevation in bing point when when mity of the solution is Unity this is how you remember it this is how you remember it this is how you remember it right now if I particularly talk about this KB a little more this KB has got certain Expressions as well KB is equal R t² / 1000 lb or M1 RT squide how Delta vaporization now what are these terms first of all what are these terms first of all look here first of all this t t is the boiling point of pure solvent LV is the latent heat of vaporization for 1 G of solvent Delta H vaporization is latent heat of vaporation for one mole of solvent the difference between LV and Delta H LV lat vaporization this is for this is for one g of solent this is for one mole of solvent and R is what you call as gas constant or you can call it a solution constant you know its value 8.314 right units of KB any one of these two and every solvent has got a particular KB KB for water is this right KB for Benzene is this Kelvin kg per mole okay let's try to solve a few questions on the basis of elevation in boiling point but before solving the question let me quickly summarize it when you put a nonvolatile solute in a pure solvent okay when you put a nonvolatile solute into P pure solvent you are going to answer me you are going to answer me when you put a nonvolatile solute into pure solvent what happens to VOR pressure does vapor pressure increase or decrease quickly does vapor pressure increase or decrease quickly I'm talking about vapor pressure I'm talking about vapor pressure when you introduce a nonvolatile solute into pure volatile liquid does vapor pressure increase or decrease vapor pressure decreases boiling point balling Point balling Point increases balling Point increases that increase in the Bing point is called as elevation in Boiling Point represented by Delta TB which is equal KB multipli mity or you can write it as you can write it as boiling point of solution minus boiling point of pure solvent right perfect how do we calculate KB these are the results RT squ divid 1000 LV or M1 RT squ divid Delta / 1,000 delt vaporization what is M1 M1 is the m mass of solvent by the way M1 that is basically the molar mass of solvent M1 is basically the M mass of pure solvent correct okay actually tell me one more thing if mity of the solution is more if mity of the solution is more elevation and boiling point will be more if elevation and boiling point will be more can I say basically boiling point of the solution will be more simple right can I say balling point of the solution will be more throat is gone is done and dusted yeah perfect let's try to solve see right now I won't be able to give you some good good questions because good equations on elevation and boiling point and all the C Properties will be done once we do the V Factor first after V of factor you can solve a lot of questions but right now we can solve some basic basic questions for example this is one basic question this is one base question which we have this is one base question which we have what weight of sucrose what weight of sucrose must be added to 100 G of water to have boiling point of solution as 100.5 de simple and basic question simple and basic question can you solve this so basically if I give you the exact scenario of this particular equation as per this question we have taken water imagine this is water in the container which is your solvent which your pure solvent correct we have taken 100 G of pure water we have taken 100 G of pure water you know boiling point of water boiling point of solvent you know that is your 100° Centigrade correct now as per the question you are adding a nonvolatile solute sucros into it after the addition of nonvolatile solute sucros into it what do we get we get a solution this is your solution basically this is your solution basically the solution which we got as for the question boiling point of the solution is given to me as 10.5° Centigrade right so I'm supposed to calculate what weight of nonvolatile solute sucros has been added what of what weight of nonvolatile solute sucros has been added right so I'll be using elevation balling point so Delta TB is equal KB * m Delta TB elevation in Boiling Point initially it was 100 100 to 100.5 means elevation in boiling point is 0.5° right is equal KB value that's given to us 0.52 mity of the solution mity of the solution will be mass of solute in G multipli th000 molar mass of solute mass of solvent in Gams right now people this this is going to be 0.52 is equal to 0.52 multiplied by this is 0.5 here sorry 0.5 is equal to 0.52 multiplied by mass of nonvolatile solute sucrose has to be calculated multip by th000 molar mass of sucrose you should be knowing it is 342 G per mole mass of solvent mass of solvent is 100 so look at this particular equation one equation one unknown one equation one unknown you can easily calculate the mass of nonvolatile solute here right WB is is calculated done I mean this is a basic question which easily you can do which easily you can do similarly you have got a question find KB of a liquid whose balling point is given and LV is given balling point is given LV is given balling point is given LV is given right so boiling point is given LV is given right R value will be 8.314 that understood put the values in this equation get the value of KB so these are some basic questions guys right but here LV is given in terms of calories calories so r value which r value you'll be using here R has got three values I hope you know R has got three values one is 8.314 jewles per Kelvin per mole one is 2 calories per Kelvin per mole and one more is 0.0821 ATM L per Calin per mole these are the three values of R basically now which value of R you'll be using in this question LV is given in calories so you'll be taking this value of R Bing b is 60° Centigrade so convert it in Calin it will be 333 Kelvin correct and KB you can get easily corre guys there is no need to think a lot over this there is one more that is depression in freezing point that is one more cative property well it is similar I'm just going to write the statement and the equation let's not going into the details remember write the expression and apply the expression in the questions see as for the name is concerned depression in freezing point see on addition on addition of a nonvolatile salute in a pure volatile solvent in a pure volatile solvent freezing point decreases freezing point decreases and that decrease in the freezing point and that decrease in the freezing point is called as depression in freezing point that is called as depression in phrasing point which is represented by Delta TF Delta TF represented by Delta TF so for example let me give you an example over here let me give you one example over here imagine this is a container in this container you have got a pure solvent let's assume its freezing point is 0° Centigrade now if you'll be adding a nonvolatile solute here you'll be getting a solution what you'll observe you'll observe fre freezing point of this solution will be less than this so its freezing point will be for example- 3° C right perfect so how much freezing point has decreased how much freezing point has decreased initially it was Zero now it is minus 3 how much is the decrease in freezing point write it like this higher minus lower higher is solvent so it is freezing point of solvent minus freezing point of solution minus freezing point of solution now in the questions do remember freezing point depression is also directly proportional to marity and it is equal to KF MTI M where KF is a constant right or you can represent it like this Delta TF is equal KF multip mity which is mass of solute in G multip by th000 divide M mass of solute and and this is mass of solvent in GRS correct you can remember them and at the same time this KF this KF is called as mol depression constant molar depression constant or you can call it as the croscopic constant you can call it as the croscopic constant it is again solvent dependent so every solvent has got a particular value of KF every solvent has got a particular value of KF you need to remember that as well you need to remember that as well every solvent has got a particular value of KF and this particular KF value can be calculated with the help of these two results right where T is what you call as freezing point of pure solvent freezing point of pure solvent right this M1 is what you call as molar mass of pure solvent right what do you have LF LF is the latent heat of fusion latent heat of fusion per gram of solvent per gram of solvent you have got Delta HF this is the latent heat of fusion per mole of solvent per mole of solvent correct so by using any of these Expressions you can easily get the value of KF right yes it's croscopic right it is croscopic constant I don't remember the spelling exactly whether it is CR y or CR I think is CR y right something like that CR Y is that perfect so solve this question solve this question people solve this question once I told you only basic basic questions can be solved for now till we do the vanous vector quickly guys find the freezing point of the solution obtained by mixing 18 G of glucose and 90 G of water so the solvent is water mass of solvent is 90 G molar mass of solvent that means molar mass of water that is 18 G per mole the nonvolatile salute which you are adding that is glucose how many grams of nonvolatile salute 18 G right and molar mass of this nonvolatile solute glucose is 180 g per mole perfect solvent is your water correct and freezing point of solvent freezing point of water pure water you should be knowing it it is 0° Centigrade again I'll be using the basic equation nothing else the basic equation says that a freezing point of solvent minus freezing point of solution is equal to KF multiplied by mity of the solution which is mass of solute in gam multipli 1000 molar mass of solute mass of solvent in G phasing point of solvent is zero right so that minus comes on that side it goes on that side so you can get the freezing point of solution is equal to minus KF value is 1.86 mass of nonvolatile solute is given as 18 * by, 000 molar mass of nonvolatile solute is 180 and mass of solvent is again 18 so solve this and get the phasing point of the solution that is something which you were supposed to calculate right so these are some basic basic equations as I told you already right similarly you can solve this question as well if 2.5 g of ethylene glycol is added to 20 G of water find the freezing point of solution can you do it on your own because it is similar question it is similar question can you do it on your own let me know once can you do it on your own people quickly yeah can it be done again the same procedure you have to do the question is based on the same procedure you know it Delta TF which is TF of solvent minus TF of solution is equal AF multiplied mity all the parameters are given all the parameters are given now let's talk about the last ctive property and after that there is one interesting topic that is your that is your van of vector right basically there are only few slides which we have to discuss now only some 10 to 12 slides are there right some 10 to 12 slides are there which we have to discuss good okay okay are you sleeping or what are you sleeping or what are you alive everyone of you are you guys alive are you all alive say say it in the chats I want you to be alive till the end yes you guys are still breathing that's great that's great all right so first of all what is osmotic pressure look at look at this particular scenario my dear students imagine this is the container which I have a bigger container okay I'm dividing this container into two equal parts with the help of a semi-permeable membrane this is a semi-permeable membrane which I'm using right and I'm assuming the semi- permeable membrane which I'm using over here it allows only the moment of solvent molecules it only allows the moment of solvent molecules solvent particles I would say it allows the moment of only solvent particle right now first of all on one side of this chamber on this one side of the chamber I'm keeping a solution a solution on another side I'm keeping a pure solvent pure solent on one side I have kept a solution whose concentration is for example c m on another side I have kept a pure solvent now people what you'll observe here this is for example the Piston which is fitted at the top right let me just let me just increase the height of this container okay I increased the height of this container try to understand what exactly I'm going to say so first of all you have got a solution and the solvent in direct contact with each other with the help of semi- permal membrane whenever you see this kind of the scenario in which in which a solution and the solvent or I can say more concentr solution and less concentr solution are in contact with the help of semi-permeable membrane what you will find you'll find the net moment of solvent molecules you'll find the net moment of solvent molecules you will find the net moment of solvent molecules net moment of solvent molecules from from less concentrated side to more concentrated side that means from solvent side to solution side from solvent side to solution side or I can say less concentrated side to more concentrated side the more concent decide whenever you'll be having pure solvent and a solution in direct contact with each other with the help of semi permal membrane or if I say it like this whenever you have got more concentrated and less concentrated solution in direct contact with each other with the help of semi peral membrane there will be net moment of solvent molecules from solvent side to the solution side from less concentrate side to more concentr side and this phenomenon is something which is what you call as OS osis this phenomenon is what you call as osmosis so what is osmosis it is the net moment of solvent molecules from solvent side to solution side or less concentrated side to more concentrated side right perfect the finon is what you call as osmosis now my dear students due to osmosis due to osmosis due to osmosis can I say this piston will start going upwards right due to osmosis I will say this piston it will start going up upwards this this piston it will start going upwards perfect now in order to make sure that piston does not move in order to make sure that piston does not move do we have to apply some excess pressure from the top yes that minimum excess pressure that minimum excess pressure that has to be applied on the solution side the minimum excess pressure that has to be applied on the solution side that this piston does not show any moment the minimum excess pressure that has to be applied on the solution side to prevent the moment of solvent molecules to prevent the net moment of solvent molecules into the solution side right that minimum excess pressure is something which you call as osmotic pressure and this osmotic pressure it is represented by what it is represented by pi so it is pretty much simple guys see basically this a it's a very huge topic if this is taught in detail it's a very huge topic there are a lot of terminologies involved here right but whatever is needed for our need 2024 that is some this is something which you need to remember nothing else okay whatever I'm telling you that's something which you have to remember nothing else otherwise it's a very very very vast topic a lot of new new terminologies different types of pressures right maybe you would have studied in biology as well right we can involve free energy here we can involve entropy here correct okay so basically again I'm telling you whenever a whenever a pure solvent and a solution is in contact with the help of semi membrane there'll be net flow of solvent molecules from solvent side to solution side or from less concentrated side to more concentrated side right this fenon is called as osmosis due to osmosis this piston should go up now in order to stop the moment of piston right in order to make sure that piston should not go up minimum excess pressure has to be given from the top or I'll say the minimum excess pressure that has to be given from the tops on the solution side so as to stop the net moment of solvent molecules from solvent side to solution side that minimum excess pressure is something which you call as which you call as osmotic pressure right which what you call as osmotic pressure and people people the pressure which you are applying from the Top If the applied pressure if if the applied pressure if the applied pressure on the solution side is more than that of osmotic pressure is more than that of osmotic tic pressure if the applied pressure from this particular side is more than that of osmotic pressure what will happen then I'll say there'll be net moment of solvent molecules from solution side to solvent side at that point of time there will be net moment of solvent from solution to solvent side and this phenomenon is what you call as rever reverse osmosis this phenomenon is called as reverse osmosis and this reverse osmosis is used where it is used for desalination of sea water this reverse osmosis it is a technique by means of which you desalinate the sea water perfect so guys if the pressure higher than osmotic pressure if the pressure higher than osmotic pressure is applied on the is applied on the solution side the solvent will flow from solution side into pure solvent side through SPM right and that's what you call as reverse osmosis for example you have got desalination of sea water correct now there are few terminologies which you need to know then I can give you certain expressions for osmotic pressure and we can solve some of its questions I there is a term called as isotonic Solutions what are isotonic Solutions isotonic Solutions are the ones which have got same osmotic isotonic Solutions are the ones which have got which have got same osmotic pressure or you can call them as you can call those Solutions as isoosmotic Solutions as well right since we are talking about the osmotic pressure that is pi osmotic pressure is directly proportional to the concentration of solution it is directly proportional to the temperature as well So eventually what you get you get pi is equal to CR RT this is the expression by means of which you calculate the osmotic pressure of any particular solution now remember the terminologies what is C over here C is what you call as concentration or marity of the solution which is number of moles of solute divided by volume of solution in liters right R is what you call a solution constant right which has got three values you know it and T is the temperature and T is the temperature perfect T is the temperature right so basically those Solutions which have got same osmotic pressure those two solutions which have got same osmotic pressure you call them as isotonic Solutions you call them as isotonic Solutions so isotonic Solutions have got same osmotic pressure that's why they are called as isoosmotic as well how do you calculate this osmotic pressure there is one General expression which is used to calculate this osmotic pressure you can write it as number of moles of solute in the solution divide by volume of solution in liters multip R * T right this is the expression by means of which you can calculate the osmotic pressure basically correct for example you have got a question over here a simple question calculate the osmotic pressure at 273 Kelvin of a 5% solution over here let me tell you this is 5% weight by volume 5% weight by volume solution of UA you have you have got 5% weight by volume per of UA what does that mean it means that 5 G of Uria are present in 100 ml of solution are present in 100 ml of solution what do I have to calculate I need to calculate its osmotic pressure osmotic pressure Pi is equal CRT or you can say MRT the choice is all worse C is concentration concentration is nothing that's marity right so Pi is equal C RT so Pi has to be equal concentration means marity number of moles of solute and the solute is Ura here so number of moles of Ura divided by volume of solution in lers multili R MTI T So Pi has to be equal number of moles of Ura will be equal mass of UA divided by mol mass of UA divided by volume of solution liters this is an mL convert it in lit so it is 0.1 liter R value which I should use I have to calculate osmotic pressure in ATM so I'll be using 0.0821 as the value of R temperature is given to me as 273 kin correct it's a matter of calculation solve it and get the osmotic pressure in ATM yes this is how you can easily calculate solve these sort of questions I believe you can do that now there is one more question which is on your screen a 4% solution of sucrose is isotonic isotonic means means their osmotic pressure is same is isotonic with 3% solution of an unknown organic substance so you have got a solution of sucrose on one side and the solution of some organic substance on another side that osmotic pressure is same that osmotic pressure is same right that osmotic pressure is same means their concentrations their concentrations as well as their temperatures will be same their osmotic pressure same means their concentrations as well as their temperatures will be same what is Pi Pi is CT so it is C1 R T1 it has to be equal C2 R T2 now if concentration temperature is same so C1 C2 cancel right or just a second just a second first of all let's write the statement you have got a 4% solution of sucrose what does it mean 4% solution of sucrose what does it mean all these percentage which will be given here they'll be weight by volume percentage right now guys you can easily I mean decode this statement what does it mean 4% it means 4 G 4 G of sucrose are present in 100 ml of solution similarly organic substance it is weight by volume is 3% that means 3 G of organic substance is present in 100 ml of solution Can You Solve IT accordingly guys can you solve it accordingly can it be done it is just equate the osmotic pressures right equate cosmotic pressures you are done can it be solved so I'm giving you this question as the homework question do let me know its answer in the comment section at the end now the most important topic right that is Introduction to vanov factor that is Introduction to vanov factor I think let's take a break for 10 to 15 minutes then we'll start the V of factor right okay the time is uh 23:59 but I want everyone of you to be back because this is something which is very important guys you have to understand this properly so session I'll be resuming at I'll be resuming at 1220 okay be back on time everyone of you be back on time okay it will take I think it will take one and a half hour more one and a half more more it will take that's it but I want you guys to be back are you planning to be back or you will study this topic on your own if you are not back so then I'll be ending the session then this topic is your homework then okay see you all then see you all see you all see you all at 12:20 is everyone back yes is everyone back right people yes back to business huh back to business thank you shirui thank you good morning guys good morning no need it's completely your choice whether you want to join now or not okay so let's complete the last topic let's complete the last topic of the session that is the vanov factor this is again one very important topic guys okay and this might I mean a lot of students they find it confusing right so I'll try to decode it from the complete Basics itself yeah okay so first of all what this vanov factor is all about why this vanov factor is introduced right what was the need for this particular V of factor everything we are going to discuss in detail okay first of all before understanding this V of factor till now if you remember we have discussed four cative properties one was your relative lowering in vapor pressure and as per the formula it is p minus PS / p is equal mole fraction of solute mole fraction of nonvolatile solute okay this was the first cative property which we have discussed second one was elevation in boiling point which was Delta TB as per the formula it used to be KB * m mity of the solution the next one was depression in phrasing point right which was basically your Delta TF that used to be equal KF multiplied M and at the end it was the osmotic pressure right right osmotic pressure Pi which is equal C * R * T where C is the concentration of the solution right so till now we have discussed four cative properties guys if you if you have a look on all these formulas this is Sky B mole fraction of solute moles of solute divided by total moles present in the solution if you elaborate all the formulas you you will understand one thing see first of all first of all this particular term it is a cative property this particular term it is a cative property this one is a cative property even this one is a cative property right if you elaborate their formulas what you'll observe you'll observe in all the colligative properties what you will see you will see all the colligative properties are inversely proportional to molar mass of solute all the cative properties you'll find inversely proportional to mol mass of solute for example if I take the last one for example if I if I take the last one your Pi is equal CRT right so Pi is equal what is C basically concentration marity number of moles of solute number of moles of solute divided by volume of solution in liters volume of solution in liters and this is R this is T right if you further want to elaborate it number of moles of solute can be written as mass of solute in G divided by molar mass of solute correct in the denominator already you have got volume of solution in lers in the numerator there is r and there is T correct as you can see this osmotic pressure and this molar mass of solute are they directly proportional or inversely they're inversely proportional they are inversely proportional so my dear students if all these are inversely proportional if all these are inversely proportional whatever whatever whatever cative property you take out of these four right all these cative properties you'll find one thing all these ctive properties are inversely proportional to the molar mass of solute all these colligative properties are inversely proportional to mol mass of solute whatever colligative property you take you'll find the same observation cative properties they are inversely proportional to what molar mass of solute this is one point which I want you guys to take a note of correct number one number two guys number two try to understand what exactly I'm going to say try to understand what exactly I'm going to say let me tell you we have exactly got two ways to calculate the value of the cative properties we have got two ways to calculate the cative properties be it any cative property two ways are there one is one is one is you will carry out C certain experiments right one is you'll carry out certain experiments by means of which you will calculate the observed value of colligative property you'll call it as at the observes observed value of colligative property or you can call it as the experimental value of colligative property this is the first way perfect you'll carry out certain experiments and by the experiments you'll get to know the value of any cative property for example I need to calculate relative lowering wer pressure I've got two ways to calculate relative lowering either experimentally either experimentally and whenever I calculate it experimentally whenever I calculate it experimentally I'll be calling the value of the colligative property as The observed value of colligative property understand again what I'm saying understand again what I'm saying I have got two ways to calculate the value of colligative properties for example I need to calculate elevation in bing point there are two ways by means of which I can calculate elevation and boiling point number one I can do some experiment and calculate the value of elevation in boiling point that elevation in boiling point which I calculate with the help of experiments that is something which I call as observed value of elevation and boiling point I hope you're getting it right now there is one more way of calculating the cative property that is with the help of these results that is with the help of these results right whenever you use these results to calculate the value of cig properties you call that particular value as the cative property calculated or you call it as the theoretical cative property theoretical you call it as the cative property theoretical right so basically again I'm repeating the same thing because this is something which is super important every colligative property can be calculated by two ways one is with the help of experiments when you use the experiment when you do the experiment and calculate the colligative property that colligative property which is calculated with the help of experiment is called as is called as observed value of colligative property or experimental value in the similar way if you use the for formas to calculate the ctive property you call them directly as calculated ctive property or thetical ctive property right perfect now guys as I told you few minutes back your cative properties they're inversely proportional to molar mass of solute they are inversely proportional to M mass of salute this is point number one point number two which you need to understand Point number two which you need to understand for example for example I'm taking two containers over here this is container number one this is container number two these are two containers which I have okay two containers assume that in this particular container I have got for example 100 ml of solvent or for example let me take let me take 100 mL of water as the solvent in this particular container also what I'm doing exactly I'm keeping 100 ml of solvent for example 100 mL of water perfect in both the containers I'm keeping 100 mL of water now for example people imagine I'm introducing one particle of Ura imagine I'm introducing one particle of Ura over here as you know your UA is a nonvolatile solute in the second container what I'm assume what I'm doing exactly I'm introducing one particle of NaCl I'm introducing one particle of NAC which is again a nonvolatile salute right what will happen since you are introducing a nonvolatile salute in a pure volatile liquid in a pure volatile solvent what will happen there will be relative lowering pressure there will be elevation in boiling point there will be depression in freezing point right perfect now my dear students if you look at the first scenario if I want to calculate the value of for example elevation in volume Point that's a cative property how many ways I have to calculate the value of elevation in Boiling Point I've got two ways either I can do some experiment or I can calculate it with the help of formula so basically I have got two ways to calculate the colligative properties one will be one will be observed colligative property which is which is done which is calculated with the help of experiments one is one is theoretical or you can call it as the calcul ated value of cative property perfect imagine in this particular scenario I need to calculate elevation and Bing point I can calculate elevation in Boiling Point through experiment or I can calculate elevation in Boiling Point through the formula or for example let's say I need to calculate to lowering in vapor pressure I can calculate it with the help of experiment or I can calculate it with the help of formula perfect similarly in the second scenario let's say I need to calculate the cative property one I can do with the help of with the help of experiment which is called as observed and I can calculate the same thing with the help of formula which I call as calculated correct what you will observe my dear students you will observe in the first scenario The observed value of colligative property will be same as that of calculated value of cative property but in the second case you will not observe the same The observed value of cuga property will be different than that of calculated why is that that why is that why in the second case these two things came out to be different why in the second case these two things came out to be different see guys you can understand it like this Ura is a salute that neither dissociates nor Associates it is a salute which neither dissociates nor Associates so I have taken one particle I've introduced this one particle into this liquid right so basically basically I've introd just one particle over here in the second case I was expecting I was thinking that I had introduced one particle but you know when NaCl gets into water it dissociates as na posi plus CL Nega it dissociates as na POS plus CL Nega it dissociates as na POS plus CL Nega so whatever I was I was expecting that I introduced one partic I over here perfect but the same na it dissociates into two particles basically it dissociates into two particles what is the ctive property that property of the solution which depends on number of particles of solute so I was expecting that I I was imagining right I was considering that I introduced one particle but in reality it got converted into two particles so whatever colligative property I was expecting with the help of formula will it be same when you see the scenario experimentally they'll be different because the value of calculated C property will come due to one particle but in reality that one particle got dissociated into two particles right so so experimentally actually the value of cative property here will be due to two particles correct that is the reason why here why here the calculated and observed cative property comes out to be different yes right is this clear is this point clear to you let me know once in the chats let me make it a little more simple for you let me make a little more simple for you see this particular value this calculate value of cative property we calculate with the help of what with the help of the formula this one is calculate experimentally perfect now what I have done over here I've introduced NAC after introducing one particle of NAC I calculated the value of any cative property with the help of formula I got some value now I'm seeing the same solution in the lab and trying to calculate its cative property experimentally which I'm calling as cative property observed now I'm finding these two are coming out to be different what is the logic what is the reason the reason is I did not take into consideration that this one particle will break into two particles I did not take into consideration but when we saw it experimentally when you saw it experimentally when you analyzed it experimentally this one particle has got dissociated into two particles right so I can say in reality the number of particles of solute they have increased and colligative property depends on number of particles of solute right cative property depends on the number of particles of solute so in reality the value of colligative property will come out to be due to two particles but I had only one particle that is the reason why these two are coming out to be different why here these two are not coming out to be different because it neither dissociates not Associates I had introduced one particle right perfect after introducing one particle of uh this Ura I calculated any cative property perfect I got some value now I analyze the same solution experimentally I got the same thing I got the same value perfect because it neither dissociates nor Associates perfect it neither dissociates not Associates here here it is getting dissociated it is getting dissociated so for those for those solutes which we introduce into the solution which we introduce into the solvent if the salute dissociates or Associates the value of the observed and calculated cig property it comes out to be different right perfect those solutes which either dissociates into solvent or Associates into solvent for then the observed and calculated value of colligative properties comes out to be different that is the reason why in this particular case they came out to be different now people since they came out to be different since they came out to be different now if I if I if I want to equalize left hand side and right hand side right now they are not equal but if I want to make them equal in order to make them equal I'll be introducing a factor over here which I represent with I just to make LHS side and Rh side equal and this factor which we are introducing to make observed and calculated cative properties equal right this factor is something which is what you call as vops Factor so how do I Define the vops factor I I is equal observed value of cative property divided by calculated value of cative property perfect I hope you got to know why this concept of vano factor was introduced yeah perfect right guys so I can say your V of factor I is basically equal it is basically equal observed value of cative property divided by calculated value of cative property perfect perfect okay if I ask you one simple thing over here if I ask you one simple thing over here see all these formulas which we got whether you will be getting correct result always from these formulas tell me that whether you'll be getting correct result from these formulas always no if the salute dissociates or Associates if the salute dissociates or Associates the value which you get by solving these formulas that will be different than experimental that will be different then experimental so from these formulas I won't be always getting the correct value of I won't be always getting the correct value of actual value of cative properties right so in order to get the actual value of colligative properties from these formulas itself I have to multiply by the factor I over here after multiplying by a factor I now I can say that from here I'll be getting the exact actual cative property values yes right is this clear to everyone is this clear to everyone I hope you got to know what was the need what was the need to introduce this fanto Factor so that we always get the exact the correct value of cative properties with the help of these formulas yeah clear to everyone now tell me one thing tell me one thing till now I told you few things let me summarize them let me summarize them first thing I told you cative property is inversely proportional to M mass of salute number one correct number two I told you number two I told you your van of factor your V Factor it is equal observed value of cative property divided by calculated value of cative property correct if cative property is inversely prop M mass of solute can I say can I say cative property observed divide by cative property calculated will be equal in terms of first relation can you say it will be equal molar mass of solute that is calculated divided by molar mass of solute that's observed right is the this particular statement clear to you is this particular statement clear to you is this particular statement clear to you yeah I is equal MC / M because your cative property that's inversely proportional to M mass of salute right inversely proportional inversely proportional I hope this particular statement is again clear to you yeah now tell me one thing tell me one thing that salute which neither dissociates nor Associates I'm calling that as non-electrolyte if you have got a non-electrolyte the salute which is non- electrolyte which neither dissociates nor Associates tell me for that particular solute what will be the value of V of vector I I which neither dissociates not Associates so calculate and observed will be same I value will be equal one I value will be equal one correct I value will be equal to one now tell me one thing that particular salute that particular electrolyte which dissociates if the electrolyte dissociates I value will it be equal to one greater than one or less than one you tell me in the chat what do you think that particular salute which dissociates which dissociates that means in reality you get more particles right and cative property depends on particles more the particles more the value of cative property right right people so observed value of colligative property at that time I'll say observed value of colligative property at that time will be greater than what will be greater than that of calculated perfect which clearly tells you that I value will be greater than one because it is a division of two if numerator is more then denominator I value will be greater than one yes now if you'll be having that electrolyte which Associates which Associates that electrolyte which Associates into the solvent for that cative property observed will be less than that of cative property calculated so I value here will be less than one if you have got the solute which Associates for example you have got a solute which dimerizes so you are introducing two particles of solute and you're expecting the value of colligative property will be there of two part will be there due to two particles but in reality it Associates into one particle right so number of particles in the solution are decreasing I was expecting that I was introducing two particles but in reality it is becoming one particle so calculated value of cative property will be less than that of what will be less than that of the observed value of colligative property due to the reason y i value will be less than one here so do remember that particular non- electrolyte I value is one here it's greater than one here it is less than one I believe it's clear I believe it's clear clear guys is it clear properly so the these are the few things which you need to remember these are the few things which you need to remember okay perfect say yes or no in the chats then I can move ahead now guys try to understand few more things what are those few more things have a look that electrolyte which under goes dissociation that electrolyte which under goes dissociation how do I calculate van of factor for that particular electrolyte have a look how do I calculate I'm giving you direct result and I'll show you how to apply it I'm not deriving the result here okay see for example I've got the electrolyte like this a n is it gives n * a stric coefficient here is one here the stric coent is n and I'm assuming n value I'm assuming n value is greater than one if n value is greater than one so consider it as two so one particle is giving two particles from one particle you are getting two particles right or from one particle you'll be getting three particles from one particle you'll be getting four particles let's say you got a solute which is undergoing dissociation so this solute in the solvent is basically undergoing dissociation correct it is undergoing dissociation how do I calculate this V of factor I will be equal 1 + n minus1 multiplied by Alpha where Alpha is the degree of dision of this particular solute right this is one of the results which you need to remember and it is valid only in case of dissociation whenever you'll be having a salute which under goes Association which under goes Association whenever you have got a solute which under goes Association for example you have got na a it gives a n to Geometry coecient here is n here it is one I'm assuming n value is greater than one so imagine n value is two so two particles of a are giving one particle three particles will be giving one particle four particles will be giving one particle so more than one particles are assembling and giving one particle so this is the case of Association and this Alpha over here I'm calling as degree of associ how do I calculate V factor in this particular case I will be equal 1 + 1 / N - 1 into Alpha this is the result which you have to remember to calculate what to calculate I when the salute under goes Association when the salute under goes Association perfect right guys I hope these things are absolutely clear to you right before doing the questions before doing the questions there are a few more things which you need to remember see guys for example if I talk about if I talk about elevation in bing Point Delta TB in reality Delta TB is I MTI KB MTI M correct yes and what is this Delta TB Delta TB is elevation in boiling point which is boiling point of solution minus boiling point of solent minus boiling point of solent correct do you see Delta TB is directly proportional to I yes it is directly proportional to I Delta TB is directly proportional to I yeah do you agree with that it is directly proportional to I over here let's say I have got that solute which dissociates let's say I have got that salute which dissociates perfect for that I'll say Delta TB will be directly proportional to I and for that I will be directly proportional to n number of ions which are generated in the solution perfect so if vanov factor for the electrolyte is more Delta TB is more if Delta TB is more the difference between them is more and when can be the difference between them more when when boiling point of the solution will be more when boiling point of the solution will be more right so do remember one simple thing when the electrolyte under goes disor at that point of time the boiling point of the solution boiling point of the solution is directly proportional to the vof vector or you can say it is directly proportional to what the number of ions generated if the concentration are kept same right over here I'm mentioning same concentration same concentration what is the meaning of this way do we use this particular statement that is something which I'll let you know in some time that's something which I'll let you know in some time similarly if I talk about freezing point if I talk about freezing point if I talk about freezing point you know your Delta TF is equal I * KF * m which is equal freezing point of freezing point of solvent minus freezing point of solution freezing point of solution correct let's say I taken that electrolyte which under goes dissociation which under goes dissociation right I'll say Delta TF I'll say Delta TF is directly proportional to I right perfect is directly proportional to I and I is directly proportional to number of ions generated so more the I more Delta TF more I more Delta TF more I more Delta TF more Delta TF means more is the difference between the two when can be difference between the two more if freezing point of the solution is less if freezing point of the solution is less so do remember one simple thing your freezing point of solution is inversely proportional to vof Vector right when you have used the electrolyte which under goes dissociation yeah perfect perfect guys and this is also valid when the concentration is kept same what is meant by the same concentration you'll get the idea in some time but before that before solving those sort of questions which are based on these statements let me show you some basic questions let me show you some basic question so that you can properly interpret all these things so I hope all these things are absolutely clear right all these things we have discussed in detail now it is a time to do some questions which will make you understand all these things in detail have a look on this particular question the values of observed and calculated molecular weights of AG NO3 are this and this calculate degree of dissociation of ag3 first of all tell me one thing how ag3 under goes dissociation how this ag3 will undergo dissociation I have to calculate degree of dissociation right how ag3 would have undergone dissociation AG positive plus NO3 negative right so first of all is is the case of dissociation absolutely one particle is giving two particles so n value is two one particle is giving two particles particle so n value here is two from one particle you're getting two particles right now my dear students what do I have to calculate I have to calculate Alpha degree of disso how to calculate degree of disso first of all I have told you already how to calculate I there was one formula which is M calculated divid M observed what is M calculated M calculated is 170 and M observed is how much 92. 64 can you solve this and let me know the exact value which you'll be getting after solving this can you let me know quickly I believe it'll be 1.9 or 1.8 something like that so you got the I value which was understood because we know in case of disso I value is always greater than one right yeah now people try to understand one more thing you got the I value since it's a case of diso instead of I I can write 1 + n -1 into Alpha is equal to 1.9 9 so I can write 1 plus or leave one aside take one on that side n minus 1 n is 2 2 - 1 is 1 so Alpha is equal to 1.9 - 1 comes out be 0.9 this is something which you are supposed to calculate the value of Alpha and you have calculated you have calculated it I'm giving the approximate value okay this approximate 1.9 yeah this approximate this not exact a lot of people are saying the value is coming out to be 1.83 if the value is coming out to be 1.83 then Alpha will be 0 .83 right I hope this is clear I hope this particular question is clear to you all right look at this particular question look at this particular question look at this particular question carefully solve this see it looks difficult because there is something degree of hydrolysis etc etc that's given some marity is given right volume is somewhere given this is all just to confuse you nothing else just to confuse you we have to calculate V factor for nh4cl tell me how this nh4cl under goes dissociation it will undergo dissociation like this nh4 positive plus CL Nega one particle is giving two particles so n value is equal to two now how do you calculate I I is equal 1 + n -1 into Alpha correct right now I we have to calculate right I is equal to 1 + n Valu is 2 - 1 and Alpha is given to me as 0.8 nothing else we need to do so the value directly comes out to be 1.8 so all the parameters which were given in the question they were just given to to confuse you nothing else right they were just given to confuse you okay let's have a look on one more a Millar solution of potassium feride a Millar solution of what potassium feride is 70% dissociated 70% dissociated means Alpha value is 0.7 right find the osmotic pressure at 27° Cen in ATM first of all how this K3 fn6 will undergo disor you tell me that it will be 3 * K POS Plus f e cn6 TR negative right now one particle is giving 3 + 3 + 1 four particles so n value over here is four perfect if n is four can I calculate I I will be 1 + N - 1 into Alpha right which will be 1 + n Val is 4 - 1 Alpha is how much 0.7 so 3 721 so it is 3.1 so you got the V Factor you got the V of vector right you got the V of vector after getting the V of factor we have to calculate osmotic pressure Pi is equal I * C RT right so Pi we have to calculate I value already we have gotten 3.1 concentration concentration is given Millar that means 10 ra^ minus 3 mol right we have to calculate osmotic pressure in ATM so the value of R has to be taken as 0.0821 ATM L per kin per mole temperature is how much 27° Cen which is 300 kin solve this and get the answer exactly in ATM in atmosphere I believe this is again clear to everyone yeah perfect yes Nathan you can do that you can do that so these sort of questions you can solve like this let's see some more questions let's see some more questions let's see some more questions look at this particular question let's see what exactly it means acetic acid dimerizes in Benzene acidic acid dimerizes in Benzene that means two acidic acid molecules two acidic acid molecules right they're dimerizing they're dimerizing so two particles are associating and forming one particle and in case of Association in case of Association the number of particles associating to form one particle that gives you the value of n this is the case of assocition n value is two correct n value is two so first thing what do we have to calculate acidic acid diiz in Benzene to form a solution when this this this is dissolved in this boiling point raised by 0. 36° calculate I and Alpha all right so first of all Delta TB it is based on elevation Bing Point Delta TB is equal i m KB MTI M correct I MIP KB MIP M okay I think there is one thing that is missing in the question that is the value of KB yes the value of KB should be given the value of KB should be given in the question let me check the value of KB in the question exactly just a second people just a second yes KB is equal KB is given in the equation as 2.57 right KB is given as 2.57 now first of all Delta TB that is the elevation in bing Point how much the balling Point has raised 0.36 de so elevation in bing point is 0.36 is equal I multiplied by what is KB KB is 2.57 mity mity is equal mass of solute mass of solute is 1.65 G multip th000 divide by m mass of solute M mass of acidic acid which will be 60 multiplied by mass of solvent mass of solvent Benzene 100 G so from this particular expression you can get the value of I from this particular expression you can get the value of I once you get the I value larika is saying I value is equal to 1.65 okay CH I believe you I value is equal to 1.65 all right right if I value is equal 1.65 Then I then in order to calculate Alpha since it is a case of Association in case of Association I is equal 1 + 1 / nus 1 into Alpha right so I value we have calculated 1.65 is equal to 1 + 1 / n Val is 2 minus 1 multip Alpha from here you can easily calculate Alpha which is a degree of Association so these were the two things which were supposed to be calculated right these were two things which were by the way I value won't be 1.65 this is the case of assocition in case of assocition I value will be less than one right I value will be less than one so find the exact value of I which will be absolutely less than one and put that value of I over here right and get the value of alpha yeah perfect right people okay look at one more question look at one more question look at one more question we have got two solutions one is potassium iodide solution one is sros solution well sacros it is a non- electrolyte it neither dissociates nor Associates right it neither dissociates nor Associates potassium iodide potassium iodide will dissociate for sure into what K positive plus I negative so here n value is equal to two here n value is equal two sross it doesn't it it neither Associates nor dissociates right it neither Associates nor dissociates both are having same concentration 0.1 M here 0.1 mol here both are having same concentration right we are given with their osmotic pressures as well osmotic pressures as well right Pi for potassium iodide is given to me as is given to me as 0.46 ATM and Pi for Cru this is given to me as 0.245 at they have got same osmotic pressure sorry they have not same osmotic pressure they have got they have got what I mean their osmotic pressures are given their osmotic pressures are given find the value of I and Alpha for KI can you do it can you do this question can you do this question can you do this question one thing which I would want to add over here assuming temperature of both the solutions to be same right assuming temperature and assuming temperature of both the solutions to be same now if I talk about Pi for KI Pi for KI is equal C R and T correct by I is equal C rnt and people Pi for srose is given to me as 0.245 uh just a second just a second Pi for sros is going to be equal sorry this is I multiplied this is I for potassium iodide multiplied with CRT this will be I for srose multiplied by C R and T perfect now if I divide these two equations it's going to be Pi for potassium iodide divided Pi for sucrose will be equal I for potassium iodide divided by I for sucrose that's one right rest all the other things will get canceled because C is same R is same a is same okay well people from this particular expression what do I get I'll be getting I for potassium iodide is equal Pi for kiid pi for srose what is pi for KI Pi for k is 0.46 5 for sros is 0.245 right from here you'll be getting the value how much approximately 0.2 something no how much is the value exactly approximately it'll be it'll be 1 point something 1.8 something yeah how much will be the value exactly 1.8 so you got the first question the I value which is 1.8 if you got the I value since you have to calculate Alpha for this potassium iodide as well since in case of dissociation I is equal 1 + N - 1 into Alpha I value for potassium iodide is 1.8 is equal to 1 + n value is 2 - 1 into Alpha so Alpha value will come out to be 0.8 okay so Alpha for potassium iodide came out to be 0.8 this sort of equation again you should be able to solve okay one more type of the question look at these four scenarios 0.1 M UA 0.1 M this 0.1 M this 0 so concentration everywhere is same so leave the concentration part aside I want you guys to arrange them on the basis of their boiling points I want you guys to arrange them on the basis of their boiling points will you be able to do will you be able to arrange them on the base of their balling points you should be in a position to arrange them you should be see V factor for Ura is one imagine 100% disor its V Vector I will be equal it will be giving 3 + 2 5 ions this will be giving three ions this will be giving two ions few minutes back only we have discussed few minutes back only we have discussed few minutes back only we have discussed I've given you one simple statement when the electrolyte under goes disso at that point of time I is directly proportional to boing point of the solution so more the ions generated more is the boiling point so maximum Bing point will be for B followed by what B will have maximum balling Point followed by your C followed by your D and followed by your a followed by your a this is the order of the balling points right this is the order of the balling point so more the ions getting generated right more the elevation in balling Point more the elevation in balling Point means more is the balling point of the solution few minutes back only we have discussed when the concentration is same when the concentration is same at that point of time if you remember this particular slide just have a look just have a look this particular slide I'm talking about boiling point of the solution is directly proportional to I directly proportional to n in case of the electrolyte which under goes dissociation right yeah similarly people similarly if you look at one more case here the concentrations are different here the concentration is different when the concentrations are different at that point of time you have to see I multiplied by C value more the I multiplied by C value you have to c i multiplied by C value V Factor multiplied by concentration I multipied by C value you have to see here right and let me tell you more the I multiplied by C value do remember more is going to be the elevation and Bing Point more is going to be the elevation in balling point and if elevation in balling point is more what does that mean that means TB solution balling point solution minus boiling point of solvent is more right when can be this term more this term will be more only when boiling point of the solution will be more so basically more the IC value more will be the boiling point of solution this answer of this question you can let me know in the chats arrange them on the base of their balling points okay arrange them on the base of their balling points similarly questions can be asked based on the freezing points as well and when the solute under goes dision at that point of of time freezing point of solution freezing point of solution see it is simple you know Delta TF is directly proportional to V of vector and V of vector is directly proportional to number of ions right V of vector is directly proportion number of ions which gets generated during dissociation correct so more the ions produce during dissociation more is going to be the Delta TF Delta DF means freezing point of solvent minus freezing point of solution right more I generated means more depression phasing point that means more is the difference between these two right when can be difference between these two more when freezing point of the solution will be less correct so you can understand it like this you can understand it like this freezing point of the solution a in case of dissociation will be inversely proportional to the V of factor or you can say the number of ions generated this is correct for dissociation when the concentrations are same when the concentrations are same and if the concentrations are different then you have to see I multiplied by C value right so at that time I multiplied by c will be inversely proportional to freezing point of solution I MTI c will be inversal freeing part of solution right and your in case of different concentration I multiplied by c will be directly proportional to Bing point of solution right I'm talking about the dissociation cases perfect so how did you find the I value for na2 SO3 it is na2 SO3 right how does it dissociate 2 * na positive plus s SO3 D negative right so imagine 100% dissociation so 1 particle is get 2 + 1 3 so n value is equal to 3 I value will be 1 + 3 - 1 into Alpha so 2 + 1 is 3 correct that's how I did it perfect and people with this with this your chapter solution is done anded so these points are valid when the concentrations are given same and they are valid for dissociation purpose right when the electrolyte under goes dissociation if the concentration is different then I multiplied by C value you have to check then then instead of I here you'll be writing i c then instead of I here you'll be writing i c okay perfect so with this your one more chapter Bites the Dust right I hope you got this chapter properly is it clear guys is it completely clear tell me in the chats quickly all the things are completely clear I hope whatever we have discussed today I hope you remember all the things right I hope you remember all the things I hope these sort of questions you can easily solve yeah these sort of question questions you can easily solve now and these sort of questions are frequently Asked guys so you have to check the disso part more the ions right more the ions more is going to be elevation and Bing Point more is going to be de pressure and freezing point more is going to be the related to lowering of vapor pressure right more is going to be the osmotic pressure because all the cative properties they are directly proportional to your vant of factor right when the electrolyte under goes dissociation uh next chapter I think it is either going to be isomerism either I'll be doing isomerism the next chapter or thermodynamics or kinetics or ionic equilibrium I'll see I'll see right I'll see hell do let me know in the comment section of this particular video how you exactly liked it was it beneficial or not yeah hello take care guys God bless you all see you see you guys take care