and this is our chemistry channel it's called easy chemistry for all um it would be great if you can share it with all your friends and and classmates so that everyone benefits already many students are benefiting from these videos so that would be great for you and for them to subscribe and share uh the channel uh with all the students that you know on social media or your friends and so on okay so so we are starting in uh or We are continuing module 13 mixtures and solutions from The Inspired mystery book and this is listen to solution concentration part two here's a reminder about the learning outcomes of this lesson first of all we will learn how to solve problems related to the concentration of solutions by performing calculations involving moles and express the results in various units for example mole per liter molarity mole per kilogram molality parts per million percent by mass and percent by volume in Parts 1 we already covered molarity percent by mass and percent by volume also we will learn how to prepare practically Solutions with specific concentrations by dissolving a solid solute in a solvent or by diluting the concentrated solution today's lesson is about solution concentration part two we already covered solution concentration part 1 in the previous lesson so here's a reminder of the main idea the main idea was that concentration can be explained in terms of percent or in terms of moles so what I would like you to do now is stop the video and you and use the knowledge you gained from part one to complete the K and the W parts of the kwl chart what you know about concentration and what you would like to find out so for the keywords we already covered the definition of concentration and the definition of molarity in part one in part two we are going to learn about molality molality it is the number of moles of a solute dissolved per kilogram of solvent also we are going to learn about mole fraction fraction the ratio of moles of a solute or a solvent in a solution to the total number of moles in the solution now that we have learned how to calculate the molarity using the equation we can learn how we can prepare a molar solution with a certain molarity so how can we prepare a one liter or one decimeter Cube solution of NaCl that has a concentration or molarity of 0.5 molar so how can we do that so the first step that we need to do is to calculate the mass of the solute or measure it all right because we don't have a balance for moles we have a balance for Mass so the first thing that we need to do is find the moles of the solute using the equation of molarity because we have the liters and we have the molarity so using this equation all right we can find the moles of NaCl multiply the volume by molarity and now we can multiply the moles by the molar mass to get the mass of NaCl and the mass of nacla is 29.25 grams now we can weigh this 29.25 grams of NaCl and after that we can paste it in a volumetric flask with the correct volume a one liter volumetric flask add a little bit of water dissolve it and then keep adding water until we reach the one liter mark because now we have a solution of one liter let's check your understanding so here's a quick fire question which procedure or technique is correct for preparing a 0.5 gram NaCl solution of one liter is it adding one liter of water to 0.5 grams of NaCl or is it adding 0.5 liter of water to 0.5 gram NaCl and then keep adding water until it reaches the one liter Mark which one is the correct one you can stop the video and think about it all right so as we explained in the previous Slide the correct answer is going to be the second one because the total volume here is going to be 1 liter but imagine if you add one liter to a 0.5 grams so it's going to be larger or the volume is going to be larger than 1 liter so we are no longer in the one liter there is another way in which we can prepare Solutions with certain molarity and that is diluting molar Solutions here's an example your chemistry teacher challenged you with this question during the molarity lesson how would you decrease the concentration or dilute a stock solution of 4 molar NaCl to two molar in ACL you can stop the video use the hint and try to answer the question alright thank you for your efforts so let's begin with the equation molarity equals moles of solute over liters of solution so we know that this these two relationships are there molarity is directly proportional to the number of moles and indirectly proportional to the volume so if we reduce if we reduce the number of moles of NaCl we will lower the molarity and if we add more water we will lower the molarity because we will have more volume right all right keep this in your mind especially the second point and let's watch the video we often prepare a solution of a given concentration by diluting a more concentrated solution the addition of water to a solution reduces its concentration but does not change the number of moles of solute the number of moles of solute before dilution is the same as the number of moles of solute after dilution molarity is defined as the number of moles of solute divided by the volume of the solution hence the number of moles of solute is the product of molarity and volume since both Solutions the concentrated one and the dilute one contain the same number of moles of solute we can calculate the molarity of the dilute solution from the relationship PSI 1 V1 equals PSI 2 B2 [Music] all right so this is the relationship that we get from the video N1 equals N2 so we can use this relationship to figure out how much of water do we need to add to reduce the molarity let's solve an example on the equation that we studied in the last slide so how much water should be added to 125 centimeter cube of a 1.2 molar solution of Koh potassium hydroxide to obtain a 0.5 molar solution of Koh you can stop the video here try to fill in the gaps and try to find the final solution okay thank you so let's do it together so this is a dilution problem because you have volume and the concentration and then another lower concentration so you know you are going to use this equation okay so C1 times V1 equals C2 times V2 so let's fill in the parts of the equation so what is the first concentration or C1 it's given to you here 1.2 all right and what's the volume or V1 125 centimeter Cube deposit here is going to equal what is the second concentration or the new concentration that you want Your solution to have 0.5 molar multiplied by V2 which is the unknown we don't know it yet so try to find V2 using your math skills so here you have 1.2 multiplied by 125 over 0.5 and the answer is going to be 300 centimeter Cube did we finish the question no because the question is asking you how much water should be added to so this is the final volume but how much should I add to 125 to reach 300 it is simple just subtract volume 1 from volume 2. so the volume of water that I need to add is going to be V2 which is 300 minus V1 125 and that is going to equal to 75 centimeter Cube is that correct no it is going to be equal to 175 centimeter Cube so far we have discussed three ways to calculate the concentration using mathematical methods the first one is percent by mass second one is percent by volume and the third one is molarity now in ideal Solutions we will be using molarity but sometimes we have changes in temperature and that affects the volume so what would happen to the volume when the temperature increases usually the volume is going to increase right because when the temperature increases the distance between atoms increases and thus the volume increases right so the molarity is going to decrease in order to avoid such problems we use or chemists use another way to find concentration and that is molality it's the ratio of the number of moles of a solute dissolved in one kilogram of solvent and we find that using the using this equation molality equals moles of solute over kilograms of solvent pay attention here it's kilograms of solvent not solution okay so here's an example what is the molality of a solution made from dissolving 0.5 mole of cacl2 in 500 grams of water so you know the moles of cacl2 and the mass of water and you don't know the molality if you write down the equation molality equals moles over the mass in kilogram just plug in the numbers so it's 0.5 mole of the solute cacl2 over 500 grams of the solvent which is water but we need it in kilograms so we convert the grams into kilograms by dividing by 1000 and finally this give us or gives us the solution as one molal or one M referring to the molality the final method we use to calculate concentration using a mathematical method is using the mole fractions so if you know the moles of a solute and the solvent you can express the concentration of a solution using the mole fractions and here is an example so x a or a move fraction which is X the mole fraction for substance a is equal to n a over n a plus and B so the moles of the solute over the moles of the total solution okay and similarly here if I have b as the solvent so the mole fraction of the solvent is going to be moles of solvent over the total moles of the solution okay so here's an example a solution of HCL is made from 0.99 mole of HCL the solute and 3.6 mole of water and solvent find the mole fractions so we have two substances here all right we have HCL and we have water so we just use the examples here so the moles of HCL over the moles of HCL plus the moles of water is going to be 0.99 over 0.99 plus 3.6 so the mole fraction of HCL is 0.22 and if we do the same for water we will get 0.78 if you add both you will find that it results in 1. so that means that 0.22 of the moles are for HCL and 0.78 of the moles are for water let's revise the concepts that we studied using true or false questions so a diluted solution of NaCl contains more solute than concentrated solution of NaCl of the same volume and the correct answer is false because it should contain less solute concentration can be described relatively and quantitatively numerically that is true molarity is defined as the moles of solute that are dissolved per liter of solvent you might think it is true but it is false because we said per liter of solution so this is wrong molality is used to calculate concentration because temperature alterations can affect molarity that is true because temperature affects volume and molarity depends on volume finally the sum of mole fractions of a certain solution must equal to one that is true as we have seen in the last example we have reached the final summary of part 1 and part 2 of solution concentration what I would like you to do now is to try and fill out the L part of the KW chart it is about what you have learned all right thank you very much for your efforts so we have covered the following in Parts one and part two concentrations can be measured qualitatively and quantitatively molarity is the number of moles of a solute in one liter of solution molality small m is the number of moles of solute dissolved in one kilogram of a solvent finally the number of moles of a solute does not change in a dilution process