in this video we're going to talk about colligative properties now you might be wondering what is a colligative property a colligative property is a property that is dependent on the concentration of solu particles and not the identity of those particles now this four colligative properties we're going to talk about boiling point elevation freezing point depression osmotic pressure and vapor pressure so let's start with the first one boiling point elevation so what does that phrase tell you well when you add solute particles to a solution the boiling point will elevate or it will go up let's use water as an example here pure water has a boiling point of 100 degrees celsius at sea level that is at an atmospheric pressure one atm the boiling point does depend on pressure as you go up a mountain the boiling point of water decreases but at one atm let me do that again the boiling point is 100 degrees celsius now what happens if we add salt to water as you add salt to water the boiling point of the solution goes up salt is the solute water is the solvent but combined they make up the solution now just to give you some numbers here this is going to be the molality of the solution and this is going to be the boiling point if you use a 1 molar nacl solution if you have that the boiling point is going to be approximately 101 degrees celsius if you have a 5m nacl solution the boiling point will be approximately 105 degrees celsius technically it's like 105.5 and if you use a 10m nacl solution the boiling point is going to be 110 degrees celsius to calculate the boiling point you could use this formula the change in temperature is equal to kb times m times i kb is the boiling point elevation constant m is the molality i is the vent hot factor by the way i think i said 105.5 for 5m solution it should be like 105.1 the kb for water is 0.51 and if the molality is five and the vet hawk factor is two the reason why the vet have factors two is because when sodium chloride dissolves in water it creates two solu particles na plus and cl minus so when you multiply these three numbers out you're going to get 5.1 and that's the change in temperature so the original boiling point was 100 thus the boiling point is going to increase by 5.1 so it's going to be 105.1 so thus you can see why boiling point elevation is a colligative property it depends on the concentration of the solute particles as the concentration of the solu particles go up the boiling point of the solution goes up and also it depends on the vent hot factor as well the more ions that are dissolved in a solution the boiling point will go up as well now the next type of colligative property that you need to be familiar with is something known as freezing point depression so think about what this expression tells you the word depression means something low something that's down so as you add salt to water the freezing point it goes down it decreases the formula for the freezing point depression it's equal to negative kf times the molality times the van gaal factor the negative sign is to help you to see that the freezing point goes down if you don't use the negative sign then kf is going to have a negative value so if you have a positive sign kf will be negative in the case of water it's going to be negative 1.86 degrees celsius per molality by the way molality is equal to the moles of the solute particles divided by the kilograms of solvent so that's how you calculate the molality of the solution for those of you who might be wondering so as more salt is added to water the freezing point let's put fp for freezing point it goes down and this property is very useful particularly in the wintertime perhaps if let's say for those of you who live up north you probably notice that during the winter time salt is added to the roads in order to prevent freezing and when you add salt to water it makes it difficult for the liquid water molecules to freeze and so that's why adding salt to ice it helps ice to melt because the freezing point goes down and that's why saw is typically added to the roads in the winter time it makes it difficult for ice to form but easy for ice to melt now let's add some numbers to this situation so let's see the relationship between molality and freezing point if the concentration is zero that is the concentration of the solute we have pure water water freezes at zero degrees celsius now if we have a 1m sodium chloride solution if you multiply negative 1.86 times 1 times the two ions in the sodium chloride solution you're going to get 3.72 but if you apply the negative sign it will be negative 3.72 so that means the freezing point will decrease by 3.72 units from zero so the new freezing point of the solution will be negative 3.72 degrees celsius if we were to increase the solute concentration by a factor of five the freezing point will increase by a factor of five that is the change in the freezing point so it's going to be negative 18.6 degrees celsius now let's say if we were to use a 5m solution but of a different substance alcl3 the identity of the substance doesn't matter but what matters is the quantity of the solute in the substance aluminum chloride can break up into four ions we can get the aluminum three plus cation so that's one ion and we can get three chloride ions so one formula unit generates four ions which means our i value is four so if we multiply negative 1.86 times five times four that will give you a freezing point or the change in freezing point which will be negative 37.2 degrees celsius but starting from zero the new freezing point will be that value so now you can see why freezing point depression is a colligative property because this property depends on the concentration of solu particles as the concentration goes up the freezing point depresses it goes down it doesn't depend on the identity of the solute particles but it depends on the concentration for instance if i were to use a 1m solution of ki and both of these have two ions per formal unit the identity is different but the concentration is the same thus the freezing point will be the same and that's what makes it a colligative property it depends on the concentration of the solute particles and not the identity of the solu particles now let's move on to the next colligative property the next one is vapor pressure this property also depends on the concentration of the solute particles now let's briefly talk about why the vapor pressure of the solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the solvent now the mole fraction of the solvent let's say if we're using a saltwater solution the water is going to be the solvent so this is going to be the moles of h2o divided by the total moles of the solution which will be the moles of the salt that we have let's use sodium chloride plus the moles of water so looking at this equation we could see that the vapor pressure of the solution depends on the moles of solu particles so if we were to increase the moles of the solu particles what's going to happen to the vape pressure will it go up or will it go down well as we increase the moles of solute particles we are increasing the value of the denominator of the fraction a fraction has a numerator and it has a denominator whenever you increase the value of the denominator of a fraction the value of the whole fraction goes down there's an inverse relationship between the two so as we increase the moles of solu particles in the solution the mole fraction of the solvent that is this entire fraction goes down in value as the mole fraction of the solvent goes down in value the vapor pressure of the solution goes down in value and so we see an inverse relationship between solute concentration and the vapor pressure of the solution as the moles of the solute goes up the vapor pressure of the solution goes down making it a colligative property now let's consider the fourth one the fourth colligative property we're going to talk about today is known as osmotic pressure here's the formula for osmotic pressure it's pi is equal to m rt times the velha factor i capital m is the molarity r is the gas constant 0.08206 and its liters times atm divided by moles times kelvin so because r has the units kelvin the temperature has to be in kelvin keep in mind the kelvin temperature is the celsius temperature plus 273.15 the molarity is different than molality keep in mind molarity is the moles of solute divided by the liters of solution and let's contrast that with molality molality is equal to the moles of solute divided by the kilograms of solvent so both of these quantities are different forms of concentration but the result is the same as we increase the concentration of the solute particles the osmotic pressure is affected in this case there is a direct relationship between the two the osmotic pressure goes up so let's review the four colligative properties number one boiling point elevation as you increase the concentration of the solute particles boiling point goes up number four osmotic pressure as you increase the molarity osmotic pressure goes up so for those two there's like a direct relationship now for number two and three freezing point depression and vapor pressure there's like an inverse relationship if you increase the solute concentration the freezing point goes down and if you increase the solid concentration the vape pressure goes down so let's just write that so you can see a summary so as we increase the concentration of the solute or the salt the boiling point goes up and the osmotic pressure goes up the vapor pressure goes down and what was the other one freezing point the freezing point goes down as well so that's the relationship between the cy concentration and the four colligative properties now let's work on a few problems associated with boiling point elevation and freezing point depression so in this example problem we have 20 grams of sodium hydroxide dissolved in 200 grams of water our goal is to calculate the boiling point of the solution and we're given the boiling point elevation constant kb it's 0.51 so what can we do here this is the formula that we need the change in the boiling point is going to be equal to kb times the molality of the solution times the van hauf factor the delta t is the difference between the boiling point of the solution and the boiling point of pure water so i'm going to move this to the other side of the equation so the boiling point of the solution is going to be the boiling point of pure water plus kb times m times i so we know the value of kb and for the van hauf factor sodium hydroxide consists of two ions the na plus ion and the hydroxyl ion so when sodium hydroxide is dissolved that one formula unit becomes two ions thus the van't hoff factor is two for this problem now the only thing that we're missing is the molality of the solution molality is moles of solute divided by kilograms of solvent so we're given the grams of the solute 20 grams of sodium hydroxide what we need to do is convert it to moles to do that we need the molar mass of naoh so we need the periodic table the atomic weight of sodium is 22.99 let me get my periodic table just to make sure i have it correct and yes that's it 22.99 for oxygen is 16 and for hydrogen it's 1.008 so this is 39.998 which i'm going to round to 40 because 20 and 40 it's like just half of each other so one mole of sodium hydroxide has a mass of approximately 40 grams of sodium hydroxide so the unit grams cancel and now we have moles of solu we need to divide this by the kilograms of the solvent the solute is naoh the solvent is water so we need to convert 200 grams to kilograms one kilogram is a thousand grams so anytime you need to convert from grams to kilograms simply divide by a thousand two hundred divided by a thousand is point two so we're going to divide the moles of solute by point two kilograms of solvent so this here is going to give us the molality of the solution now let's get rid of that and let's plug in the numbers so 20 divided by 40 that's 0.5 0.5 divided by 0.2 is 2.5 so that is the molality of the solution now let's plug in everything into this formula so we have the boiling point of water or at least we know what it is the boiling point of pure water at atmospheric pressure of let's say at sea level at an atmospheric pressure of 1 atm that is 100 degrees celsius now kb is 0.51 with the units celsius per molality times a 2.5 mole solution times the vector factor of 2. so 0.51 times 2.5 times 2 that's 2.55 and then once we add that to 100 we're going to get the boiling point of the solution which is 102.55 degrees celsius so that's how you can calculate the boiling point of a solution when you dissolve a solute in water now let's move on to the next problem number two determine the freezing point of a solution if 400 grams of aluminum chloride was dissolved in 1600 grams of water and we're given the kf for water is 1.86 so the boiling point of the solution is going to be the boiling point of the pure solvent which is water again but this time it's going to be plus the kf value times the molality times the velha factor now we're dealing with aluminum chloride when you dissolve aluminum chloride in water it's going to break up into the aluminum three plus cation and we're also going to get three chloride ions so notice that we get a total of four ions per formula unit thus the vet half factor is going to be four in this example now let's calculate the molality so let's start with the grams of solute just like we did in the last problem so we need to convert grams of aluminum chloride into moles which means we need the molar mass of alcl3 so we have one aluminum atom and three chlorine atoms using the periodic table the atomic weight of aluminum is 26.98 and the atomic weight of chlorine is 35.45 times three so let's perform the computation so this gives us a molar mass of 133.33 grams per mole so this tells us that one mole of alcl3 equates to a mass of 133.33 grams now our next step is to divide the moles of solute by the kilograms of solvent so we have 1600 grams of water if we divide 1600 by a thousand it's going to give us 1.6 kilograms of solvent so remember the molality is equal to the moles of solute divided by the kilograms of the solvent so this is going to be 400 divided by 133.33 which is essentially 3 and 3 divided by 1.6 gives us this answer 1.875 m so that's the molality of the solution now let's plug in everything to get the final answer so the boiling point i mean rather the freezing point of the solution is going to be the freezing point of water which is zero degrees celsius plus kf now whenever you add a solute to water the freezing point goes down the topic is freezing point depression when you add a solute to water the boiling point goes up thus the expression boiling point elevation so in this problem the kf for water is going to be negative because it depresses or decreases the freezing point so this is going to be negative 1.86 times a molality of 1.875 times the valhalla factor of 4. and so we're going to get negative 13.95 degrees celsius so that is the freezing point of the solution now let's move on to the last problem in this video which of the following solutions has the highest boiling point is it a b c or d well let's write the equation we know that the change in temperature is going to be equal to kb times m times i now kb is going to be constant for all of the solutions we're assuming that the solvent is the same most likely water so the only part that's different is the mi product because for each of the four solutions i mean the four answers the molality is different and the velha factor is different so we've got to find out which of these has the highest mi product value so let's consider answer choice a the molality is 0.35 and aluminum bromide has four ions one aluminum ion and three bromide ions so if we multiply point three five by four that's going to be 1.4 now let's move on to answer choice b the molality is 0.75 but glucose doesn't ionize into ions glucose is it's just a solute particle that doesn't ionize so it has a vent how factor of one now moving on part c we have a molality of 0.50 and calcium chloride has three ions ca and the two cl minus ions so 0.5 times 3 is 1.5 finally answer choice d the molality is 0.8 sodium chloride breaks up into two ions so 0.8 times 2 is 1.6 so therefore answer choice d is the answer because we have the highest mi value for that solution you