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### Colligatives Properties

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Colligative Properties 1:07
• Boiling Point Elevation
• Example 1: Question
• Example 1: Solution
• Freezing Point Depression
• Example 2: Question
• Example 2: Solution
• Osmotic Pressure
• Example 3: Question
• Example 3: Solution

### Transcription: Colligatives Properties

Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.0000

Last lesson, we discussed the vapor pressure of solutions; today, we are actually going to continue along that same theme, and we are going to talk about something that is classified as the colligative properties.0004

Traditionally, the colligative properties have been three of them: boiling point elevation, freezing point depression, and the existence of osmotic pressure.0016

Sometimes, the vapor pressure of solutions--the things that we discussed last time--they put it under colligative properties.0024

You know what, it really doesn't matter; they are just names; what is important is the actual things, like vapor pressure, like boiling point/elevation, freezing point/depression.0030

But again, it is good to know what it is that they are called: the reason that they are called the colligative properties is because, again, it doesn't have to do with the identity of the species involved in the solution that you are forming; it just has to do with the number of particles in solution.0039

In other words, how much of something do I drop in the solvent?0054

Identity is irrelevant: it is the collection--that what this means--it is the collection of particles--how many particles.0058

So, having said that, let's go ahead and get started.0064

The first of the colligative properties that we are going to discuss is something called boiling point elevation.0068

Boiling point elevation: all right, adding a solute to a solvent (now, this can be a solid solute; it can be a liquid solute; in this case, it doesn't really matter) raises the temperature at which the solution boils--at which the solvent boils.0075

This is kind of interesting: here is where you get into a little bit of a problem, in terms of nomenclature--in terms of the words that we are using, the vocabulary.0114

When we take a solvent (like water--pure water), when we drop some salt into it, well, yes, it is a solution; so now, we can talk about the boiling point of the solution; but really, what we are boiling is the water.0123

I mean, we are not boiling off the salt; so again, when we say solvent, solution...I wouldn't get too picky about "Well, is it this, or is it this? Is it a solvent? Is it a solution?"--the problem is one of vocabulary.0135

It is not one of science: we know what it is we are talking about--we either have a pure solvent, or we have something that is mixed in with the solvent, which we call a solution.0150

Because we like to give names to things, often the names that we give them tend to create a little bit of a stumbling block, as we are trying to explain it; so don't let that get in the way of the understanding of the material; the material is what is important, not the vocabulary.0159

OK, so if we add a solute to a solvent, it raises the temperature at which the solvent boils.0174

We know that water boils at 100 degrees Celsius; well, if I drop in 30 grams of salt or 30 grams of sugar into that solution and dissolve it, now I'm going to actually have to raise the temperature a little bit higher than 100 degrees, depending on how much salt or sugar I put in there.0180

Now, it's not going to boil until it reaches maybe 100.5, maybe 101, maybe 101.5; that is what that means.0198

It raises the temperature at which the liquid boils off into solution.0205

Now, the change in temperature--the difference between a normal boiling point and the boiling point that it rises to--is given by the following equation.0212

ΔT is equal to KB times m of the solute.0221

Now, notice: this is molality of the solute: not molarity; not mole fraction; this is molality of the solute, this time.0230

OK, this is very, very clear; what is important is to know what species we are discussing.0244

This is something called the boiling point elevation constant.0250

It's just a constant, and there are lists of these that are available in the CRC handbook, in your book; so this is called the boiling point elevation constant, and it is characteristic of the particular solvent we are talking about.0254

It is characteristic of a given solvent.0276

There you go; OK, so, well, let's just do an example; that is the only way that this is actually going to make sense.0287

Once again, when you add a solute to a solvent, you raise the temperature at which the solvent boils (boiling point elevation--one of the colligative properties), and the temperature difference at which it boils--the ΔT--is equal to this boiling point elevation constant, times the molality of the solute--the thing that you put into the solvent.0298

Let's do Example #1: OK, a solution is made by dissolving 27 grams of glucose in 110.0 grams of water (so just a nice glucose solution).0320

The solution was found to have a boiling point of 101.18 degrees Celsius.0353

So notice, in this case, the boiling point went up by 1.18 degrees Celsius, which is actually a lot--because, you remember, the heat capacity of water is 4.18 Joules per degree Celsius per gram; so that is actually kind of a lot.0370

Here, what we are asking for is: Calculate the molar mass of the glucose.0385

One of the nice things about the colligative properties is: they can actually be used to calculate molar masses of molecules, and they are still done like this.0397

Mostly osmotic pressure is used, and we will discuss why in a minute; but it's a perfectly valid experimental procedure.0406

OK, well, let's recall what we mean by molar mass; so the molar mass of something is the mass of something, divided by the number of moles that is in that mass.0413

If I have a mass of 10 grams, and it occupies 2 moles, well, that is 10 grams per 2 moles, which is 5 grams per mole.0424

Mass in grams: well, we know what the mass in grams is; we just said that we have 27 grams of this glucose.0432

27.0 grams of this glucose--what we need is how many moles of glucose this is.0441

When we can find the number of moles, we put it in the denominator; we do our division; and we get our molar mass.0448

It's a pretty common question; and again, this is the kind of question that you will often see on the experimental section of the AP exam, because a lot of the experiments that are done in the AP often will rearrange equations to find molar mass.0455

So, be aware of this; this could actually be an actual, specific question that you will see on the experimental part of the AP exam.0469

OK, so let's go ahead and write down our equation: ΔT equals KB times m of solute (and again, m is going to be molality).0479

Well, that is equal to: KB...well, molality is the moles of solute divided by the kilograms of solvent (right?--that is the definition of molality); therefore, if I want to find the number of moles of solute (which is the moles of glucose), I just rearrange this equation.0491

I write (so let me rewrite it again: so it's): 1 is KB times mol of glucose, divided by kilogram of water (because water is our solvent and glucose is our solute).0516

When I rearrange this, I get: moles of glucose equals kilograms of H2O times ΔT divided by KB.0535

OK, well, that equals...well, kilograms of water--we said we had 110 grams of water, which is 0.110 kilograms.0554

Well, ΔT is 101.18-100.0, divided by KB, which for water is 0.51.0571

It is just...for water, that is the boiling point elevation constant.0584

And now, when we run this, we end up with 0.2545 moles of glucose: that is it--we just rearranged an equation that was given to us.0590

We have to give you the equations in some form, but it's up to you to arrange the equation in a way that you need to solve for what you need.0602

That is the whole idea; this is the real difference in science.0610

Equations are given in a certain form; you need to be able to manipulate those equations to get what you want from the equation.0615

That is the real test of understanding.0622

Let's go ahead and go back--molar mass: we said we had 27 grams of glucose; we know we have 0.2545 moles of glucose; when we do this division, we get 106 grams per mole, and I sure hope that is the right molar mass for glucose.0625

I'm pretty confident that it is.0643

That is it: OK, now let's discuss why the boiling point actually goes up.0645

It is the same reason for why the vapor pressure goes up: well, that is what boiling point is--when the vapor pressure of a solution (remember, vapor pressure changes with temperature--well), as I keep raising the temperature, at some point the vapor pressure of the solution is going to match the atmospheric pressure, which means that, when that happens, all the liquid has enough energy to jump up into the vapor phase.0652

If I have solute floating around mixed up with the water, those solute molecules are going to get in the way of the water rising to the surface of the water-air interface; they can't jump up to the vapor phase.0678

So, it takes them longer to get up there; they need more energy to climb through all of those excess solute molecules--work their way through to get to the surface before they jump up.0694

More energy is required; because more energy is required to boil it off, boiling point elevation; that is what is happening--the exact same thing.0704

It is the same for all of it: freezing point depression, osmotic pressure--it is because of the presence of solute particles that get in the way of the solvent.0712

That is all; OK.0720

So now, the second colligative property is going to be freezing point depression.0724

Let's write: freezing point depression; OK, this time, adding a solute causes the freezing point (now I'm going to write this a little bit differently)--it causes the temperature at which a solvent freezes to drop.0729

In other words, in order to freeze the solution, you must make it colder.0774

That is what it means.0790

So, if I take a cup of water, and I stick it in the freezer, at 0 degrees Celsius it's going to freeze.0794

If I take that cup of water, and I drop in 50 grams of salt, and dissolve it, and I stick it in there into a 0-degree freezer, and I pull it out, it's still going to be liquid.0801

It is not going to freeze; in order to freeze it, I have to drop it below 0 degrees Celsius--maybe -1, maybe -5, maybe -10--it depends on how much salt I actually put in there (generally, how much solute I put into my solvent).0816

It lowers the temperature at which a solution freezes, which means I have to make it colder.0830

You notice: opposite ends--the freezing point drops; the boiling point increases; so what I actually do whenever I add a solute to a solvent--I increase the temperature range at which something stays a liquid.0836

That is what I am doing.0849

OK, now, this equation is actually the same, except just a different solvent: the change in temperature between the normal freezing point and the depleted, the dropped, freezing point, is equal to--this time it's another constant, but this time it's KF, times...well, you have it: the molality of the solute.0851

And again, molality of the solute--this time, this is called the freezing point depression constant.0873

It also depends on the particular solvent that we are talking about.0878

So, let's see: in the case of water, it's going to be 1.86; so let's just do an example.0881

Example #2: What mass of ethylene glycol must be added to 15.0 liters of water to produce a solution which freezes at -25 degrees Celsius?0891

So, I want to take water, but I don't want it to freeze at 0 degrees Celsius; I need for it to freeze at -25 degrees Celsius.0942

How much ethylene glycol (how much solute) do I have to add to that water so that the freezing point drops by 25 degrees Celsius?0951

That is a huge drop; OK, this ethylene glycol--this is what they add to antifreeze; this is what gives antifreeze its green color.0957

Antifreeze is exactly what it means; under normal conditions across the country, often the temperatures drop below 0; well, you can't have water in your engine trying to cool it at 0 degrees Celsius; it's going to freeze.0965

So, we need to drop the temperature: we put ethylene glycol in it as antifreeze that keeps it from actually freezing.0978

The temperatures can drop to -5, -10, -15, -20, and the water in your engine won't freeze and crack the cylinder block and things like that.0985

OK, so let's see: we said 15 liters of water, to produce a solution which freezes at -25 degrees Celsius.0995

OK, so we have our equation; so now, if I can find the moles of solute--if I can find the moles of the solute--I can use the molar mass of ethylene glycol to give me an actual mass.1002

That is what I am going to do.1022

Let's go ahead and write: ΔT equals the freezing point, times the molality.1025

Well, we know what the molality is: by definition, molality is moles of solute over kilograms of solvent; it is the same thing as before.1031

Therefore, I have...let's see: moles of solute (which, in this case, is the ethylene glycol) is equal to ΔT times the kilograms of water, divided by this freezing point depression thing; and now, let's go ahead and calculate some of these values.1045

I'm going to do these over here on the right.1068

ΔT is equal to the absolute value of -25 degrees Celsius, minus 0 degrees Celsius; and again, ΔT, the temperature difference, is always given as a positive, so we take the absolute value.1070

It is 25 degrees Celsius.1087

In other words, we want the difference in temperature, between 0 and -25--it's 25 degrees.1089

Now, let's see: OK, so let's put it in now.1095

ΔT: so we have 25 degrees Celsius; well, how much water do we have--what is the mass of water?1103

Well, we have 15 liters; 15 liters is 15,000 milliliters; 15,000 milliliters is...well, the density of water is 1 gram per milliliter; it's 15,000 grams, so it's 15 kilograms; there you go, right?1109

1 gram is 1 milliliter; 1 kilogram is 1 liter of water.1128

We have 15.0 kilograms (right?--because mass has to be in kilograms here; that is molality), divided by 1.86 (which is the freezing point depression constant for water), and when I do this, I end up getting 201.6 moles.1133

Well, OK: so now, let me go ahead and take 201.6 moles, multiplied by 62.1 grams per mole (which is the molar mass of our ethylene glycol).1152

Moles cancel, and my final number is 12,520 grams, which is equivalent to 12.5 kilograms.1172

So, if I have 15 kilograms of water (in other words, 15 liters of water), I am going to add to it 12.5 kilograms of ethylene glycol (almost the same amount--it's going to be roughly 13 or 14 liters of ethylene glycol), and that will actually drop the freezing temperature of that solution by 25 degrees Celsius.1185

So now, I can use it as a coolant in my car, and I don't have to worry if the temperature drops--very nice.1212

OK, now, for the third colligative property: this is going to be osmotic pressure.1220

OK, osmotic pressure: this always seems to be a bit of a problem, discussing this one; it's always kind of strange for kids to sort of wrap their mind around.1226

And understandably so: it's kind of odd; hopefully, I can dispel the difficulty; so let me draw a picture and see if I can make sense of this for you.1236

I have this U-tube, and at the bottom of that U-tube, I have this semipermeable membrane.1248

I say semipermeable, because certain things pass and certain things don't.1254

What passes is solvent molecules--OK, let's just say water is our solvent; so solvent molecules pass.1260

What doesn't pass is solute; so if I have sugar or salt or potassium chloride, potassium sulfate, anything else, it will not pass.1267

Water can pass; that is what it means; OK.1275

On this side, on the left, I have pure solvent; so this is pure water on the left.1278

On the right-hand side, I have solvent, with a whole bunch of solute in it: it could be anything--sugar, salt, whatever--it doesn't really matter; it's just a bunch of free particles floating around.1285

This is solution.1297

Now, if you were to just see something like this, your natural instinct--again, just from your experience, without any sort of scientific training at all--you would know that what this solution wants to do is: it wants to mix with the pure solvent.1301

In other words, these solute particles--they want to distribute themselves evenly (and you know this, also, from your study of thermodynamics a couple of lessons past).1317

It wants to distribute itself evenly, but we said that the solute particles can't pass this way; they can't pass through the membrane.1327

Well, there is another way for these solute particles to participate in all of this excess solvent.1335

What they can do is: they can literally pull solvent across that membrane; and that is exactly what they do.1343

In order to mix...you have this concentrated solution, and you have this dilute solution; what is going to end up happening is: these solute particles want to pull water in to become more dilute.1352

That is what they want: they want to distribute themselves evenly across all the solvent that is available to it.1364

Since they can't pass this way, they pull the water this way.1370

Well, in the process of pulling water this way--so now, water molecules are flowing across this membrane--well, now the water level is rising here.1374

The water level is rising, rising, rising; here it's falling, falling, falling, falling; at a certain point, this excess amount of water that has been pulled into it...well, it has weight, right?1384

That weight is going to be pushing down on this, and yet, these solute particles are still going to want to pull water into it.1398

At some point, the pull this way and the push this way balance out, and it reaches an equilibrium, and there is this certain height difference between this water level and this water level.1406

That right there--that is the osmotic pressure.1419

The best way to think about it is: this, by definition, is the osmotic pressure of this solution--in other words, it is the pull that it is creating; but pressure--we don't think of pressure as a pull.1422

You could think of a pull as a negative pressure, which is fine; now let me give you another way of thinking about this.1439

If I were to put this apparatus together, but if I were to put a little piston (so now let me erase this) on top of that, and apply a pressure to keep the water from actually flowing in--to keep things exactly where they are--that pressure that I apply here...that is equal to the osmotic pressure.1444

Osmotic pressure is the amount of pressure that I have to apply to the top of a solution to keep free solvent from flowing into that solution to dilute it.1465

That is what is going on.1476

Let me write that down: so, osmotic pressure is the minimum pressure exerted at the top of the solution (right here) to prevent osmosis.1478

This pull across the semipermeable membrane--that is osmosis.1518

In other words, it is the pressure that I have to apply to prevent the osmosis from taking place.1525

That is what is happening; that is defined as the osmotic pressure.1532

The equation for this is...osmotic pressure--the symbol for it is pi (the normal mathematical symbol)...equals M (molarity), times R, times T.1538

OK, this is the molarity this time; so, with boiling point elevation and freezing point depression, we had molality; with osmotic pressure, we have molarity--moles per liter, not moles of solute per kilograms of solvent.1551

It is moles of solute per liter of solution.1564

It is the molarity of the solution.1567

R is the gas constant, 0.08206 liter-atmosphere per mole-Kelvin; and T is the temperature in Kelvin (so you definitely have to watch: not degrees Celsius; not degrees Fahrenheit; that is in Kelvin; and this R--it's the .08206, not the 8.35 Joules per mole-Kelvin, the one that we used for the thermodynamics).1571

OK, nice; so that is the equation; now, here is what is nice about osmotic pressure.1604

We'll do an example: Osmotic pressure is valuable (let's write that with a v) because a small concentration of solute produces a high osmotic pressure.1611

In other words, you don't have to put a lot of solute into a solution to create osmotic pressure.1643

OK, boiling point elevation and freezing point depression--they require large concentrations.1653

You saw: in order to get a 25-degree difference, I had to add virtually...almost the same amount of solute as there was solvent (about 13 kilograms of ethylene glycol to 15 kilograms of water).1658

I don't have to do that with osmotic pressure; very little solute actually creates a huge osmotic pressure difference.1672

OK, so here is Example 3; the statement of the question is actually a bit long, but it's not a problem--it's a pretty standard experimental procedure, and it's something that you may run across.1680

A 20.0-milligram sample of a biological macromolecule (and by "biological macromolecule," I mean a protein; it could be a nucleic acid--RNA, DNA; it could be a very large sugar: macromolecule) was dissolved in enough water to make 25.0 (let's write that a little bit better) milliliters of solution.1693

OK, so I have 20 milligrams of this biological macromolecule (let's just say it's a protein); we dissolve it in enough water; we bring the volume to 25 milliliters; OK.1750

The osmotic pressure was measured (which--we can measure it) to be 0.56 atmospheres (that is actually pretty high for a solution) at 25 degrees Celsius.1762

My question is: what is the molar mass of the protein?1787

This is a standard experimental procedure for determining the molar mass of a protein.1790

Well, it used to be; nowadays, we just use mass spectrometry; but that is fine.1794

What is the molar mass (again, if you don't have a mass spectrometer in your lab, this is what you do) of the molecule?1800

Well, what is our equation?--our equation is: osmotic pressure equals molarity, times R, times temperature (Kelvin); it equals the moles of solute (by definition, the molarity is moles of solute over the volume), times R, times the temperature.1818

Therefore...again, I have the mass of this thing; I have the mass--what I need to do is find the moles, and if I take the mass divided by the moles, that is molar mass--grams per mole.1841

I'm going to rearrange this equation; I'm going to solve...moles of solute equals the osmotic pressure, times the volume, divided by R, and divided by T.1854

Does this kind of look familiar: PV/RT? PV=nRT? number of moles equals pressure times volume over R times temperature?1872

Notice: it is the same form as the gas law, except this is not a gas pressure; this is an osmotic pressure, but it's still a unit of pressure.1880

It's pretty extraordinary, isn't it, that for a solution and a gas--things that you wouldn't expect to behave the same--they actually have the same form of the equation.1888

Wow, that is actually pretty amazing.1895

OK, 0.56 atmospheres, times 0.025 liters...right?--because whenever we use the gas constant, .08206, our pressure has to be in atmospheres and our volume has to be in liters, so 25 milliliters--I have to call it .025 liters...1898

So again, watch your units: there is a lot to keep track of--it's not difficult; it's just tedious.1922

There are a lot of little things, a lot of detail, in these.1927

...over 0.08206 liter-atmosphere per mole-Kelvin (I'll leave that off), and 25 degrees Celsius...yes, 25 degrees Celsius; in Kelvin, it is 298 (is that right?...yes, 298 is correct)...1931

So, the moles of solute equals 7.53x10-7 mol--a very, very, very tiny amount.1948

Well, now that we have the moles and we have the milligrams, let's go ahead and do some division.1960

Molar mass (which is what we want): it's equal to 20 milligrams (which is 20.0 times 10-3 grams--just do the definition of milligrams instead of doing the conversion--I am perfectly free to do that), divided by 7.53, times 10-7 mol.1967

When we do that division, we get 2.66x104 grams per mole; that is huge--2,660 grams per mole.1990

That is actually pretty small for a protein.2002

There you go: osmotic pressure was used by using the definition of part of it, molarity (moles per liter); this is the kind of stuff that you will see on the experimental section of the AP exam.2005

Take a look at the equation: if the equation has something like molarity or molality, it's defined by something: molality is moles of solute over kilograms of solvent.2018

Write out the entire equation with each component of that particular thing.2028

And then, see what you need, and you will get what you want.2034

2.66x104 grams per mole--that is the molar mass of our particular protein, using a colligative property--in this case, osmotic pressure.2037

Thank you for joining us here at Educator.com.2048

We'll see you next time; goodbye.2050