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Let's think about what might happen to the boiling point or

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the freezing point of any solution if we start adding

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particles, or we start adding solute to it.

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For our visualization, let's just think about water again.

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It doesn't have to be water.

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It can be any solvent, but let's just think about water

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in its liquid state.

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The particles are reasonably disorganized because of their

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kinetic energy, but they still have that hydrogen bonds that

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wants to make them be near each other.

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So this is in the liquid state, and they have a

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reasonable amount of kinetic energy.

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You know, each of these particles is moving in some

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direction, rubbing against each other, bouncing off of

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each other.

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Now, to move it into the solid state, or to freeze it, what

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has to happen?

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The ice has to enter kind of a crystalline structure.

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It has to get pretty organized, so let's say it has

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to look something like this.

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The water molecules are going to have a regular structure

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where the hydrogen bonds dominate any kind of kinetic

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movement they want to do, and all the kinetic movement,

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they're just vibrating in place.

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So you have to get a little bit orderly

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right there, right?

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And then, obviously, this lattice structure goes on and

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on with a gazillion water molecules.

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But the interesting thing is that this

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somehow has to get organized.

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And what happens if we start introducing molecules into

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this water?

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Let's say the example of sodium-- actually, I won't do

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any example.

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Let's just say some arbitrary molecule, if I were to

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introduce it there, if I were to put something--

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let me draw it again.

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So now I'll just use that same-- I'll introduce some

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molecules, and let's say they're pretty large, so they

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push all of these water molecules out of the way.

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So the water molecules are now on the outside of that, and

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let's have another one that's over here, some relatively

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large molecules of solute relative to water, and this is

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because a water molecule really isn't that big.

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Now, do you think it's going to be easier or harder to

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freeze this?

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Are you going to have to remove more or less energy to

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get to a frozen state?

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Well, because these molecules, they're not going to be part

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of this lattice structure because frankly, they wouldn't

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even fit into it.

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They're actually going to make it harder for these water

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molecules to get organized because to get organized, they

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have to get at the right distance for the hydrogen

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bonds to form.

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But in this case, even as you start removing heat from the

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system, maybe the ones that aren't near the solute

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particles, they'll start to organize with each other.

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But then when you introduce a solute particle, let's say a

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solute particle is sitting right here.

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It's going to be very hard for someone to organize with this

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guy, to get near enough for the hydrogen bond to start

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taking hold.

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This distance would make it very difficult.

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And so the way I think about it is that these solute

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particles make the structure irregular, or they add more

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disorder, and we'll eventually talk about

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entropy and all of that.

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But they make it more irregular, and it's making it

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harder to get into a regular form.

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And so the intuition is is that this should lower the

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boiling point or make it-- oh, sorry,

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lower the melting point.

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So solute particles make you have a lower boiling point.

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Let's say if we're talking about water at standard

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temperature and pressure or at one atmosphere then instead of

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going to 0 degrees, you might have to go to negative 1 or

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negative 2 degrees, and we're going to talk a little bit

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about what that is.

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Now, what's the intuition of what this will do when you

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want to go into a gaseous state, when you

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want to boil it?

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So my initial gut was, hey, I'm already in a disordered

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state, which is closer to what a gas is, so wouldn't that

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make it easier to boil?

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But it turns out it also makes it harder to boil, and this is

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how I think about it.

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Remember, everything with boiling deals with what's

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happening at the surface, and we talked about that in our

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vapor pressure.

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So at the surface, we said if I have a bunch of water

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molecules in the liquid state, we knew that although the

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average temperature might not be high enough for the water

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molecules to evaporate, that there's a distribution of

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kinetic energies.

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And some of these water molecules on the surface

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because the surface ones might be going

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fast enough to escape.

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And when they escape into vapor, then they create a

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vapor pressure above here.

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And if that vapor pressure is high enough, you can almost

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view them as linemen blocking the way for more molecules to

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kind of run behind them as they block all of the other

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ambient air pressure above them.

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So if there's enough of them and they have enough energy,

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they can start to push back or to push outward is the way I

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think about it, so that more guys can come in behind them.

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So I hope that lineman analogy doesn't completely lose you.

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Now, what happens if you were to introduce solute into it?

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Some of the solute particle might be down here.

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It probably doesn't have much of an effect down here, but

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some of it's going to be bouncing on the surface, so

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they're going to be taking up some of the surface area.

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And because, and this is at least how I think of it, since

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they're going to be taking up some of the surface area,

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you're going to have less surface area exposed to the

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solvent particle or to the solution or the stuff that'll

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actually vaporize.

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You're going to have a lower vapor pressure.

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And remember, your boiling point is when the vapor

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pressure, when you have enough particles with enough kinetic

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energy out here to start pushing against the

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atmospheric pressure, when the vapor pressure is equal to the

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atmospheric pressure, you start boiling.

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But because of these guys, I have a lower vapor pressure.

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So I'm going to have to add even more kinetic energy, more

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heat to the system in order to get enough vapor pressure up

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here to start pushing back the atmospheric pressure.

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So solute also raises the boiling point.

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So the way that you can think about it is solute, when you

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add something to a solution, it's going to make it want to

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be in the liquid state more.

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Whether you lower the temperature, it's going to

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want to stay in liquid as opposed to ice, and if you

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raise the temperature, it's going to want to stay in

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liquid as opposed to gas.

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I found this neat-- hopefully, it shows up

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well on this video.

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I have to give due credit, this is from chem.purdue.edu/

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gchelp/solutions/eboil.html, but I thought it was a pretty

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neat graphic, or at least a visualization.

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This is just the surface of water molecules, and it gives

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you a sense of just how things vaporize as well.

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There's some things on the surface that just bounce off.

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And here's an example where they visualized sodium

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chloride at the surface.

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And because the sodium chloride is kind of bouncing

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around on the surface with the water molecules, fewer of

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those water molecules kind of have the room to escape, so

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the boiling point gets elevated.

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Now, the question is by how much does it get elevated?

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And this is one of the neat things in life is that the

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answer is actually quite simple.

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The change in boiling or freezing point, so the change

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in temperature of vaporization, is equal to some

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constant times the number of moles, or at least the mole

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concentration, the molality, times the molality of the

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solute that you're putting into your solution.

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So, for example, let's say I have 1 kilogram of-- so let's

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say my solvent is water.

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I'll switch colors.

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And I have 1 kilogram of water, and let's say we're

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just at atmospheric pressure.

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And let's say I have some sodium chloride, NaCl.

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And let's say I have 2 moles of NaCl.

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I'll have 2 moles.

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The question is how much will this raise the boiling point

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of this water?

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So first of all, you just have to figure out the molality,

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which is just equal to the number of moles of solute,

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this 2 moles, divided by the number of

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kilograms of solvent.

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So let's say we have 1 kilogram of solvent.

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This was, of course, moles.

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So our molality is 2 moles per kilogram.

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So we just have to figure out what this constant is, and

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then we'll know the temperature elevation.

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And actually, that same Purdue site, they

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gave a list of tables.

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I haven't run the experiments myself.

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They have some neat charts here.

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But they say, OK water, normal boiling point is 100 degrees

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Celsius at standard atmospheric pressure.

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And then they say that the constant is 0.512 Celsius

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degrees per mole.

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So let's just say 0.5.

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So it equals 0.5.

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So k is equal to 0.5.

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And I want to be very clear here because this is a very--

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I won't say a subtle point, but it's an interesting point.

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So I said that there's 2-- the molality of-- I just realized

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I made a mistake.

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I said the molality of sodium chloride is 2.

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2 moles per kilograms. But that would be if sodium

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chloride stayed in this molecular state, if it stayed

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together, right?

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But what happens is that the sodium chloride actually

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disassociates, and we learned all about it in

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that previous video.

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Each molecule or each sodium chloride pair disassociates

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into two molecules, into a sodium ion

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and a chlorine anion.

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And because of that, because this disassociates into two,

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the molality is actually going to be two times the number of

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moles of sodium chloride I have. So it's going to be two

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times this.

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So my molality will actually be 4.

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And this is an interesting point.

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If I was dealing with-- and I wrote it here.

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So this right here is glucose, and this is sodium chloride,

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or at least sodium chloride in its crystal form.

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One molecule, I guess you can view it, or one salt of it.

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I guess you could just view it as one of these little pairs

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right here.

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But the interesting thing is is you could have the same

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number of moles of sodium chloride when you view it as a

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compound and glucose.

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But glucose, when it goes into water, it just stays as one

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molecule of glucose.

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So a mole of glucose will disassociate into a mole of

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glucose in water.

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Well, I guess it won't disassociate.

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It'll just stay as one mole, while a mole of sodium

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chloride will turn into two moles because it

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disassociates.

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It turns into two separate particles.

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So in my example, when I start with a mole of this, I end

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up-- actually, once I dissolve it in water, I ended up with 4

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moles per kilogram of molality, because this turns

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into two particles.

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So given that the molality is 4 moles.

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2 moles of sodium, 2 moles of chloride per kilogram.

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So I just use that constant that I just got from Purdue.

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And I get the change in temperature is equal to that

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constant, 0.5, times 4, which is equal to 2 degrees.

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So my boiling point will be elevated by 2 degrees.

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Now, if I had the same number of moles, if I had 2 moles of

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glucose dissolved into my water, I'd only get half as

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much, half as much of an increase.

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Because the molality would be half as much.

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Because it doesn't turn into two particles.

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In some textbooks, you'll actually see it

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00:12:16,720 --> 00:12:18,090
written like this.

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You'll actually see the same formula written like change in

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boiling temperature, or vapor temperature, or whatever you

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00:12:25,860 --> 00:12:30,250
want to think, is equal to k times m times i, where they'll

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00:12:30,250 --> 00:12:34,269
say this is the molality of the compound

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00:12:34,269 --> 00:12:35,110
you're talking about.

256
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In this case, this number would be 2, and i is the

257
00:12:39,879 --> 00:12:43,330
number of molecules or the number of things that it

258
00:12:43,330 --> 00:12:44,639
disassociates into.

259
00:12:44,639 --> 00:12:46,909
So in this case, this would have been 2.

260
00:12:46,909 --> 00:12:49,519
And that's where we would have gotten 4 times k, which is

261
00:12:49,519 --> 00:12:51,019
0.5, which is 2.

262
00:12:51,019 --> 00:12:53,179
In the case of water, this would be-- oh, sorry, in the

263
00:12:53,179 --> 00:12:55,389
case of the glucose, this would still be 2.

264
00:12:55,389 --> 00:12:57,399
But it only turns into one particle when it goes in the

265
00:12:57,399 --> 00:12:58,639
water, so that would be 1.

266
00:12:58,639 --> 00:13:02,370
So you would only have a 1 degree increase in the boiling

267
00:13:02,370 --> 00:13:03,370
point of water.

268
00:13:03,370 --> 00:13:06,460
Now, freezing point is the same thing.

269
00:13:06,460 --> 00:13:10,500
Change in freezing point is also

270
00:13:10,500 --> 00:13:13,179
proportional to the molality.

271
00:13:13,179 --> 00:13:15,799
And you can either say the molality of the original

272
00:13:15,799 --> 00:13:18,689
non-in-water compound times the number of compounds it

273
00:13:18,690 --> 00:13:22,290
disassociates into, although this k is going to be

274
00:13:22,289 --> 00:13:26,439
different for freezing than it is for boiling.

275
00:13:26,440 --> 00:13:29,660
Of course, this k changes at different pressures and for

276
00:13:29,659 --> 00:13:30,370
different elements.

277
00:13:30,370 --> 00:13:33,950
But the really big takeaway is just to realize that even if

278
00:13:33,950 --> 00:13:37,160
you have a mole of this and a mole of that, and they're

279
00:13:37,159 --> 00:13:39,809
going to be dissolved into the same amount of water, because

280
00:13:39,809 --> 00:13:42,159
this dissociates into two particles and this

281
00:13:42,159 --> 00:13:45,169
disassociates into only one for every-- or this

282
00:13:45,169 --> 00:13:47,620
disassociates into two moles for every mole of the crystal

283
00:13:47,620 --> 00:13:50,610
you have-- this doesn't disassociate; it just stays as

284
00:13:50,610 --> 00:13:53,810
one-- this'll have twice as large of an effect on the

285
00:13:53,809 --> 00:13:57,299
freezing point change or on the boiling point elevation

286
00:13:57,299 --> 00:13:59,319
than the glucose will.

287
00:13:59,320 --> 00:13:59,737