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In the previous lecture, we began our

study of Solutions and their Physical properties.

In particular, we observed the effect that a solute in a

liquid solvent can have on the Vapor Pressure of that liquid solvent.

And we observe something call Raoult's Law and then we developed a

model of dynamic equilibrium to account for the changes that we observed.

To review quickly, Raoult's Law simply tells

us that the vapor pressure of the solution,

as depicted here, is equal to the vapor pressure of

the pure solvent multiplied by the mole fraction of the solvent.

As the mole fraction of the solvent decreases due to the presence

of solute, the vapor pressure decreases as is shown in this graph.

We developed an understanding of this, based upon Dynamic equilibrium.

Specifically, the presence of the solute reduces the rate of evaporation

in order to continue to have equilibrium

that must also reduce the rate of condensation.

A lower rate of condensation requires a lower

rate of lower vapor pressure, because it lowers

the rate at which gas molecules colleed, collide

with the surface of the liquid and thereby condense.

Let's now think about what the impact of this has on boiling of that solvent.

Remember, we discussed before, for pure

solvents, for pure liquids.

In order to boil, the liquid's vapor

pressure must be equal to the applied pressure.

Here we have again an applied pressure pushing down on the syringe.

And there is some pressure of the

gas inside the syringe arising from the equilibrium.

In order for us to have equilibrium, the applied pressure and

the pressure inside the syringe must be equal to each other.

If the applied pressure exceeds

the pressure inside the syringe, the syringe will

simply collapse all of the gas into the liquid.

Therefore, to boil, to be at equilibrium,

the vapor pressure must equal the applied pressure.

Therefore, the boiling point temperature is the temperature at which

the liquid's vapor pressure is equal to the applied pressure.

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Let's think about what that means, if we

were to put a solute into the liquid below.

Imagine we have added now a solute here. Then we know the vapor pressure goes down.

The pressure above here is now lower than it used to be.

If it is lower than it used to be then it is now lower than the applied pressure.

The syringe now collapses all of the gas into the liquid form.

We're clearly no longer at the boiling point.

So when we insert the solute into the solution we change the

boiling point of the liquid.

That actually makes a lot of sense if we

think about the dynamic equalibrium which is taking place.

Because it has changed the rates of evaporation,

and so it would seem to make sense

that it would change the temperature at which

we would expect the boiling point to occur.

What do we think we need to do to be able

to restore the boiling point once we've added the solute and disrupted

the boiling?

To figure that out, let's go back and

look at our phase diagram that describes the relationship

between the vapor pressure and the temperature or

alternatively between the boiling point and the applied pressure.

Remember this was our graph from before.

According to Raoult's Law, at each temperature when we add the solute

we will lower the vapor pressure. So, on this particular graph,

add each temperature. Let's imagine that we have added eh,

a solute that low, that is about 5% mole fraction.

That should lower the vapor pressure, let's make it 10% so I can draw it.

That should lower the vapor pressure of

the liquid by about 10% at each temperature.

So, at each

temperature, we've gone down by about 10%. If we've gone down by 10% this

curve has suddenly dropped to look something like this.

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What does that mean?

It means at each temperature, the vapor pressure is

lower so, correspondingly, we will be below the applied pressure.

And points which used to correspond to equilibrium such as here are

now firmly in the range of where the liquid

is because we're above the liquid vapor equilibrium line.

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At a particular pressure, in order to restore boiling,

clearly, we're going to have to go to a higher temperature.

Therefore, there will be an elevation of the

boiling point because of the lowering of the

vapor pressure.

If we lower the vapor pressure, we'll need to

increase the temperature to restore a higher vapor pressure.

To match the applied pressure, and therefore

the temperature of the boiling point goes

up by the amount that we commonly called delta Tb, where b stands for boiling.

Turns out experimentally, we will observe this relationship here.

The amount by which the boiling point goes

up is proportional to the molality, remember we defined that in the previous

lecture, the molality of the solute, in fact it is proportional to it.

This proportionality constant here, is a number

which is unique for each particular liquid.

Kb depends upon the properties of the solvent.

It does not, it turns out, depends upon the properties of the solute.

That we have put into that solvent.

Interestingly, that means, in fact, well, here's an

example that I'll get to before I mention it.

For water that proportionality constant is about a half a degree centigrade, per,

mole of solute per kilogram of solution, in other words per mole a unit.

We have a one molal solution, one mole per kilogram,

the boiling point of that solution or, or, of that water

solution, will go up by, about a half a degree centigrade.

Notice it does not matter what that solute was, the change is dependent

only upon the number of moles of the solute and not on what the solute is.

And that's a surprising outcome. It tells us, that in this diagram, that we

have drawn before, when we think about the effect of the dynamic

equilibrium, between the liquid and the gas.

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That the disruption of the evaporation process

by the presence of solute molecules does not

actually depend upon what those solute molecules are,

but only how many there are in concentration.

That's consistent with the idea that each

of the solute molecules complexes by solvation.

A certain number of solvent molecules and also that the solute

molecules are essentially blocking some fraction of the

surface by complexing the solvent molecules near the surface.

Correspondingly then, the reduction, I'm sorry, the

elevation of the boiling point increases proportionally to

the number of moles of solute we

have added, regardless of what those solutes are.

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But we could actually mix together two liquids each of which

is volatile and each of which therefore is capable of evaporating.

So we might imagine then forming Liquid

Solutions rather than a solid and a liquid.

In this case we may dissolve one liquid into another

where both are volatile.

And in general, we will remember the rule that in order for a solution

to form, the two liquid solutions need to be similar for example in their polarity.

That often means that they will have similar kinds of molecular structures.

For example as described in your text, we consider the comparison

of benzene to toluene the two molecules which are shown here.

The difference between these two is just

the addition of the carbon and three hydrogens

to the benzene ring which is shown.

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Let's imagine now that we mix

together different components, or different compositions,

different ratios of benzene and toluene and measure the vapor pressure above.

So again, we're doing exactly the same thing we've done before.

Except now below, we have a solution that contains both benzene and toluene.

And in the gas phase, we expect

to see both benzene and toluene as well, as both benzene

molecules are capable of evaporating and

toluene molecules are capable of evaporating.

The vapor pressure we observed, will be a combination of the pressures of the

benzene in the vapor phase and the toluene which is in the vapor phase.

What do we observe experimentally?

When we measured the total pressure above the liquid,

as a function of the composition of that liquid in terms of benzene,

we wind up seeing something that looks in fact like a remarkable straight line.

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And for comparison sake here I have given

you, the vapor pressure of the pure toluene.

Now, remember if we are talking about pure

toluene the mole fraction of benzene is zero.

And if it's zero, then the vapor pressure should be exactly equal

to the vapor pressure of toluene and it is.

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If in fact we have pure benzene, then the mole fraction of benzene is 1.

And if we look on the graph where the mole fraction is one, we should arrive

at a vapor pressure which is exactly equal to the vapor pressure of pure benzene.

And it is.

And it turns out that there is a straight line relationship, as we increase from one

to the other.

That's actually perhaps not all that surprising.

Let's take a look at that relationship in a bit more detail over here on the pad.

Remember, from Raoult's Law, we would think if

we had a non-volatile solute in the benzene solution.

That the vapor pressure of the benzene, would be equal to the

vapor pressure of pure benzene multiplied by the mole fraction of the benzene.

That's just Raoult's law applied to

the benzene.

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But we might also expect that Raoult's law would

apply to the vapor pressure of the toluene, as well.

So, the vapor pressure of the toluene, we would expect to be the vapor pressure of

pure toluene times the mole fraction, oops of the toluene.

Going back to our discussion of Dalton's law, we

know that the total pressure of two gasses mixed together

is equal to the pressure of each gas, the

partial pressure of each gas and we sum them together.

This result we called Dalton's Law.

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If we combine these two together, by inserting the vapor pressure

of benzene and the vapor pressure of toluene together we get.

That the total vapor pressure, should be equal to

the vapor pressure of pure benzene, multiplied by the

mole fraction of benzene, plus the vapor pressure of

pure toluene, times the mole fraction of pure toluene.

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You get the linear relationship that we give

to you on the screen in the experimental data.

We can also take

advantage of the fact, that we know that for

some of the mole fractions must be equal to 1.

We showed that in the previous lecture.

Let's insert the mole fraction of toluene as

being 1 minus the mole fraction of benzene.

Then we have the total vapor is the vapor

pressure of pure benzene times the mole fraction of benzene.

Plus the vapor pressure

of pure toluene multiplied now by 1 minus the mole fraction of benzene.

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Let's put the benzene mole fraction components together here.

We can rewrite this as the vapor pressure of pure benzene,

minus the vapor pressure of pure toluene, multiplied by

the mole fraction of benzene plus the

vapor pressure pure toluene.

This equation is exactly the graph that we have observed back here.

It's exactly that, because the Y intercept is the pressure of pure toluene.

Here it is.

On the low end and if we insert the vapor

pressure, I'm sorry the mole fraction of benzene being one.

The vapor pressures of the pure toluene cancel

out and we wind up with the vapor pressure

of pure benzene.

And we notice as well, that the pressure is

a linear function of the mole fraction of the benzene.

So really the data we are observing, can be

understood by simply combining Dalton's law, with Raoult's law.

Now there's an interesting consequence of these equations.

Let's consider a particular circumstance here.

Let's imagine that the mole

fraction of benzene is equal to one half, in

other words, half of our liquid solution, is benzene.

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Then the other half is toluene, so the mole

fraction of the toluene is also equal to a half.

What do we think the vapor pressure of that corresponds to?

Well let's see, it's here right?

Half and if we go across we should be able to calculate that.

We could also calculate it from the above equation.

If we calculate the pressure,

when we take the vapor pressure pure benzene, minus the vapor pressure of pure

toluene times a half and add to it the vapor pressure of pure toluene.

That number turns out to be 61.9 torr.

Okay.

What can we do with that information?

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Let's actually compare that now to the

vapor pressure of the benzene in this solution.

Remember, we have this equation here.

The vapor pressure of the benzene and the

vapor above the solution is equal to the vapor

pressure of pure benzene, which is 95.1 times

the mole fraction of benzene, which is one half.

And if we multiply those two numbers together, we get

47.6 torr. Well, what else might we say about that?

Remember, from Dalton's law, one of the ways we

wrote Dalton's law is that the pressure of one

component can be equal to the pressure of the

total, multiplied by the mole fraction of that component.

And now, I'm going to be careful to say that that mole fraction is now the

mole fraction in the vapor phase for the benzene.

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The vapor pressure of the benzene is now 47.6 torr.

The total vapor pressure is 61.9 torr. And, that must be

multiplied now by the mole fraction in the vapor phase of the benzene.

We can solve that equation

quite easily, right? Just with a calculator.

The mole fraction of the vapor in the benzene, turns out

to be equal to, where's my calculation, 0.77.

Well, why is that significant?

The vapor pressure of the benzene in the vapor phase is 0.77.

But remember, the vapor pressure of the benzene in the liquid phase was one half.

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It therefore also has a lower mole fraction of toluene.

Why more benzene than toluene? Well, look at the data.

The data here tell

us, that benzene has a higher vapor pressure and is therefore more volatile.

Toluene has a lower vapor pressure and is therefore less volatile.

Therefore, when we take the vapor which is above a

liquid solution, that vapor is richer in the more volatile component.

That aspect actually would allow us to then take that

vapor away, condense it and we would have a solution

which is now richer in the

more volatile component, higher concentration of benzene.

And that is the fundamental basis of what is called fractional distillation.

The ability to separate two liquid components by

taking advantage of the difference of their vapor pressures.

We are going to continue this study, the properties of solutions

in terms of other physical properties in the next lecture.