>> In this module we're talking about colligative properties,
specifically boiling point elevation,
freezing point depression, and osmotic pressure.
First of all, what a colligative property is.
It's a property of a solution
that depends only upon how many solute particles there are
and not their identity.
So, boiling point elevation.
First of all, this is the important equation
that you're going to memorize.
It says that the change in the boiling point
of a solvent is equal
to something called the van't Hoff factor times the molality
of that solution, times the --
what's called the boiling point elevation constant.
So, first of all, the parts of this equation.
The change in the boiling point is the boiling point
of the solution minus the boiling point
of the pure solvent, before you add something in.
Now, it ends up that for any liquid,
if you add a non-volatile solute,
you will increase the boiling point,
that's what this is describing of that --
of the resulting solution compared to the pure solvent.
Because, this number's always bigger than this,
this will always be a positive number.
This i, the van't Hoff factor, this is the definition,
it's the mols of particles in the solution divided
by the mols of the solute.
If it's a molecular compound, then i ,
the van't Hoff factor is just 1, because for every particle
for that molecular --
that molecule you put in there you just get, you know,
it's the same particle inside.
However, if it's an ionic compound,
then it can dissociate in that solution.
And when it does, you create more particles.
For example, if you put sodium chloride into water,
it dissociates into sodium ions and chloride ions.
So, it would be about 2 over 1.
So, 1 -- for every 1 mol of sodium chloride,
we get about two mols of particles roughly.
Come back to that in a second.
Molality is the same as we saw before.
Remember the definition of molality,
it's mols of solute divided by kilograms of solvent.
And this boiling point elevation constant, you look it up,
you don't have to memorize any of them.
Just, the only thing -- the important thing to remember is
that you look it up based upon what the solvent is, that's all.
So, this van't Hoff factor.
So, like we were saying, if it's sodium chloride,
we'd expect one formula unit of sodium chloride to break
up into two particles, the sodium ion and the chloride ion,
we'd expect it to be -- the van't Hoff factor to be 2.0.
In fact, it's about 1.9.
The reason for this is something called ion pairing.
So, when we put that sodium ion in the water, okay,
most of the sodium ions and the chloride ions separate
from each other, and they're surrounded by a sea
of solvation by water molecules.
However, some of them still say --
stay associated with each other.
And that means instead of getting two particles,
this whole thing here only counts as one particle.
Magnesium chloride, we'd expect to get three particles,
two chloride and one magnesium.
So, we'd expect that van't Hoff factor
to be three, in fact it's 2.7.
And here, just for comparison notice that glucose,
which is a molecular compound does not dissociate.
So, we'd expect it's van't Hoff factor
to be 1, and it actually is.
So, freezing point depression works
like boiling point elevation, except that the freezing point
of a solution is always lower than the freezing point
of the pure solvent when you add a non-volatile solute to it.
So, here delta T F, the change in freezing point is equal
to the freezing point of the solution,
minus the freezing point of the solvent -- pure solvent.
Now, because this would be lower than this,
this will be a negative number delta T F.
When you're putting it here in this equation to use it,
just make it positive, it'll work out better that way.
Same van't Hoff factor, same volatility.
The freezing point depression constant,
again it depends only upon what the solvent is.
You look it up, you don't have to memorize any of them.
So, why does this happen?
Well this is a Phase diagram for water.
Please pardon my attempt to draw curves here.
But, the idea is this, we know from the last module
that by adding a solute
to a solvent we lower the vapor pressure
of the resulting solution.
And so, in this Phase diagram, remember it's solid over here,
it's liquid here, and its gas over here.
So, this is pressure versus temperature.
As we go along this curve,
the pressure along this curve here will be the vapor pressure
of that liquid, in this case water.
So, by adding -- so, we know by adding the solute we go
over the vapor pressure.
That's why the green curve which it the vapor pressure
of the solution, is lower than the white curve,
which is the vapor pressure of pure water.
So, green is the vapor pressure of the solution,
white is the vapor pressure of water.
And so, if we look at, okay, this is one atmosphere,
the pressure -- the external pressure, right?
Let's say that the atmospheric pressure was one atmosphere.
The solution or the solvent will boil
when its vapor pressure is equal to the external pressure.
So, when the vapor pressure reaches one atmosphere,
that temperature is the boiling point.
Because this is -- this curve is higher than this curve,
we reach that point at a lower temperature for the pure solvent
than we do for the solution.
This difference here is the change in the boiling point --
the boiling point elevation going from here pure solvent
to here , the solution Same thing works
with the freezing point, only the opposite way.
This is a temperature at which, you go from liquid to solid
or solid to liquid, the freezing point.
So, from pure solvent, in this case water,
it would be this temperature right here.
The lowering of the vapor pressure of the solution means
that it freezes at a lower temperature here.
And this difference here would be the change
in the freezing point.
Now, we -- remember when we talked
about vapor pressure lowering by an addition of the solute.
We -- and talked about how entropy is involved,
it still applies here, it's the same reasoning.
All right.
So, the other colligative property is osmotic pressure.
This is supposed to be a pi, we call it pi capital.
And osmotic pressure is equal
to the van't Hoff factor, same as before.
Now this M is the Molarity not the molality,
watch out it's a capital M, times the gas law constant,
which we know from before from the gas factor.
And T is the temperature in kelvin.
R, remember is .08206 liters atmospheres per kelvin mol.
And that's only four sig figs.
So, what is osmotic pressure?
Well, it ends up if you put a solution and separate it
from some of its solvent
by what's called a semi-permeable membrane,
the column of liquid on the side of the solution will rise,
and -- but you can stop it from rising
by exerting an -- a pressure on it.
And that pressure for that solution is the osmotic pressure
of that solution.
So, if we look at it, this --
what this semi-permeable solution does is it allows
solvent particles to pass through
but not solute particles.
So, what the -- what'll here is just left to its own,
the pure solvent side is going to try
to dilute the solution side.
And it's -- what it's trying to do is
to get the concentration the same on both sides, zero,
it can never get there.
But what happens is, as the solvent particles pass
through the semi-permeable membrane,
this column of liquid starts increasing in height.
And just like with the mercury barometer we talked
about in the gas chapter, as it increases --
as it goes up higher and higher in this column,
gravity is exerting a force down,
and thus creating a pressure at this barrier.
And when it reaches a height such that the pressure exerted
down balances the pressure
of the solvent coming through, it stops rising.
That difference is a measure of the osmotic pressure.
So, let's do an example.
This one happens to deal with freezing point depression,
but boiling point elevation
and osmotic pressure calculations are pretty similar,
just remember those equations.
So, we have 1 gram of quinine -- quinine, it's a molecule,
a molecular compound, is dissolved
in 10 grams of cyclohexane.
So, quinine is a molecular compound
as is cyclohexane an enzyme.
The freezing point
of the resulting solution is .24 degrees Celsius,
almost 0 not quite.
What's the normal mass of quinine?
If we have pure cyclohexane it ends up that it freezes
at 6.47 degrees Celsius.
And the freezing point depression constant
for the solvent which is cyclohexane is this,
20.2 kelvins per molal.
So, here our approach is going to be,
we're going to use the freezing point depression equation.
And we're going to solve it for the molality.
Because -- okay, so, the freeze -- first the freezing point,
it's the freezing point
of the solution minus the freezing point
of the pure solid.
Which is, .24 minus 6.47 degrees,
it's negative 6.23 degrees Celsius.
And because, the difference in the temperature
in Celsius is the same as the difference in kelvin,
this is also the difference in the freezing point,
the change in the freezing point is also negative 6.23 Kelvin.
Think about that if you're not sure why
that is, but that's true.
And so, now we have the freezing point depression equation,
we're going to solve for the molality.
Here, because quinine is a molecular compound, i is 1.
So, we don't have to worry about it.
If we solve for the molality, just divide the change
in freezing point, we're going to make it positive, by the,
well i, the van't Hoff factor times the freezing point
depression constant, i is 1.
We saw that k is -- k sub F is 20.2,
I gave you that T in the problem.
Do the division, we get .308, three sig figs,
first insignificant figure is a 4, molality.
So, that's the molality of the solution But,
we want the molar mass of the quinine.
So, what we're going to do is, because the definition
of molality is mols of solute over kilograms of solvent.
And we know how many kilograms
of solvent, which is cyclohexane.
We had 10.0 grams, which is 0.1000 kilograms --
two zeros after that.
If we multiply the molality that we just figured
out times the kilograms of solvent, we get mols of solute.
So, this is mols of quinine, the solute.
And all we have to do now to get the molar mass is take the mass
of the quinine in grams divided by the mols, grams per mol.
So, the molality was 1 gram divided by that many mols.
When we do the division we get to three sig figs,
and you are 3.24 grams per mol.
And there you go.