Chem102ColligativeProperties

Colligative properties of solutions: boiling point elevation, freezing point depression, and osmotic pressure.

>> 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.