>> host: Welcome back
to "Chemistry Matters"
and our study of solutions.
Solutions often possess
different properties
than the solvent
by itself would possess.
For example, properties
such as boiling point
and freezing point
are different for pure water
than they are for salt water.
We've all seen footage
of winter storms with trucks
putting salt on the road.
This is because the salt
will lower the freezing point
of pure water and make
the roads safer for cars.
Salt water will also boil
at a higher temperature
than pure water by itself would.
These are two examples of what
we call colligative properties.
Colligative properties
are properties of a solution
that are different than those
of a pure solvent by itself.
These properties are also
dependent on the concentration
of the solute particles,
not just their identity.
There are four types
of colligative properties,
but the two types we'll explore
in this unit are
freezing point depression
and boiling point elevation.
In order to calculate
colligative properties,
we'll need to look
at a third way to describe
how concentrated a solution is:
molality, which is abbreviated
with a lowercase "m".
The word molality looks
and sounds a lot like molarity,
but there's an important
distinction between the two.
While both expressions
have moles of solute written
in the numerator,
Molarity has volume of
the solution in Liters
in the denominator,
whereas molality has kilograms
of solvent in the denominator.
Since both moles and mass
do not change with temperature,
molality, unlike molarity,
is not temperature dependent,
so we'll use this measure
of concentration
when we study solutions
at varying temperatures.
The reason molarity
can change slightly
if an experiment is run
at multiple temperatures
is because the volume
of the solution can change
with temperature.
As a solution cools, the volume
can constrict slightly.
As the temperature increases,
the volume can expand slightly.
This concept is called
thermal expansion,
and it's why engineers
have to place joints in roads
when they build our highways.
Anytime a substance is heated,
there's thermal expansion.
In other words, the volume
that it occupies increases.
The same thing happens
when a solution is heated:
the volume that it occupies
will increase.
That's why chemists have
this third way to express
the concentration of solutions.
Chemists will use molality
to express concentration
when running experiments at
widely varying temperatures.
Molality is also
used in calculations
when we're investigating
colligative properties,
and the effects that
various solutes have on the
properties of solutions, such
as freezing point depression
and boiling point elevation.
Freezing point depression
is a colligative property
that describes how
the freezing point of
a solution is lowered
compared to the freezing point
of the pure solvent.
Boiling point elevation
occurs when the boiling point
of a solution is higher
than the boiling point
of the pure solvent alone.
The amount that a solution's
freezing point or boiling point
can be altered, can be
calculated with a formula.
We'll learn about that formula
in just a bit.
But first let's get out
to our classroom for
some practice investigating
colligative properties
that I'm pretty sure
you're going to like.
>> professor: For this activity
we're actually going to conduct
a delicious investigation
of colligative properties
by creating some
homemade ice cream.
Colligative properties
are properties of a solution,
like the freezing point,
that are different than what
the property would be if it
were just a solvent alone.
For this activity, I'd like
for you and your partner
to follow the instructions
I've put on the monitor.
Each team will place
one cup of milk,
and two tablespoons of sugar,
and one half teaspoon
of vanilla extract
into a plastic zipper bag.
Get as much air out
of the bag as possible
and then seal it tightly.
Move the contents of the bag
around with your fingers until
the sugar dissolves.
Next, you're going to add
some more crushed ice
to the cylinder
and add a generous amount
of the salt.
Then you'll take your
plastic bag that's sealed
and you'll add it
on top of the ice.
And then you'll add
more ice and more salt.
Once the ice is filled
to the top, you're going
to close the jar and make sure
that it's sealed.
Okay, so you'll roll the jar
on the table back and forth
between the two of you
until the mixture turns solid.
This will become evident
when air bubbles
inside the plastic bag
stop moving around.
If too much ice melts
inside the jar
before that happens,
add more ice and salt,
and then continue rolling it.
Once you've finished,
open the jar,
take out the bag,
and scoop out the contents
of your ice-cream,
dividing it equally
between two of you.
And then you get
to enjoy the ice cream!
>> student: Mmm, it's so good.
[laughter]
>> professor: Okay, now
that you're all enjoying
your solutions,
let's attach some chemistry
to what just happened.
First of all I'd like
both teams to take a quick
temperature reading of the
salt, ice, and water solution
that's still in your cylinders.
>> student: Whoa, that's weird.
The freezing point of water
is zero degrees Celsius.
But this says -12,
and it's still going down.
>> student: Yeah, mine says
-7.9 degrees Celsius.
How is it still liquid water
at that temperature?
>> professor: Those are
great observations,
and that was really
the whole point of the lab.
You see, when solutes
are dissolved into solvents
they alter the properties
of that solvent.
So when you dissolve
salt into the water,
you essentially make it
more difficult
for the water molecules
to come together
to solidify into ice,
because there are all these
dissolved solute particles
that are in the way.
This means that
the temperature of the water
has to get colder than
it normally would be
so that it could freeze.
So, since you lowered
the freezing point
of the salt and water and ice
mixture in the container,
your milk and sugar,
and vanilla mixture froze.
>> host: What our students
just got a taste of,
aside from delicious ice cream,
was a real life example
of how colligative properties
work in solutions.
We can do calculations
to determine the degree
to which freezing point
or boiling point will be
altered due to the presence
of solute particles.
And that's an important
part of chemistry.
The formula for calculating
freezing point depression
and boiling point elevation
can be seen here:
delta T equals I Times K
times molality.
As you can see, it requires
three variables.
Delta T represents
the change in temperature
in degrees Celsius
and it's by this amount
that the freezing point
will be lowered
for the solution relative
to the pure solvent alone.
K is a constant that is
dependent on the solvent
being affected.
These solvent constants
are well known by chemists
and are available in tables
like the chart
you can get in the
"Chemistry Matters" Toolkit.
So that part is easy.
M represents Molality.
I represents the number
of particles dissolved
in the solution.
Remember the models of
sugar and salt dissolving
our students drew earlier
in this unit?
Sugar, like other
nonelectrolyte, covalent
substances, will remain
intact when dissolved,
and so it will contribute
one particle.
If the substance is
an electrolyte that will
dissociate into ions
when dissolved,
like the salt, or NaCl,
our students drew,
it has an I value of 2
for the 2 ions that formed.
A salt, like sodium phosphate,
is made of 4 ions:
3 sodiums and 1 phosphate,
and will have an I value of 4.
Similarly, for boiling point
elevation, the formula
would be the same, but you
must use the boiling point
solvent constant for K.
You can get a chart
for boiling point constants in
the "Chemistry Matters" Toolkit
as well.
Let's head back to our
classroom where our students
have finished their ice cream
and are now getting
some practice with colligative
property calculations.
>> professor: Let's get started.
Team 1, I'd like you to solve
a practice problem involving
boiling point elevation.
So first, you're going to
have a solution of
0.73 moles of glucose
in 225 grams of phenol.
And you'll figure out
what the new boiling point
of the phenol is.
The boiling constant or the Kb,
is 3.60 degrees Celsius
per molal, and the normal
boiling point of phenol
is 182 degrees Celsius.
Team 2, I want you guys
to work on a freezing point
depression calculation.
Okay, so in your problem,
you're going to find
the freezing point of
a saturated solution
of sodium chloride,
where you've got 28 grams
of sodium chloride,
and 100 grams of water.
The freezing constant,
or the Kf, is 1.86 degrees
Celsius per molal.
And as you know, the normal
freezing point of water
is zero degrees Celsius.
So once you've finished
your work on your whiteboards,
we'll discuss it.
[students whispering]
Team 1, walk us through
your calculations?
>> student: Okay, molality
is moles of solute divided
by kilograms of solvent.
In this case, we have
225 grams of solvent, phenol,
which equals 0.225 kilograms.
So 0.73 moles of glucose
divided by 0.225 kilograms
of phenol is 3.2 molal.
And I is going to be 1,
since the solute is glucose,
which is a covalently bonded
compound that stays as
one molecule when it dissolves.
And from our chart,
the boiling constant Kb of
phenol is 3.60 degrees Celsius
per molal.
>> professor: Very good.
>> student: Solving for delta T,
we found an answer
of 12 degrees Celsius,
which means the boiling point
of this solution would
actually be 194 degrees Celsius,
12 degrees higher than what
the normal boiling point
of phenol alone would be,
182 degrees.
>> professor: Outstanding!
Team 2?
How did your freezing point
depression problem go?
>> student: So we took
28 grams of salt
and converted that to moles.
We then divided that
by the kilograms of solvent,
which was the water.
Since we had 100 milliliters,
and 100 milliliters of water
is equivalent to 100 grams,
we knew we had .1 kilograms
of water.
>> student: We then divided
the moles of salt by .1 kilogram
and found the molality
to be 4.8 molal.
I in this case would be two,
since we have an ionic compound
that's breaking up to form
two ions in the solution.
And from our chart,
the freezing constant Kf
for water is
1.86 degrees Celsius per molal.
>> professor: Very good
catch there, noticing
the ionic compound.
>> student: Solving for Delta T
we found 18 degrees Celsius.
The freezing point of
this solution would be
-18 degrees Celsius,
which is 18 degrees colder
than the normal freezing point
of pure water alone.
>> professor: Great work,
everyone.
Sometimes these colligative
property calculations
are tricky.
Remember when you calculate
your delta T, that's not
your new temperature.
It's just the amount
by which the normal boiling
or freezing point is changed.
So you still have
to subtract or add the number
that you calculate for delta T
from the freezing
or the boiling point
of the solvent alone is.
But nice work everyone!
>> host: Indeed!
Some impressive work
on colligative properties
from our students.
Now that we've spent some time
on colligative properties,
when you think back to
the beginning of this unit,
you'll understand why we
chose the King of Pops
to start things off.
The pops that King of Pops
make are frozen,
but they don't get as hard
as pure ice because of
the sugar and salt dissolved
in the water of the pop mixture.
The freezing point of the water
is depressed by the sugar
and salt making the pop
a more enjoyable texture,
and delicious.
We'll take a break here
so you can practice
colligative property
calculations, and test
your skills by making
some ice cream of your own.
You can find instructions
for the freezing point
ice cream lab, as well as
practice problems in our
"Chemistry Matters" Toolkit.
And there's even a Closer Look
video that can help.
When you're ready,
I'll see you at the top
of the next video
of the Unit 7 playlist, where
we'll begin to investigate
some extremely important
classes of solutions,
acids and bases.
See you there.