COLLIGATIVE PROPERTIES
Physical properties of a solution that depend on the amount of solute in the
solution rather than what the solute is, are called colligative properties.
Properties that depend on the amount of solute in a solution include boiling
point elevation, freezing point depression, and osmotic pressure. We will be
taking a closer look at the first two of these properties.
Boiling Point Elevation
Whenever a solute is dissolved in a pure solvent it raises the boiling point of
the solvent. This phenomenom is called boiling point elevation. The more
solute that is dissolved in the solution, the higher the boiling point becomes.
There is an equation that helps us predict what the change in the boiling point
of a solvent will be if we know the amount of solute dissolved. That equation
is shown below.
Tb = kbm
Tb = the change in the boiling point temperature (°C)
kb = the boiling point constant for the solvent used (°C/m)
m = the molality of the solution (moles solute/kg solvent)
Every solvent has its own boiling point constant, so the boiling points of
different solvents will change by varying amounts, depending on what the
solvent is. The kb for water is 0.520 °C/m. This number tells us that if we have
a 1.00 molal solution with water as the solvent, the boiling point of the solution
would be 0.520 °C higher than the boiling point for pure water. Below are
some sample problems.
Example 1
What would the boiling point be of a 3.50 molal NaCl solution with water as
the solvent?
Tb = kbm
kb = 0.520 °C/m
m = 3.50 m
Tb = (0.520 °C/m)(3.50 m) = 1.82 °C
+ 100 °C
101.82 °C
Since water normally boils at 100 °C, we would add the change in boiling point
temperature to the normal boiling point and have the new boiling point
temperature.
Example 2
How many grams of NaCl would have to be added to 1400 grams of water in
order to increase the boiling point to 102.5 °C?
Since we know that the normal boiling point of water is 100 °C, we can tell that
the Tb is 2.5 °C. By using the change in the boiling point temperature and the
boiling point constant for water
(0.52 °C/m) we can calculate the molality of the solution needed. Once we
know the molality of the solution we can calculate the number of moles of
NaCl needed and the number of grams needed.
2.5 °C = (0.520 °C/m)(m)
2.5 °C
= m = 4.8 m = ? moles solute
moles solute
0.520 °C/m
1.400 kg solvent
(4.8 m)(1.400 kg) =
6.7 moles NaCl x 58.5 grams = 3.9 x 102
grams NaCl
1 mole
Freezing Point Depression
Whenever a solute is dissolved in a pure solvent it lowers the solvents freezing
point. This is called freezing point depression. The more solute that is
dissolved, the lower the freezing point of the solution. There is an equation that
helps us predict the change in the freezing point of the solution if we know the
amount of solute dissolved. This equation is the same as the one for boiling
point elevation, except the b (for boiling) is replaced with an f (for freezing).
There are kf values for different solvents but they are not the same as the kb
constant. For example, the kb for water is
0.520 °C/m while the kf for water is 1.86 °C/m. The difference in constants
shows that the freezing point will be changed more than the boiling point for an
aqueous solution. Below is a sample problem for calculating freezing point
depression.
Tf = kfm
Example 3
What would the new freezing point be for a solution with 126 grams of KNO 3
dissolved in 750 grams of water?
126 g KNO3 x 1 mole = 1.25 moles KNO3
101.1 g
m = 1.25 moles = 1.67 m
0.750 kg
Tf = (1.86 °C/m)(1.67 m) = 3.11 °C
Since water normally freezes at 0 °C, the new freezing point would be lowered
to -3.11 °C