Chemistry 11
Unit V
Problem Set 5.8 – Colligative Properties
Colligative properties are four specific properties that are only affected by the molality of the solution.
The four properties are:
1. Freezing Point Depression: The freezing temperature of a solvent gets lower as more solute is dissolved.
2. Vapour Pressure Depression: The vapour pressure of a liquid gets lower as more solute is dissolved.
3. Boiling Point Elevation:
The boiling point of a solvent gets higher as more solute is added.
4. Osmotic Pressure Elevation: The osmotic pressure of a system with a semipermeable membrane gets higher
as more solute is added to one side of the membrane
These properties are independent of the nature of the solute. In other words, it does not matter what the solute is, only
how much is added to the system.
1.
Freezing Point Depression
When a solvent freezes, the particles begin to organize into a regular, repeating crystal structure. The presence of
solute in the solvent disrupts this structure and means that the solvent particles must be moving more slowly (less
kinetic energy) in order to become ordered or organized. Because of this, the solvent solidifies or freezes at a lower
temperature with solute present than it does without solute present.
The amount the freezing point will drop can be calculated from the following formula.
∆𝑇𝑓 = 𝑘𝑓 𝑚 𝑖
Where:
T is the change in freezing point temperature (freezing Tpure solvent - freezing Tsolution)
kf is the freezing point depression constant, which is a specific value based on the solvent. The freezing point
depression constant for water is 1.853 K.kg/mole.
i is the van’t Hoff factor which is a multiplier that accounts for the total number of dissolved particles. So, for
example, NaCl has a van’t Hoff factor of 2, since 1 mole of NaCl dissociates into 1 mole of Na+ and 1 mole of Clfor a total of 2 moles of ions. For covalent compounds that form molecular solutions (like sugar) the van’t Hoff
factor is 1. Consider the following examples:
Reaction
van’t Hoff factor
+
NaCl(s)  Na + Cl
2
MgCl2(s)  Mg2+ + 2Cl3
Al(NO3)3(s)  Al3+ + 3NO34
C6H12O6(s) (sugar)  C6H12O6(aq) 1
Al2(SO4)3  2Al3+ + 3SO425
2.
Vapour Pressure Depression
Vapour pressure occurs because all liquids evaporate to some extent. Some of the liquid particles have enough kinetic
energy to escape from the liquid and fly off into the gaseous state. The ease with which a liquid evaporates depends on
a few things. Firstly, it depends on how strongly one liquid molecule is attracted to its neighboring liquid molecule, or in
other words how strong the intermolecular forces are between solvent particles. The more strongly they are attracted
to each other, the less likely they are to become free of the liquid. Solvents with strong intermolecular forces (like
water) have low vapour pressures and those with weak intermolecular forces (like rubbing alcohol or acetone) have high
vapour pressures and therefore evaporate quickly. The vapour pressure also depends on temperature. The higher the
temperature, the more likely the liquid particles can escape into the gas phase.
Chemistry 11
Unit V
Solute particles, since they are generally solid, do not readily evaporate into the gas state. The solute particles compete
for space at the surface of the liquid, thereby preventing the solvent molecules from escaping into the gas phase.
Consider the following diagram of closed systems with pure solvent (left) and solution (right):
Pure solvent system
Solute dissolved in solvent
higher
vapour
pressure
lower
vapour
pressure
solution
pure liquid
This decrease in vapour pressure is called Raoult’s Law. In its simplest form, Raoult’s law says that the vapour pressure
of a solution is equal to the vapour pressure of the solvent multiplied by the mole fraction of the solvent.
In equation form, this is
𝑃𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑋𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑥 𝑃𝑠𝑜𝑙𝑣𝑒𝑛𝑡
The mole fraction of the solvent is simply the moles of solvent divided by the total moles (solvent + solute). So, for
example, if the solvent is 99% of the solution, the vapour pressure of the solution will be 99% of what it would be at that
temperature for the pure solvent.
There are some significant limitations of Raoult’s law: it only works for ideal solutions. Ideal solutions contain solutes
which do not easily evaporate (like solids) and are solutions which are very dilute. They must be dilute because the
solvent – solute interactions must be about the same strength as solvent – solvent interactions for this law to hold true.
This is not normally the case, but if there are very few solute molecules, it will not significantly change the number of
solvent molecules that can escape at the surface of the liquid.
3.
Boiling Point Elevation
Temp
Press.
0
0.6
5
0.9
10
1.2
15
1.8
20
2.3
25
3.2
30
4.2
35
5.6
40
7.4
50
12.3
60
19.9
70
31.2
80
47.3
90
70.1
100
101.3
A very similar equation to the one used for freezing point depression is used for boiling point
elevation. Like freezing point depression there is a specific constant for each solvent, and molality
must also be used. i is the van’t Hoff factor as before.
∆𝑇𝑏 = 𝑘𝑏 𝑚 𝑖
Boiling point elevation is conceptually connected with vapour pressure depression because:
the boiling point of a liquid is achieved when the liquid’s vapour pressure is equal to the
atmospheric pressure.
For water at sea level, when the temperature is 1000C, the vapour pressure is 101.3 kPa, the same as
the atmospheric pressure, so the water boils. As you may have experienced, the boiling point of water
at very high elevations is lower than at sea level, since the atmospheric pressure is lower as you rise
above sea level. There is less air pressing down, so the pressure is less. For example, at about 95oC,
the vapour pressure of water is 80 kPa, which is what the atmospheric pressure is at the peak of
Whistler. This means that water will boil at about 95oC at the peak. See the temperature vs. vapour
pressure chart to the left.
Chemistry 11
Unit V
As the vapour pressure drops due to the presence of solute, the temperature has to be increased to bring the vapour
pressure back to what it would be as a pure liquid. So, water that boils at 100oC will not boil if it has solute in it, but if it
is heated to something above 100oC, the vapour pressure will get back to 101.3 kPa and the solution will boil.
4.
Osmotic Pressure Elevation
Osmosis is the process of water moving across a semipermeable membrane. The semipermeable membrane allows the
water to pass through, but it does not allow the larger solute molecules to pass through. As a result, the water moves
from an area of high concentration (pure water) to lower concentration (when mixed with solute).
water
side
solution
side
(higher
concentration
of solute)
water
side
solution
side
(lower
concentration
of solute)
semipermeable
membrane
Water moves from left to right across this membrane, creating a water level difference on the two sides of the
membrane. The higher the concentration, the more water will move and the higher the difference in levels will be.
The pressure difference created by the different solution levels can be found from:
𝜋=𝑀𝑖𝑅𝑇
Where:
is the pressure due to the difference in solution levels
M is the molarity (moles solute / L of solution) of the side with dissolved substances
R is a constant (8.31 kPa L/oC mole)
T is temperature.
In essence, though, this means that at a given temperature, doubling the concentration will double the solution level
difference. There is a direct relationship.
Chemistry 11
Unit V
Questions:
Freezing Point Depression
1
What is the freezing point of water if 10.0 g of NaCl are dissolved into 100.0 g of water?
2
What is the freezing point of water if 10.0 g of LiF are dissolved into 100.0 g of water?
3
A solution of AlF3 has a freezing point of -1.56 oC. If 56.0 g of solvent present, how many moles of AlF3 are
present in the solution?
4
A solution of sugar, C6H12O6, (does not dissociate when it dissolves, so i = 1) has a freezing point of -4.6 oC. If
450 mL of water are present, how many moles of sugar are dissolved? What mass of sugar is this?
5
3.5 g of a compound with a van’t Hoff factor of 3 is mixed with 46.0 g of water. The freezing point of this
mixture is -4.68 oC. What is the molar mass of the substance?
6*
A compound contains Lithium and one other element and has a van’t Hoff factor of 2. When 10.0 g of the
substance are dissolved into 100.0 g of water, the freezing point is -14.3 oC. What is the other element present?
7
8
9
10
11
12
13
14
15
16
Vapour Pressure Depression
Use the vapour pressure table above to help answer these questions.
A water solution at 90.0 oC has a vapour pressure of 65.0 kPa. What is the mole fraction of the solvent in the
solution?
A water solution at 100.0 oC has a vapour pressure of 95.0 kPa. If there are 100.0 g of water present, how many
moles of solute are present?
A water solution at 50.0 oC has 10.0 g of NaCl dissolved into 65.0 g of water. What will the vapour pressure of
the mixture be?
Boiling Point Elevation
Note: The boiling point elevation constant kb for water is 0.521 oC.kg/mole. The density of water is 1.0 g/mL
What will the boiling point be for a solution which has 1.0 moles of LiF dissolved into 1.0 kg of water?
What will the boiling point be for a solution which has 5.0 g of Sr(OH)2 dissolved into 100 mL of water?
A solution has a boiling point of 105 oC and has a van’t Hoff factor of 3. What is the molality of the solution?
A solution of TiCl3 has a boiling point of 108.4 oC. If 100.0 g of water are present, how many grams of TiCl3 are
dissolved into the solution?
A solution has a boiling point of 102.4 oC. If 3.5 g of solute (which is of AB3 type) are added to 72 g of water,
what is the molar mass of the compound?
Osmotic Pressure
A semipermeable membrane separates two liquids such that after diffusion, the water side is 8 cm high and the
solution side is 12 cm high. If the concentration of the solution was doubled, what would the liquid levels be?
How would the difference of liquid levels compare in the following two situations:
a) one side of a semipermeable membrane has water the other 1.0 M NaCl
b) one side of a the same semipermeable membrane has water and the other has 1.0 M AlCl3.
Answers:
1. -6.34 oC
5. 9.0 x 101 g/mole
9. 11.7 kPa
13. 62.23 g
2. -14.3 oC
6. Fluorine (LiF)
10. 101 oC
14. 42.2 g/mole
3. 0.0118 moles
7. 0.927
11. 100.6 oC
15. 6 cm and 14 cm
4. 1.1 moles sugar, 201 g
8. 0.368 moles solute
12. 3.20 mole/kg
16. b) would have double the
difference that a) has