1. COLLIGATIVE PROPERTIES
of SOLUTIONS

2. Why do we need to add salt to ice to make ice cream?
Ice and Salt

3. The salt added to the ice lowers the freezing point of the ice so that the temperature surrounding the ice cream is even colder than freezing.
The lowering of the freezing point of water is a colligative property.
A colligative property is a property of solutions that depends only upon the number of solute particles, not upon their identity.
Three important colligative properties of solutions are:
Freezing-point depression
Vapor-pressure lowering
Boiling-point elevation

4. Freezing-Point Depression
When a liquid is cooled, the average kinetic energy of its particles decreases.
As the particles slow down, the intermolecular attractive forces draw the particles close together, and the liquid freezes to a solid.

5. Pure water Water + solute
In a solution, the solute particles interfere with the attractive forces among the solvent particles. This prevents the solvent from entering the solid state at its normal freezing point. The temp of the solution must be lower further until the particles draw close together and the solution freezes.
Pure water
Water + solute

6. The freezing point of a solution is lower than that of a pure solvent.
Salt lowers the freezing point of water.
Antifreeze lowers the freezing point of water.

7. The greater the number of solute particles in a
solvent, the lower the freezing point of the solution.
The more solute particles, the more interference in the attractive forces among the solvent particles, and the lower the freezing point.
Why?
1 mol of solute particles
(in 1000 g of water) lowers
the freezing point by 1.86°C.
2 mol of solute particles
(in 1000 g of water) lowers
the freezing point by 3.72°C.

9. Vapor-Pressure Lowering
Recall that vapor pressure is the pressure exerted by a vapor that is in dynamic equilibrium with its liquid in a closed system.

10. Why does adding a solute lowering the vapor pressure?
A solution that contains a solute that is nonvolatile (not easily vaporized) always has a lower vapor pressure that the pure solvent.
Glucose and sodium chloride are nonvolatile solutes.
The greater the number of solute particles in a solvent, the lower the resulting vapor pressure.
Why does adding a solute lowering the vapor pressure?
Normal vapor
pressure
Lower vapor
pressure
However, when the solvent contains solute, a mix of solute and solvent particles occupies the surface area. With fewer solvent particles at the surface, fewer solvent particles enter the gaseous state, and the vapor pressure is lowered.
The particles that produce vapor pressure escape the liquid phase at its surface. When a solvent is pure, its particles occupy the entire surface area.
Pure Solvent
Solution

11. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent.
SKIP

12. The greater the number of solute particles in a solvent,
the lower the resulting vapor pressure.
Which solution will have the lowest vapor pressure?
CaCl2
Produces 6 mol of particles
because each formula unit
of NaCl dissociates into two ions.
H2O
3NaCl(s)
3Na+(aq) + 3Cl-(aq)
3 mol C6H12O6
3 mol NaCl
3 mol CaCl2
H2O
H2O
3C6H12O6(s)
3C6H12O6(aq)
3CaCl2(s)
3Ca2+(aq) + 6Cl-(aq)
Produces 3 mol of particles
because glucose
does not dissociate.
Produces 9 mol of particles
because each formula unit
of CaCl2 dissociates into three ions.

13. The greater the number of solute particles in a solvent,
the lower the resulting vapor pressure.
Ionic compounds that dissociate have a greater effect on vapor pressure than molecular compounds because…....
They produce more particles when they dissolve.
3 mol C6H12O6
3 mol NaCl
3 mol CaCl2

15. Boiling-Point Elevation
In a solution containing a nonvolatile solute, its vapor pressure is lower.
A liquid boils when its
vapor pressure equals
atmospheric pressure.
When this solution is heated and the temp is raised to the boiling point of the pure solvent, the resulting vapor pressure is still less than the atmospheric pressure and the solution will not boil.
Solution
The solution must be heated to a higher temp to raise the vapor pressure to atmospheric pressure. The temp difference between a solution’s boiling point and a pure solvent’s boiling point is called the boiling point elevation.

16. The fluid circulating through a car’s cooling system is a solution of water and ethylene glycol, or antifreeze. Ethylene glycol raising the boiling point of the solution to above the boiling point of water. This helps protect the car from overheating in the summer.

18. Molality moles of solute (mol) Molality (m) = kilogram of solvent(kg)
Molality (m) is the number of moles of solute dissolved in 1 kg (1000g) of solvent.
moles of solute (mol)
Molality (m) =
kilogram of solvent(kg)

19. How many grams of potassium iodide must be dissolved
Example 1:
How many grams of potassium iodide must be dissolved
in g of water to produce a molal solution?
Given
Formula
moles of solute (mol)
0.060 m
Molality =
kg of solvent (kg)
500.0 g water x
Calculate:
0.060 m =
0.500 kg
(0.060 mol/kg)
(0.500 kg) =
0.030 mol
166.0
g KI
0.030 mol X =
4.98 =
5.0 g of KI 1
mol KI

20. The magnitudes of the freezing-point depression (∆Tf) and the boiling-point elevation (∆Tb) of a solution are directly proportional to the molal conc (m), assuming the solute is molecular, not ionic.
∆Tf = Kf X m
∆Tb = Kb X m

21. Example 1:
Antifreeze protects a car from freezing and overheating. Calculate the freezing-point depression of a solution containing 1.00 x 102 g of ethylene glycol (C2H6O2) in kg of water.
Formula
Given
∆Tf = Kf X m
100 g of C2H6O2
moles of solute (mol)
Molality =
0.500 kg water
kg of solvent (kg)
Calculate: 1
mol C2H6O2
100 g of C2H6O2 X =
1.61 mol
62.0
g of C2H6O2
1.61 mol C2H6O2
m = =
3.22 m
0.500 kg of water
∆Tf =
(1.86 °C/m)
(3.226 m) =
5.9892 =
5.99 °C