1. Colligative Properties
Consider three beakers:
50.0 g of ice
50.0 g of ice moles NaCl
50.0 g of ice moles sugar (sucrose)
What will the temperature of each beaker be?
Beaker 1:
Beaker 2:
Beaker 3:

2. Colligative Properties
The reduction of the freezing point of a substance is an example of a colligative property:
A property of a solvent that depends on the total number of solute particles present
There are four colligative properties to consider:
Vapor pressure lowering (Raoult’s Law)
Freezing point depression
Boiling point elevation
Osmotic pressure

3. Colligative Properties – Vapor Pressure
A solvent in a closed container reaches a state of dynamic equilibrium.
The pressure exerted by the vapor in the headspace is referred to as the vapor pressure of the solvent.
The addition of any nonvolatile solute (one with no measurable vapor pressure) to any solvent reduces the vapor pressure of the solvent.

4. Colligative Properties – Vapor Pressure
Nonvolatile solutes reduce the ability of the surface solvent molecules to escape the liquid.
Vapor pressure is reduced.
The extent of vapor pressure lowering depends on the amount of solute.
Raoult’s Law quantifies the amount of vapor pressure lowering that is observed.

5. Colligative Properties – Vapor Pressure
Raoult’s Law:
PA = XAPOA
where PA = partial pressure of the solvent (A) vapor above the solution (ie with the solute)
XA = mole fraction of the solvent (A)
PoA = vapor pressure of the pure solvent (A)

6. Colligative Properties – Vapor Pressure
Example: The vapor pressure of water at 20oC is 17.5 torr. Calculate the vapor pressure of an aqueous solution prepared by adding 36.0 g of glucose (C6H12O6) to 14.4 g of water.

7. Colligative Properties – Vapor Pressure
Answer: torr

8. Colligative Properties – Vapor Pressure
Example: The vapor pressure of pure water at 110oC is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.10 atm at the same temperature. What is the mole fraction of ethylene glycol in the solution?
Both ethylene glycol and water are liquids. How do you know which one is the solvent and which one is the solute?

9. Colligative Properties – Vapor Pressure
Answer: XEG = 0.219

10. Colligative Properties – Vapor Pressure
Ideal solutions are those that obey Raoult’s Law.
Real solutions show approximately ideal behavior when:
The solution concentration is low
The solute and solvent have similarly sized molecules
The solute and solvent have similar types of intermolecular forces.

11. Colligative Properties – Vapor Pressure
Raoult’s Law breaks down when solvent-solvent and solute-solute intermolecular forces of attraction are much stronger or weaker than solute-solvent intermolecular forces.

12. Colligative Properties – BP Elevation
The addition of a nonvolatile solute causes solutions to have higher boiling points than the pure solvent.
Vapor pressure decreases with addition of non-volatile solute.
Higher temperature is needed in order for vapor pressure to equal 1 atm.

13. Colligative Properties- BP Elevation
The change in boiling point is proportional to the number of solute particles present and can be related to the molality of the solution:
DTb = Kb.m
where DTb = boiling point elevation
Kb = molal boiling point elevation constant
m = molality of solution
The value of Kb depends only on the identity of the solvent (see Table 13.4).

14. Colligative Properties - BP Elevation
Example: Calculate the boiling point of an aqueous solution that contains 20.0 mass % ethylene glycol (C2H6O2, a nonvolatile liquid).
Solute =
Solvent =
Kb (solvent) =
Tb = Kb  m

15. Colligative Properties - BP Elevation
Molality of solute:
DTb =
BP = oC

16. Colligative Properties - BP Elevation
Example: The boiling point of an aqueous solution that is 1.0 m in NaCl is oC whereas the boiling point of an aqueous solution that is 1.0 m in glucose (C6H12O6) is oC. Explain why.

17. Colligative Properties - BP Elevation
Example: A solution containing 4.5 g of glycerol, a nonvolatile nonelectrolyte, in g of ethanol has a boiling point of 79.0oC. If the normal BP of ethanol is 78.4oC, calculate the molar mass of glycerol.
Given: DTb =
mass solute =
mass solvent =
Kb = 1.22oC/m (Table 13.4)
Find: molar mass (g/mol)
Tb = Kb  m

18. Colligative Properties - BP Elevation
Step 1: Calculate molality of solution
Step 2: Calculate moles of solute present
Step 3: Calculate molar mass

19. Colligative Properties - Freezing Pt Depression
The addition of a nonvolatile solute causes solutions to have lower freezing points than the pure solvent.
Solid-liquid equilibrium line rises ~ vertically from the triple point, which is lower than that of pure solvent.
Freezing point of the solution is lower than that of the pure solvent.

20. Colligative Properties - Freezing Pt Depression
The magnitude of the freezing point depression is proportional to the number of solute particles and can be related to the molality of the solution.
Tf = Kf  m
where DTf = freezing point depression
Kf = molal freezing point depression constant
m = molality of solution
The value of Kf depends only on the identity of the solvent (see Table 13.4).

21. Colligative Properties - Freezing Pt Depression
Example: Calculate the freezing point depression of a solution that contains 5.15 g of benzene (C6H6) dissolved in 50.0 g of CCl4.
Given: mass solute =
mass solvent =
Kf solvent =
Find: DTf
Tf = Kf  m

22. Colligative Properties - Freezing Pt Depression
Molality of solution:
DTf =

23. Colligative Properties - Freezing Pt Depression
Example: Which of the following will give the lowest freezing point when added to 1 kg of water: 1 mol of Co(C2H3O2)2, 2 mol KCl, or 3 mol of ethylene glycol (C2H6O2)? Explain why.

24. Colligative Properties - Freezing Pt Depression
Reminder: You should be able to do the following as well:
Calculate the freezing point of any solution given enough information to calculate the molality of the solution and the value of Kf
Calculate the molar mass of a solution given the value of Kf and the freezing point depression (or the freezing points of the solution and the pure solvent).

25. Colligative Properties - Osmosis
Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking other larger particles.
In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so.
If two solutions with identical concentration (isotonic solutions) are separated by a semipermeable membrane, no net movement of solvent occurs.

26. Colligative Properties - Osmosis
Osmosis: the net movement of a solvent through a semipermeable membrane toward the solution with greater solute concentration.
In osmosis, there is net movement of solvent from the area of lower solute concentration to the area of higher solute concentration.
Movement of solvent from high solvent concentration to low solvent concentration

27. Colligative Properties - Osmosis
Osmosis plays an important role in living systems:
Membranes of red blood cells are semipermeable.
Placing a red blood cell in a hypertonic solution (solute concentration outside the cell is greater than inside the cell) causes water to flow out of the cell in a process called crenation.

28. Colligative Properties
Placing a red blood cell in a hypotonic solution (solute concentration outside the cell is less than that inside the cell) causes water to flow into the cell.
The cell ruptures in a process called hemolysis.

29. Colligative Properties - Osmosis
Other everyday examples of osmosis:
A cucumber placed in brine solution loses water and becomes a pickle.
A limp carrot placed in water becomes firm because water enters by osmosis.
Eating large quantities of salty food causes retention of water and swelling of tissues (edema).