1. V. Colligative Properties
Ch Solutions
V. Colligative Properties

2. A. Definition Colligative Property
property that depends on the concentration (# of moles) of solute particles, not their chemical nature (identity).
For example: NaCl and MgSO4

3. B. Types 1. Freezing point depression (lowering)
2. Boiling point elevation (raising)
3. Vapor-pressure reduction
4. Osmotic pressure

4. 1. Freezing Point Depression
B. Types
1. Freezing Point Depression
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5. B. Types Freezing Point Depression (tf)
f.p. of a solution is lower than f.p. of the pure solvent
tf -difference between the freezing point of the solution and the freezing point of the pure solvent

6. B. Types Applications salting icy roads melts ice
making ice cream- lowers temp.
Antifreeze (ethylene glycol)
cars (-64°C to 136°C)

7. 2. Boiling Point Elevation
B. Types
2. Boiling Point Elevation
Solute particles weaken IMF in the solvent.

8. B. Types 2. Boiling Point Elevation (tb)
b.p. of a solution is higher than b.p. of the pure solvent
tb - difference between the boiling point of the solution and the boiling point of the pure solvent

9. C. Calculations t: change in temperature (°C)
tf(b) = kf(b) · m · n
t: change in temperature (°C)
k: constant based on the solvent (°C·kg/mol)
m: molality (m)
n: # of particles
0.51 oC/m Boiling point k
1.86oC/m freezing point k

10. C. Calculations NaCl Na+ + Cl- n=2 MgCl2 Mg+2 + 2Cl- n=3
# of Particles:
Nonelectrolytes (covalent)
remain intact when dissolved
C6H12O C6H12O n=1
Electrolytes (ionic)
dissociate into ions when dissolved
2 or more particles
NaCl Na Cl n=2
MgCl Mg Cl- n=3

11. C. Calculations
At what temperature will a solution that is composed of 0.73 moles of glucose (C6H12O6) in 225 g of water boil?
GIVEN:
b.p. = ?
tb = ?
kb = 0.51°C/ m
WORK:
m = 0.73mol ÷ 0.225kg
tb = (0.51°C/ m)(3.2 m)(1)
tb = 1.6°C
b.p. = 100. °C + 1.6°C
b.p. = 101.6°C
m = 3.2 m
n = 1
tb = kb · m · n

12. C. Calculations
Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water.
GIVEN:
f.p. = ?
tf = ?
kf = 1.86°C/ m
WORK:
m = 0.48mol ÷ 0.100kg
tf = (1.86°C/ m)(4.8 m)(2)
tf = 18°C
f.p. = 0.00°C - 18°C
f.p. = -18°C
m = 4.8 m
n = 2
tf = kf · m · n

13. D. Molar Mass determination
Any colligative property can be used to determine the molar mass of an unknown (Exp. #24)
STRATEGY: Calculate number of moles required to produce observed change in colligative property

14. D. Molar Mass determination
Use: tb = kb · m · n to find # of moles
tb = kb · x moles/kgsolv · n
Molar Mass ( M )= mass solute/x moles

15. D. Molar Mass determination
Or use this formula:
mass(g)solu · Kb(f)
tb(f) · kgsolv
Molar Mass (M) =

16. D. Molar Mass determination
A 10.0 g sample of an unknown compound that does not dissociate is dissolved in g of water. The boiling point of the solution is elevated C above the normal boiling point. What is the molar mass of the unknown sample?