Colligative Properties of Solutions
Colligative properties = physical
properties of solutions that depend
on the # of particles dissolved, not
the kind of particle.
Colligative Properties




Lowering vapor pressure
Raising boiling point
Lowering freezing point
Generating an osmotic pressure
Colligative Properties




Lowering vapor pressure
Raising boiling point
Lowering freezing point
Generating an osmotic pressure
Boiling Point Elevation

a solution that contains a nonvolatile
solute has a higher boiling pt than the
pure solvent; the boiling pt elevation is
proportional to the # of moles of solute
dissolved in a given mass of solvent.
Boiling Point Elevation
 Tb
where:
= kb m
Tb = elevation of boiling pt
m = molality of solute
kb = the molal boiling pt elevation constant

kb values are constants; see table 15-4, p. 472
(honors text)

kb for water = 0.52 °C/m
Ex: What is the normal boiling pt of a 2.50
m glucose, C6H12O6, solution?

“normal” implies 1 atm of pressure
Tb = kbm
Tb = (0.52 C/m)(2.50 m)
Tb = 1.3 C

Tb = 100.0 C + 1.3 C = 101.3 C



Ex: How many grams of glucose, C6H12O6,
would need to be dissolved in 535.5 g of water
to produce a solution that boils at 101.5°C?



Tb = kbm
1.5 C= (0.52 C/m)(m)
m = 2.885
x
2.885 m 
0.5355 kg
x  1.5449 mol  180 g  278.1 g  280 g
1 mol
Freezing/Melting Point Depression

The freezing point of a solution is
always lower than that of the pure
solvent.
Freezing/Melting Point
Depression
 Tf
= kfm
where:
Tf = lowering of freezing point
m = molality of solute
kf = the freezing pt depression constant

kf for water = 1.86 °C/m

kf values are constants;
see table 15-5, p. 474 (honors text)
Ex: Calculate the freezing pt of a
2.50 m glucose solution.

Tf = kfm
Tf = (1.86 C/m)(2.50 m)
Tf = 4.65 C

Tf = 0.00C - 4.65 C = -4.65C


Ex: When 15.0 g of ethyl alcohol, C2H5OH, is
dissolved in 750 grams of formic acid, the freezing
pt of the solution is 7.20°C. The freezing pt of pure
formic acid is 8.40°C. Determine Kf for formic acid.
15.0 g C2H5OH 1mol

 0.3261mol
46 g
0.3261 mol
 0.4348 m
0.75 kg
Tf = kfm
1.20 C= (kf)( 0.4348 m)
kf = 2.8 C/m
Ex: An antifreeze solution is prepared containing 50.0 cm3
of ethylene glycol, C2H6O2, (d = 1.12 g/cm3), in 50.0 g
water. Calculate the freezing point of this 50-50 mixture.
Would this antifreeze protect a car in Chicago on a day
when the temperature gets as low as –10° F?
(-10 °F = -23.3° C)
50.0 cm3 C2H6O2
1.12 g
1mol


 0.90323 mol
3
62.0 g
cm
0.90323 mol
 18.06 m
0.050 kg
Tf = kfm
Tf = (1.86C/m)(18.06 m)
Tf = 33.6 C
Tf = 0 C – 33.6 C = -33.6 C
YES!
Electrolytes and Colligative
Properties
• Colligative properties depend on the # of
particles present in solution.
• Because ionic solutes dissociate into ions,
they have a greater effect on freezing pt and
boiling pt than molecular solids of the same
molal conc.
Electrolytes and Colligative
Properties

For example, the freezing pt of water is
lowered by 1.86°C with the addition of any
molecular solute at a concentration of 1 m.
– Such as C6H12O6, or any other covalent
compound

However, a 1 m NaCl solution contains 2
molal conc. of IONS. Thus, the freezing pt
depression for NaCl is 3.72°C…double that of
a molecular solute.
– NaCl  Na+ + Cl- (2 particles)
Electrolytes - Boiling Point Elevation and
Freezing Point Depression
The relationships are given by the following equations:

Tf = kf ·m·n
or
Tb = kb·m·n
Tf/b = f.p. depression/elevation of b.p.
m = molality of solute
kf/b = b.p. elevation/f.p depression constant
n = # particles formed from the dissociation of
each formula unit of the solute
Ex: What is the freezing pt of:
a) a 1.15 m sodium chloride solution?

NaCl  Na+ + Cl-
n=2

Tf = kf·m·n
Tf = (1.86 C/m)(1.15 m)(2)
Tf = 4.28 C

Tf = 0.00C - 4.28 C = -4.28C


Ex: What is the freezing pt of:
b) a 1.15 m calcium chloride solution?

CaCl2  Ca2+ + 2Cl-
n=3

Tf = kf·m·n
Tf = (1.86 C/m)(1.15 m)(3)
Tf = 6.42 C

Tf = 0.00C – 6.42 C = -6.42C


Ex: What is the freezing pt of:
c) a 1.15 m calcium phosphate solution?


Ca3(PO4)2  3Ca2+ + 2PO43n=5

Tf = kf·m·n
Tf = (1.86 C/m)(1.15 m)(5)
Tf = 10.7 C

Tf = 0.0C – 10.7 C = -10.7C


Determining Molecular Weights
by Freezing Point Depression
Ex: A 1.20 g sample of an unknown molecular compound is dissolved in
50.0 g of benzene. The solution freezes at 4.92°C. Determine the
molecular weight of the compound. The freezing pt of pure benzene
is 5.48°C and the Kf for benzene is 5.12°C/m.




Tf = 0.56°C
Tf = kf·m
0.56°C = (5.12°C/m)(m)
m = 0.1094
0.1094m 
x mol
0.050 kg
x  0.00547mol
1.20g
Molar Mass 
 219 g mol
0.00547mol
Ex: A 37.0 g sample of a new covalent compound was dissolved in 200.0
g of water. The resulting solution froze at –5.58°C. What is the
molecular weight of the compound?




Tf = 5.58°C
Tf = kf·m
5.58°C = (1.86°C/m)(m)
m = 3.00 m
3.00m 
x mol
0.200 kg
x  0.60mol
37.0g
Molar Mass 
 61.7 g mol
0.60mol