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Colligative Properties Calculations
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Boiling-Point Elevation and Freezing-Point Depression
 Because the addition of a nonvolatile solute to a volatile solvent
lowers the vapor pressure of the solution relative to the pure solvent,
the vapor pressure of the solution will be lower than the pure solvent
at any temperature.
 If the normal boiling point of a liquid occurs when the vapor
pressure equals 1 atm, and the addition of solute will lower the
vapor pressure of a solution from that of the pure solvent, then it
makes sense that a temperature higher than the normal BP of the
pure solvent will be required to make the vapor pressure of the
solution be 1 atm. The BP of the solution will therefore be higher
than the normal BP.
 If we denote the increase in BP as Tb, the molality of the solution
as m, then there is a molal boiling-point-elevation constant, Kb,
that is the constant of proportionality between the two:
Tb = Kbm
Some typical values for Kb are found in the table below. Once again,
it is important to recognize that the concentration of solute particles
in solution can vary depending on whether or not the solute is an
electrolyte.
 Similar explanation can be given for the depression of the freezing
point, so that the following relationship holds:
Tf = Kfm
where Tf is the change in the freezing point, and Kf is the molal
freezing-point-depression constant. Kb values for common
solvents are also shown in the table below:
Molal Freezing-Point and Boiling-Point Constants
Solvent
Normal FP (C)
Molal FP
C
constant, Kf (
)
m
Normal BP (C)
Molal BP
constant, Kb
(
C
)
m
Acetic Acid
16.6
3.90
117.9
3.07
Camphor
178.8
39.7

207.4
5.61
Ether
116.3
1.79
34.6
Naphthalene
80.2
6.94
217.7

2.02
5.80
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Colligative Properties Calculations
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Phenol
40.9
7.40
181.8
3.60
Water
0.00
1.86
100.0
0.51
Practice Exercise #1:
Calculate the freezing point depression of a solution containing
17.1 g of sucrose, C12H22O11 and 200. g of water. What is the
actual freezing point of the solution?
Calculate the molar mass of eucalyptol:
C: 12.0 g  12.0 mol = 144.0 g
H: 1.00 g  12.0 mol = 12.0 g
O: 16.0 g  11.0 mol = 176.0 g
1 mol C12H22O11 = 342.0 g C12H22O11
17.1g _ C12H 22O11


1
1.00 _ mol _ C12H 22O11 

 0.0500 _ mol _ C10H18O
 342g _ C12H 22O11 
0.0500 _ mol _ C12H 22O11 
m  
 0.250m
0.200_kg_H 2O


1.86C 
Tf  
 0.250m  0.465C
 m 
Tf  0.000C  (  .465C)  0.465C

Practice Exercise #2
Which of the following solutes will produce the largest increase in
boiling point upon addition of 1 kg of water: 1 mol Co(NO3)2, 2
mol KCl, or 3 mol of ethylene glycol, C2H6O2?
The correct answer is 2 mol KCl because it produces 4 mol of
particles when added to water, while the other two only provide 3
mol of particles each when dissolved in water.
This characteristic of electrolytes to dissociate (break apart) when
dissolved is described by the van’t Hoff factor, i, which is the
measure of the extent of electrolyte dissociation. It is defined as
follows:
i
T f (measured )
T f (measured _ for _ nonelectrolyte)
,
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Colligative Properties Calculations
Page 3 of 3
so that the true way of writing the depression and elevation
equations is:
Tf = iKfm and Tb = iKbm
Use of Colligative Properties to Determine Molar Mass
Any of the colligative properties can be used to determine molar
mass. Here is an example using freezing point depression
information
Practice Exercise 3
Camphor, C10H16O, melts at 178.8C; it has a particularly large
freezing-point depression constant, Kf = 39.7C/m. When 0.186 g
of an organic substance of unknown molar mass is dissolved in
22.01 g of liquid camphor, the freezing point of the mixture is
found to be 176.7C. What is the approximate molar mass of the
solvent.
Tf = iKfm, where
Tf = 176.70C  178.80C = 2.1C,
22.01 g camphor = 0.02201 kg solvent, and
Kf = 39.7 C , and
m
i= 1 (unknown substance is organic, and is dissolving in a nonpolar
solvent, so we assume it is a nonelecrolyte)
m

Tf 2.10C

 0.0529m
Kf
39.7 Cm
mole _ solute
 moles _ solvent  (m)(kg_ solvent)
kg_ solvent
0.0529 _ mol _ solute 
 
0.02201kg_ solvent 
kg_ solvent


 0.0011643_ mol _ solute  0.186g _ solute
m

M

0.186g
g
 159.8
0.0011643_ mol
mol