Colligative Properties Worksheet
The amount of the elevation of the boiling point of a solution is given by Tb = ikbm where kb is the
molal boiling point elevation constant of the solvent and m is the molal concentration of the solute.
The amount of the depression of freezing point is given by Tf = -ikfm where kf is the molal freezing
point depression constant of the solvent and m is the molal concentration of the solute.
When working with ionic solutes in water and other polar solvents, one must be careful to take into
consideration the fact that the ion concentration is higher than the concentration of the solute. For example,
table salt, NaCl dissolves in water to form ions NaCl(s) -> Na+(aq) + Cl-(aq). Each mole of NaCl makes
two moles of ions, so if we have a 1 molal solution of NaCl in water, the ion concentration is 2 molal.
Remember, nonelectrolyte solutes do not produce any ions.
1. The normal freezing point of water is 0.00 ˚C and it's freezing point depression constant is 1.86 ˚C/m. If
we make up a 1.00 molal solution of Na2SO4, what is the freezing point of the mixture.
2.
For a 0.222 m aqueous solution of sucrose, C12H22O11, What is the boiling point? kb = 0.512˚C/m
3.
What is the normal boiling point of a 0.122 m aqueous solution of KI?
4. Automotive antifreeze consists of ethylene glycol, C2H6O2 , a nonvolatile nonelectrolyte. Calculate
the boiling point and freezing point of a 25.0 mass percent solution of ethylene glycol in water.
5.
Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol
(C10H18O), a fragrant substance found in the leaves of eucalyptus tree. K f for CHCl3 is 4.68 ˚C/molal and
normal freezing point is -63.5°C.
6. List the following aqueous solutions in order of their expected freezing points (from lowest to highest):
0.050 m CaCl2; 0.15 m NaCl; 0.10 m HCl; 0.050 m HC2H3O2; 0.10 m C12H22O11
7.
A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g in 40.0 g
CCl4. The normal boiling point of the resultant solution increased by 0.357˚C. Calculate the molar mass of
the solute. Kb= 5.02°C/molal
8. Camphor, C10H16O, melts at 179.8˚C, it has particularly large freezing point depression constant, Kf
= 40.˚C/molal. 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 solute? Tf= 179.8° C and Kf= 40. °C/molal
9. A compound called pyrene has the empirical formula C8H5 . When 4.04 g of pyrene is dissolved in 10.00 g of
benzene, the boiling point of the solution is 85.1°C. Calculate the molar mass of pyrene and determine its
molecular formula. The molal boiling-point constant for benzene is 2.53°C/m. Its normal boiling point is 80.1°C
10. A mass of CaCl2 , when dissolved in 100.00 g of water, gives an expected freezing point of -5.0°C. What
mass of glucose would give the same result?

Answers
1)
f =kf•m•i=(1.86˚C/m)(1.00m)(3)= -5.58˚C
2)
Bp=100.000˚C+[(0.222m)(.513°C/m)(1)] = 100.114˚C
3)
Kb=100.000˚C+[(0.122m)(.513°C/m)(2)] = 100.125°C
4)
Mass% 
Mass solute
mass solution to convert to m assume 1.00kg of solution
250g
moles solute
moles solute
62.0g/mole
m
subtract solute to get

 5.377m
kg solvent
solution  solute 1.00kg  .25kg
boiling point = 102.72 °C
freezing point = -9.87 °C
 42.0g

154g/mole
Fp  63.5ÞC 
 0.600kg



5.
6)




 x 4.68ÞC/m x 1 65.6ÞC






NaCl, HCl, CaCl2, C12H22O11, HC2H3O2
7)
0.250g
Tb
0.357ÞC
m

 0.07112m  MM
kb x i
5.02ÞC/m x 1
.0400kg
8)
MM  87.9g/mole
110 g/mol (same method as # 7)
9) Use formula in #7. MF=204.4g/mol. EM=101 so MF= C16H10
10) Use formula in # 7 but solve for grams:
Tf x Kg Solvent
g
5.0C x 0.100kg 110.99g
x

x
 9.9g CaCl 2
Kf
mole
3 x 1.86C  m
mole
T x Kg Solvent
g
5.0C x 0.100kg 180.16g
 f
x

x
 49g glucose
Kf
mole
1.86C m
mole
CaCl2:
Mass grams 
Sugar:
Mass grams
