Honors Chemistry Name______
Chapter 15.1: Colligative Properties Worksheet III Date _____/_____/_____ Period _____
Use the chart below as needed.
Solvent / Boiling Point / Kb / Freezing Point / Kfwater / 100.00 oC / 0.52 oC/m / 0.000 oC / 1.86 oC/m
benzene / 80.00 oC / 2.53 oC/m / 5.50 oC / 5.10 oC/m
camphor / 208.00 oC / 5.95 oC/m / 179.80 oC / 40.0 oC/m
1. The equilibrium vapor pressure of pure water at 30.oC is 31.6 torr. When 22.4 g of a nonvolatile, nonelectrolyte solute
is dissolved in 50.0 g of water, the vapor pressure of the solution is 2.50 torr lower than that of the pure water at the
same temperature.
a. Calculate the mole fraction of solute.
b. Calculate the molecular weight of the solute.
2. A 28.0% (w/w) aqueous glucose solution at 20. oC Also, the equilibrium vapor pressure of pure water at 20. oC is
17.4 torr. (Assume the density of the solution is 1.00 g/mL)
a. Estimate the vapor pressure of the glucose solution
28.0 % = 28.0 g solute x 100
100.0 g solution
28.0 g C6H12O6 / 1 mole C6H12O6 / = 0.155 mole C6H12O6180.2 g C6H12O6
72.0 g H2O / 1 mole H2O / = 4.00 mole H2O
18.02 g H2O
17.4 torr x 4.00 mole H2O = 16.7 torr or 0.0220 atm
4.16 mole solution
b. Estimate the boiling point of the solution
m = 0.155 mole = 2.15 m
0.0720 kg
DTb = Kb m
DTb = 0.52 oC x 2.15 m = 1.1 oC
m
BP solution = BP solvent + DTb
BP solution = 100.00 oC + 1.1 oC = 101.1 oC
c. Estimate the osmotic pressure of the solution
M = 0.155 mole = 1.55 M
0.100 L
p = MRT
p = (1.55 M) (0.0821 L * atm) (293 K) = 37.3 atm
(mole * K)
3. 3.42 g of a nonvolatile, nonelectrolyte solute of unknown molecular weight is dissolved in 40.0 g of camphor, and the
freezing point of the solution is measured to be 163.7 oC. What is the molecular weight of the solute?
DTf = 179.80 oC – 163.7 oC = 16.1 oC
DTf = Kf m
DTf = Kf ( x )
kg of solvent
16.1 oC = (40.0 oC)( x )
m 0.0400 kg
x = 0.0161 mole
molecular weight = 3.42 g = 212 g/mole
0.0161 mole
4. Assuming complete ionization of the solute, what would one expect the boiling point of a solution of 50.0 g of NaI in
125 g of water to be?
50.0 g NaI / 1 mole NaI / = 0.334 mole NaI149.89 g NaI
m = 0.334 mole NaI = 2.67 m
0.125 kg solvent
DTb = Kb mi
DTb = 0.52 oC x 2.67 m x 2 = 2.8 oC
m
BP solution = BP solvent + DTb
BP solution = 100.00 oC + 2.8 oC = 102.8 oC
5. The osmotic pressure of blood serum is 7.65 atm at body temperature, 37 oC. How many grams of glucose, C6H12O6,
should be used per liter of solution to prepare an IV solution that is isotonic with blood serum, meaning, having the
same osmotic pressure as blood serum?
p = MRT
M = p = (mole * K)(7.65 atm) = 0.301 M
RT (0.0821 L * atm)(310. K)
0.301 mole C6H12O6 / 180.2 g C6H12O6 / = 54.2 g1 L / 1 mole / 1 L
6. Which of the following aqueous solutions should have the lowest freezing point?
a. 100 mL of 0.5 m NaCl
DTf = I Kf m
DTf = (2)(1.86 oC/m)(0.5 m)
DTf = 2 oC
b. 200 mL of 0.10 m CaCl2
DTf = i Kf m
DTf = (3)(1.86 oC/m)(0.10 m)
DTf = 0.56 oC
c. 500 mL of 0.10 m sucrose, C12H22O11
DTf = i Kf m
DTf = (1)(1.86 oC/m)(0.10 m)
DTf = 0.19 oC
d. 100 mL of 0.20 m sucrose
DTf = i Kf m
DTf = (1)(1.86 oC/m)(0.20 m)
DTf = 0.37 oC
7. 0.00550 g of a highly purified protein was dissolved in 25.0 mL of water. At 50.0 oC, the solution exhibited an osmotic
pressure of 0.00157 atm. What is the molecular weight of the protein? (Assume the solution volume is 25.0 mL.)
p = iMRT
0.00157 atm = (1)( x )(0.0821 L * atm)(323 K)
0.0250 L mole * K
x = 0.00000148 moles
molecular weight = 0.00550 g = 3720 g/mole
0.00000148 moles