Chem 400
Lecture Problems for Colligative Properties
Nuss
1. The vapor pressure of water at 35 oC is 42.175 mm Hg. The vapor pressure of ethyl alcohol
(C2H5OH) at 35 C is 100.5 mm Hg. What is the vapor pressure of a solution prepared by
dissolving 250 g of C2H5OH in 375 g of H2O? Ans: 48.3 mm Hg
2. A student needs to prepare an aqueous solution of sucrose at a temperature of 20o C with a
vapor pressure of 15.0 mm Hg. How many grams of sucrose (mm = 342 g/mol) does she
need if she uses 375 g H2O? (The vapor pressure of water at 20o C is 17.5 mm Hg.)
Ans: 1190 g
3. What is the vapor pressure in mm Hg of a solution prepared by dissolving 18.3 g of NaCl in
500.0g H20 at 70 C. (The v.p. of water at 70C is 233.7 mm Hg.) Ans: 228.5 mm Hg
4. What will be the freezing point and boiling point of an aqueous solution containing 55.0 g of
glycerol, C3H5(OH)3, and 250 g of water? Kb(H2O) = 0.51 oC/m and Kf = 1.86o C/m.
Ans: -4.45 o C and 101.22 o C.
5. How many grams of (NH4)3PO4 need to be added to 500. g of H2O so that the freezing point
of the solution is lowered to –8.3o C? Assume that the ammonium phosphate completely
dissociates. (Kf = 1.86o C/m) Ans: 91 g
6. What is the osmotic pressure of a 0.075 M solution of aspartic acid at 18.5o C. ans:1.80 atm
7. What is the osmotic pressure in mm Hg of 6.00L of a 0.108 M solution of barium chloride at
30 C? Ans: 6.13 x 103 mm Hg
8. An isotonic solution will produce an osmotic pressure of 7.84 atm measured against pure
water at human body temperature (37.0 C). How many g of sodium chloride must be
dissolved in a liter of water to produce an isotonic solution? Ans: 9.01 g NaCl/L soln.
9. A solution is prepared from 25.0 g of benzene, C6H6, and 2.50 g of an unknown compound.
The freezing point of this solution is 4.3oC. The normal freezing point of benzene is 5.5 oC
and the freezing point depression constant for benzene is –5.12 oC/m . Determine the molar
mass of the compound.
Ans: 427 g/mol
Chem 400
Lecture Problems for Colligative Properties
Nuss
10. 0.0800 g of a compound with the empirical formula C2H2N is dissolved in 10.00 g of
benzene (C6H6). The resulting solution freezes at 4.99 C. Benzene freezes at 5.48 C and has
a freezing point depression constant kf = 4.90 C/m. What is the molecular formula for the
compound? ans: C4H4N2
11. When 2.0 g of an electrolyte was dissolved in 25.0 g of water, the resulting solution was
found to have a M of 0.815 M and a m of 0.800 m. The freezing point of the solution was
determined to be –3.72 C. the freezing point depression constant for water is –1.86 C/m.
Calculate the MW of the electrolyte and the density of the aqueous solution.
ans: MW = 100. g/mol, d = 1.10 g/mL.
12. The freezing point of pure benzene was measured as 5.49 C. The freezing point of a solution
prepared by dissolving 5.782 g of an unknown substance (sometimes sold in stores as “moth
balls”) in 100.2 g of benzene was found to be 3.48 C. Kf for benzene is –5.12C/m.
Calculate the MW of the unknown.
In a separate analysis of the unknown, it was found
to contain 49.0% carbon, 48.2% chlorine, and 2.75% hydrogen. What is the molecular
formula of the unknown?
ans: MW = 147 g/mol Formula = C6H4Cl2
13. If a solution contains 4 mols of A and 6 mols of B and has a normal bp of 85 C and the
vapor pressure of pure A is 500. mm Hg, what is the vapor pressure of pure B at this
temperature?
(Hint: what is the pressure at the “normal bp” for any substance?) Ans: 933 mm Hg
14. 3.2 g of a compound with a molecular weight of 96.0 g/mol dissolved in 50.0 g of water gave
a solution which freezes at –1.50 C. Kf for water = –1.86 C/m. Describe the electrolyte as
strong, weak, or nonelectrolyte based on the van’t Hoff factor. ans: weak electrolyte; van’t
Hoff factor = 1.21.