CH302 Spring 2009 Answer Key—
10 Questions on Colligative Properties and 10 Questions to Introduce Chemical Equilibria
All of this is intended to be done without the aid of a calculator. All of the calculations are designed such
that approximating should be straight-forward and produce a correct result.
1. Based on the physical constants involved, which colligative property has the greatest magnitude for a
solution of a given concentration? Which can't be compared in this way? Why?
Osmotic pressure - except at extremely low temperatures - the RT term is going to be much larger than
the value of kb or kf. It is difficult to compare Raoult's Law in these terms because it uses no physical
constants.
2. Which colligative properties have a linear concentration dependence? Write their equations.
All colligative properties have a linear concentration dependence.
Osmotic pressure (Π = i·M·R·T)
Boiling point elevation (ΔTb = i·m·Kb)
Freezing point depression (ΔTf = -i·m·Kf)
Vapor pressure lowering (P = P°·x)
3. Rank the following aqueous solutions in terms of increasing boiling point: 3 m sugar, 2 m NaCl, 0.5 m
Mg(OH)2, 5 m AlN, 1 m urea.
5 m AlN < 1 m urea < 0.5 m Mg(OH)2 < 3 m sugar < 2 m NaCl
4. Assuming a cell wall can withstand an osmotic pressure of 1 atmosphere and the concentration of Na+
in a cell is 50 mM, approximate the [Na+] outside the cell that would cause lysis.
The product of 0.0821 L·atm·mole-1·K-1 and 298 K is roughly 25 L·atm·mol-1 (or atm·M-1).
1 atmosphere of pressure would thus be produced by roughly 40 mM Na+.
Therefore an exterior concentration of approximately 10 mM Na+ (or less) would be required.
5. If you dissolved 58 grams of NaCl in 90 grams of pure H20 hot enough to have a vapor pressure of 30
torr, what will the new vapor pressure be?
The molecular weight of NaCl is roughly 58 g·mol-1, so we have about 1 mol of NaCl.
The molecular weight of H20 is roughly 18 g·mol-1, so we have about 5 mol of water.
The mol fraction of water is therefore 5/6, times a vapor pressure 30 torr for pure water, is roughly 25
torr for the new solution.
6. Assuming standard conditions and a Kf = 0.2 K·m-1 and a Kb = 0.5 K·m-1 for water, what would be
the freezing point of a solution that boiled at 375.5 K? Express your answer in both K and °C.
The normal boiling point of water is 373 K, so we have a ΔTb = 2.5 K, which equates to an effective
molality of m = 5, resulting in a ΔTf = -1 K, so our solution will freeze at 272 K, or -1 °C.
7. Based on the question above and assuming 1 kg of water, how many moles of NaCl would be needed to
produce this effect? What about sugar?
Accounting for the van't hoff coefficient, 2.5 moles of NaCl would produce a 5 m solution.
It would take 5 moles of sugar to produce the same effect.
8. Based on you understanding of boiling point elevation, why doesn't salting water help food to cook
faster?
The effects involved are very small. Given the magnitude of the boiling point elevation constant as well
as the amount of salt required to achieve even a 1 m solution, the change in boiling point resulting from
salting food is too small to substantially effect cooking times.
9. Vapor pressure is often described as a "surface phenomenon." Define this term in your own words to
the best of your ability.
