Question

A solution is made by dissolving 0.541 mol of nonelectrolyte solute in 865 g of benzene. Calculate the freezing point, T f , and boiling point, T b , of the solution.

Answer 1

Answer:

T f = (Normal freezing point of benzene - Molality of the solution × freezing point depression per molal of benzene) = 5.5 ° C - .625 m × 5.12° C = 2.30° C.

T f = 2.30° C.

T b = (Normal boiling point of benzene - Molality of the solution × boiling point elevation per molal of benzene) = 80.1 ° C. - .625 × 2.53° C = 81.68 ° C.

T b = 81.68 ° C.

Explanation:

The molality of the solution is .541 ÷ .865 = .625 m.

Ideally benzene freezes at 5.5 ° C and the depression of the freezing point is 5.12° C per molal.

At a molality of 0.625 the freezing point will be depressed by (molality of solution × freezing point depression per molal of benzene) or .625 × 5.12° C = 3.20° C.  So the resulting freezing point will be (Ideal freezing point - freezing point depression) or T f =5.5 ° C. - 3.20 ° C. = 2.30° C.

Also ideally benzene boils at a temperature of 80.1° C and the elevation of the boiling point per molal is 2.53° C .  So the boiling point of benzene will be elevated by (molality of solution × boiling point elevation per molal of benzene) .625 × 2.53° C = 1.58° C.  So the boiling point of the current solution, T b, will be 80.1 ° C. + 1.58 ° C. = 81.68 ° C.