Respuesta :
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.
