Question

What are the boiling point and freezing point of a 1.47 m solution of naphthalene in benzene? (The boiling point and freezing point of benzene are 80.1°C and 5.5°C respectively. The boiling point elevation constant for benzene is 2.53°C/m, and the freezing point depression constant for benzene is 5.12°C/m.) Boiling point = °C Freezing point = °C

Answer 1

We know ∆T_b = K_b* m

Where ∆T_b is boiling point elevation = T_b(solution) - T_{b(pure solvent)}

K_b is boiling point elevation constant, m is molality of solution

Given , m = 1.47 m, K_b = 2.53 ^oC/m

∆T_b = 1.47 m * 2.53 ^oC/m = 3.72 ^oC

Boiling point of solution = ∆T_b+T_{b(pure solvent)} = 3.72 ^oC + 80.1 ^oC = 83.82 ^oC

Boiling point of solution = 83.82 ^oC ---------Answer

We know ∆T_f = T_f(solvent) - T_f(solution) = K_f * m

∆T_f is freezing point depression, K_f = freezing point depression constant, m is Molality

∆T_f = 5.12 ^oC/m * 1.47 m = 7.526 ^oC

Freezing point of solution = 5.5 ^oC - 7.526 ^oC = -2.026 ^oC -------Answer