Question

A solution is made by dissolving 0.636 mol of nonelectrolyte solute in 841 g of benzene. Calculate the freezing point, T_f and boiling point T_b of the solution.

Answer 1

Given the mass of benzene (solvent) = 841 g

moles of the nonelectrolyte solute = 0.636 mol

Molality of a solution is defined as the moles of the solute present in 1000 g of the solvent.

Hence molality of the solution of benzene and the nonelectrolyte,

m = (0.636 mol / 841 g benzene) * (1000 g benzene / 1 kg benzene) = 0.756molKg^-1

molal boiling point elevation constant of benzene, Kb = 2.65 degC*Kg*mol^-1

Boiling point of benzene, Tb = 80.1 DegC

Let the boiling point of the solution be T^'b, which is higher than normal boiling point Tb

Now elevation in boiling point of the solution can be calculated as,

DeltaTb = Kb*m

=> T^'b - Tb = (2.65 degC*Kg*mol^-1) *  (0.756molKg^-1 ) = 2 degC

=> T^'b - 80.1 degC = 2 degC

=> T^'b = 82.1 degC (answer)

molal freezing point depression constant of benzene, Kf = 5.12 degC*Kg*mol^-1

Freezing point of benzene, Tf = 5.5 DegC

Let the freezing point of the solution be T^'f, which is lower than normal freezing point Tf

Now depression in freezing point of the solution can be calculated as,

DeltaTf = Kf*m

=> Tf - T^'f -  = (5.12 degC*Kg*mol^-1) *  (0.756molKg^-1 ) = 3.87 degC

=> 5.5 degC - T^'f = 3.87 degC

=> T^'f = 5.5 degC - 3.87 degC = 1.63 degC (answer)

Note: use the value of Kb, Kf, Tb and Tf given in yotur text book to get exact result