Question

A solution is made by dissolving 0.647 mol of nonelectrolyte solute in 771 g of benzene. Calculate the freezing point, Tf, and boiling point, Tb, of the solution. Constants may be found here.

Answer 1

molality = number of moles of solute / mass of solution in kg

molality = 0.647 / 0.771 kg = 0.839 m

deltaTf = Kf * m

delta Tf = 4.90 * 0.839 = 4.11 ⁰C

freezing point of solution = freezing point of pure solvent - deltaTf

freezing point of solution = 5.5 - 4.11 = 1.39 ⁰C

Therefore, freezing point of solution = 1.39 ⁰C

deltaTb = Kb * m

Kb of benzene = 2.53 ⁰C/m

delta Tb = 2.53 * 0.839 = 2.12 ⁰C

boiling point of solution = boiling point of pure solvent + delta Tb

boiling point of solution = 80.100 +2.12 = 82.2 ⁰C

Therefore, boiling point of solution = 82.2 ⁰C