Question

A solution is made by dissolving 0.656 mol of nonelectrolyte solute in 869 g of benzene. calculate the freezing point, tf, and boiling point, tb, of the solution. constants may be found here.

Answer 1

Answer is: the freezing point is 1.63°C and boiling point is 82.01°C..

1) n(nonelectrolyte solute) = 0.656 mol.
m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
m(C₆H₆) = 0.869 kg.
b(solution) = n(nonelectrolyte solute) ÷ m(C₆H₆).
b(solution) = 0.656 mol ÷ 0.869 kg.
b(solution) = 0.754 mol/kg.

2) ΔT = Kf(benzene) · b(solution).
ΔT = 5.12°C/m · 0.754 m.
ΔT = 3.865°C.
Tf = 5.50°C - 3.865°C.
Tf = 1.63°C.

3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
ΔTb = 1.91°C.
Tb = 80.1°C + 1.91°C.
Tb = 82.01°C.