Question

When 3.24 g of p-dichlorobenzene was dissolved in 25.40 g of benzene, the solution froze at –1.9˚C. The freezing point of pure benzene is 5.4 ˚C. Calculate the K_f of benzene.

Answer 1

Mass of p-dichlorobenzene = 3.24 g, Molar mass of p-dichlorobenzene = 147 g/mol

Number of moles = mass in gram / Molar mass

Number of moles of p-dichlorobenzene = 3.24 g / 147 g/mol = 0.022 mol

Weight of solvent = 25.40 g = 25.40 Kg / 1000 = 0.0254 Kg

molality = number of moles of solute / weight of solvent in Kg

molality = 0.022 mol / 0.0254 Kg = 0.866 mol/Kg = 0.866 m

We know from freezing point of depression

∆Tf = Kf * b ----(1)

∆Tf = Freezing point of depression = Tf(pure solvent) - Tf(solution) = 5.4 oC - (-1.9 oC) = 7.3 oC

b = molality of solution

From(1)

Kf = ∆Tf / b = 7.3 oC / 0.866 m = 8.43 oC/m

Kf of benzene = 8.43 oC