Question

The molal freezing point constant of benzene is 4.90 and its melting point is 5.51°C. A solution of 0.816 g of a compound A when dissolved in 7.5 g of benzene freezes at 1.59°C. Determine the molecular weight of A.

Answer 1

Given: Mass of solute (A) = w₂ = 0.816 g

Mass of solvent (benzene) = w₁ = 7.5 g = 0.0075 kg

Freezing point of pure benzene = 5.51°C

Freezing point of solution = 1.59°C

Depression in freezing point:

ΔT_f = 5.51−1.59 = 3.92°C

K_f (benzene) = 4.90 K kg mol⁻¹

`Delta T_f = K_f * w_2/(M * w_1)`

⇒ `3.92 = 4.90 * 0.816/(M * 0.0075)`

`3.92 = (4.90 xx 0.816)/(0.0075 xx M)`

⇒ `3.92 = 3.9984/(0.0075 * M)`

⇒ `0.0075 xx M = 3.9984/3.92`

⇒ 0.0075 × M = 1.02

⇒ `M = 1.02/0.0075`

M = 136 g/mol

∴ The molecular weight of compound A is 136 g/mol.