Question

The latent heat of fusion of benzene is 30.25 cal/g. 0.0802 g of a substance when dissolved in 13.03 g of benzene lower the freezing point by 0.490°C. If pure benzene freezes at 5.4°C, calculate the molecular mass of the substance.

Answer 1

Given: Mass of solute = 0.0802 g

Mass of benzene (solvent) = 13.03 g = 0.01303 kg

Depression in freezing point (ΔT_f) = 0.490°C

Latent heat of fusion of benzene = 30.25 cal/g

Freezing point of pure benzene = 5.4°C

K_f​ for benzene = 5.12 K kg mol⁻¹  ...(Known value)

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

`0.490 = 5.12 xx 0.0802/(M * 0.01303)`

`0.490 = 0.4106/(M * 0.01303)`

`0.01303 xx M = 0.4106/0.490`

0.01303 × M = 0.838

`M = 0.838/0.01303`

M = 64.3 g/mol

∴ The molecular mass of the substance is 64.3 g/mol.