$1.22 g$ of benzoic acid (molar mass $122 g mo l^{- 1}$ ) dissolved in $50.0 g$ of benzene lowers the freezing point by $0.640 K$ ( $K_{f} = 5.12 K kg mo l^{- 1}$ ). Benzoic acid dimerises in benzene as $2 C X_{6} H X_{5} COOH ⇌ (C X_{6} H X_{5} COOH)_{2}$ . Find the percentage of benzoic acid that is dimerised, to the nearest integer.

Numerical answer type, answer in %, tolerance 1.

Answer: 75 %

Without association,

m = (1.22/122) /0.0500 = 0.200 mol k g^{- 1}

and

Δ T_{f} = 5.12 \times 0.200 = 1.024 K

. So

i = 0.640/1.024 = 0.625

. For dimerisation

i = 1 - α /2

, giving

α = 2 (1 - 0.625) = 0.75

, i.e.

75%