A carboxylic acid HA (0.50 mol kg^-1 formal molality) is dissolved in a solvent with K_f=5.0 K kg mol^-1 in which it trimerises: 3 HAleftharpoons(HA)_3 with degree of association alpha. The observed freezing point depression is 1.40 K. The value of alpha is

JEE Advanced levelSingle correct optionDifficulty 3 of 3so191

A carboxylic acid HA ( $0.50 mol k g^{- 1}$ formal molality) is dissolved in a solvent with $K_{f} = 5.0 K kg mo l^{- 1}$ in which it trimerises: $3 HA ⇌ (HA)_{3}$ with degree of association $α$ . The observed freezing point depression is $1.40 K$ . The value of $α$ is

Without association

Δ T_{f} = 5.0 \times 0.50 = 2.50 K

, so

i = 1.40/2.50 = 0.56

. Starting with 1 mol:

(1 - α)

monomer plus

α /3

trimer, so

i = 1 - \frac{2 α}{3}

. Thus

α = \frac{3 ( 1 - 0.56 )}{2} = 0.66

. The dimer formula

i = 1 - α /2

would wrongly give

0.88