$2 \,g$ of benzoic acid $(C_{6}H_{5}COOH)$ dissolved in $25 \,g$ of benzene shows a depression in freezing point equal to $1.62 \,K$. The molal depression constant for benzene is $4.9 \,K \,kg \,mol^{-1}$. What is the percentage association of the acid if it forms a dimer in the solution?

(D) Given: $w_{2} = 2 \,g$,$K_{f} = 4.9 \,K \,kg \,mol^{-1}$,$w_{1} = 25 \,g$,$\Delta T_{f} = 1.62 \,K$.
The molar mass of benzoic acid $(C_{6}H_{5}COOH)$ is $M_{normal} = (7 \times 12) + (6 \times 1) + (2 \times 16) = 122 \,g \,mol^{-1}$.
Using the formula for depression in freezing point: $\Delta T_{f} = K_{f} \times m = K_{f} \times \frac{w_{2} \times 1000}{M_{observed} \times w_{1}}$.
$M_{observed} = \frac{K_{f} \times w_{2} \times 1000}{\Delta T_{f} \times w_{1}} = \frac{4.9 \times 2 \times 1000}{1.62 \times 25} = 241.98 \,g \,mol^{-1}$.
Van't Hoff factor $i = \frac{M_{normal}}{M_{observed}} = \frac{122}{241.98} \approx 0.504$.
For dimerization: $2A \rightleftharpoons A_{2}$.
$i = 1 - x + \frac{x}{2} = 1 - \frac{x}{2}$,where $x$ is the degree of association.
$0.504 = 1 - \frac{x}{2} \implies \frac{x}{2} = 0.496 \implies x = 0.992$.
Percentage association = $0.992 \times 100 = 99.2 \%$.