Pure benzene freezes at 5.3°C . A solution of 0.223 g of phenylacetic acid $\left({\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{COOH}\right)$  in 4.4 g of benzene

$\left(K_f=5.12K\mathrm{kg}{\mathrm{mol}}^{-1}\right)$ freezes at 4.47°C. From this 

observation, one can conclude that

$\mathrm{\Delta }{T}_f=T_0-T_f=5.3-4.47=0.83$

molality of solvent $m=\frac{0.223}{136}\times \frac{1000}{4.4}=0.373$

(Molecular weight of phenylacetic acid = 136)

$\begin{array}{l}{K}_f=5.12\\ \mathrm{\Delta }{T}_f=i.K_{f}m\\ i=\frac{\mathrm{\Delta }{T}_f}{K_{f}m}=\frac{0.83}{5.12\times 0.373}=0.45\end{array}$

'i' is nearly equal to 0.5 it means phenylacetic acid undergoes dimerization in benzene.