75.2 ~g of C_6 H_5 OH (phenol) is dissolved in a solvent of K_f=14...

Phenol is dimerised as follows At $t=0$ 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \rightleftharpoons\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)_2 At equilibrium (1-$\alpha) \quad \alpha$/ 2 Total moles after dimerisation $=1-\alpha+\alpha / 2=1-\alpha / 2$$\begin{aligned}\text{ Van't Hoff factor (i) }& =\frac{\text{ Normal molecular wt.\text{ of } phenol }}{\text{ Abnormal molecular wt. of phenol }}=\frac{94}{m_{a b}}\\ m_{a b} & =\frac{1000 \times K_f \times w}{\Delta T_f \times W} \end{aligned}$$$Given that$$\begin{aligned}& K_f=14 \\ & w=72.5 \mathrm{gm}\\ & W=1000 \mathrm{gm}\\ & \therefore \quad m_{a b}=\frac{1000 \times 14 \times 72.5}{7 \times 1000}=145 \\ & \end{aligned}\therefore \quad i=$\frac{94}{145} \text{ (i) }=\frac{\text{ No.\text{ of } particles after dimerisation }}{\text{ No. of particles before dimerisation }}=\frac{1-\alpha / 2}{1}$From eq. (i) and (ii)\begin{aligned}1-\alpha/ 2 & =\frac{94}{145}=0.65 \\ \frac{\alpha}{2}& =1-0.65=0.35 \end{aligned}$$Hence,$35 \%$ phenol is present in dimeric form.