• question_answer $75.2g$ of ${{C}_{6}}{{H}_{5}}OH$ (phenol) is dissolved in a solvent of ${{K}_{f}}=14$ .If the depression in freezing point is 7 K, then find the percentage of phenol that dimerises. A) 62% B) 98% C) 55% D) 75%

 Molar mass of solute $\left( {{M}_{B}} \right)=\frac{1000\times {{K}_{f}}\times {{W}_{B}}}{{{W}_{A}}\times \Delta {{T}_{f}}}$$\Rightarrow \,\,\,\,{{M}_{B}}=\frac{1000\times 14\times 75.2}{1000\times 7}$ ${{M}_{B}}=150.4g\,per\,mol$ Actual molar mass of phenol $=94g/mol$ Now, van't Hoff factor, $i=\frac{calculated\,molar\,mass}{observed\,molar\,mass}$ $\therefore \,\,\,\,i=\frac{94}{150.4}0.625$ Dimerisation of phenol can be shown as:
 $\begin{matrix} {} & 2{{C}_{6}}{{H}_{5}}OH & & {{({{C}_{6}}{{H}_{5}}OH)}_{2}} \\ Initial & 1 & {} & 0 \\ At\,equilibrium & 1-\alpha & {} & \frac{\alpha }{2} \\ \end{matrix}$ Total number of moles at equilibrium, $i=1-\alpha +\frac{\alpha }{2}$ $i=1-\frac{\alpha }{2}$ But, $i=0.625$, thus, $0.625=1-\frac{\alpha }{2}$ $\frac{\alpha }{2}=1-0.625$ $\alpha =0.75$ Thus, the percentage of phenol that dimerises is 75%.