Chemistry - Liquid Solution Question with Solution | TestHub

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 $$(1-\alpha )\alpha /2$$ Total moles after dimerisation $=1-\alpha +\alpha /2=1-\alpha /2$ $$\begin{array}{rl}\text{ Van’t Hoff factor (i) }& =\frac{\text{ Normal molecular wt. of phenol }}{\text{ Abnormal molecular wt. of phenol }}=\frac{94}{m_{ab}}\\ m_{ab}& =\frac{1000\times K_f\times w}{\Delta T_f\times W}\end{array}$$ Given that $$\begin{array}{rl}& K_f=14\\ & w=72.5\mathrm{gm}\\ & W=1000\mathrm{gm}\\ & \therefore m_{ab}=\frac{1000\times 14\times 72.5}{7\times 1000}=145\\ & \end{array}$$ $$\therefore i=\frac{94}{145}$$ $$\text{ (i) }=\frac{\text{ No. of particles after dimerisation }}{\text{ No. of particles before dimerisation }}=\frac{1-\alpha /2}{1}$$ From eq. (i) and (ii) $$\begin{array}{rl}1-\alpha /2& =\frac{94}{145}=0.65\\ \frac{\alpha }{2}& =1-0.65=0.35\end{array}$$ Hence, $35\%$ phenol is present in dimeric form.