Chemistry - Liquid Solution Question with Solution | TestHub

Given : $w_2=0.2\mathrm{g},w_1=20\mathrm{g},\Delta T_f=0.45^{\circ }\mathrm{C}$ $\Delta T_f=\frac{1000\times K_f\times w_2}{w_1\times M}\Rightarrow 0.45=\frac{1000\times 5.12\times 0.2}{20\times M}$ $\therefore M_{\text{(observed) }}=113.78$ (acetic acid) $2{\mathrm{CH}}_3\mathrm{COOH}\rightleftharpoons {\left({\mathrm{CH}}_3\mathrm{COOH}\right)}_2$ Before association After association $1-\alpha$ $\alpha /2$ (where $\alpha$ is degree of association) Molecular weight of acetic acid $=60$ $\begin{array}{rl}& i=\frac{\text{ Normal molecular mass }}{\text{ Observed molecular mass }}\\ & \therefore \frac{M_{(\text{normal })}}{M_{(\text{observed) }}}=1-\alpha +\frac{\alpha }{2}\\ & \text{ or, }\frac{60}{113.78}=1-\alpha +\frac{\alpha }{2}\therefore \alpha =0.945\text{ or }94.5\%\end{array}$