Chemistry - Electrochemistry Question with Solution | TestHub

Hint: Firstly we find ${\lambda }_{\mathrm{m}}\left(\mathrm{aq}{\mathrm{CH}}_3\mathrm{COOH}\right)$ ${\lambda }_{\mathrm{m}}=\frac{\mathrm{K}\times 1000}{\mathrm{C}(\mathrm{M})}=\frac{1.65\times 10^{-4}\times 1000}{0.02}=8.25{\mathrm{Scm}}^2{\mathrm{mol}}^{-1}$ ${\lambda }_{\mathrm{m}}({\mathrm{CH}}_3\mathrm{COOH}$ at infinite dilution $)={\lambda }_{\mathrm{m}}^{\infty }\left({\mathrm{H}}^{+}\right)+{\lambda }_{\mathrm{m}}^{\infty }\left({\mathrm{CH}}_3{\mathrm{COO}}^{-}\right)$ Now degree of dissociation $\begin{array}{rl}& \alpha =\frac{{\lambda }_{\mathrm{m}}\left[{\mathrm{CH}}_3\mathrm{COOH}\text{ at }0.02(\mathrm{M})\right]}{{\lambda }_{\mathrm{m}}\left[{\mathrm{CH}}_3\mathrm{COOH}\text{ at infinite dilution }\right]}\\ & \alpha =\frac{8.25}{349.1+40.9}=0.021\end{array}$