Question

Specific conductance of $0.1MC{H}_{3}COOHat{25}^{\circ }$ is $3.9\times {10}^{-4}oh{m}^{-1}c{m}^{-1}$. If ${\lambda }^{\infty }\left(H_3{O}^{+}\right)and{\lambda }^{\infty }\left(V{H}_{3}CO{O}^{-}\right)$ at ${25}^{\circ }C$ are 349.0 and 41.0 $oh{m}^{-1}c{m}^{2}mo{l}^{-1}$ respectively, degree of ionization of $C{H}_{3}COOH$ at the given concentration is :

• A. $2.0\%$
• B. $1.0\%$
• C. $4.0\%$
• D. $5.0\%$

Answer 1

The correct option is B

$1.0\%$

${\lambda }_m\left(C{H}_{3}COOH\right)=\frac{1000k}{0.1}=\frac{1000\times 3.9\times {10}^{-4}}{0.1}=3.9oh{m}^{-1}c{m}^{2}mo{l}^{-1}{\alpha }_{C{H}_{3}COOH}=\frac{{\lambda }_m}{{\lambda }_m^{\circ }}=\frac{3.9}{{\lambda }_{C{H}_{3}CO{O}^{-}+{\lambda }_{H^{+}}^{\circ }}^{\circ }}=\frac{3.9}{41+349}=1.0\times {10}^{-2}=1\%$