Q24 / 90 · KVPY 2016 (SB/SX) · Chemistry
The specific conductance ( $\kappa$ ) of $0.02 \mathrm{M}$ aqueous acetic acid solution at $298 \mathrm{~K}$ is $1.65 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}$. The degree of dissociation of acetic acid is
[Given : equivalent conductance at infinite dilution of $\mathrm{H}^{+}=349.1 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ and $\mathrm{CH}_{3} \mathrm{COO}^{-}=40.9 \mathrm{~S} \mathrm{~cm}^{2}$ $\left.\mathrm{mol}^{-1}\right]$
- Correct answerA$0.021$
- B$0.21$
- C$0.012$
- D$0.12$
Solution
$\lambda_{\mathrm{M}}=\frac{1000 \times \mathrm{K}}{\mathrm{M}}$
$=\frac{1000 \times 1.65 \times 10^{-4}}{0.2}$
$=8.25$
$\lambda_{\mathrm{H}}^{\infty}=\lambda_{\mathrm{M}}^{\infty}\left(\mathrm{H}^{+}\right)+\lambda_{\mathrm{M}}^{\infty}\left(\mathrm{CH}_{3} \mathrm{COO}^{-}\right)$
$=349.1+40.9$
$=390$
$\propto=\frac{8.25}{390}=.0211$
Hence, the correct answer is option: 'A'.