Q34 / 155 · WBJEE 2024 · Chemistry

The specific conductance \((\mathrm{k})\) of \(0.02(\mathrm{M})\) aqueous acetic acid solution at 298 K is \(1.65 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}\). The degree of dissociation of acetic acid is \(\left[\lambda_{\mathrm{H}^{+}}=349.1 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\right.\) and \(\left.\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}=40.9 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\right]\)

  • Correct answer
    A
    0.021
  • B
    0.21
  • C
    0.012
  • D
    0.12

Solution

Hint: Firstly we find $\lambda_{\mathrm{m}}\left(\mathrm{aq} \mathrm{CH}_3 \mathrm{COOH}\right)$ $\lambda_{\mathrm{m}}=\frac{\mathrm{K} \times 1000}{\mathrm{C}(\mathrm{M})}=\frac{1.65 \times 10^{-4} \times 1000}{0.02}=8.25 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$ $\lambda_{\mathrm{m}}\left(\mathrm{CH}_3 \mathrm{COOH}\right.$ 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 $\begin{aligned} & \alpha=\frac{\lambda_{\mathrm{m}}\left[\mathrm{CH}_3 \mathrm{COOH} \text { at } 0.02(\mathrm{M})\right]}{\lambda_{\mathrm{m}}\left[\mathrm{CH}_3 \mathrm{COOH} \text { at infinite dilution }\right]} \\ & \alpha=\frac{8.25}{349.1+40.9}=0.021\end{aligned}$

Hence, the correct answer is option: 'A'.

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