Q18 / 110 · VITEEE 2022 · Chemistry

Equivalent conductances of \(\mathrm{NaCl}, \mathrm{HCl}\) and \(\mathrm{CH}_3 \mathrm{COONa}\) at infinitedilution are \(126.45,426.16 \mathrm{and}\) \(91 \mathrm{dhm}^{-1} \mathrm{~cm}^2\) respectively. Theequivalent conductance of \(\mathrm{CH}_3 \mathrm{COOH}\) at infinite dilution would be

  • A
    \(101.38 \mathrm{ohm}^{-1} \mathrm{~cm}^2\)
  • B
    \(253.62 \mathrm{ohm}^{-1} \mathrm{~cm}^2\)
  • Correct answer
    C
    \(390.71 \mathrm{ohm}^{-1} \mathrm{~cm}^2\)
  • D
    \(678.90 \mathrm{ohm}^{-1} \mathrm{~cm}^2\)

Solution

By Kohlraush's law, \(\wedge_{\text {eq }}^0 \mathrm{NaCl}=126.45\) \(\begin{aligned} & \lambda_{\mathrm{Na}^{+}}^0+\lambda_{\mathrm{Cl}^{-}}^0=126.45 \\ & \lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{Cl}^{-}}^0=426.16 . \\ & \lambda_{\mathrm{CH}_3 \mathrm{OO}^{-}}^0+\lambda_{\mathrm{Na}^{+}}^0=91 \end{aligned}\) on adding (2) and (3) then subtract (1) from it \(\begin{aligned} & \lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^0+\lambda_{\mathrm{Na}^{+}}^0=517.16-126.45 \\ & \wedge_{\left(\mathrm{CH}_3 \mathrm{COOH}\right)}^0=390.71 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \end{aligned}\)

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

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