Q46 / 120 · NEET 2012 (Mains) · Chemistry

Molar conductivities $\left(\Lambda_m^{\circ}\right)$ at infinite dilution of $\mathrm{NaCl}, \mathrm{HCl}$ and $\mathrm{CH}_3 \mathrm{COONa}$ are 126.4, 425.9 and $91.0 \mathrm{~S} \mathrm{~cm} \mathrm{cmol}^{-1}$ respectively. $\Lambda^{\circ}{ }_m$ for $\mathrm{CH}_3 \mathrm{COOH}$ will be

  • A
    $425.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  • B
    $180.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  • C
    $290.8 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  • Correct answer
    D
    $390.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Solution

$$ \begin{aligned} & \text { } \mathrm{CH}_3 \mathrm{COONa}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{CH}_3 \mathrm{COOH} \\ & 91+425.9=126.4+x \\ & \therefore \quad x=516.9-126.4 \\ & =390.5 \mathrm{~S} \mathrm{~cm}{ }^2 \mathrm{~mol}^{-1} \\ & \end{aligned} $$

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

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