Q27 / 120 · VITEEE 2008 · Chemistry

The molar conductivities of $\mathrm{KCl}$, $\mathrm{NaCl}$ and $\mathrm{KNO}_{3}$ are 152,128 and $111 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ respectively. What is the molar conductivity of $\mathrm{NaNO}_{3} ?$

  • A
    $101 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
  • Correct answer
    B
    $-101 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
  • C
    $87 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
  • D
    $-391 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

Solution

Since $\mathrm{NaNO}_{3}$ is formed by the reaction $\mathrm{Nacl}+\mathrm{KNO}_{3} \rightarrow \mathrm{NaNO}_{3}+\mathrm{KCl}$ hence, using Kohlrausch's law $\begin{aligned} \stackrel{o}{\wedge} \mathrm{m}_{\mathrm{NaNO}_{3}} &=\stackrel{o}{\wedge}_{\mathrm{NaCl}}+\stackrel{o}{\wedge}_{\mathrm{KNO}_{3}}-\stackrel{\circ}{\wedge}_{\mathrm{KCl}} \\ &=128+111-152=87 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \end{aligned}$

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

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