Q48 / 180 · JIPMER 2018 · Chemistry

If molar conductivity of $\mathrm{Ca}^{2+}$ and $\mathrm{Cl}^{-}$ions are 119 and $71 \mathrm{Sm}^2 \mathrm{~mol}^{-1}$ respectively, then the molar conductivity of $\mathrm{CaCl}_2$ at infinite dilution is

  • A
    $215 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  • B
    $340 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  • C
    $126 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  • Correct answer
    D
    $261 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Solution

Given, $\begin{array}{r}\Lambda^{\circ} \mathrm{Ca}^{2+}=119 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ \Lambda^{\circ} \mathrm{Cl}^{-}=71 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{array}$ As we know that, $\begin{array}{ll} & \Lambda^{\circ} \mathrm{CaCl}_2=\Lambda_{\mathrm{Ca}^2+}^{\circ}+2 \times \Lambda^{\circ}{ }_{\mathrm{Cl}^{-}} \\ \therefore \quad & \Lambda^{\circ} \mathrm{CaCl}_2=119+2 \times 71 \\ & \Lambda^{\circ} \mathrm{CaCl}_2=261 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{array}$

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

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