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