Question

If molar conductivity of $C{a}^{2+}$ and $C{l}^{-}$ ions are $119$ and $71Sc{m}^{2}mo{l}^{-1}$ respectively, then the molar conductivity of $CaC{l}_2$ at infinite dilution is
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• A. $126Sc{m}^{2}mo{l}^{-1}$
• B. $261Sc{m}^{2}mo{l}^{-1}$
• C. $215Sc{m}^{2}mo{l}^{-1}$
• D. $340Sc{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is B $261Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_{C{a}^{2+}}^0=119Sc{m}^{2}mo{l}^{-1}$ ${\Lambda }_{C{l}^{-}}^0=71Sc{m}^{2}mo{l}^{-1}$ For a strong electrolyte $CaC{l}_2$, the limiting molar conductivity is the sum of the individual ionic conductivities. Thus, ${\Lambda }_{CaC{l}_2}^0={\Lambda }_{C{a}^{2+}}^0+2\times {\Lambda }_{C{l}^{-}}^0$ $=119+2\times 71$ $=261Sc{m}^{2}mo{l}^{-1}$