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

• A. $126Sc{m}^{2}mo{l}^{-1}$
• B. $340Sc{m}^{2}mo{l}^{-1}$
• C. $215Sc{m}^{2}mo{l}^{-1}$
• D. $261Sc{m}^{2}mo{l}^{-1}$

Answer 1

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

${\Lambda }_{C{a}^{2+}}^0=119Sc{m}^{2}mo{l}^{-1}$

${\Lambda }_{C{l}^{-}}^0=71Sc{m}^{2}mo{l}^{-1}$

For a strong electrolyte $NaBr$, the limiting molar conductivity is the sum of the individual ionic conductivities.

Thus,
${\Lambda }_{CaC{l}_2}^0={\Lambda }_{C{a}^{2+}}^0+2\times {\Lambda }_{C{l}^{-}}^0$
$=119+2\times 71$

$=261Sc{m}^{2}mo{l}^{-1}$