If the molar conductance values of ${\mathrm{Ca}}^{2+}\text{ and }{\mathrm{Cl}}^{-}$ at infinite dilution are respectively 118.88 × 10^-4  ${\mathrm{m}}^2 \mathrm{mho} {\mathrm{mol}}^{-1}\text{ and }77.33\times {10}^{-4}{\mathrm{m}}^2 \mathrm{mho} {\mathrm{mol}}^{-1}$ $\text{ then that of }{\mathrm{CaCl}}_2\text{ is (in }{\mathrm{m}}^2 \mathrm{mho} {\mathrm{mol}}^{-1}\text{ ) }$ 

According to Kholrausch's law 

${\wedge }_{{\mathrm{m}}_{{\mathrm{CaCl}}_2}}^{\mathrm{o}}={\mathrm{\lambda }}_{{\mathrm{Ca}}^{2+}}^{\mathrm{o}}+\hspace{0.17em}2{\mathrm{\lambda }}_{\mathrm{Cl}-}^{\mathrm{o}}$

given that , 

$$\begin{gathered} {\mathrm{\lambda }}_{{\mathrm{Ca}}^{2+}}^{\mathrm{o}}= 118.88\mathrm{X} {10}^{-4}{\mathrm{m}}^2 \mathrm{mho} {\mathrm{mol}}^{-1} \\ {\mathrm{\lambda }}_{{\mathrm{Cl}}^{-}}^{\mathrm{o}} =77.33 \mathrm{X}\hspace{0.17em}{10}^{-4} {\mathrm{m}}^{2 } \mathrm{mho} {\mathrm{mol}}^{-1} \end{gathered}$$

${\wedge }_{CaCl2}^o=( 188.88 \times \hspace{0.17em}{10}^{-4} ) +\hspace{0.17em} 2(77.33\hspace{0.17em}\times {10}^{-4})=\mathbf{ }\mathbf{273}\mathbf{.}\mathbf{54}\mathbf{ }\mathbf{\times }\mathbf{\hspace{0.17em}}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$