Question

If the molar conductance values of $\ce{Ca^{2+}}$ and $\ce{Cl^{-}}$ at infinite dilution are respectively $\ce{118.88 \times 10^{-4} m^2 \; mho \; mol^{-1}}$ and $\ce{77.33 \times 10^{-4} \; m^2 \; mho \; mol^{-1}}$ then that of $\ce{CaCl_2}$ is (in $\ce{m^2 \; mho \; mol^{-1}}$)

• A. $118.88 \times 10^{-4}$
• B. $154.66 \times 10^{-4}$
• C. $273.54 \times 10^{-4}$
• D. $196.21 \times 10^{-4}$

Answer 1

Answer: (C)

From Kohlrausch's law :
$\Lambda_{m}^{\infty}=v_{+} \lambda_{+}^{\infty}+v_{-} \lambda_{-}^{\infty}$
For $CaCl _{2}:$
$\Lambda_{m}^{\infty}\left( CaCl _{2}\right)=\lambda_{ Ca ^{2+}}^{\infty}+2 \lambda_{ Cl ^{-}}^{\infty}$
$=118.88 \times 10^{-4}+2 \times 77.33 \times 10^{-4}$
$=118.88 \times 10^{-4}+154.66 \times 10^{-4}$
$=273.54 \times 10^{-4} m ^{2} \,mho\, mol ^{-1}$