Question

At $25^{\circ} \,C$, the molar conductances at infinite dilution for the strong electrolytes $NaOH , NaCl$ and $BaCl _{2}$ are $248 \times 10^{-4}, 126 \times 10^{-4}$ and $280 \times 10^{-4} \,Sm ^{2}\, mol ^{-1}$ respectively, $\lambda_{m}^{\circ} Ba ( OH )_{2}$ in $Sm ^{2} \,mol ^{-1}$ is

• A. $52.4 \times 10^{-4}$
• B. $524 \times 10^{-4}$
• C. $402 \times 10^{-4}$
• D. $262 \times 10^{-4}$

Answer 1

Answer: (B)

$BaCl _{2}+2 NaOH \rightarrow Ba ( OH )_{2}+2 NaCl$
$\lambda_{m}^{\infty} Ba ( OH )_{2}=\lambda_{m}^{\infty} BaCl _{2}+2 \lambda_{m}^{\infty} NaOH -2 \lambda_{m}^{\infty} NaCl$
$=280 \times 10^{-4}+2 \times 248 \times 10^{-4}-2 \times 126 \times 10^{-4}$
$=(280+496-252) \times 10^{-4}$
$=524 \times 10^{-4} \,Sm^ 2\, mol ^{-1}$