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At $25^{\circ} \mathrm{C}$, the molar conductances at infinite dilution for the strong electrolytes $\mathrm{NaOH}, \mathrm{NaCl}$ and $\mathrm{BaCl}_{2}$ are $248 \times 10^{-4}, 126 \times 10^{-4}$ and $280 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ respectively, $\lambda_{\mathrm{m}}^{\circ} \mathrm{Ba}(\mathrm{OH})_{2}$ in $\mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ is
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The correct answer is:
$524 \times 10^{-4}$
$$
\begin{aligned}
\mathrm{BaCl}_{2} &+2 \mathrm{NaOH} \longrightarrow \mathrm{Ba}(\mathrm{OH})_{2}+2 \mathrm{NaCl} \\
\lambda_{m \mathrm{Ba}(\mathrm{OH})_{2}}^{\infty} &=\lambda_{m}^{\infty} \mathrm{BaCl}_{2}+2 \lambda_{m}^{\infty} \mathrm{NaOH}-2 \lambda_{m}^{\infty} \mathrm{NaCl} \\
&=280 \times 10^{-4}+2 \times 248 \times 10^{-4} \\
&=(280+496-252) \times 10^{-4} \\
&=524 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}
\end{aligned}
$$
\begin{aligned}
\mathrm{BaCl}_{2} &+2 \mathrm{NaOH} \longrightarrow \mathrm{Ba}(\mathrm{OH})_{2}+2 \mathrm{NaCl} \\
\lambda_{m \mathrm{Ba}(\mathrm{OH})_{2}}^{\infty} &=\lambda_{m}^{\infty} \mathrm{BaCl}_{2}+2 \lambda_{m}^{\infty} \mathrm{NaOH}-2 \lambda_{m}^{\infty} \mathrm{NaCl} \\
&=280 \times 10^{-4}+2 \times 248 \times 10^{-4} \\
&=(280+496-252) \times 10^{-4} \\
&=524 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}
\end{aligned}
$$
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