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Molar conductivities $\left(\Lambda_m^{\circ}\right)$ at infinite dilution of $\mathrm{NaCl}, \mathrm{HCl}$ and $\mathrm{CH}_3 \mathrm{COONa}$ are 126.4, 425.9 and $91.0 \mathrm{~S} \mathrm{~cm} \mathrm{cmol}^{-1}$ respectively. $\Lambda^{\circ}{ }_m$ for $\mathrm{CH}_3 \mathrm{COOH}$ will be
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The correct answer is:
$390.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
$$
\begin{aligned}
& \text { } \mathrm{CH}_3 \mathrm{COONa}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{CH}_3 \mathrm{COOH} \\
& 91+425.9=126.4+x \\
& \therefore \quad x=516.9-126.4 \\
& =390.5 \mathrm{~S} \mathrm{~cm}{ }^2 \mathrm{~mol}^{-1} \\
&
\end{aligned}
$$
\begin{aligned}
& \text { } \mathrm{CH}_3 \mathrm{COONa}+\mathrm{HCl} \longrightarrow \mathrm{NaCl}+\mathrm{CH}_3 \mathrm{COOH} \\
& 91+425.9=126.4+x \\
& \therefore \quad x=516.9-126.4 \\
& =390.5 \mathrm{~S} \mathrm{~cm}{ }^2 \mathrm{~mol}^{-1} \\
&
\end{aligned}
$$
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