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The molar conductivities $\wedge_{\mathrm{NaOAC}}^{\circ}$ and $\wedge_{\mathrm{HCl}}^{\circ}$ at infinite dilution in water at $25^{\circ} \mathrm{C}$ are $91.0$ and $426.2 \mathrm{~S} \mathrm{~cm}^2 / \mathrm{mol}$ respectively. To calculate $\wedge_{\mathrm{HOAc}}^O$, the additional value required is
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Verified Answer
The correct answer is:
$\wedge_{\mathrm{NaCl}}^{\circ}$
$\wedge_{\mathrm{NaCl}}^{\circ}$
$$
\begin{aligned}
& \lambda_{\mathrm{CH}_3 \mathrm{COONa}}^0=\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^0+\lambda_{\mathrm{Na}^{+}}^0 \\
& \lambda_{\mathrm{HCl}}^0=\lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{C}}^0 \cdots \cdots \cdots \cdots \cdot \\
& \lambda_{\mathrm{NaCl}}^0=\lambda_{\mathrm{Na}}^0+\lambda_{\mathrm{Cl}}^0 \cdots \cdots \cdots \cdots \\
& \lambda_{\mathrm{CH}_3 \mathrm{COOH}}^0=(1)+(2)-(3)
\end{aligned}
$$
\begin{aligned}
& \lambda_{\mathrm{CH}_3 \mathrm{COONa}}^0=\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}^0+\lambda_{\mathrm{Na}^{+}}^0 \\
& \lambda_{\mathrm{HCl}}^0=\lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{C}}^0 \cdots \cdots \cdots \cdots \cdot \\
& \lambda_{\mathrm{NaCl}}^0=\lambda_{\mathrm{Na}}^0+\lambda_{\mathrm{Cl}}^0 \cdots \cdots \cdots \cdots \\
& \lambda_{\mathrm{CH}_3 \mathrm{COOH}}^0=(1)+(2)-(3)
\end{aligned}
$$
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