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Calculate $\Lambda_{H O A C}^{\infty}$ using appropriate molar conductances of the electrolytes listed above at infinite dilution in $\mathrm{H}_2 \mathrm{O}$ at $25^{\circ} \mathrm{C}$.
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1758 Upvotes
Verified Answer
The correct answer is:
390.7
$$
\begin{aligned}
& \text { } \Lambda_{\mathrm{HOAC}}^{\infty}=\Lambda_{\mathrm{NaOAC}}^{\infty}+\Lambda_{\mathrm{HCl}}^{\infty}-\Lambda_{\mathrm{NaCl}}^{\infty} \\
& =91.0+426.2-126.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\
& =390.7 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
\end{aligned}
$$
\begin{aligned}
& \text { } \Lambda_{\mathrm{HOAC}}^{\infty}=\Lambda_{\mathrm{NaOAC}}^{\infty}+\Lambda_{\mathrm{HCl}}^{\infty}-\Lambda_{\mathrm{NaCl}}^{\infty} \\
& =91.0+426.2-126.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\
& =390.7 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
\end{aligned}
$$
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