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If the values of $\Lambda_{\infty}$ of $\mathrm{NH}_4 \mathrm{Cl}, \mathrm{NaOH}$ and $\mathrm{NaCl}$ are 130, 217 and $109 \mathrm{ohm}^{-1} \mathrm{~cm}^2$ equiv $^{-1}$ respectively, the $\Lambda_{\infty}$ of $\mathrm{NH}_4 \mathrm{OH}$ in ohm ${ }^{-1} \mathrm{~cm}^2$ equiv $^{-1}$ is
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The correct answer is:
238
On adding Eq. (i) and (ii) and subtracting the Eq. (iii)
$\begin{aligned}
\Lambda_{\infty} \mathrm{NH}_4^{+}+\Lambda_{\infty} \mathrm{OH}^{-} & =130+217-109 \\
\Lambda_{\infty} \mathrm{NH}_4 \mathrm{OH} & =347-109 \\
& =238 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \text { equiv }^{-1}
\end{aligned}$
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