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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Aicharya · Accepted Answer

On adding Eq. (i) and (ii) and subtracting the Eq. (iii) $\begin{aligned} \Lambda_{\infty} \mathrm{NH}_4^{+}+\Lambda_{\infty} \mathrm{OH}^{-} & =130+217-109 \\ \Lambda_{\infty} \mathrm{NH}_4 \mathrm{OH} & =347-109 \\ & =238 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \text { equiv }^{-1} \end{aligned}$

If the values of $\Lambda_{\infty}$ of $\mathrm{NH}_4 \mathrm{Cl}, \mathrm{NaOH}$ and $\mathrm{NaCl}$ are…

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