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At $298 \mathrm{~K}$ the molar conductivities at infinite dilution $\left(\Lambda_m^{\circ}\right)$ of $\mathrm{NH}_4 \mathrm{Cl}, \mathrm{KOH}$ and $\mathrm{KCl}$ are 152.8, 272.6 and $149.8 \mathrm{~S} \mathrm{~cm} \mathrm{cmol}^{-1}$ respectively. The $\Lambda_m^{\circ}$ of $\mathrm{NH}_4 \mathrm{OH}$ in $\mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-1}$ and $\%$ dissociation of $0.01 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$ with $\Lambda_m=25.1 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ at the same temperature are
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Verified Answer
The correct answer is:
$275.6,9.1$
$$
\begin{aligned}
\Lambda_m^{\circ} \mathrm{NH}_4 \mathrm{OH} & =\Lambda_m^{\circ}\left(\mathrm{NH}_4 \mathrm{Cl}+\mathrm{KOH}\right)-\Lambda_m^{\circ}(\mathrm{KCl}) \\
& =152.8+272.6-149.8 \\
& =275.6 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
\end{aligned}
$$
Degree of dissociation
$$
\begin{aligned}
(\alpha)=\frac{\Lambda_m}{\Lambda_m^{\circ}} & =\frac{25.1}{275.6} \\
& =0.091
\end{aligned}
$$
$\therefore \%$ degree of dissociation $=9.1 \%$
\begin{aligned}
\Lambda_m^{\circ} \mathrm{NH}_4 \mathrm{OH} & =\Lambda_m^{\circ}\left(\mathrm{NH}_4 \mathrm{Cl}+\mathrm{KOH}\right)-\Lambda_m^{\circ}(\mathrm{KCl}) \\
& =152.8+272.6-149.8 \\
& =275.6 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
\end{aligned}
$$
Degree of dissociation
$$
\begin{aligned}
(\alpha)=\frac{\Lambda_m}{\Lambda_m^{\circ}} & =\frac{25.1}{275.6} \\
& =0.091
\end{aligned}
$$
$\therefore \%$ degree of dissociation $=9.1 \%$
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