At $298 \mathrm{~K}$ the molar conductivities at infinite dilution $\left(\Lambda_m^{\circ}\right)$ of…

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
  1. $275.6,0.91$
  2. $275.6,9.1$
  3. $266.6,9.6$
  4. 30,84

Solution

$ \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 $=9.1 \%$

Asked in: AP EAMCET 2014

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