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Calculate molar conductivity of $\mathrm{NH}_4 \mathrm{OH}$ at infinite dilution if molar conductivities of $\mathrm{Ba}(\mathrm{OH})_2 \mathrm{BaCl}_2$ and $\mathrm{NH}_4 \mathrm{Cl}$ at infinite dilution are $520,280,129 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively.
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The correct answer is:
$249.0 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
According to Kohlrausch law,
i. $\quad \wedge_0\left(\mathrm{Ba}(\mathrm{OH})_2\right)=\lambda_{\mathrm{Ba}^{2+}}^0+2 \lambda_{\mathrm{OH}^{-}}^0$
ii. $\wedge_0\left(\mathrm{BaCl}_2\right)=\lambda_{\mathrm{Ba}^{2+}}^0+2 \lambda_{\mathrm{Cl}^{-}}^0$
iii. $\quad \wedge_0\left(\mathrm{NH}_4 \mathrm{Cl}\right)=\lambda_{\mathrm{NH}_4^{+}}^0+\lambda_{\mathrm{Cl}^{-}}^0$
$\mathrm{Eq}(\mathrm{i})+\frac{1}{2} \mathrm{Eq}$ (ii) $-\frac{1}{2} \mathrm{Eq}$ (iii) gives
$\begin{aligned} \wedge_0\left(\mathrm{NH}_4 \mathrm{OH}\right)=\wedge_0\left(\mathrm{NH}_4 \mathrm{Cl}\right)+ & \frac{1}{2} \wedge_0\left(\mathrm{Ba}(\mathrm{OH})_2\right) \\ & -\frac{1}{2} \wedge_0\left(\mathrm{BaCl}_2\right)\end{aligned}$
$\wedge_0\left(\mathrm{NH}_4 \mathrm{OH}\right)=129+\frac{1}{2} 520-\frac{1}{2} 280$
$=249.0 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
i. $\quad \wedge_0\left(\mathrm{Ba}(\mathrm{OH})_2\right)=\lambda_{\mathrm{Ba}^{2+}}^0+2 \lambda_{\mathrm{OH}^{-}}^0$
ii. $\wedge_0\left(\mathrm{BaCl}_2\right)=\lambda_{\mathrm{Ba}^{2+}}^0+2 \lambda_{\mathrm{Cl}^{-}}^0$
iii. $\quad \wedge_0\left(\mathrm{NH}_4 \mathrm{Cl}\right)=\lambda_{\mathrm{NH}_4^{+}}^0+\lambda_{\mathrm{Cl}^{-}}^0$
$\mathrm{Eq}(\mathrm{i})+\frac{1}{2} \mathrm{Eq}$ (ii) $-\frac{1}{2} \mathrm{Eq}$ (iii) gives
$\begin{aligned} \wedge_0\left(\mathrm{NH}_4 \mathrm{OH}\right)=\wedge_0\left(\mathrm{NH}_4 \mathrm{Cl}\right)+ & \frac{1}{2} \wedge_0\left(\mathrm{Ba}(\mathrm{OH})_2\right) \\ & -\frac{1}{2} \wedge_0\left(\mathrm{BaCl}_2\right)\end{aligned}$
$\wedge_0\left(\mathrm{NH}_4 \mathrm{OH}\right)=129+\frac{1}{2} 520-\frac{1}{2} 280$
$=249.0 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
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