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.

According to Kohlrausch law,
i. \(\wedge_0\left(\mathrm{Ba}(\mathrm{OH})_2\right)=\lambda_{\mathrm{Ba}^{2+}}^0+2 \lambda_{\mathrm{OH}^{-}}^0\)
ii. \(\wedge_0\left(\mathrm{BaCl}_2\right)=\lambda_{\mathrm{Ba}^{2+}}^0+2 \lambda_{\mathrm{Cl}^{-}}^0\)
iii. \(\wedge_0\left(\mathrm{NH}_4 \mathrm{Cl}\right)=\lambda_{\mathrm{NH}_4^{+}}^0+\lambda_{\mathrm{Cl}^{-}}^0\)
\(\mathrm{Eq}(\mathrm{i})+\frac{1}{2} \mathrm{Eq}\) (ii) \(-\frac{1}{2} \mathrm{Eq}\) (iii) gives
\(\wedge_0\left(\mathrm{NH}_4 \mathrm{OH}\right)=\wedge_0\left(\mathrm{NH}_4 \mathrm{Cl}\right)+ \frac{1}{2} \wedge_0\) \(\left(\mathrm{Ba}(\mathrm{OH})_2\right) \) \( -\frac{1}{2} \wedge_0\left(\mathrm{BaCl}_2\right)\)
\(\wedge_0\left(\mathrm{NH}_4 \mathrm{OH}\right)=129+\frac{1}{2} 520-\frac{1}{2} 280\)
\(=249.0 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)