Chemistry - Electrochemistry Question with Solution | TestHub

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 $\begin{array}{rl}{\wedge }_0\left({\mathrm{NH}}_4\mathrm{OH}\right)={\wedge }_0\left({\mathrm{NH}}_4\mathrm{Cl}\right)+& \frac{1}{2}{\wedge }_0\left(\mathrm{Ba}(\mathrm{OH}{)}_2\right)\\ & -\frac{1}{2}{\wedge }_0\left({\mathrm{BaCl}}_2\right)\end{array}$ ${\wedge }_0\left({\mathrm{NH}}_4\mathrm{OH}\right)=129+\frac{1}{2}520-\frac{1}{2}280$ $=249.0{\Omega }^{-1}{\mathrm{cm}}^2{\mathrm{mol}}^{-1}$