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(D) Given: $\lambda_{m}^{\circ}(NH_4Cl) = 185 \ S \ cm^2 \ mol^{-1}$,$\lambda_{m}^{\circ}(OH^-) = 170 \ S \ cm^2 \ mol^{-1}$,$\lambda_{m}^{\circ}(Cl^-

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$3$

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(D) Given: $\lambda_{m}^{\circ}(NH_4Cl) = 185 \ S \ cm^2 \ mol^{-1}$,$\lambda_{m}^{\circ}(OH^-) = 170 \ S \ cm^2 \ mol^{-1}$,$\lambda_{m}^{\circ}(Cl^-) = 70 \ S \ cm^2 \ mol^{-1}$.Using Kohlrausch's law: $\lambda_{m}^{\circ}(NH_4Cl) = \lambda_{m}^{\circ}(NH_4^+) + \lambda_{m}^{\circ}(Cl^-) = 185$.Therefore,$\lambda_{m}^{\circ}(NH_4^+) = 185 - 70 = 115 \ S \ cm^2 \ mol^{-1}$.Now,$\lambda_{m}^{\circ}(NH_4OH) = \lambda_{m}^{\circ}(NH_4^+) + \lambda_{m}^{\circ}(OH^-) = 115 + 170 = 285 \ S \ cm^2 \ mol^{-1}$.The degree of dissociation $(\alpha)$ is given by $\alpha = \frac{\lambda_{m}}{\lambda_{m}^{\circ}} = \frac{85.5}{285} = 0.3$.Since $\alpha = x 	imes 10^{-1}$,we have $0.3 = x 	imes 10^{-1}$,which gives $x = 3$.

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