What is percent dissociation of $\mathrm{NH}_4 \mathrm{OH}$ if molar conductance at zero concentration for $\mathrm{NH}_4 \mathrm{Cl}, \mathrm{NaCl}$ and NaOH are 130,109 and $213 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively and molar conductivity of $0.01 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$ is $9.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ ?

The percent dissociation of $\mathrm{NH}_4\mathrm{OH}$ is determined using conductivity measurements. Kohlrausch's law gives the limiting molar conductivity as $\Lambda_0(\mathrm{NH}_4\mathrm{OH}) = \Lambda_0(\mathrm{NH}_4\mathrm{Cl}) + \Lambda_0(\mathrm{NaOH}) - \Lambda_0(\mathrm{NaCl}) = 130 + 213 - 109 = 234\ \mathrm{S\ cm}^2\ \mathrm{mol}^{-1}$.

The degree of dissociation $\alpha$ relates measured conductivity to infinite dilution: $\alpha = \frac{\Lambda_m}{\Lambda_0} = \frac{9.0}{234}$. Percent dissociation is $\alpha \times 100 = \frac{900}{234} = \frac{100}{26}\%$.

This matches option D, $\frac{100}{26}\%$.