The degree of dissociation of ${\mathrm{NH}}_4\mathrm{OH}$ in a solution which is $0.2 \mathrm{M}$ with respect to

${\mathrm{NH}}_4\mathrm{Cl}\text{ and }0.1\mathrm{M}$ with respect to ${\mathrm{NH}}_4\mathrm{OH}$ is:

                                    ${\mathrm{NH}}_4\mathrm{OH}\rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{-}$

Conc. . at start           $\begin{array}{l}\mathrm{C} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{C\alpha }\hspace{0.25em} \hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\mathrm{C\alpha }\end{array}$

Conc. at equilibrium     $\mathrm{C}=(1-\mathrm{\alpha } ) \mathrm{C} \mathrm{\alpha }$

                                        ${\mathrm{NH}}_4\mathrm{Cl}\rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{Cl}}^{-}$

Conc.at equilibrium                           $0.2$

$\left[{\mathrm{NH}}_4\mathrm{OH}\right]=C(1-\mathrm{\alpha })=C$ (assuming $\mathrm{\alpha }$ to be very small).

$\left[{\mathrm{NH}}_4^{+}\right]=\mathrm{C}\alpha +0.2;\left[{\mathrm{OH}}^{-}\right]=\mathrm{C}\alpha .$

$\begin{array}{l}\mathit{ }\mathit{ }\mathit{\therefore }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }\mathit{ }{K}_b=\frac{\left[{\mathrm{NH}}_4^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_4\mathrm{OH}\right]}\\ 1.8\times {10}^{-5}=\frac{(\mathrm{C}\alpha +0.2)\times \mathrm{C}\alpha }{C}\end{array}$

or   $1.8\times {10}^{-5}=0.2\alpha$

          $[\because \alpha \text{ is very small, }C\alpha +0.2=0.2]$

$\alpha =\frac{1.8\times {10}^{-5}}{0.2}=9\times {10}^{-5}$

or            $\alpha \mathit{ }= 9 \times {10}^{-5} \times 100 = 9 \times {10}^{-3}\mathrm{\%}$

So, the correct answer is (a).