Volume of strong mono basic acid of normality 10 is needed to prepare 1L of a buffer solution of p^H9, using 1 mole of NH₃ is”X” ml. Find $\frac{\mathrm{X}}{16}$. 
$\left[{\mathrm{K}}_{\mathrm{b}}\text{ for }{\mathrm{NH}}_3=1.8\times {10}^{-5};{\mathrm{p}}^{{\mathrm{K}}_{\mathrm{b}}}=4.7447\right]$

NH₄Cl is a salt of S.A-W.B
Since W.B (NH₃) is left and NH₄Cl is formed. So a mixture of W. Base & Salt is formed Which is a basic buffer.
$\begin{array}{l} {\mathrm{NH}}_3+\mathrm{HCl}\to {\mathrm{NH}}_4\mathrm{Cl}\\ \text{ Initial }\\ \begin{array}{lll} 1& \mathrm{x}& 0\end{array}\\ \text{ Final }\\ 1-\mathrm{x}\mathrm{x}\mathrm{x}\\ {\mathrm{p}}^{\mathrm{OH}}=14-{\mathrm{p}}^{\mathrm{H}}\\ {\mathrm{p}}^{\mathrm{OH}}=14-9=5\\ {\mathrm{p}}^{\text{OH }}={\mathrm{p}}^{{\mathrm{k}}_{\mathrm{b}}}+{\log }_{10}\frac{\left[\text{ salt }\right]}{\left[\text{ base }\right]}\\ 5=4.7447+{\log }_{10}\frac{\mathrm{x}}{1-\mathrm{x}}\\ 5-4.7447={\log }_{10}\frac{\mathrm{x}}{1-\mathrm{x}}\Rightarrow \mathrm{x}=0.64\\ {\mathrm{N}}_1{\mathrm{V}}_1=0.64\text{; }\\ \Rightarrow {\mathrm{V}}_1=\frac{0.64}{{\mathrm{N}}_1}\Rightarrow \frac{0.64}{10}=0.064\mathrm{L}=64\mathrm{ml}=\mathrm{X}\\ \frac{\mathrm{X}}{16}=4\\ \end{array}$