Calculate amount of $N{H}_{4}Cl$ required to be dissolved in $500mL$ water to have $pH=5.[K_{b}N{H}_{4}OH=1.8\times {10}^{-5}].$
(Molar mass of $N{H}_{4}Cl=53.5gmo{l}^{-1}$)

The correct option is C 4.815 g
$\left[H^{+}\right]={10}^{-pH}={10}^{-5}M$
$\underset{C(1-h)}{N{H}_4^{+}+H_{2}O}\rightleftharpoons \underset{Ch}{N{H}_{4}OH}+\underset{Ch}{H^{+}}$
$K_h=\frac{\left[N{H}_{4}OH\right]\left[H^{+}\right]}{\left[N{H}_4^{+}\right]}=\frac{(Ch{)}^2}{C(1-h)}$
Since weak base, $h<0.11-h\approx 1$

$N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}{K}_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}$

$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}{K}_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
$K_h=C{h}^2=\frac{K_w}{K_b}=\frac{{10}^{-14}}{1.8\times {10}^{-5}}=5.5\times {10}^{-10}$

$h=\frac{K_h}{Ch}=\frac{K_h}{\left[H^{+}\right]}=\frac{5.5\times {10}^{-10}}{{10}^{-5}}=5.5\times {10}^{-5}$
$C=\frac{\left[H^{+}\right]}{h}=\frac{{10}^{-5}}{5.5\times {10}^{-5}}=0.18mol/L$
500 ml of $N{H}_{4}Cl$ contains $\frac{0.18}{2}=0.9mol$
Mass in g $=0.9\times 53.5=48.15g$