(આપેલ : $pK _{ b }\left( NH _3\right)=4.74,NH _3$ નું મોલર દળ $=17\, g\, mol ^{-1},NH _4 Cl$નું મોલર દળ $= 53.5\, g\, mol ^{-1}$
$= pK _{ b }+\log \frac{\left[ NH _{4}^{+}\right]}{\left[ NH _{3}\right]}$
$=5.74=4.74+\log \frac{\left[ NH _{4}^{+}\right]}{0.2} \Rightarrow\left[ NH _{4}^{+}\right]=2$
Hence,$NH _{4} Cl =2 \times 53.5=107\,g$