107 g of a ammonium chloride sample decomposes on heating to give HCl(g) and 25.5 g of NH₃(g). The percentage purity of ammonium chloride sample is

$$\begin{gathered} \begin{array}{lc}{\mathrm{NH}}_4\mathrm{Cl}\left(\mathrm{s}\right)\stackrel{\mathrm{\Delta }}{⟶}{\mathrm{NH}}_3\left(\mathrm{g}\right)+\mathrm{HCl}\left(\mathrm{g}\right) \\ 53.5 \mathrm{g} 17 \mathrm{g} 36.5 \mathrm{g}& \end{array} \end{gathered}$$
17 g NH₃ is obtained by 53.5 g NH₄Cl
So, 25.5 g NH₃ will be obtained by
$=\frac{53.5}{17}\times 25.50=80.25 \mathrm{g} {\mathrm{NH}}_4\mathrm{Cl}$
$\therefore \mathrm{\%}\text{ purity of sample }=\frac{80.25}{107}\times 100=75\mathrm{\%}$