7g of a sample of NaCl on treatment with excess of silver nitrate gave 14.35g of AgCl. NaCl in the sample is

$$\begin{gathered} \mathrm{NaCl}+{\mathrm{AgNO}}_3\to \mathrm{AgCl}+{\mathrm{NaNO}}_3 \\ 58.5\mathrm{g} 143.5\mathrm{g} \\ ? 14.35\mathrm{g} \\ \mathrm{Mass} \mathrm{of} \mathrm{NaCl}=\frac{14.35}{143.5}\times 58.5=5.85\mathrm{g} \\ \%\mathrm{purity} \mathrm{of} \mathrm{NaCl}=\frac{5.87}{7}\times 100=83.5\% \end{gathered}$$