The impure 6 g of NaCl is dissolved in water and then treated with excess of silver nitrate solution. The white of precipitate of silver chloride is found to be 14 g. The % purity of NaCl solution would be:

The reaction that takes place is
$\mathrm{NaCl}+{\mathrm{AgNO}}_3⟶\mathrm{AgCl}\downarrow +{\mathrm{NaNO}}_3$
$\because$ 143.5 g of AgCl is produced from 58.5 g NaCl
$\because$ 14g of AgCl will produce from
$\frac{58.5\times 14}{143.5}=5.70\mathrm{g}$
This is the amount of NaCl in common salt;
$\% \mathrm{purity} =\frac{5.70}{6}\times 100=95\%$