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

Solution :
The reaction that takes place is
$$NaCl + AgN{O_3} \to AgCl \downarrow + NaN{O_3}$$
$$\therefore \,\,143.5\,g$$   of $$AgCl$$  is produced from $$58.5$$  $$g$$ $$NaCl$$
$$\therefore \,\,14\,g$$   of $$AgCl$$  will be produced from $$\frac{{58.5 \times 14}}{{143.5}} = 5.70\,g\,NaCl$$
This is the amount of $$NaCl$$  in common salt; $$\% \,{\text{purity}} = \frac{{5.70}}{6} \times 100 = 95\% $$