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

Question

Aicharya · Accepted Answer

The reaction that takes place is
$\mathrm{NaCl}+\mathrm{AgNO}_{3} \longrightarrow \mathrm{AgCl} \downarrow+\mathrm{NaNO}_{3}$
$\therefore 143.5 \mathrm{~g}$ of $\mathrm{AgCl}$ is produced from $58.5$ $\mathrm{g} \mathrm{NaCl}$
$\therefore 14 \mathrm{~g}$ of $\mathrm{AgCl}$ will be produced from
$\frac{58.5 \times 14}{143.5}=5.70 \mathrm{~g} \mathrm{NaCl}$
This is the amount of $\mathrm{NaCl}$ in common salt:
$\%$ purity $=\frac{5.70}{6} \times 100=95 \%$

The impure $6 \mathrm{~g}$ of $\mathrm{NaCl}$ is dissolved in water and then treated with excess of silver…

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