Question

The impure $6g$ of $NaCl$ is dissolved in water and then treated with excess of silver nitrate solution. The weight of precipitate of silver chloride is found to be $14.35g$. The $\%$ purity of $NaCl$ solution would be :

• A. $97.5\%$
• B. $87.5\%$
• C. $77.5\%$
• D. $67.5\%$

Answer 1

The correct option is A $97.5\%$

The reaction that takes place is
$NaCl+AgN{O}_3\to AgCl\downarrow +NaN{O}_3$
$\because 143.5ofAgCl\text{is produced from}58.5gNaCl\therefore 14g\text{of AgCl will produce from}\frac{58.5\times 14.35}{143.5}=5.85g$
$\text{This is the amount of NaCl in common salt.}\%purity=\frac{5.85}{6}\times 100=97.5\%$