Question

The impure $6$g of $NaCl$ is dissolved in water and then treated with an excess of silver nitrate solution. The weight of precipitate of silver chloride is found to be $14$g. The percentage purity of $NaCl$ sample(approx.) would be:

• A. $95\%$
• B. $85\%$
• C. $75\%$
• D. $65\%$

Answer 1

The correct option is A $95\%$

$NaCl+AgN{O}_3$ (excess) $⟶NaN{O}_3+AgCl$

$1$ mole of $NaCl⟶1$ mole of $AgCl$

Molar mass: $(23+35.5)g$ of $NaCl⟶(108+35.5)g$ of $AgCl$

$58.5g$ of $NaCl⟶143.5g$ of $AgCl$

$6g$ of $NaCl⟶\frac{143.5}{58.5}\times 6g$ of $AgCl$

$=14.71g$ of $AgCl$

But in question $14g$ of $AgCl$ precipitate

So, impurity= $14.71-14=0.71g$

% of impurity=$0.71/14.71\times 100=4.82$%

So, % purity= $100-4.82=95.18$%