Question

An impure sample of sodium chloride that weighed $0.50$ g gave $0.90$ g of silver chloride as precipitate on treatment with excess of silver nitrate solution. Calculate the percentage purity of the sample:

• A. $72\%$
• B. $78\%$
• C. $80\%$
• D. None of the above

Answer 1

The correct option is A $72\%$

The balanced reaction is as follows:

$NaCl+AgN{O}_3\to AgCl+NaN{O}_3$

$143.5$ g of $AgCl$ is obtained from $58.5$ g of pure $NaCl$.

$0.9$ g of $AgCl$ is obtained from $\frac{58.5\times 0.9}{143.5}$g $=0.36$ g of pure $NaCl$.

Percentage purity $=\frac{0.36\times 100}{0.5}=72\%$