Question

$3.92$ g of ferrous ammonium sulphate (FAS) reacts completely with $50$ ml $\frac{N}{10}KMn{O}_4$ solution. The percentage purity of the sample is:

• A. $50$
• B. $78.4$
• C. $80$
• D. $28$

Answer 1

The correct option is A $50$

$50$ ml of $0.1N$ potassium permanganate corresponds to $\frac{50\times 0.1}{1000}=0.005eq.KMn{O}_4$.
Thus, $0.005$ eq. $FAS$ has reacted.
The equivalent weight of $FAS$ is $392$ g.
$0.005$ equivalents of $FAS$ corresponds to $0.005\times 392=1.96g$ of $FAS$.

Hence, the percentage purity of $FAS$ sample is $\frac{1.96}{3.92}\times 100$ or $50\%$.