Question

$25.0$ g of $FeS{O}_4.7H_{2}O$ was dissolved in water containing dilute $H_{2}S{O}_4$ and the volume was made up to $1.0$ L. $25$ mL of this solution required $20$ mL of an $\frac{N}{10}KMn{O}_4$ solution for complete oxidation. The percentage of $FeS{O}_4.7H_{2}O$ in the acid solution is:

• A. $78\%$
• B. $98\%$
• C. $89\%$
• D. $79\%$

Answer 1

The correct option is C $89\%$

As given,
$\Rightarrow \frac{\left(W_{FeS{O}_4\cdot 7H_{2}O}\right)\times 1000}{278/1}=80$
$\Rightarrow W_{FeS{O}_4\cdot 7H_{2}O}=22.24$
$\Rightarrow \%$ of $FeS{O}_4\cdot 7H_{2}O=\frac{22.24}{25}\times 100\%=88.96\%$