Question

$0.222g$ of iron ore was brought into solution, $F{e}^{3+}$ is reduced to $F{e}^{2+}$ with $SnC{l}_2$. The reduced solution required $20mL$ of $0.1NKMn{O}_4$ solution. The percentage of iron present in the ore is:

[Equivalent weight of iron is $55.5$]

• A. $55.5$%
• B. $45.0$%
• C. $50.0$%
• D. $40.0$%

Answer 1

Answer: (C)

Number of equivalents of $Fe$ $=$ Number of equivalents of $KMn{O}_4$

$\frac{NV}{1000}=\frac{0.1\times 20}{1000}=0.002$

Mass of iron (pure) $=0.002\times 55.5=0.111g$

% $purity=\frac{\text{Mass of pure iron}}{\text{Mass of iron ore}}\times 100=\frac{0.111}{0.222}\times 100=50$%.

Hence, option $C$ is correct.