Question

A sample $(5.6 \,g )$ containing iron is completely dissolved in cold dilute $HCl$ to prepare a $250\, mL$ of solution. Titration of $25.0 \,mL$ of this solution requires $12.5\, mL$ of $0.03 \,M \,KMnO _{4}$ solution to reach the end point. Number of moles of $Fe ^{2+}$ present in $250\, mL$ solution is $x \times 10^{-2}$ (consider complete dissolution of $FeCl _{2}$ ). The amount of iron present in the sample is $y \%$ by weight.
(Assume: $KMnO _{4}$ reacts only with $Fe ^{2+}$ in the solution
Use: Molar mass of iron as $56 \,g \,mol ^{-1}$ )
The value of $y$ is _______.

Answer 1

Moles of $Fe ^{+2}= x \times 10^{-2}=1.875 \times 10^{-2}$
wt. of $Fe ^{+2}=1.875 \times 10^{-2} \times 56$
Hence percentage of $Fe ^{+2}$
$=\frac{1.875 \times 10^{-2} \times 56}{5.6} \times 100 \%$
$=18.75 \%$