A sample $(5.6 \mathrm{~g})$ containing iron is completely dissolved in cold dilute $\mathrm{HCl}$ to…

Moles of ${\mathrm{Fe}}^{2+}$ present in $250 \mathrm{ml}$ solution $=\mathrm{x}\times {10}^{-2}$

Moles of ${\mathrm{Fe}}^{2+}$ present in $25 \mathrm{ml}$ solution $= \mathrm{x}\times {10}^{-3} \mathrm{mole}$

At equivalence point

eq. of ${\mathrm{Fe}}^{2+}$ = eq. of ${\mathrm{KMnO}}_4$

${\left(\mathrm{Moles}\right)}_{{\mathrm{Fe}}^{2+}}\times {\left(\mathrm{vf}\right)}_{{\mathrm{Fe}}^{2+}}={\left(\mathrm{Moles}\right)}_{{\mathrm{KMnO}}_4}\times {\left(\mathrm{vf}\right)}_{{\mathrm{KMnO}}_4}$

$\mathrm{x}\times {10}^{-3}\times 1=0.03\times 12.5\times {10}^{-3}\times 5$

$\mathrm{x}=1.875 \mathrm{mole}$

Moles of ${\mathrm{Fe}}^{2+}$ present in $250 \mathrm{ml}$ solution $=1.875\times {10}^{-2}$

Mass of ${\mathrm{Fe}}^{2+}$ present in $250 \mathrm{ml}$ solution

$=1.875\times {10}^{-2}\times 56 \mathrm{gm}$

$=1.05 \mathrm{gm}$

$\%$ of $\mathrm{Fe}\left(\mathrm{y}\right)=\frac{1.05}{5.6}\times 100$

$=18.75\%$