$0.2828 \mathrm{g}$ of iron wire was dissolved in excess dilute $H_{2}S{O}_4$ and the solution was made up to $100 \mathrm{mL}. 20 \mathrm{mL}$. of this solution required $30 \mathrm{ml}$of $\frac{N}{30}{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ solution for exact oxidation. Calculate percent purity of Fe in wire.

$\mathrm{Fe}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{FeSO}}_4+{\mathrm{H}}_2$

Milliequivalents of Fe= Milliequivalents of

${\mathrm{FeSO}}_4$

= Milliequivalents for second reaction

${\mathrm{k}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$

${\mathrm{N}}_1{\mathrm{v}}_1={\mathrm{N}}_2{\mathrm{V}}_2$

$\mathrm{M}\times \frac{1000}{100}\times 20=\frac{\mathrm{N}}{30}\times 30$ M= moles of Iron

$M=\frac{1}{200}$

amount of iron in sample $=\frac{56}{200}=0.28$

 So, percentage purity $=\frac{0.28}{0.2828}\times 100$

$=99\mathrm{\%}$

 Hence, answer 99% is correct.