The purity of $H_2{O}_2$ in a given sample is $85\%$. The weight of impure sample of $H_2{O}_2$ which requires $10 ml$ of $M/5{KMnO}_4$ solution in a titration in acidic medium is gram

 In acidic medium ${KMnO}_4$ undergoes reduction.

Milliequivalents of ​${\mathrm{MnO}}_4^{-}$= Milliequivalents of ${\mathrm{H}}_2{\mathrm{O}}_2$
$10\times \frac{\mathrm{M}}{5}\times 5$=Milliequivalents of  ${\mathrm{H}}_2{\mathrm{O}}_2$
⇒ Milliequivalents of ​${\mathrm{H}}_2{\mathrm{O}}_2$=10

Weight of $H_2{O}_2=10\times {10}^{-3}\times \frac{32}{2}=0.17 g$

Weight of 85% impure $H_2{O}_2=\frac{100}{85}\times 0.17=0.2 g$

Therefore, the weight of impure $H_2{O}_2$ is 0.2g.