A 20 . 0mL solution containing 0 . 2g impure H _ 2 O _ 2 reacts...

Let mass of pure ${H}_{2}{O}_{2}$ is $x$ gram

$$\begin{array}{ll}+ 7 \\ MnO{4}^{-}\end{array}$$$$\begin{array}{ll}+\end{array}$$$$\begin{array}{ll}- 1 \\ {H}_{2}{O}_{2} \\ vf = 2\end{array}$$$$\begin{array}{ll}\to\end{array}$$$$\begin{array}{ll}+ 2 \\ {Mn}^{2 +}\end{array}$$$$\begin{array}{ll}+\end{array}$$$$\begin{array}{ll}0 \\ {O}_{2}\end{array}$$

Eq. of ${H}_{2}{O}_{2} = Eq .$ of $MnO{4}^{-}$

\left[\dfrac{x}{34}\right]2 = \left[\dfrac{0 .316}{158}\right]5$</p><p>$x = 0 . 17$</p><p>So,$%$purity of${H}_{2}{O}_{2}$solution$= \dfrac{0 .17}{0 .2} \times 100 = 85 %$