No of moles of KMnO_{4} present in 20ml of 0.1M

KMnOH solution = \(\frac{20}{1000}\) x 0.1

= 2 x 10^{-3}

The balanced equation for the redox reactions is:

2KMnO_{4} + 5H_{2}O_{2} + 3H_{2}SO_{4} → K_{2}SO_{4} + 2MnSO_{4} + 8H_{2}O + 5O_{2} …(1)

From the equation, 2 moles of KMnO_{4} = 5 moles of H_{2}O_{2}

2 x 10^{-3} moles of KMnO_{4} will react with H_{2}O_{2} = \(\frac{5}{2}\) x 2 x 10^{-3} = 5 x 10^{-3} moles Molecular wt. of H_{2}O_{2} = 34

Amount of H_{2}O_{2} actually present

= 34 x 5 x 10^{-3}

= 0.17g

Percentage purity of in 20 ml is

= \(\frac{0.17}{0.20}\) x 100 = 85