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

KMnO_{4} solution = \( \frac{20}{1000}L\times 0.1\,M\)

= 2 x 10^{-3} mol.

**The balanced chemical equation for the redox reaction is : **

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

**According to above balance chemical equation -**

∵ We have required 5 mole pure H_{2}O_{2} to reduce 2 mol KMnO_{4}.

∴ Number of moles of pure H_{2}O_{2} required to reduce 2 x 10^{-3} mol KMnO_{4}

= \(\frac{5}{2}\) x 2 x 10^{-3} mol

= 5 x 10^{-3} mol of pure H_{2}O_{2}

Molecular weight of H_{2}O_{2} = 34 g/mol

∴ Mass of pure H_{2}O_{2} = 5 x 10^{-3} mol x 34g/mol

= 0.17 g

**But,**

Amount of H_{2}O_{2} actually precent = 0.2g

∴ Percentage purity of H_{2}O_{2} = \(\frac{0.17}{0.20}\times 100\)

= 85

**Hence,**

The purity of H_{2}O_{2} was 85%.