No of moles of KMnO4 present in 20 mL of 0.1M
KMnO4 solution = \( \frac{20}{1000}L\times 0.1\,M\)
= 2 x 10-3 mol.
The balanced chemical equation for the redox reaction is :
2KMnO4 + 5H2O2 + 3H2SO4 → 2MnSO4 + K2SO4 + 8H2O + 5O2 .....(1)
According to above balance chemical equation -
∵ We have required 5 mole pure H2O2 to reduce 2 mol KMnO4.
∴ Number of moles of pure H2O2 required to reduce 2 x 10-3 mol KMnO4
= \(\frac{5}{2}\) x 2 x 10-3 mol
= 5 x 10-3 mol of pure H2O2
Molecular weight of H2O2 = 34 g/mol
∴ Mass of pure H2O2 = 5 x 10-3 mol x 34g/mol
= 0.17 g
But,
Amount of H2O2 actually precent = 0.2g
∴ Percentage purity of H2O2 = \(\frac{0.17}{0.20}\times 100\)
= 85
Hence,
The purity of H2O2 was 85%.