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0.2g of a sample of H2O2 reduced 20 ml of 0.1M KMnO4 solution in acidic medium. What is the percentage purity of H2O2?

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No of moles of KMnO4 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:

2KMnO4 + 5H2O2 + 3H2SO4 → K2SO4 + 2MnSO4 + 8H2O + 5O2 …(1)

From the equation, 2 moles of KMnO4 = 5 moles of H2O2

2 x 10-3 moles of KMnO4 will react with H2O2\(\frac{5}{2}\) x 2 x 10-3 = 5 x 10-3 moles Molecular wt. of H2O2 = 34

Amount of H2O2 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

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