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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?

2 Answers

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by (64.4k points)
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Best answer

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

by (57 points)
Plz solve it in some other simple way
+2 votes
by (26.4k points)

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%.

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