Verified Solution • 2025–26 Session moderate

QJEEM20Chaa79: JEE Main 2020 Chemistry - Class 11 JEEM20Chaa79 Solution: A 20.0 mL solution containing 0.2 g impure H 2 O 2 reacts…

A 20.0 mL solution containing 0.2 g impure H2O2 reacts completely with 0.316 g of KMnO4 in acid solution. The purity of H2O2 (in %) is _____________
(mol. wt. of H2O2 = 34; mol. wt. of KMnO4 = 158)

Step-by-Step Verified Solution

5H2O2 + 2MnO4- + 6H+ →\to 2Mn2+ + 5O2 + 8H2O

Moles of KMnO4 = 0.316158{{0.316} \over {158}} = 2 ×\times 10-3

Equivalents of H2O2 = Equivalent of KMnO4

= 2 × 10–3 × 5 = 0.01

Moles of H2O2 = 0.012{{0.01} \over 2} = 0.005

Mass of pure H2O2 = 0.005 × 34 = 0.170 gm

Percentage purity = 0.170.2{{0.17} \over {0.2}} ×\times 100 = 85 %

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