JEE MAIN - Chemistry (2020 - 4th September Morning Slot - No. 7)
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)
(mol. wt. of H2O2 = 34; mol. wt. of KMnO4 = 158)
Answer
85
Explanation
5H2O2 + 2MnO4- + 6H+ $$ \to $$ 2Mn2+ + 5O2 + 8H2O
Moles of KMnO4 = $${{0.316} \over {158}}$$ = 2 $$ \times $$ 10-3
Equivalents of H2O2 = Equivalent of KMnO4
= 2 × 10–3 × 5 = 0.01
Moles of H2O2 = $${{0.01} \over 2}$$ = 0.005
Mass of pure H2O2 = 0.005 × 34 = 0.170 gm
Percentage purity = $${{0.17} \over {0.2}}$$ $$ \times $$ 100 = 85 %
Moles of KMnO4 = $${{0.316} \over {158}}$$ = 2 $$ \times $$ 10-3
Equivalents of H2O2 = Equivalent of KMnO4
= 2 × 10–3 × 5 = 0.01
Moles of H2O2 = $${{0.01} \over 2}$$ = 0.005
Mass of pure H2O2 = 0.005 × 34 = 0.170 gm
Percentage purity = $${{0.17} \over {0.2}}$$ $$ \times $$ 100 = 85 %
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