20 ml of solution having 0.2 g of impure H2O2 reacts with 0.316 of KMn

1.	20 ml of solution having 0.2 g of impure H2O2 reacts with 0.316 of KMnO4 in acidic medium. Calculate (a) Purity of H2O2 (b) Volume of dry oxygen evolved at 0°C and 750 mm pressure.
Answer» 2MnO₄^-(aq) + 6H⁺(aq) + 5H₂O₂(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 50(g) 158 x 2 g of KMnO₄ reacts with 34 × 5 g of H₂O₂ 0.316 g of KMnO₄ will react with \(\frac{170}{0.2}\) x 0.316 g of H₂O₂ = 0.169 g of H₂O₂ ⇒ Percentage purity of H₂O₂ = \(\frac{0.169}{0.2}\) x 100 = 85% Also 316 g of KMnO₄ at STP produces 5 × 22 L of O₂ ⇒ 0.316g of KMnO₄ at STP will produce \(\frac{5\times 22.4}{316}\) x 0.316 = 0.112 L of O₂ Conversion to 750 mm pressure 760 x 0.112 = x × 750x = 0.113 L

20 ml of solution having 0.2 g of impure H2O2 reacts with 0.316 of KMnO4 in acidic medium. Calculate (a) Purity of H2O2 (b) Volume of dry oxygen evolved at 0°C and 750 mm pressure.

Answer»

2MnO₄^-(aq) + 6H⁺(aq) + 5H₂O₂(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 50(g)

158 x 2 g of KMnO₄ reacts with 34 × 5 g of H₂O₂

0.316 g of KMnO₄ will react with \(\frac{170}{0.2}\) x 0.316 g of H₂O₂ = 0.169 g of H₂O₂

⇒ Percentage purity of H₂O₂ = \(\frac{0.169}{0.2}\) x 100 = 85%

Also 316 g of KMnO₄ at STP produces 5 × 22 L of O₂

⇒ 0.316g of KMnO₄ at STP will produce \(\frac{5\times 22.4}{316}\) x 0.316

= 0.112 L of O₂

Conversion to 750 mm pressure 760 x 0.112 = x × 750x = 0.113 L

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20 ml of solution having 0.2 g of impure H2O2 reacts with 0.316 of KMnO4 in acidic medium. Calculate (a) Purity of H2O2 (b) Volume of dry oxygen evolved at 0°C and 750 mm pressure.

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