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H 2 O 2 + 2KI → 40% yield I 2 + 2 KOH H 2 O 2 + 2 KMnO 4 + 3 H 2 SO 4 → 50% yield K 2 SO 4 + 2 MnSO 4 + 3 O 2 + 4 H 2 O 150 mL of H 2 O 2 sample was divided into two parts. First part was treated with KI and formed KOH required 200 mL of M/2 H 2 SO 4 for neutralisation. Other part was treated with KMnO 4 yielding 6.74 litre of O 2 at 1 atm and 273 K. Using % yield indicated, the volume strength of H 2 O 2

Options

  1. A5.04
  2. B10.08
  3. C3.36
  4. D33.6

Correct answer

D. 33.6

Step-by-step solution

2 K O H   +   H 2 S O 4     → K 2 S O 4   +   2 H 2 O 2 mole of KOH react with 1 mole of H 2 SO 4 . Mole of H 2 SO 4 = 0.1 ; mole of KOH = 0.2 whereas in first reaction 1 mole H 2 O 2 form 2 mole KOH. mole of H 2 O 2 used in first reaction = 0.2 2 × 1 0.4 = 0.25 mole in second reaction. mole of produced O 2 = 6.74 22.4 = 0.3 mole ∴ mole of H 2 O 2 used in second reaction = 0.3 3 × 0.5 = 0.2 mole Total mole of consumed H 2 O 2 = 0.25 + 0.2 =0.45 mole Molarity of H

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