One litre of a sample of ozonized oxygen at 0°C and 1 atm on passing through a KI solution, liberated iodine which required 9 ml of a thiosulphate solution. A volume of 12 ml of a '5.6 volume

Question

One litre of a sample of ozonized oxygen at 0°C and 1 atm on passing through a KI solution, liberated iodine which required 9 ml of a thiosulphate solution. A volume of 12 ml of a '5.6 volume' hydrogen peroxide solution liberated iodine from another iodide solution, which required 24 ml of the same thiosulphate solution. The volume per cent of ozone in the ozonized oxygen sample is

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'5.6 volume' H₂O₂ = 5.6/11.2 = 0.5 M; 12 ml holds 6 mmol H₂O₂, liberating 6 mmol I₂, matched by 24 ml thiosulphate. So 9 ml thiosulphate \equiv 9/24 × 6 = 2.25 mmol I₂ = 2.25 mmol O₃ (O₃ liberates I₂ 1:1). O₃ volume at STP = 2.25 × 22.4 = 50.4 ml per litre = 5.04% ≈ 5. Ans: 5.

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