One litre of a mixture of O 2 and O 3 at 1 atm and 300 K was allowed to react with an excess of acidified solution of KI. The iodine liberated required 40 mL of M /10 sodium thiosulphate solution

Question

One litre of a mixture of O 2 ​ and O 3 ​ at 1 atm and 300 K was allowed to react with an excess of acidified solution of KI. The iodine liberated required 40 mL of M /10 sodium thiosulphate solution for titration. What is the mass percent of ozone in the mixture?

TestHub · Accepted Answer

Na₂S₂O₃ used = 40 mL of M/10 = 0.04 × 0.1 = 0.004 m o l I₂ formed: I 2 ​ = 2 0.004 ​ = 0.002 m o l From reaction: Ozone liberates 1 mol I₂ per 1 mol O₃ ⇒ moles O₃ = 0.002 mol Mass of O₃: 0.002 × 48 = 0.096 g Now total gas mass (1 L at 1 atm, 300 K): Using density of air ≈ 1.29 g / L ⇒ total mass ≈ 1.29 g Mass % of O₃: 1.29 0.096 ​ × 100 = 7.32%

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