On subjecting 10 ml mixture of N 2 and CO to repeated electric spark, 3 ml of O 2 was required for combustion. What was the mole percentage of CO in the mixture under identical conditions?

Question

On subjecting 10 ml mixture of N 2 ​ and CO to repeated electric spark, 3 ml of O 2 ​ was required for combustion. What was the mole percentage of CO in the mixture under identical conditions?

TestHub · Accepted Answer

Let the mixture have x mL of CO and ( 10 − x ) mL of N₂. CO + 2 1 ​ O 2 ​ → CO 2 ​ x 2 x ​ = 3 x ∴ x = 6 m l At similar conditions of temperature ( T ) and pressure ( P ), mole percentage is the same as volume percentage. Therefore, the mole percentage of CO is: 10 6 ​ × 100 = 60%

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