100 ml of $C H_{4}$ and $C_{2} H_{2}$ were exploded with excess of $O_{2}$ . After explosion and cooling, the mixture was treated with KOH, where a reduction of 165 ml was observed. Therefore the composition of the mixture is -

C H_{4} = 35 m l; C_{2} H_{2} = 65 m l

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C H_{4} = 65 m l; C_{2} H_{2} = 35 m l

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C H_{4} = 75 m l; C_{2} H_{2} = 25 m l

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C H_{4} = 25 m l; C_{2} H_{2} = 75 m l

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Solution

The correct option is A $C H_{4} = 35 m l; C_{2} H_{2} = 65 m l$
Let vol. of $C H_{4} = x m l$
$∴ v o l . o f C_{2} H_{2} = (100 - x)$
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (ℓ)$
x ml. V ml - -
Finally - (V - 2x) ml x ml -
- 3(100 – x)
$C_{2} H_{2} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (ℓ)$
(100 – x) V ml - -
Finally - V - 3(100 - x) 2(100 - x)
- 2x
After combustion, the total vol. of gas
= x + 2 (100 – x) ml
$∴ x + 2 (100 - x) = 165$
200 - x = 165
or x = 35 ml

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