7.5 ml of a gaseous hydrocarbon was exploded with 36 ml of $O_{2}$ . On cooling, the volume of gases was found to be 28.5 ml, 15 ml of which was absorbed by $K O H$ and the rest was absorbed in a solution of alkaline pyrogallol. The formula of hydrocarbon is:

C_{2} H_{6}

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C_{2} H_{4}

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C_{3} H_{8}

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C_{3} H_{6}

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Solution

The correct option is B $C_{2} H_{4}$
$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} ⟶ x C O_{2} + \frac{y}{2} H_{2} O$

$7.5 m l 7.5 (x + \frac{y}{4}) m l 7.5 x m l$

Volume of $C O_{2} = 15 m l =$ volume absorbed by KOH
$∴ 7.5 x = 15 ⟹ x = 2$

Volume left $= 28.5 - 15 = 13.5 m l$

Volume of $O_{2}$ (used) $= 36 - 13.5 = 22.5 m l$

$∴ 7.5 (x + \frac{y}{4}) = 22.5$

Solve for y, putting $x = 2, y = 4$

Hence, the formula $= C_{2} H_{4}$

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