Ten milliliters of a gaseous hydrocarbon was burnt completely in 80 ml of $O_{2}$ at STP. The Volume of the remaining gas is 70 ml. The volume became 50 ml on treatment with NaOH. The formula of the 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 $C_{2} H_{4}$
$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} ⟶ x C O_{2} + \frac{y}{2} H_{2} O$

$1 m l (x + \frac{y}{4}) m l x m l$

$10 m l 10 (x + \frac{y}{4}) m l 10 x m l$
volume of $C O_{2} = (70 - 50) = 20 m l$

$10 x = 20$ , therefore, $x = 2$

Volume of $C O_{2} + V o l u m e o f O_{2} (l e f t) = 70 m l$

Volume of $O_{2}$ (left) $= 70 - 20 = 50 m l$

Volume of $O_{2}$ (used) $= 80 - 50 = 30 m l$

$∴ 10 (x + \frac{y}{2}) = 30$

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

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

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