5 ml of a gaseous hydrocarbon was burnt in 30ml of oxygen gas. The remaining gases on cooling to room temperature measured 25ml. When the remaining gases were passed through KOH solution, the volume of the gases left over to be 15ml. Find the formula of the hydrocarbon if all the measurements were made under identical conditions

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$C_{2} H_{6}$

$C_{3} H_{8}$

$C_{3} H_{6}$

$C_{2} H_{4}$

answer is D.

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Detailed Solution

$C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x C O_{2} + y_{2} H_{2} O$

5 ml $5 (x + \frac{y}{4}) O_{2} 5 x C O_{2}$
Volume of $C O_{2} = 5 x = 10$
X=2
Volume of $O_{2}$ taken =30 volume of $O_{2}$ left =15
Volume of $O_{2}$ used = 30-15 = 15
$5 (x + \frac{y}{4}) = 15$

$\begin{array}{l} x + \frac{y}{4} = 3 \\ \frac{y}{4} = 3 - x = 3 - 2 = 1 \\ y = 4 \\ ∴ f o r m u l a : C_{2} H_{4} x + \frac{y}{4} = 3 \\ \frac{y}{4} = 3 - x = 3 - 2 = 1 \\ y = 4 \\ ∴ f o r m u l a : C_{2} H_{4} \end{array}$