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Q.

7 .5 ml of a gaseous hydrocarbon was exploded with 36 ml of oxygen. The volume of gases on cooling was found to be 28.5 ml, 15 ml of which was absorbed by KOH and the rest was absorbed in a solution of alkaline pyrogallol. ff all volumes are measured under same conditions, the formula of hydrocarbon is:

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a

$C_{3} H_{4}$

b

$C_{2} H_{4}$

c

$C_{2} H_{6}$

d

$C_{3} H_{6}$

answer is B.

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Detailed Solution

Here, in the reaction,

$\begin{array}{l} C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \to x {CO}_{2} + \frac{y}{2} H_{2} O \\ \begin{array}{ccc} 7.5 & 36 & - \\ - 36 - 15 (x + \frac{y}{4}) & 7.5 x \\ = 15 \end{array} \\ \Rightarrow x = 2 \\ 36 - 7.5 [1 + \frac{y}{4}] + 15 = 28.5 \\ \Rightarrow y = 4 \\ C_{2} H_{4} \end{array}$

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7 .5 ml of a gaseous hydrocarbon was exploded with 36 ml of oxygen. The volume of gases on cooling was found to be 28.5 ml, 15 ml of which was absorbed by KOH and the rest was absorbed in a solution of alkaline pyrogallol. ff all volumes are measured under same conditions, the formula of hydrocarbon is: