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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, 15mL of which was absorbed by $K O H$ & the rest was absorbed in a solution of alkaline pyrogallol. If all volumes are measured under same conditions the formula of hydrocarbon is?

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a

$C_{2} H_{2}$

b

$C_{2} H_{4}$

c

$C_{2} H_{8}$

d

None

answer is C.

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Detailed Solution

$\underset{(g)}{C_{x} H_{y}} + \underset{(g)}{(x + \frac{y}{4})} O_{2} \to_{}^{} x \underset{(g)}{C O_{2}} + \frac{y}{2} \underset{(g)}{H_{2} O}$
$\underset{\underset{0}{7.5}}{C_{x}} + \underset{\underset{36 - 15 [x + \frac{y}{4}]}{36}}{(x + \frac{y}{4}) O_{2}} \to \underset{\underset{7.5 x = 15}{O}}{x C O_{2}} + \underset{0}{\frac{y}{2}} H_{2} O$
$∴ x = 2$
$36 - 7.5 [1 + \frac{y}{4}] + 15 = 28.5 \Rightarrow y = 4$
Hydrocarbon is $C_{2} H_{4}$

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