A $10 L$ flask contains $41.4 g$ of a mixture of gases $C_{x} H_{8}$ and $C_{x} H_{12} .$ The total pressure at 44°C in flask is 1.56 atm. Analysis revealed that the gas mixture has 87% total $C$ and 13% total $H$ . Find out that value of x .
[Use R= 0.082 L atm $m o l^{- 1} K^{- 1}$ ]

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answer is 5.

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Detailed Solution

Given, $P = 1.56 a t m; V = 10 L$
$T = 317 K, R = 0.082 T o t a l m o l e s (n) = \frac{P V}{R T} = \frac{1.56 \times 10}{0.082 \times 317} = 0.6 m o l$

Let $C_{x} H_{8}$ be a mol, therefore moles of $C_{x} H_{12} = (0.6 - a) m o l;$ mass of $C$ in a mol of
$C_{x} H_{12} = 12 a x g; m a s s o f C i n (0.6 - a) m o l o f C_{x} H_{12} = 12 \times (0.6 - a) g$
Total mass of $C$ in mixture = $12 a x + 12 x (0.6 - a) g$
$= 7.2 x g$
$% o f C in mixture = \frac{7.2 x}{41.4} \times 100$
Given $% o f C = 87 %$ or $\frac{720 x}{41.4} = 87 o r x = 5$