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Concentration Terms

A mixture of ...

Question

A mixture of $C_{3} H_{8}$ and $C H_{4}$ exerts a pressure of $320$ mm Hg at temperature $T (K)$ in a $V$ litre flask. On complete combustion, gaseous mixture contains $C O_{2}$ , only and exerts a pressure of $448$ mm Hg under identical condition. Hence, mole fraction of $C_{3} H_{8}$ in the mixture is :

A

0.2

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B

0.8

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C

0.25

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D

0.75

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Solution

The correct option is A $0.2$
Let $V$ litre at $T$ k and $320 m m H g$ represents $1 m o l$ then $V$ litre at $T$ k and $448 m m H g$ represents $= \frac{448}{320} m o l = 1.4 m o l$

$C_{3} H_{8} (g) x m o l + S O_{2} (g) \to 3 C O_{2} 3 x m o l (g) + 4 H_{2} O (l)$

$C H_{4} (g) (1 - x) m o l + 2 S O_{2} (g) \to 3 C O_{2} (g) (1 - x) m o l) + 4 H_{2} O (l)$

Let moles of $C_{3} H_{8} = x$

Moles of $C H_{4} = 1 - x$

Moles of $C O_{2}$ produced $= 3 x + 1 - x = 1 + 2 x$
$1 + 2 x = 1.4$
$2 x = 0.4$
$x = 0.2$

Mole fraction of $C_{3} H_{8} = \frac{x}{1} = 0.2$

$∴$ option A is correct.

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