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Dalton's Law of Partial Pressures

A mixture of ...

Question

A mixture of $C_{2} H_{2}$ and $C_{3} H_{8}$ occupied a certain volume at a pressure of 80 mm Hg. The mixture was completely burnt to form $C O_{2}$ and $H_{2} O (l)$ . If the pressure of $C O_{2}$ was found to be 230 mm Hg at the same temperature and volume, then the mole fraction of $C_{3} H_{8}$ in the mixture is:

A

0.125

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B

0.875

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C

0.6

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D

0.8

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Solution

The correct option is B 0.875
Reactions are :
$(i) C_{2} H_{2} + 2.5 O_{2} \to 2 C O_{2} + H_{2} O (l)$
$(ii) C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O (l)$
Since, $V$ and $T$ are constants, $P \propto n$ .
Say, $P_{C_{2} H_{2}} = P_{m m}$ , then, $P_{C_{3} H_{8}} = (80 - P) mm$
now, $P_{C O_{2}} (i) = 2 \times P_{C_{2} H_{2}} = 2 P$
$P_{C O_{2}} (i i) = 3 P_{C_{3} H_{8}} = 3 (80 - P)$
$P_{C O_{2} (t o t a l)} = 2 P + 3 (80 - P) = 230$ (Given)
$∴ 240 - P = 230 \Rightarrow P = 10$
$∴ P_{C_{3} H_{8}} = 80 - 10 = 70 mm$
$∴$ mole fraction $= \frac{70}{80} = 0.875$

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