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Concentration Terms

A mixture of ...

Question

A mixture of $a$ mol of $C_{3} H_{8}$ and $b$ mol of $C_{2} H_{4}$ was kept is a container of $V$ L which exerts a pressure of $4.93$ atm at temperature T. Mixture was burnt in presence of $O_{2}$ to convert $C_{3} H_{8}$ and $C_{2} H_{4}$ into $C O_{2}$ in the container at the same temperature. The pressure of gases after the reaction and attaining the thermal equilibrium with atmosphere at temperature T was found to be $11.08$ atm.

The mole fraction of $C_{3} H_{8}$ in the mixture is :

A

0.25

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B

0.75

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C

0.45

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D

0.55

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Solution

The correct option is A $0.25$
The balanced reactions are as follows:

$C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O (l)$
$C_{3} H_{4} + 3 O_{2} \to 2 C O_{2} + 4 H_{2} O (l)$
Initially,
$P V = n R T$
$4.93 \times V = (a + b) R T$ .....(i)
After combustion, pressure is due to the total moles of $C O_{2}$ .
$11.8 \times V = (3 a + 2 b) R T$ .....(ii)
Divide equation (ii) by equation (i), we get
$\frac{11.08}{4.93} = 2.25 = \frac{3 a + 2 b}{a + b}$
$2.25 a + 2.25 b = 3 a + 2 b$
$0.25 b = 0.75 a$
$b = 3 a$
$x_{C_{3} H_{8}}$ or $x_{a} = \frac{a}{a + b} = \frac{a}{a + 3 a} = \frac{1}{4} = 0.25$

Hence, the mole fraction of $C_{3} H_{8}$ in the mixture is $0.25$ .

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