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Dalton's Law of Partial Pressures

Equal masses ...

Question

Equal masses of $S O_{2}$ and $O_{2}$ are kept in a vessel at $27^{o} C$ . The total pressure of the mixture is 2.1 $a t m$ . The partial pressure of $S O_{2}$ is:

A

1.4

a t m

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B

7

a t m

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C

0.7

a t m

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D

14

a t m

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Solution

The correct option is B 0.7 $a t m$
The molecular weight of $S O_{2}$ is twice that of $O_{2}$ .
Mole fraction of $S O_{2} = \frac{w / 64}{w / 64 + w / 32} = 1 / 3$

Mole fraction of $S O_{2} = 1 : 3$
$p_{S O_{2}} = X_{S O_{2}} \times P_{T o t a l} = 1 / 3 \times 2.1$

Therefore, partial pressure of $S O_{2}$ is equal to $\frac{2.1}{3} = 0.7$ $a t m$ .

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