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Dalton's Law of Partial Pressures

Equal masses ...

Question

Equal masses of $S O_{2}, C H_{4}$ and $O_{2}$ are mixed in empty container at $298$ K, where total pressure is $2.1$ atm. The partial pressures of $C H_{4}$ in the mixture is:

A

0.5

atm

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B

0.75

atm

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C

1.2

atm

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D

0.6

atm

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Solution

The correct option is C $1.2$ atm
Partial pressure is directly proportional to the mole fraction of the gas
Partial pressure $=$ Total pressure $\times$ Mole fraction

Let the mass of each constituent $= x$
Moles of $S O_{2}$ $=$ $\frac{x}{64}$

Moles of $C H_{4}$ $=$ $\frac{x}{16}$

Moles of $O_{2}$ $=$ $\frac{x}{32}$

Total Moles $=$ $\frac{7 x}{64}$

Partial pressure $=mole fraction×total pressure=$ $\frac{\frac{x}{16}}{\frac{7 x}{64}}$ x21

$= 1.2 a t m$

Hence partial pressure of $C H_{4} = 1.2 a t m$

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