Question

The pressure of the mixture of equal weights of two gases of molecular weight 4 and 40 is 1.1 atm. The partial pressure of the lighter gas in the gas mixture is:

• A. 0.55 atm
• B. 0.11 atm
• C. 1 atm
• D. 0.1 atm

Answer 1

Answer: (C)

1 atm

Suppose 40 $g$ each of the lighter and heavier gas is present.

The molar masses are 4 and 40 $g/mol$ respectively.

The total number of moles are $\frac{40}{4}=10$ moles for lighter gas

and $\frac{40}{40}=1$ mole for heavier gas.

The mole fraction of the lighter gas is $\frac{10}{10+1}=\frac{10}{11}$

The partial pressure of the lighter gas is $\frac{10}{11}\times 1.1atm=1atm$