A mixture of 40g of oxygen gas and 2g of helium gas exerts a total pressure of 2.4 atm, what is the mole fraction of oxygen gas in the mixture

The molar mass of $\text{O}_2$ is $2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol}$.

$$n_{\text{O}_2} = \frac{\text{mass of O}_2}{\text{molar mass of O}_2} = \frac{40 \text{ g}}{32.00 \text{ g/mol}} = 1.25 \text{ mol}$$

The molar mass of $\text{He}$ is $4.00 \text{ g/mol}$.

$$n_{\text{He}} = \frac{\text{mass of He}}{\text{molar mass of He}} = \frac{2 \text{ g}}{4.00 \text{ g/mol}} = 0.50 \text{ mol}$$

The mole fraction of a component is the ratio of the moles of that component to the total moles in the mixture.

$$\chi_{\text{O}_2} = \frac{n_{\text{O}_2}}{n_{\text{total}}} = \frac{1.25 \text{ mol}}{1.75 \text{ mol}}$$ $$\chi_{\text{O}_2} \approx 0.714$$

The total pressure information (2.4 atm) is not needed to calculate the mole fraction.

The mole fraction of oxygen gas in the mixture is approximately 0.714.

$$0.714$$