Equal masses of methane and oxygen are mixed in an empty container at $$25^\circ C$$. The fraction of the total pressure exerted by oxygen is

Dalton’s law of partial pressures states that in a gas mixture each component exerts a pressure proportional to its number of moles.
Thus, the fraction of the total pressure contributed by a gas equals its mole fraction:

$$\frac{P_{\text{O}_2}}{P_{\text{total}}}= \frac{n_{\text{O}_2}}{n_{\text{total}}}$$

Let the mass of methane mixed be $$m$$ g. The same mass $$m$$ g of oxygen is taken.

Moles of methane:
$$n_{\text{CH}_4}= \frac{m}{M_{\text{CH}_4}} = \frac{m}{16}$$

Moles of oxygen:
$$n_{\text{O}_2}= \frac{m}{M_{\text{O}_2}} = \frac{m}{32}$$

Total moles:
$$n_{\text{total}} = n_{\text{CH}_4}+n_{\text{O}_2}= \frac{m}{16}+\frac{m}{32}= \frac{2m+m}{32}= \frac{3m}{32}$$

Mole fraction of oxygen (hence pressure fraction):
$$\frac{n_{\text{O}_2}}{n_{\text{total}}}= \frac{\dfrac{m}{32}}{\dfrac{3m}{32}}= \frac{1}{3}$$

Therefore, the fraction of the total pressure exerted by oxygen is $$\dfrac{1}{3}$$.

Option C which is: $$\frac{1}{3}$$