To find the fraction of the total pressure exerted by oxygen, we need to consider the mole fractions of methane (CH₄) and oxygen (O₂) in the mixture. Here are the steps to solve this problem:

1. Let us assume equal weights, say 'W' grams, of methane and oxygen are mixed.

2. The molecular weights of methane and oxygen are 16 g/mol and 32 g/mol, respectively. The moles of methane (n_CH4) and moles of oxygen (n_O2) in the mixture are then calculated as follows:

• $$ n_{CH_4} = \frac{\text{Mass of CH}_4}{\text{Molar mass of CH}_4} = \frac{W}{16} \text{ moles} $$
• $$ n_{O_2} = \frac{\text{Mass of O}_2}{\text{Molar mass of O}_2} = \frac{W}{32} \text{ moles} $$

3. For W grams of each gas, ratio of moles becomes:

• $$ \text{The mole ratio of CH}_4 \text{ to O}_2 = \frac{W/16}{W/32} = \frac{32}{16} = 2 \text{ to } 1 $$

4. With 2 moles of CH₄ for every 1 mole of O₂, the total moles in the mixture are 3 moles (2 moles CH₄ + 1 mole O₂).

5. According to Dalton's law of partial pressures, the total pressure P in the container is the sum of the partial pressures of the components. Each component's partial pressure is proportional to its mole fraction:

• Mole fraction of O₂, $$ X_{O_2} = \frac{n_{O_2}}{n_{total}} = \frac{1}{3} $$

6. Hence, the fraction of the total pressure exerted by oxygen would be:

$$ \text{Fraction exerted by O}_2 = X_{O_2} = \frac{1}{3} $$

Therefore, the correct answer is: Option A - 1/3.