Equal masses of methane and oxygen are mixed in an empty container at `25^(@)C`. The fraction of the total pressure exerted by oxygen is:

Options

A. `1/2`
B. `2/3`
C. `1/3 xx (273)/(298)`
D. `1/3`

Explanation

To find the fraction of the total pressure exerted by oxygen when equal masses of methane (CH₄) and oxygen (O₂) are mixed in a container at 25°C, we can follow these steps:

### Step 1: Define the masses of methane and oxygen
Let the mass of methane (CH₄) be \( x \) grams and the mass of oxygen (O₂) also be \( x \) grams.

### Step 2: Calculate the number of moles of each gas
- For methane (CH₄):
\[ \text{Molar mass of CH₄} = 16 \, \text{g/mol} \]
\[ \text{Moles of CH₄} = \frac{x}{16} \]

- For oxygen (O₂):
\[ \text{Molar mass of O₂} = 32 \, \text{g/mol} \]
\[ \text{Moles of O₂} = \frac{x}{32} \]

### Step 3: Calculate the total number of moles
The total number of moles in the mixture is:
\[ \text{Total moles} = \text{Moles of CH₄} + \text{Moles of O₂} = \frac{x}{16} + \frac{x}{32} \]

To add these fractions, find a common denominator:
\[ \frac{x}{16} = \frac{2x}{32} \]
So,
\[ \text{Total moles} = \frac{2x}{32} + \frac{x}{32} = \frac{3x}{32} \]

### Step 4: Calculate the mole fraction of oxygen (O₂)
The mole fraction of oxygen is given by:
\[ \chi_{O₂} = \frac{\text{Moles of O₂}}{\text{Total moles}} = \frac{\frac{x}{32}}{\frac{3x}{32}} = \frac{1}{3} \]

### Step 5: Calculate the fraction of total pressure exerted by oxygen
According to Dalton's Law of Partial Pressures, the partial pressure of oxygen is:
\[ P_{O₂} = \chi_{O₂} \times P_{\text{total}} \]
Thus, the fraction of the total pressure exerted by oxygen is:
\[ \text{Fraction of pressure by O₂} = \chi_{O₂} = \frac{1}{3} \]

### Final Answer
The fraction of the total pressure exerted by oxygen is \( \frac{1}{3} \).

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