Calculate relative rate of effusion of `O_(2)` to `CH_(4)` from a container container containing `O_(2)` and `CH_(4)` in 3 :2 mass ratio.

Options

A. `(3sqrt(2))/(4)`
B. `(3)/(4sqrt(2))`
C. `(3)/(2sqrt(2))`
D. none of these

Explanation

To calculate the relative rate of effusion of \( O_2 \) to \( CH_4 \) from a container containing \( O_2 \) and \( CH_4 \) in a 3:2 mass ratio, we can use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Identify the Molar Masses**:
- The molar mass of \( O_2 \) (oxygen) is \( 32 \, g/mol \).
- The molar mass of \( CH_4 \) (methane) is \( 16 \, g/mol \).

2. **Determine the Masses in the Given Ratio**:
- Let the mass of \( O_2 \) be \( 3m \) and the mass of \( CH_4 \) be \( 2m \).

3. **Calculate the Number of Moles**:
- Moles of \( O_2 \) = \( \frac{\text{mass}}{\text{molar mass}} = \frac{3m}{32} \)
- Moles of \( CH_4 \) = \( \frac{\text{mass}}{\text{molar mass}} = \frac{2m}{16} = \frac{m}{8} \)

4. **Apply Graham's Law**:
- According to Graham's law, the relative rate of effusion of \( O_2 \) to \( CH_4 \) is given by:
\[ \frac{\text{Rate of } O_2}{\text{Rate of } CH_4} = \sqrt{\frac{M_{CH_4}}{M_{O_2}}} \]
- Substituting the molar masses:
\[ \frac{\text{Rate of } O_2}{\text{Rate of } CH_4} = \sqrt{\frac{16}{32}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

5. **Final Calculation**:
- The relative rate of effusion of \( O_2 \) to \( CH_4 \) can be expressed as:
\[ \frac{\text{Rate of } O_2}{\text{Rate of } CH_4} = \frac{1}{\sqrt{2}} \approx 0.707 \]

### Conclusion:
The relative rate of effusion of \( O_2 \) to \( CH_4 \) is \( \frac{1}{\sqrt{2}} \) or approximately \( 0.707 \).