`SO_2 and CH_4` are introduced in a vessel in the molar ratio 1:2.The ratio of molecules of two gases present in the container when their rate of effusion becomes equal is :

Options

A. `1:2`
B. `2:1`
C. `4:1`
D. `1:4`

Explanation

To solve the problem, we need to determine the ratio of molecules of \( SO_2 \) and \( CH_4 \) present in a container when their rates of effusion become equal. We will 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 Ratio**:
We are given that \( SO_2 \) and \( CH_4 \) are introduced in a molar ratio of \( 1:2 \). This means:
\[ n_{SO_2} : n_{CH_4} = 1 : 2 \]

2. **Use Graham's Law of Effusion**:
According to Graham's law, the rates of effusion of two gases are given by:
\[ \frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}} \]
where \( R_1 \) and \( R_2 \) are the rates of effusion of \( SO_2 \) and \( CH_4 \) respectively, and \( M_1 \) and \( M_2 \) are their molar masses.

3. **Molar Masses**:
The molar masses of the gases are:
- \( M_{SO_2} = 64 \, g/mol \)
- \( M_{CH_4} = 16 \, g/mol \)

4. **Set Up the Equation for Equal Rates of Effusion**:
When the rates of effusion are equal, we can set \( R_{SO_2} = R_{CH_4} \). Therefore, we have:
\[ \frac{n_{SO_2}}{n_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_{SO_2}}} \]

5. **Substitute the Molar Masses**:
Plugging in the molar masses into the equation gives:
\[ \frac{n_{SO_2}}{n_{CH_4}} = \sqrt{\frac{16}{64}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

6. **Final Ratio of Molecules**:
Thus, the ratio of the number of molecules of \( SO_2 \) to \( CH_4 \) when their rates of effusion become equal is:
\[ n_{SO_2} : n_{CH_4} = 1 : 2 \]

### Conclusion:
The ratio of molecules of \( SO_2 \) to \( CH_4 \) when their rates of effusion become equal is \( 1:2 \).

---