If the molecules of `SO_(2)` effuse a distance of 150cm in a certain period of time, the distance travelled by the molecules of `CH_(4)` effusing in the same time is

Options

A. 300 cm
B. 600 cm
C. 37.5 cm
D. 75 cm

Answer

To solve the problem, 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 Masses**:
- Molar mass of \( CH_4 \) (methane):
- Carbon (C) = 12 g/mol
- Hydrogen (H) = 1 g/mol (4 H atoms)
- Total = \( 12 + (4 \times 1) = 16 \) g/mol
- Molar mass of \( SO_2 \) (sulfur dioxide):
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol (2 O atoms)
- Total = \( 32 + (2 \times 16) = 64 \) g/mol

2. **Apply Graham's Law**:
- According to Graham's law:
\[ \frac{\text{Rate of } SO_2}{\text{Rate of } CH_4} = \frac{\sqrt{\text{Molar mass of } CH_4}}{\sqrt{\text{Molar mass of } SO_2}} \]
- Substituting the molar masses:
\[ \frac{\text{Rate of } SO_2}{\text{Rate of } CH_4} = \frac{\sqrt{16}}{\sqrt{64}} = \frac{4}{8} = \frac{1}{2} \]

3. **Set Up the Distance Relationship**:
- Let \( D \) be the distance traveled by \( CH_4 \) in the same time that \( SO_2 \) travels 150 cm.
- According to the relationship of rates and distances:
\[ \frac{150 \text{ cm}}{D} = \frac{1}{2} \]

4. **Solve for \( D \)**:
- Cross-multiplying gives:
\[ 150 \text{ cm} = \frac{1}{2} D \]
- Therefore:
\[ D = 150 \text{ cm} \times 2 = 300 \text{ cm} \]

5. **Conclusion**:
- The distance traveled by the molecules of \( CH_4 \) in the same time is **300 cm**.