The rate of diffusion of methane is twice that of X. The molecular mass of X is

Options

A. 16
B. 32
C. 80
D. 64

Explanation

To solve the problem, we will use Graham's law of effusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Understanding the relationship**: According to Graham's law, the rate of diffusion (R) of gas is inversely proportional to the square root of its molecular mass (M). This can be expressed mathematically as:
\[ R \propto \frac{1}{\sqrt{M}} \]

2. **Setting up the equation**: We are given that the rate of diffusion of methane (CH₄) is twice that of gas X. If we denote the rate of diffusion of methane as \( R_{CH_4} \) and that of gas X as \( R_X \), we can write:
\[ R_{CH_4} = 2 R_X \]

3. **Expressing rates in terms of molecular masses**: From Graham's law, we can express the rates in terms of their molecular masses:
\[ R_{CH_4} = \frac{1}{\sqrt{M_{CH_4}}} \quad \text{and} \quad R_X = \frac{1}{\sqrt{M_X}} \]
Here, \( M_{CH_4} \) is the molecular mass of methane, which is 16 g/mol.

4. **Substituting the rates into the equation**: We can substitute these expressions into the equation from step 2:
\[ \frac{1}{\sqrt{M_{CH_4}}} = 2 \cdot \frac{1}{\sqrt{M_X}} \]

5. **Rearranging the equation**: Rearranging gives:
\[ \sqrt{M_X} = 2 \cdot \sqrt{M_{CH_4}} \]

6. **Substituting the molecular mass of methane**: Now, substituting \( M_{CH_4} = 16 \):
\[ \sqrt{M_X} = 2 \cdot \sqrt{16} \]
\[ \sqrt{M_X} = 2 \cdot 4 = 8 \]

7. **Squaring both sides**: To find \( M_X \), we square both sides:
\[ M_X = 8^2 = 64 \]

8. **Conclusion**: Therefore, the molecular mass of gas X is 64 g/mol.

### Final Answer:
The molecular mass of X is **64 g/mol**.