Rate of diffusion of `NH_(3)` is twice that of X. What is the molecular mass of X

Options

A. 68
B. 48
C. 12
D. 8

Explanation

To solve the problem, we will use Graham's law of effusion, which states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass. The formula can be expressed as:

\[ \frac{\text{Rate of diffusion of } A}{\text{Rate of diffusion of } B} = \sqrt{\frac{M_B}{M_A}} \]

Where:
- \( A \) is ammonia (\( NH_3 \))
- \( B \) is gas \( X \)
- \( M_A \) is the molar mass of ammonia
- \( M_B \) is the molar mass of gas \( X \)

### Step-by-Step Solution:

1. **Identify the given information:**
- The rate of diffusion of \( NH_3 \) is twice that of gas \( X \).
- This can be expressed as:
\[ \frac{\text{Rate of diffusion of } NH_3}{\text{Rate of diffusion of } X} = 2 \]

2. **Set up the equation using Graham's law:**
- According to Graham's law, we can write:
\[ \frac{\text{Rate of diffusion of } NH_3}{\text{Rate of diffusion of } X} = \sqrt{\frac{M_X}{M_{NH_3}}} \]
- Substituting the known rate:
\[ 2 = \sqrt{\frac{M_X}{M_{NH_3}}} \]

3. **Square both sides to eliminate the square root:**
- Squaring both sides gives:
\[ 4 = \frac{M_X}{M_{NH_3}} \]

4. **Substitute the molar mass of ammonia:**
- The molar mass of ammonia (\( NH_3 \)) is approximately 17 g/mol. Therefore:
\[ 4 = \frac{M_X}{17} \]

5. **Solve for \( M_X \):**
- Rearranging the equation gives:
\[ M_X = 4 \times 17 \]
- Calculating this results in:
\[ M_X = 68 \text{ g/mol} \]

6. **Conclusion:**
- The molecular mass of gas \( X \) is 68 g/mol.

### Final Answer:
The molecular mass of \( X \) is 68 g/mol.