Ammonia gas diffuses twice as fast as gas X. The gas 'X'is

Options

A. `SO_(2)`
B. `C_(4)H_(10)`
C. `C_(5)H_(8)`
D. `Cl_(2)`

Answer

To solve the problem, we can 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. **Identify the Given Information**:
- Ammonia (NH₃) diffuses twice as fast as gas X.
- Molar mass of ammonia (NH₃) = 17 g/mol.

2. **Use Graham's Law**:
- According to Graham's law, the rate of diffusion of gas A (ammonia) and gas B (gas X) can be expressed as:
\[ \frac{\text{Rate of diffusion of NH}_3}{\text{Rate of diffusion of gas X}} = \sqrt{\frac{M_X}{M_{NH_3}}} \]
- Here, \(M_X\) is the molar mass of gas X and \(M_{NH_3}\) is the molar mass of ammonia.

3. **Set Up the Equation**:
- Since ammonia diffuses twice as fast as gas X, we can write:
\[ \frac{2}{1} = \sqrt{\frac{M_X}{17}} \]

4. **Square Both Sides**:
- Squaring both sides gives:
\[ 4 = \frac{M_X}{17} \]

5. **Solve for Molar Mass of Gas X**:
- Rearranging the equation to find \(M_X\):
\[ M_X = 4 \times 17 = 68 \text{ g/mol} \]

6. **Identify Gas X from Options**:
- Now, we need to find a gas with a molar mass of 68 g/mol from the given options.
- One of the options is C₅H₈ (pentene), which has a molar mass of:
\[ C: 12 \times 5 = 60 \text{ g/mol} \]
\[ H: 1 \times 8 = 8 \text{ g/mol} \]
\[ \text{Total} = 60 + 8 = 68 \text{ g/mol} \]

7. **Conclusion**:
- Therefore, gas X is C₅H₈, and the correct answer is option 3.