`X mL` of `H_(2)` gas effuses through a hole in a container in `5 s`. The time taken for the effusion of the same volume of the gas specified below, under identical conditions, is

Options

A. 10 sec : He
B. 20 sec : `O_(2)`
C. 25 sec : CO
D. 55 sec : `CO_(2)`

Explanation

To solve the problem of effusion of gases, we can 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 Given Data:**
- Volume of `H2` gas effused: `X mL`
- Time taken for `H2` to effuse: `T1 = 5 s`
- Molar mass of `H2`: `M1 = 2 g/mol`
- We need to find the time taken for another gas (`T2`) to effuse the same volume `X mL`.

2. **Use Graham's Law of Effusion:**
According to Graham's law:
\[ \frac{T_1}{T_2} = \frac{M_2}{M_1} \]
where:
- `T1` = time taken for `H2` to effuse
- `T2` = time taken for the other gas to effuse
- `M1` = molar mass of `H2`
- `M2` = molar mass of the other gas

3. **Rearranging the Equation:**
Rearranging the equation gives:
\[ T_2 = T_1 \cdot \frac{M_2}{M_1} \]

4. **Substituting Known Values:**
- For the first gas (let's say `O2` with a molar mass `M2 = 32 g/mol`):
\[ T_2 = 5 \cdot \frac{32}{2} \]
\[ T_2 = 5 \cdot 16 = 80 s \]

5. **Check Other Gases:**
- For `CO` (molar mass `M2 = 28 g/mol`):
\[ T_2 = 5 \cdot \frac{28}{2} = 5 \cdot 14 = 70 s \]
- For `N2` (molar mass `M2 = 28 g/mol`):
\[ T_2 = 5 \cdot \frac{28}{2} = 5 \cdot 14 = 70 s \]
- For `CH4` (molar mass `M2 = 16 g/mol`):
\[ T_2 = 5 \cdot \frac{16}{2} = 5 \cdot 8 = 40 s \]
- For `CO2` (molar mass `M2 = 44 g/mol`):
\[ T_2 = 5 \cdot \frac{44}{2} = 5 \cdot 22 = 110 s \]

6. **Conclusion:**
Based on the calculations, the time taken for the effusion of the gas can vary depending on the gas in question. The gas with the molar mass of `32 g/mol` (like `O2`) takes `80 seconds` to effuse the same volume as `H2` in `5 seconds`.