A specific volume of `H_2` requires 24 s to diffuse out of a container. The time required by an equal volume of `O_2` to diffuse out under identical conditions, is

Options

A. `24 s`
B. 96 S
C. 384 S
D. 192 S

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.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Time taken by \( H_2 \) to diffuse out, \( T_{H_2} = 24 \) seconds.
- We need to find the time taken by \( O_2 \) to diffuse out, \( T_{O_2} \).

2. **Determine the Molar Masses:**
- Molar mass of \( H_2 = 2 \) g/mol (since hydrogen has a molar mass of approximately 1 g/mol and there are two hydrogen atoms).
- Molar mass of \( O_2 = 32 \) g/mol (since oxygen has a molar mass of approximately 16 g/mol and there are two oxygen atoms).

3. **Apply Graham's Law of Effusion:**
\[ \frac{Rate_{O_2}}{Rate_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} \]
Substituting the molar masses:
\[ \frac{Rate_{O_2}}{Rate_{H_2}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]

4. **Relate Rate of Diffusion to Time:**
Since the rate of diffusion is inversely proportional to time, we can write:
\[ \frac{Rate_{O_2}}{Rate_{H_2}} = \frac{T_{H_2}}{T_{O_2}} \]
Therefore:
\[ \frac{1}{4} = \frac{T_{H_2}}{T_{O_2}} \]

5. **Rearranging to Find \( T_{O_2} \):**
\[ T_{O_2} = 4 \times T_{H_2} \]
Substituting the value of \( T_{H_2} \):
\[ T_{O_2} = 4 \times 24 \text{ seconds} = 96 \text{ seconds} \]

### Final Answer:
The time required by an equal volume of \( O_2 \) to diffuse out under identical conditions is **96 seconds**.

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