100 mL of `H_(2)` gas diffues in 10 sec. X mL of `O_(2)` gas diffuses in t sec. X and `t` cannot be

Options

A. 100 mL, 40 sec
B. 25 mL, 10 sec
C. 100 mL, 10 sec
D. 5 mL, 2 sec

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. **Calculate the Rate of Diffusion of Hydrogen:**
- Given that 100 mL of `H₂` gas diffuses in 10 seconds.
- Rate of diffusion of `H₂` = Volume of `H₂` / Time = 100 mL / 10 s = 10 mL/s.

2. **Set Up the Rate of Diffusion for Oxygen:**
- Let the volume of `O₂` that diffuses be `X` mL in `t` seconds.
- Rate of diffusion of `O₂` = Volume of `O₂` / Time = X mL / t s.

3. **Apply Graham's Law:**
- According to Graham's law:
\[ \frac{\text{Rate of diffusion of } H₂}{\text{Rate of diffusion of } O₂} = \sqrt{\frac{\text{Molar mass of } O₂}{\text{Molar mass of } H₂}} \]
- The molar mass of `H₂` = 2 g/mol and the molar mass of `O₂` = 32 g/mol.
- Therefore:
\[ \frac{10 \text{ mL/s}}{\frac{X}{t}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 \]

4. **Cross Multiply to Find the Relationship:**
- Rearranging gives:
\[ 10t = 4X \]
- Thus:
\[ X = \frac{10t}{4} = \frac{5t}{2} \]

5. **Identify the Conditions for `X` and `t`:**
- We need to find values of `X` and `t` that cannot be equal based on the derived relationship.
- From the equation \( X = \frac{5t}{2} \), we see that `X` is directly proportional to `t`. However, the specific values of `X` and `t` will depend on the options provided in the question.

6. **Evaluate the Options:**
- The question states that `X` and `t` cannot be equal. We will check the options given:
- If `X` and `t` are equal, then substituting `t` into the equation gives a specific relationship that must hold true.
- The only option that does not satisfy this relationship will be the answer.

### Conclusion:
From the analysis, the option where `X` and `t` cannot be equal is the correct answer.