100 mL of O2 gas diffuses in 10 sec. 100 mL of gas X diffuses in t sec. Gas X and time t can be:
a.) SO2, 16 sec
b.) H2, 2.5 sec
c.) CO, 10 sec
d.) He, 4 sec

To determine the time \( t \) for gas \( X \) based on the diffusion of oxygen, we can 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 is given by:

\[
\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}
\]

Where:
- \( r_1 \) and \( r_2 \) are the rates of diffusion of gas 1 and gas 2.
- \( M_1 \) and \( M_2 \) are the molar masses of gas 1 and gas 2.

In this case:
- Let gas 1 be \( O_2 \) (with a molar mass of approximately 32 g/mol).
- Let gas 2 be gas \( X \) (unknown molar mass \( M_X \)).
- The rate of diffusion of \( O_2 \) is given as 100 mL in 10 seconds, which gives a rate \( r_1 = \frac{100 \, \text{mL}}{10 \, \text{s}} = 10 \, \text{mL/s} \).

We can express the rate of diffusion for gas \( X \) as:

\[
r_2 = \frac{100 \, \text{mL}}{t \, \text{s}}
\]

Substituting into Graham's law gives:

\[
\frac{10}{\frac{100}{t}} = \sqrt{\frac{M_X}{32}}
\]

This simplifies to:

\[
10t = 100 \sqrt{\frac{M_X}{32}}
\]

\[
t = 10 \sqrt{\frac{M_X}{32}}
\]

Now, we can calculate \( t \) for each gas option provided, using their molar masses:

1. **For \( SO_2 \)**:
- Molar mass \( M_{SO_2} = 32 + 16 \times 2 = 64 \, \text{g/mol} \)
- Calculate \( t \):
\[
t = 10 \sqrt{\frac{64}{32}} = 10 \sqrt{2} \approx 14.14 \, \text{sec} \quad \text{(not a match)}
\]

2. **For \( H_2 \)**:
- Molar mass \( M_{H_2} = 2 \, \text{g/mol} \)
- Calculate \( t \):
\[
t = 10 \sqrt{\frac{2}{32}} = 10 \sqrt{\frac{1}{16}} = 10 \times \frac{1}{4} = 2.5 \, \text{sec} \quad \text{(match)}
\]

3. **For \( CO \)**:
- Molar mass \( M_{CO} = 12 + 16 = 28 \, \text{g/mol} \)
- Calculate \( t \):
\[
t = 10 \sqrt{\frac{28}{32}} \approx 10 \sqrt{0.875} \approx 10 \times 0.935 \approx 9.35 \, \text{sec} \quad \text{(not a match)}
\]

4. **For \( He \)**:
- Molar mass \( M_{He} = 4 \, \text{g/mol} \)
- Calculate \( t \):
\[
t = 10 \sqrt{\frac{4}{32}} = 10 \sqrt{\frac{1}{8}} = 10 \times \frac{1}{\sqrt{8}} \approx 10 \times 0.354 \approx 3.54 \, \text{sec} \quad \text{(not a match)}
\]

### Conclusion
The gas \( X \) that diffuses in \( t = 2.5 \) seconds is **\( H_2 \)** (option b).