At a certain temperature, the time required for the complete diffusion of 200 mL of `H_(2)` gas is 30 minutes. The time required for the complete diffusion of 50 mL of `O_(2)` gas at the same temperature will be

Options

A. 60 minutes
B. 45 minutes
C. 30 minutes
D. 15 minutes

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. **Identify the Given Data:**
- Volume of `H2` = 200 mL
- Time for `H2` to diffuse = 30 minutes
- Volume of `O2` = 50 mL
- We need to find the time for `O2` to diffuse.

2. **Calculate the Rate of Diffusion for `H2`:**
\[ \text{Rate of } H2 = \frac{\text{Volume of } H2}{\text{Time of } H2} = \frac{200 \text{ mL}}{30 \text{ min}} = \frac{20}{3} \text{ mL/min} \]

3. **Use Graham's Law:**
According to Graham's law:
\[ \frac{\text{Rate of } H2}{\text{Rate of } O2} = \sqrt{\frac{M_{O2}}{M_{H2}}} \]
Where:
- \( M_{H2} = 2 \, \text{g/mol} \) (molar mass of hydrogen)
- \( M_{O2} = 32 \, \text{g/mol} \) (molar mass of oxygen)

4. **Substituting the Molar Masses:**
\[ \frac{\frac{200}{30}}{\text{Rate of } O2} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 \]

5. **Rearranging to Find the Rate of `O2`:**
\[ \text{Rate of } O2 = \frac{200}{30 \times 4} = \frac{200}{120} = \frac{5}{3} \text{ mL/min} \]

6. **Calculate the Time Required for `O2`:**
\[ \text{Time of } O2 = \frac{\text{Volume of } O2}{\text{Rate of } O2} = \frac{50 \text{ mL}}{\frac{5}{3} \text{ mL/min}} = 50 \times \frac{3}{5} = 30 \text{ minutes} \]

### Final Answer:
The time required for the complete diffusion of 50 mL of `O2` gas at the same temperature is **30 minutes**.