50 ml of oxygen diffuses under certain conditions through a porous membrane. The volume of Hydrogen that diffuses in the same time under the same conditions is

Options

A. 12.5 ml
B. 25ml
C. 100ml
D. 200ml

Answer

To solve the problem of how much hydrogen diffuses when 50 ml of oxygen diffuses under the same conditions, we can 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 Information:**
- Volume of oxygen (V_O2) = 50 ml
- Molar mass of hydrogen (M_H2) = 2 g/mol
- Molar mass of oxygen (M_O2) = 32 g/mol

2. **Use Graham's Law of Effusion:**
Graham's law states that:
\[ \frac{\text{Rate of diffusion of } O_2}{\text{Rate of diffusion of } H_2} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} \]

3. **Express the Rates of Diffusion:**
Since the time is the same for both gases, we can express the rates in terms of volumes:
\[ \frac{V_{O_2}}{V_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}} \]

4. **Substitute the Known Values:**
Substitute the volumes and molar masses into the equation:
\[ \frac{50 \text{ ml}}{V_{H_2}} = \sqrt{\frac{2}{32}} \]

5. **Simplify the Square Root:**
Calculate the square root:
\[ \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]

6. **Set Up the Equation:**
Now we have:
\[ \frac{50}{V_{H_2}} = \frac{1}{4} \]

7. **Cross Multiply to Solve for \( V_{H_2} \):**
Cross-multiplying gives:
\[ 50 = \frac{1}{4} V_{H_2} \]
Thus,
\[ V_{H_2} = 50 \times 4 = 200 \text{ ml} \]

8. **Conclusion:**
Therefore, the volume of hydrogen that diffuses in the same time under the same conditions is **200 ml**.