A vessel having certain amount of oxygen at certain pressure and temperature develops a very small hole and 4 g of oxygen effuses out. How much hydrogen would have effused out of the same vessel had it been taken at the same pressure and temperature.

Options

A. 1g
B. 16g
C. 0.25g
D. 64g

Answer

To solve the problem, we will use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of oxygen (O₂) effused = 4 g
- Molar mass of oxygen (O₂) = 32 g/mol
- Molar mass of hydrogen (H₂) = 2 g/mol

2. **Calculate the Moles of Oxygen Effused:**
\[ \text{Moles of } O_2 = \frac{\text{Mass of } O_2}{\text{Molar mass of } O_2} = \frac{4 \text{ g}}{32 \text{ g/mol}} = 0.125 \text{ mol} \]

3. **Apply Graham's Law of Effusion:**
According to Graham's law:
\[ \frac{\text{Rate of effusion of } O_2}{\text{Rate of effusion of } H_2} = \sqrt{\frac{\text{Molar mass of } H_2}{\text{Molar mass of } O_2}} \]
Let the mass of hydrogen (H₂) that would effuse out be \( x \) grams. The moles of hydrogen can be expressed as:
\[ \text{Moles of } H_2 = \frac{x \text{ g}}{2 \text{ g/mol}} = \frac{x}{2} \text{ mol} \]

4. **Set Up the Equation Using Graham's Law:**
Since the time of effusion is the same for both gases, we can write:
\[ \frac{0.125}{\frac{x}{2}} = \sqrt{\frac{2}{32}} \]

5. **Simplify the Right Side:**
\[ \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \]

6. **Substitute and Solve for \( x \):**
\[ \frac{0.125}{\frac{x}{2}} = \frac{1}{4} \]
Cross-multiplying gives:
\[ 0.125 \cdot 4 = \frac{x}{2} \]
\[ 0.5 = \frac{x}{2} \]
Multiplying both sides by 2:
\[ x = 1 \text{ g} \]

### Conclusion:
The mass of hydrogen (H₂) that would have effused out is **1 gram**.