A pre weighed vessel was filled with oxygen at N.T.P. and weighted.It was then evacuated, filled with `SO_(2)` at the same temperature and pressure, and again weighed. The weight of oxygen will be

Options

A. Same as that of `SO_(2)`
B. `(1)/(2)` that of `SO_(2)`
C. Twice that of `SO_(2)`
D. One -fourth that of `SO_(2)`

Explanation

To solve the problem, we need to determine the weight of oxygen (O₂) when a pre-weighed vessel is filled with it at Normal Temperature and Pressure (N.T.P.) and then filled with sulfur dioxide (SO₂) under the same conditions.

### Step-by-Step Solution:

1. **Understand the Conditions**:
- The vessel is filled with oxygen (O₂) at N.T.P. and then evacuated and filled with sulfur dioxide (SO₂) at the same temperature and pressure.
- At N.T.P., 1 mole of any ideal gas occupies 22.4 liters.

2. **Molar Mass of Gases**:
- The molar mass of oxygen (O₂) is 32 g/mol (16 g/mol for each oxygen atom).
- The molar mass of sulfur dioxide (SO₂) is 64 g/mol (32 g/mol for sulfur and 16 g/mol for each of the two oxygen atoms).

3. **Number of Moles**:
- Since the conditions of temperature and pressure are the same for both gases, the number of moles of O₂ will be equal to the number of moles of SO₂ that fill the vessel.
- Let’s denote the number of moles as \( n \).

4. **Weight Calculation**:
- The weight of oxygen (O₂) can be calculated using the formula:
\[ \text{Weight of O₂} = n \times \text{Molar Mass of O₂} = n \times 32 \text{ g/mol} \]
- The weight of sulfur dioxide (SO₂) can be calculated similarly:
\[ \text{Weight of SO₂} = n \times \text{Molar Mass of SO₂} = n \times 64 \text{ g/mol} \]

5. **Relating the Weights**:
- Since we know the weights are related by the number of moles, we can express the weight of O₂ in terms of the weight of SO₂:
\[ \text{Weight of O₂} = \frac{1}{2} \times \text{Weight of SO₂} \]
- This means that the weight of oxygen is half that of sulfur dioxide.

6. **Conclusion**:
- If we denote the weight of SO₂ as \( W_{SO₂} \), then the weight of O₂ is:
\[ W_{O₂} = \frac{1}{2} W_{SO₂} \]

### Final Answer:
The weight of oxygen (O₂) in the vessel is half the weight of sulfur dioxide (SO₂) in the same vessel at N.T.P.