0.2 g of a gas X occupies a volume of 0.44 L at some pressure and temperature. Under identical conditions of P and T, 0.1 g of `CO_2` gas occupies 0.32 L. Gas X can be

Options

A. `O_2`
B. `SO_2`
C. NO
D. `C_4 H_10`

Answer

To solve the problem, we need to determine the identity of gas X based on the given information about its mass, volume, and the conditions under which it is measured.

### Step-by-Step Solution:

1. **Understanding the Given Data**:
- We have 0.2 g of gas X occupying a volume of 0.44 L.
- We also have 0.1 g of CO2 occupying a volume of 0.32 L under the same conditions of pressure and temperature.

2. **Using Avogadro's Law**:
- Avogadro's law states that under the same conditions of temperature and pressure, equal volumes of gases contain an equal number of moles. Therefore, the volume is directly proportional to the number of moles.

3. **Finding the Moles of CO2**:
- First, we calculate the number of moles of CO2 in the given volume:
\[ \text{Moles of CO2} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.1 \text{ g}}{44 \text{ g/mol}} = 0.00227 \text{ moles} \]

4. **Finding the Mass of CO2 for 0.44 L**:
- Now, we need to find the mass of CO2 that would occupy 0.44 L using the unitary method:
\[ \text{Mass of CO2 at 0.44 L} = \left(\frac{0.1 \text{ g}}{0.32 \text{ L}}\right) \times 0.44 \text{ L} = 0.1375 \text{ g} \]

5. **Finding the Moles of Gas X**:
- Since the volumes are the same and the conditions are identical, the moles of gas X will equal the moles of CO2 at 0.44 L:
\[ \text{Moles of gas X} = \text{Moles of CO2 at 0.44 L} = \frac{0.1375 \text{ g}}{M_X} \]
- Where \( M_X \) is the molar mass of gas X.

6. **Setting Up the Equation**:
- We have:
\[ \frac{0.2 \text{ g}}{M_X} = \frac{0.1375 \text{ g}}{44 \text{ g/mol}} \]

7. **Solving for Molar Mass of Gas X**:
- Rearranging the equation to find \( M_X \):
\[ M_X = \frac{0.2 \text{ g} \times 44 \text{ g/mol}}{0.1375 \text{ g}} = 64 \text{ g/mol} \]

8. **Identifying Gas X**:
- Now, we need to identify which gas has a molar mass of 64 g/mol.
- The molar mass of SO2 (sulfur dioxide) is 32 (sulfur) + 32 (two oxygens) = 64 g/mol.

### Conclusion:
Gas X is **SO2** (sulfur dioxide).