80 gm of `SO_(X)` gas occupies 14 litre at 2atm & 273K. The value of
X is :

Answer

To solve the problem, we need to determine the value of \( x \) in the compound \( SO_x \) given the mass, volume, pressure, and temperature of the gas. We will use the ideal gas law and the concept of molar mass.

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of the gas (\( m \)) = 80 g
- Volume (\( V \)) = 14 L
- Pressure (\( P \)) = 2 atm
- Temperature (\( T \)) = 273 K

2. **Write the molar mass of \( SO_x \):**
- Molar mass of sulfur (S) = 32 g/mol
- Molar mass of oxygen (O) = 16 g/mol
- Therefore, the molar mass of \( SO_x \) = \( 32 + 16x \) g/mol

3. **Calculate the number of moles (\( n \)):**
- The number of moles can be calculated using the formula:
\[ n = \frac{m}{\text{Molar mass}} = \frac{80}{32 + 16x} \]

4. **Use the ideal gas equation:**
- The ideal gas law is given by:
\[ PV = nRT \]
- Where \( R \) (ideal gas constant) = 0.08 L·atm/(K·mol)
- Substitute the known values into the equation:
\[ 2 \times 14 = n \times 0.08 \times 273 \]
- Simplifying gives:
\[ 28 = n \times 21.84 \]
- Therefore:
\[ n = \frac{28}{21.84} \approx 1.28 \text{ moles} \]

5. **Set the two expressions for \( n \) equal to each other:**
- From step 3, we have:
\[ n = \frac{80}{32 + 16x} \]
- From step 4, we have:
\[ n \approx 1.28 \]
- Setting them equal:
\[ \frac{80}{32 + 16x} = 1.28 \]

6. **Cross-multiply to solve for \( x \):**
- Cross-multiplying gives:
\[ 80 = 1.28(32 + 16x) \]
- Expanding the right side:
\[ 80 = 40.96 + 20.48x \]
- Rearranging gives:
\[ 20.48x = 80 - 40.96 \]
\[ 20.48x = 39.04 \]
- Solving for \( x \):
\[ x = \frac{39.04}{20.48} \approx 1.9 \]

7. **Round \( x \) to the nearest whole number:**
- Since \( x \) must be a whole number, we round \( 1.9 \) to \( 2 \).

### Final Answer:
The value of \( x \) is approximately **2**.