At `273K` temp, and 9 atm pressure, the compressibility fog a gas is `0.9`. The volume of 1 mill-moles of gas at this temperature and pressure is `:`

Options

A. 2.24 litre
B. 0.020 mL
C. 2.24 mL
D. 22.4 mL

Explanation

To solve the problem, we will use the compressibility factor (Z) and the ideal gas equation modified for real gases. The compressibility factor is defined as:

\[ Z = \frac{PV}{nRT} \]

Where:
- \( Z \) = compressibility factor
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant
- \( T \) = temperature

### Step 1: Identify the given values
- Temperature, \( T = 273 \, K \)
- Pressure, \( P = 9 \, atm \)
- Compressibility factor, \( Z = 0.9 \)
- Number of moles, \( n = 1 \, \text{millimole} = 1 \times 10^{-3} \, moles \)
- Ideal gas constant, \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)

### Step 2: Rearrange the compressibility factor equation to solve for volume \( V \)
We can rearrange the equation to find \( V \):

\[ V = \frac{ZnRT}{P} \]

### Step 3: Substitute the values into the equation
Now, substituting the known values into the equation:

\[ V = \frac{(0.9)(1 \times 10^{-3} \, \text{mol})(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(273 \, K)}{9 \, atm} \]

### Step 4: Calculate the volume
Now, we will perform the calculation:

1. Calculate \( nRT \):
\[ nRT = (1 \times 10^{-3})(0.0821)(273) = 22.4143 \, \text{L} \]

2. Now, substitute this value back into the volume equation:
\[ V = \frac{(0.9)(22.4143)}{9} \]

3. Calculate \( V \):
\[ V = \frac{20.17287}{9} = 2.24143 \, \text{L} \]

### Step 5: Convert volume from liters to milliliters
Since 1 L = 1000 mL, we convert the volume:

\[ V = 2.24143 \, \text{L} \times 1000 = 2241.43 \, \text{mL} \]

### Final Answer
The volume of 1 millimole of gas at 273 K and 9 atm pressure is approximately **2241.43 mL**.

---