Which of the following statements is(are) correct for a gas X having molar mass 5g and density 0.3g/litre at 0.5 atmospheric pressure at 300 K?

Options

A. The gas “X” will behave ideally
B. The force of attraction will dominate over the force of repulsion among the gas molecules
C. The force of repulsion will dominate over the force of attraction among the gas molecules
D. None of these

Explanation

To determine which statements are correct for gas X with the given properties, we will follow these steps:

### Step 1: Calculate the Volume of Gas X
Given:
- Molar mass (M) = 5 g/mol
- Density (d) = 0.3 g/L

Using the formula for density:
\[ d = \frac{m}{V} \]
Where:
- \( m \) = mass of the gas
- \( V \) = volume of the gas

We can rearrange this to find the volume:
\[ V = \frac{m}{d} \]

Assuming we have 1 mole of gas (mass = 5 g):
\[ V = \frac{5 \, \text{g}}{0.3 \, \text{g/L}} = \frac{5}{0.3} \approx 16.67 \, \text{L} \]

### Step 2: Calculate the Compressibility Factor (Z)
The compressibility factor \( Z \) is given by:
\[ Z = \frac{PV}{RT} \]

Where:
- \( P \) = pressure = 0.5 atm
- \( V \) = volume ≈ 16.67 L (from Step 1)
- \( R \) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature = 300 K

Substituting the values:
\[ Z = \frac{(0.5 \, \text{atm})(16.67 \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)})(300 \, \text{K})} \]

Calculating:
\[ Z = \frac{8.335}{24.63} \approx 0.338 \]

### Step 3: Analyze the Compressibility Factor
The value of \( Z \) is approximately 0.338, which is less than 1.

- If \( Z < 1 \), it indicates that the attractive forces between gas molecules dominate over the repulsive forces.

### Conclusion
Based on the calculations:
1. **Gas X does not behave ideally** (since \( Z < 1 \)).
2. **The attractive forces dominate over the repulsive forces** among the gas molecules.

### Final Answer
- The correct statement is: "The force of attraction will dominate over the force of repulsion among the gas molecules."