Dimer acetic acid in benzene is in equilibrium with acetic acid at a particular condition of temperature and pressure. If half of the dimer molecules are hypothetically separated out then

Options

A. Osmotic pressure of the solution reduces
B. Freezing point of the solution reduces
C. Boiling point of the solution reduces
D. Vapour pressure of the solution reduces

Answer

To solve the problem regarding the effect of hypothetically separating half of the dimer molecules of acetic acid in benzene, we will analyze the situation step by step.

### Step-by-Step Solution:

1. **Understanding the Dimerization of Acetic Acid**:
- Acetic acid (CH₃COOH) can dimerize in a solvent like benzene to form a dimer (2 CH₃COOH ↔ (CH₃COOH)₂).
- At equilibrium, there is a certain concentration of both the dimer and the monomer (acetic acid).

2. **Effect of Separating Half of the Dimer**:
- If half of the dimer molecules are hypothetically separated out, the concentration of the dimer decreases.
- This will shift the equilibrium to the left (according to Le Chatelier's principle), resulting in the formation of more monomer acetic acid.

3. **Van't Hoff Factor (i)**:
- The Van't Hoff factor (i) represents the number of particles the solute breaks into in solution. For dimeric acetic acid, i would initially be lower due to the presence of dimers.
- When half of the dimer is removed, i decreases because there are fewer dimeric units contributing to the total particle count in the solution.

4. **Calculating the Effects on Colligative Properties**:
- **Osmotic Pressure (π)**:
- π = iCRT, where C is the concentration of solute.
- As i decreases, the effect on π depends on the balance between i and C. However, since both factors are interconnected, the net effect is that π does not significantly change.
- **Freezing Point Depression (ΔTf)**:
- ΔTf = iKfM, where M is molality.
- With a decrease in i, the freezing point depression may not change significantly, as the increase in molar mass also affects M.
- **Boiling Point Elevation (ΔTb)**:
- ΔTb = iKbM, similar reasoning applies as with freezing point.
- **Vapor Pressure**:
- Vapor pressure is affected by the number of solute particles. If i decreases, the vapor pressure of the solution decreases.

5. **Conclusion**:
- The correct conclusion from the analysis is that the vapor pressure of the solution decreases when half of the dimer molecules are hypothetically separated out.

### Final Answer:
The vapor pressure of the solution reduces.