Consider the two solutions:
I: `0.5 M NaCl` aqueous solution at `25^(@)C`,
`NaCl` is complete ionized.
II:`2.0 M C_(6)H_(5)COOH` in benzene at `25^(@)C`,
`C_(6)H_(5)COOH` dimerizes to the full extent.
Which of the following statements(s) is (are) correct?

Options

A. Both the solutions display equal osmotic pressure.
B. Both have equal vapour pressure.
C. Solution II is hypertonic.
D. Solution II has greater depression in freezing point than solution I.

Explanation

To solve the problem, we need to analyze the two solutions given and determine their properties based on the information provided.

### Step 1: Analyze Solution I (0.5 M NaCl)
- **Dissociation of NaCl**: Sodium chloride (NaCl) completely dissociates in solution into sodium ions (Na⁺) and chloride ions (Cl⁻).
- **Van 't Hoff factor (i)**: For NaCl, since it dissociates into two ions, the van 't Hoff factor (i) is 2.
- **Concentration (C)**: The concentration of the NaCl solution is given as 0.5 M.

### Step 2: Calculate the osmotic pressure (π) for Solution I
Using the formula for osmotic pressure:
\[ \pi_1 = i_1 \cdot C_1 \cdot R \cdot T \]
Where:
- \(i_1 = 2\)
- \(C_1 = 0.5 \, \text{M}\)
- \(R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}\) (ideal gas constant)
- \(T = 298 \, \text{K}\) (25°C)

Substituting the values:
\[ \pi_1 = 2 \cdot 0.5 \cdot 0.0821 \cdot 298 \]
\[ \pi_1 = 24.6 \, \text{atm} \]

### Step 3: Analyze Solution II (2.0 M C₆H₅COOH)
- **Dimerization of C₆H₅COOH**: Benzoic acid (C₆H₅COOH) dimerizes in solution. This means that two molecules of benzoic acid combine to form one dimer.
- **Van 't Hoff factor (i)**: Since it dimerizes completely, the effective concentration of particles is halved, leading to \(i = 1\) (as one dimer counts as one particle).
- **Concentration (C)**: The concentration of the benzoic acid solution is given as 2.0 M.

### Step 4: Calculate the osmotic pressure (π) for Solution II
Using the same formula for osmotic pressure:
\[ \pi_2 = i_2 \cdot C_2 \cdot R \cdot T \]
Where:
- \(i_2 = 1\)
- \(C_2 = 2.0 \, \text{M}\)

Substituting the values:
\[ \pi_2 = 1 \cdot 2.0 \cdot 0.0821 \cdot 298 \]
\[ \pi_2 = 49.4 \, \text{atm} \]

### Step 5: Compare the osmotic pressures
From the calculations:
- \(\pi_1 = 24.6 \, \text{atm}\)
- \(\pi_2 = 49.4 \, \text{atm}\)

### Conclusion
1. The osmotic pressures of the two solutions are not equal; hence, the statement that they display the same osmotic pressure is incorrect.
2. Since the osmotic pressures are different, we cannot conclude that they have equal vapor pressures without additional information.
3. Solution II (benzoic acid) has a higher osmotic pressure than Solution I (NaCl).