Consider following cases:
I: `2M CH_(3)COOH` solution is benzene at `27^(@)C` where there is dimer formation to the extent of `100%`
II: `0.5M Kci` aq. Solution at `27^(@)C`, which ionises `100%`
Which is/are true statements(s):

Options

A. both are isotonic
B. I is hypertonic
C. II is hypotonic
D. none is correct

Explanation

To solve the given question, we will analyze both cases step by step.

### Case I: 2M CH₃COOH solution in benzene at 27°C (100% dimer formation)

1. **Initial Moles**:
- Let’s assume we start with 1 mole of CH₃COOH.

2. **Dimer Formation**:
- The reaction for dimer formation can be represented as:
\[ 2 \text{ CH}_3\text{COOH} \rightleftharpoons \text{(CH}_3\text{COOH)}_2 \]
- Since the dimer formation is 100%, all moles of CH₃COOH will form dimers.

3. **Moles at Equilibrium**:
- At equilibrium, the number of moles of CH₃COOH remaining will be \(1 - \alpha\) (where \(\alpha\) is the degree of association).
- Since \(\alpha = 1\) (100% association), the moles of CH₃COOH left will be:
\[ 1 - 1 = 0 \]
- The number of moles of dimer formed will be:
\[ \frac{\alpha}{2} = \frac{1}{2} = 0.5 \text{ moles} \]

4. **Total Moles After Association**:
- The total number of moles after association will be:
\[ 0 + 0.5 = 0.5 \text{ moles} \]

5. **Van't Hoff Factor (i)**:
- The Van't Hoff factor \(i\) is given by:
\[ i = 1 - \frac{\alpha}{2} \]
- Substituting \(\alpha = 1\):
\[ i = 1 - \frac{1}{2} = \frac{1}{2} \]

6. **Osmotic Pressure Calculation**:
- The formula for osmotic pressure \(\pi\) is:
\[ \pi = iCRT \]
- Substituting the values:
- \(i = \frac{1}{2}\)
- \(C = 2 \, \text{M}\)
- \(R = R\) (gas constant)
- \(T = 300 \, \text{K}\) (27°C in Kelvin)
- Therefore,
\[ \pi = \frac{1}{2} \times 2 \times R \times 300 = 300R \]

### Case II: 0.5M KCl aqueous solution at 27°C (100% ionization)

1. **Initial Moles**:
- Assume we start with 1 mole of KCl.

2. **Ionization Reaction**:
- The dissociation of KCl can be represented as:
\[ \text{KCl} \rightleftharpoons \text{K}^+ + \text{Cl}^- \]
- Since the ionization is 100%, all moles of KCl will dissociate.

3. **Moles at Equilibrium**:
- At equilibrium, the number of moles of KCl remaining will be \(1 - \alpha\) (where \(\alpha\) is the degree of dissociation).
- Since \(\alpha = 1\), the moles of KCl left will be:
\[ 1 - 1 = 0 \]
- The number of moles of K⁺ and Cl⁻ formed will be:
\[ \alpha = 1 \text{ mole each} \]

4. **Total Moles After Dissociation**:
- The total number of moles after dissociation will be:
\[ 0 + 1 + 1 = 2 \text{ moles} \]

5. **Van't Hoff Factor (i)**:
- The Van't Hoff factor \(i\) is given by:
\[ i = 1 + \alpha \]
- Substituting \(\alpha = 1\):
\[ i = 1 + 1 = 2 \]

6. **Osmotic Pressure Calculation**:
- Using the osmotic pressure formula:
\[ \pi = iCRT \]
- Substituting the values:
- \(i = 2\)
- \(C = 0.5 \, \text{M}\)
- \(R = R\)
- \(T = 300 \, \text{K}\)
- Therefore,
\[ \pi = 2 \times 0.5 \times R \times 300 = 300R \]

### Conclusion:
- The osmotic pressures for both cases are equal:
\[ \pi_{\text{Case I}} = 300R \quad \text{and} \quad \pi_{\text{Case II}} = 300R \]
- Since both solutions have the same osmotic pressure, they are isotonic.

### True Statement:
- Both solutions are isotonic, hence the correct statement is: **Both are isotonic.**