At `25^(@)C`, a `0.1m` solution of `CH_(3)COOH` is `1.35%` dissociated. Calculate the freezing point and osmotic pressure of the solution. Compare your results with those expected under conditions of no dissociation . `K_(f)` for water `=1.86^(@)Cm^(-1)`.

Explanation

To solve the problem of calculating the freezing point and osmotic pressure of a 0.1m solution of acetic acid (CH₃COOH) that is 1.35% dissociated at 25°C, we will follow these steps:

### Step 1: Calculate the degree of dissociation (α)
The degree of dissociation (α) is given as 1.35%. To convert this percentage into a fraction, we divide by 100.

\[ \alpha = \frac{1.35}{100} = 0.0135 \]

### Step 2: Calculate the van 't Hoff factor (i)
The van 't Hoff factor (i) can be calculated using the formula:

\[ i = 1 + \alpha(n - 1) \]

For acetic acid, which dissociates into one ion of acetate (CH₃COO⁻) and one ion of hydrogen (H⁺), n = 2. Thus,

\[ i = 1 + 0.0135(2 - 1) = 1 + 0.0135 = 1.0135 \]

### Step 3: Calculate the depression in freezing point (ΔTf)
Using the formula for depression in freezing point:

\[ \Delta T_f = K_f \cdot m \cdot i \]

Where:
- \( K_f = 1.86 \, ^\circ C \, kg/mol \)
- \( m = 0.1 \, mol/kg \)
- \( i = 1.0135 \)

Substituting the values:

\[ \Delta T_f = 1.86 \cdot 0.1 \cdot 1.0135 = 0.1885 \, ^\circ C \]

### Step 4: Calculate the freezing point of the solution
The freezing point of pure water is 0°C. Therefore, the freezing point of the solution is:

\[ T_f = 0 - \Delta T_f = 0 - 0.1885 = -0.1885 \, ^\circ C \]

### Step 5: Calculate the osmotic pressure (π)
The osmotic pressure can be calculated using the formula:

\[ \pi = iCRT \]

Where:
- \( C = 0.1 \, mol/L \)
- \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \)
- \( T = 25 + 273.15 = 298.15 \, K \)

Substituting the values:

\[ \pi = 1.0135 \cdot 0.1 \cdot 0.0821 \cdot 298.15 \]

Calculating this gives:

\[ \pi = 2.48 \, atm \]

### Step 6: Compare with no dissociation
If there was no dissociation, the van 't Hoff factor (i) would be 1. Thus, the osmotic pressure would be:

\[ \pi_{no \, dissociation} = 1 \cdot 0.1 \cdot 0.0821 \cdot 298.15 = 2.46 \, atm \]

### Summary of Results
- **Freezing Point of the Solution**: -0.1885°C
- **Osmotic Pressure of the Solution**: 2.48 atm
- **Osmotic Pressure without Dissociation**: 2.46 atm