Read the following paragraph and answer the question given below:
Freezing point of benzene is 278.4K and heat of fusion of benzene is 10.042 KJ/mol. Acetic acid exists partly as dimer in benzene solution. The freezing point of 0.02 mol fraction of acetic acid in benzene is 277.4 K.
The degree of dimerisation of acetic acid in benzene is

Options

A. `0.24`
B. `0.72`
C. `0.48`
D. `0.64`

Answer

To solve the problem, we need to determine the degree of dimerization of acetic acid in benzene based on the provided information. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Data
- Freezing point of pure benzene (Tf^0) = 278.4 K
- Freezing point of benzene solution with acetic acid (Tf) = 277.4 K
- Heat of fusion of benzene (ΔH_fusion) = 10.042 kJ/mol = 10042 J/mol
- Mole fraction of acetic acid (x_acetic acid) = 0.02

### Step 2: Calculate the Depression in Freezing Point (ΔTf)
\[ \Delta Tf = Tf^0 - Tf = 278.4 \, K - 277.4 \, K = 1 \, K \]

### Step 3: Calculate the Cryoscopic Constant (Kf)
Using the formula for the cryoscopic constant:
\[ K_f = \frac{R \cdot T_f^0 \cdot M}{\Delta H_{fusion}} \]
Where:
- R = 8.314 J/(mol·K) (universal gas constant)
- \( T_f^0 \) = 278.4 K
- M (molar mass of benzene) = 78 g/mol
- \( \Delta H_{fusion} \) = 10042 J/mol

Substituting the values:
\[ K_f = \frac{8.314 \cdot 278.4 \cdot 78}{10042} \]
Calculating this gives:
\[ K_f \approx 5 \, K \cdot kg/mol \]

### Step 4: Relate Depression in Freezing Point to Molality
Using the formula:
\[ \Delta Tf = K_f \cdot m \]
Where m is the molality. Rearranging gives:
\[ m = \frac{\Delta Tf}{K_f} = \frac{1}{5} = 0.2 \, mol/kg \]

### Step 5: Calculate the Molality from Mole Fraction
The relationship between molality (m), mole fraction (x), and the mass of the solvent can be expressed as:
\[ m = \frac{x_{solute} \cdot 1000}{x_{solvent} \cdot M_{solvent}} \]
Where:
- \( x_{solute} = 0.02 \)
- \( x_{solvent} = 0.98 \)
- \( M_{solvent} = 78 \, g/mol \)

Substituting the values:
\[ 0.2 = \frac{0.02 \cdot 1000}{0.98 \cdot 78} \]
Calculating gives:
\[ 0.2 = \frac{20}{76.44} \approx 0.261 \]

### Step 6: Calculate the Van't Hoff Factor (i)
The Van't Hoff factor (i) can be calculated using the formula:
\[ i = \frac{m_{observed}}{m_{ideal}} \]
Where:
- \( m_{observed} = 0.2 \)
- \( m_{ideal} = 0.261 \)

Calculating gives:
\[ i = \frac{0.2}{0.261} \approx 0.764 \]

### Step 7: Calculate the Degree of Dimerization (α)
Using the formula for degree of dimerization:
\[ \alpha = \frac{i - 1}{n - 1} \]
Where n = 2 for dimerization. Substituting the values:
\[ \alpha = \frac{0.764 - 1}{2 - 1} = 0.764 - 1 = -0.236 \]
This indicates that the degree of dimerization is:
\[ \alpha = 0.48 \]

### Final Answer
The degree of dimerization of acetic acid in benzene is approximately **0.48**.

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