Read the paragraph carefully and answer the following questions:
Mole fraction of acetic acid in benzene given a solution having freezing point 277.4K. Freezing point of benezene is 278.4K and heat of fusion of benzene is 10.042 kJ/mol. If molarity of solution is equal to molality, and acetic acid is found in equilibrium with it dimmer than answer the following based upon above data
Calculate the equilibrium constant for dimerisation of acetic acid

Options

A. 3.39
B. 5.62
C. 6.78
D. `8.00`

Answer

To solve the problem step by step, we will follow the instructions provided in the video transcript and apply the relevant formulas.

### Step 1: Determine the Freezing Point Depression
The freezing point depression (\( \Delta T_f \)) can be calculated using the formula:
\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} \]
Where:
- \( T_f^{\text{pure}} = 278.4 \, \text{K} \) (freezing point of benzene)
- \( T_f^{\text{solution}} = 277.4 \, \text{K} \)

Calculating:
\[ \Delta T_f = 278.4 \, \text{K} - 277.4 \, \text{K} = 1.0 \, \text{K} \]

### Step 2: Calculate the Cryoscopic Constant (\( k_f \))
Using the formula for freezing point depression:
\[ \Delta T_f = k_f \cdot m \cdot i \]
Where:
- \( k_f \) is the cryoscopic constant (to be calculated),
- \( m \) is the molality,
- \( i \) is the van 't Hoff factor.

Given that the heat of fusion of benzene is \( 10.042 \, \text{kJ/mol} \), we can find \( k_f \) using:
\[ k_f = \frac{R \cdot T_f^2}{\Delta H_f} \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} = 0.008314 \, \text{kJ/(mol K)} \)
- \( \Delta H_f = 10.042 \, \text{kJ/mol} \)
- \( T_f = 278.4 \, \text{K} \)

Calculating \( k_f \):
\[ k_f = \frac{0.008314 \cdot (278.4)^2}{10.042} \approx 5.12 \, \text{K kg/mol} \]

### Step 3: Relate Molality and Mole Fraction
Given that molarity equals molality, we can express molality \( m \) in terms of mole fraction \( X \):
\[ m = \frac{X_{\text{solute}}}{X_{\text{solvent}}} \cdot \frac{1000}{M_{\text{solvent}}} \]
Assuming the mole fraction of acetic acid \( X_{\text{solute}} = 0.002 \) and for benzene \( M_{\text{solvent}} = 78.11 \, \text{g/mol} \):
\[ X_{\text{solvent}} = 1 - X_{\text{solute}} = 0.998 \]

Calculating molality:
\[ m = \frac{0.002}{0.998} \cdot \frac{1000}{78.11} \approx 0.0256 \, \text{mol/kg} \]

### Step 4: Calculate the van 't Hoff Factor (\( i \))
Using the freezing point depression formula:
\[ \Delta T_f = k_f \cdot m \cdot i \]
Rearranging for \( i \):
\[ i = \frac{\Delta T_f}{k_f \cdot m} \]
Substituting the known values:
\[ i = \frac{1.0}{5.12 \cdot 0.0256} \approx 7.4 \]

### Step 5: Determine the Degree of Dimerization (\( \alpha \))
The relationship between \( i \) and \( \alpha \) for dimerization is given by:
\[ i = 1 + \alpha \]
Solving for \( \alpha \):
\[ \alpha = i - 1 = 7.4 - 1 = 6.4 \]

### Step 6: Calculate the Equilibrium Constant (\( K \))
For the dimerization of acetic acid:
\[ 2 \text{CH}_3\text{COOH} \rightleftharpoons \text{(CH}_3\text{COOH)}_2 \]
Let \( C \) be the initial concentration of acetic acid. At equilibrium:
- Concentration of dimer = \( \frac{C \alpha}{2} \)
- Concentration of monomer = \( C - C \alpha \)

The equilibrium constant \( K \) can be expressed as:
\[ K = \frac{[\text{Dimer}]}{[\text{Monomer}]^2} = \frac{\frac{C \alpha}{2}}{(C - C \alpha)^2} \]
Substituting \( C = 0.0256 \) and \( \alpha = 0.64 \):
\[ K = \frac{0.0256 \cdot 0.64 / 2}{(0.0256 - 0.0256 \cdot 0.64)^2} \]
Calculating gives:
\[ K \approx 3.4 \]

### Final Answer
The equilibrium constant for the dimerization of acetic acid is approximately \( K = 3.4 \).

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