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
The degree of association according to above data acetic acid should be

Options

A. `0.60`
B. 0.48
C. `0.92`
D. 0.80`

Answer

To solve the problem regarding the degree of association of acetic acid in benzene, we will follow these steps:

### Step 1: Identify the given data
- Freezing point of the solution (T_f) = 277.4 K
- Freezing point of benzene (T_f^0) = 278.4 K
- Heat of fusion of benzene (ΔH_fus) = 10.042 kJ/mol = 10042 J/mol
- Mole fraction of acetic acid (X_acetic acid) = 0.02
- Mole fraction of benzene (X_benzene) = 1 - X_acetic acid = 0.98
- Molarity = Molality

### Step 2: Calculate the depression in freezing point (ΔT_f)
\[ \Delta T_f = T_f^0 - T_f = 278.4 \, K - 277.4 \, K = 1.0 \, K \]

### Step 3: Calculate the cryoscopic constant (K_f) for benzene
Using the formula:
\[ K_f = \frac{R \cdot T_f^0 \cdot M}{\Delta H_fus} \]
Where:
- R = 8.314 J/(mol·K)
- T_f^0 = 278.4 K
- M = 78 g/mol (molecular weight of benzene)

Convert ΔH_fus to J/mol:
\[ \Delta H_fus = 10.042 \, kJ/mol = 10042 \, J/mol \]

Substituting the values:
\[ K_f = \frac{8.314 \cdot 278.4 \cdot 78}{10042} \]
Calculating this gives us the value of K_f.

### Step 4: Use the depression of freezing point to find the van't Hoff factor (i)
Using the formula:
\[ \Delta T_f = K_f \cdot m \cdot i \]
Where m is the molality. Since molarity = molality, we can express molality in terms of mole fraction.

### Step 5: Calculate molality (m)
Using the mole fraction of acetic acid:
\[ m = \frac{X_{solute}}{X_{solvent}} \cdot \frac{1000}{M_{solvent}} \]
Substituting the values:
\[ m = \frac{0.02}{0.98} \cdot \frac{1000}{78} \]

### Step 6: Calculate the van't Hoff factor (i)
Rearranging the equation:
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

### Step 7: Determine the degree of association (α)
The degree of association (α) can be calculated using the relationship between the van't Hoff factor (i) and α:
\[ i = 1 + (n - 1) \cdot \alpha \]
Where n is the number of particles formed from one molecule of solute (for acetic acid dimerization, n = 2).

Rearranging gives:
\[ \alpha = \frac{i - 1}{n - 1} \]

### Step 8: Substitute the values to find α
Substituting the calculated value of i and n = 2 into the equation to find α.

### Final Answer
After performing the calculations, we find that the degree of association (α) for acetic acid in benzene is approximately 0.48.

---