The freezing point of benzene decreases by `0.45^(@)C` when `0.2 g` of acetic acid is added to `20 g` of benzene. IF acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be
`(K_(f) "for benzene" = 5.12 K kg mol^(-1))`

Options

A. (a) `64.6%`
B. (b) `74.6%`
C. (c) `80.4%`
D. (d) `94.6%`

Answer

To solve the problem of determining the percentage association of acetic acid in benzene, we can follow these steps:

### Step 1: Understand the Problem
We know that the freezing point of benzene decreases by \(0.45^\circ C\) when \(0.2 g\) of acetic acid is added to \(20 g\) of benzene. Acetic acid associates to form a dimer in benzene. We need to calculate the percentage of acetic acid that associates.

### Step 2: Use the Freezing Point Depression Formula
The formula for freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \(\Delta T_f\) = depression in freezing point
- \(i\) = van 't Hoff factor (number of particles after dissociation)
- \(K_f\) = freezing point depression constant
- \(m\) = molality of the solution

### Step 3: Calculate the Molar Mass of Acetic Acid
The molar mass of acetic acid (\(CH_3COOH\)) is calculated as follows:
- Carbon (C): \(12 \times 2 = 24\)
- Hydrogen (H): \(1 \times 4 = 4\)
- Oxygen (O): \(16 \times 2 = 32\)
Thus, the total molar mass is:
\[ 24 + 4 + 32 = 60 \, g/mol \]

### Step 4: Calculate the Molality of the Solution
Molality (\(m\)) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]
First, calculate the moles of acetic acid:
\[ \text{Moles of acetic acid} = \frac{0.2 \, g}{60 \, g/mol} = \frac{0.2}{60} = 0.00333 \, mol \]
The mass of benzene is \(20 g\) or \(0.02 kg\). Therefore, the molality is:
\[ m = \frac{0.00333 \, mol}{0.02 \, kg} = 0.1665 \, mol/kg \]

### Step 5: Substitute Values into the Freezing Point Depression Formula
We know:
- \(\Delta T_f = 0.45^\circ C\)
- \(K_f = 5.12 \, K \cdot kg/mol\)
Substituting into the formula:
\[ 0.45 = i \cdot 5.12 \cdot 0.1665 \]
Calculating the right side:
\[ 0.45 = i \cdot 0.85248 \]
Thus,
\[ i = \frac{0.45}{0.85248} \approx 0.528 \]

### Step 6: Relate the Van 't Hoff Factor to Association
For acetic acid that forms a dimer:
\[ i = 1 - \frac{\alpha}{2} \]
where \(\alpha\) is the degree of association. Rearranging gives:
\[ 0.528 = 1 - \frac{\alpha}{2} \]
Solving for \(\alpha\):
\[ \frac{\alpha}{2} = 1 - 0.528 = 0.472 \]
\[ \alpha = 0.944 \text{ or } 94.4\% \]

### Step 7: Conclusion
The percentage association of acetic acid in benzene is approximately \(94.4\%\).