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. `64.6%`
B. `80/4%`
C. `74.6%`
D. `94.6%`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the molality of the solution
First, we need to find the molality (m) of the acetic acid in benzene.

Given:
- Mass of acetic acid (m_acetic_acid) = 0.2 g
- Molar mass of acetic acid (C2H4O2) = 60 g/mol
- Mass of benzene (m_benzene) = 20 g

**Calculation:**
1. Convert the mass of acetic acid to moles:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.2 \, \text{g}}{60 \, \text{g/mol}} = 0.00333 \, \text{mol} \]

2. Calculate the mass of benzene in kg:
\[ m_{\text{benzene}} = \frac{20 \, \text{g}}{1000} = 0.02 \, \text{kg} \]

3. Calculate the molality (m):
\[ m = \frac{n}{m_{\text{benzene}}} = \frac{0.00333 \, \text{mol}}{0.02 \, \text{kg}} = 0.1665 \, \text{mol/kg} \]

### Step 2: Use the freezing point depression formula
The freezing point depression (\(\Delta T_f\)) is given by:
\[ \Delta T_f = K_f \cdot m \cdot i \]
Where:
- \(\Delta T_f = 0.45 \, ^\circ C\)
- \(K_f = 5.12 \, \text{K kg mol}^{-1}\)
- \(i\) is the van 't Hoff factor

**Rearranging for \(i\):**
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

**Calculation:**
\[ i = \frac{0.45}{5.12 \cdot 0.1665} \approx 0.527 \]

### Step 3: Determine the degree of association (α)
Since acetic acid dimerizes in benzene, we can express the relationship between the initial moles and the moles at equilibrium.

Let:
- Initial moles of acetic acid = 1 (before dimerization)
- Degree of association = α (fraction that associates)
- Moles of dimer formed = α/2

At equilibrium:
- Moles of acetic acid = \(1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}\)
- Total moles at equilibrium = \(1 - \frac{\alpha}{2} + \frac{\alpha}{2} = 1\)

Thus, the van 't Hoff factor can be expressed as:
\[ i = \frac{1 - \frac{\alpha}{2}}{1} = 1 - \frac{\alpha}{2} \]

Setting this equal to the previously calculated \(i\):
\[ 1 - \frac{\alpha}{2} = 0.527 \]

**Rearranging gives:**
\[ \frac{\alpha}{2} = 1 - 0.527 = 0.473 \]
\[ \alpha = 0.946 \]

### Step 4: Calculate the percentage association
Percentage association is given by:
\[ \text{Percentage association} = \alpha \times 100 = 0.946 \times 100 = 94.6\% \]

### Final Answer:
The percentage association of acetic acid in benzene is **94.6%**.

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