The freezing point of solution containing `0.2 g` of acetic acid in `20.0 g` of benzene is lowered by `0.45^(@)C`. Calculate the degree of association of acetic acid in benzene.
`(K_(f)=5.12 K^(@) mol^(-1) kg^(-1))`

Options

A. 0.527
B. 0.8
C. 0.945
D. None of these

Answer

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

### Step 1: Identify the given data
- Mass of acetic acid (solute), \( W_b = 0.2 \, \text{g} \)
- Mass of benzene (solvent), \( W_a = 20.0 \, \text{g} \)
- Depression in freezing point, \( \Delta T_f = 0.45 \, ^\circ C \)
- Freezing point depression constant for benzene, \( K_f = 5.12 \, \text{K kg}^{-1} \text{mol}^{-1} \)

### Step 2: Calculate the number of moles of acetic acid
The molecular weight of acetic acid (CH₃COOH) is approximately \( 60 \, \text{g/mol} \).

\[ \text{Number of moles of acetic acid} = \frac{W_b}{\text{Molar mass}} = \frac{0.2 \, \text{g}}{60 \, \text{g/mol}} = \frac{0.2}{60} \approx 0.00333 \, \text{mol} \]

### Step 3: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

\[ \text{Mass of solvent in kg} = \frac{W_a}{1000} = \frac{20.0 \, \text{g}}{1000} = 0.020 \, \text{kg} \]

\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.00333 \, \text{mol}}{0.020 \, \text{kg}} = 0.1665 \, \text{mol/kg} \]

### Step 4: Use the freezing point depression formula
The formula for freezing point depression is given by:

\[ \Delta T_f = K_f \times m \times i \]

Where \( i \) is the Van't Hoff factor. Rearranging the formula to solve for \( i \):

\[ i = \frac{\Delta T_f}{K_f \times m} \]

Substituting the known values:

\[ i = \frac{0.45}{5.12 \times 0.1665} \approx \frac{0.45}{0.85248} \approx 0.527 \]

### Step 5: Relate the Van't Hoff factor to the degree of association
For acetic acid, which can associate, we can express \( i \) in terms of the degree of association \( \alpha \):

\[ i = 1 - \frac{\alpha}{2} \]

Setting the two expressions for \( i \) equal:

\[ 0.527 = 1 - \frac{\alpha}{2} \]

### Step 6: Solve for \( \alpha \)
Rearranging the equation:

\[ \frac{\alpha}{2} = 1 - 0.527 = 0.473 \]

Multiplying both sides by 2:

\[ \alpha = 0.946 \]

### Final Answer
The degree of association of acetic acid in benzene is approximately \( 0.946 \).

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