phenol associates in benzene to a certain extent in dimerisation reaction. A solution containing 0.02 kg of phenol in 1.0 kg of benzene has its freezing point depressed 0.69 k. [`K_(f)(C_(6)H_(6)) =5.12 k kg"mol"^(-1)`]. The degree of association:

Options

A. 0.63
B. 0.73
C. 0.83
D. 0.93

Answer

To solve the problem of determining the degree of association of phenol in benzene based on the given data, we can follow these steps:

### Step 1: Understand the Given Data
We are provided with:
- Mass of phenol (solute) = 0.02 kg = 20 g
- Mass of benzene (solvent) = 1.0 kg = 1000 g
- Freezing point depression (ΔTf) = 0.69 K
- Freezing point depression constant for benzene (Kf) = 5.12 K kg/mol

### Step 2: Calculate the Molality of the Solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

1. **Calculate the number of moles of phenol:**
- Molecular weight of phenol (C6H5OH) = 94.11 g/mol
- Number of moles of phenol = mass of phenol / molecular weight of phenol
\[ \text{Number of moles of phenol} = \frac{20 \text{ g}}{94.11 \text{ g/mol}} \approx 0.212 \text{ moles} \]

2. **Calculate the molality:**
\[ \text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}} = \frac{0.212 \text{ moles}}{1.0 \text{ kg}} = 0.212 \text{ mol/kg} \]

### Step 3: Use the Freezing Point Depression Formula
The freezing point depression is given by the formula:
\[ \Delta T_f = K_f \times m \times i \]
Where:
- \( \Delta T_f \) = freezing point depression
- \( K_f \) = freezing point depression constant
- \( m \) = molality
- \( i \) = Van't Hoff factor (degree of association)

Rearranging the formula to find \( i \):
\[ i = \frac{\Delta T_f}{K_f \times m} \]

### Step 4: Substitute the Values
Substituting the known values into the equation:
\[ i = \frac{0.69 \text{ K}}{5.12 \text{ K kg/mol} \times 0.212 \text{ mol/kg}} \approx \frac{0.69}{1.08544} \approx 0.636 \]

### Step 5: Relate Van't Hoff Factor to Degree of Association
For the dimerization reaction of phenol:
\[ 2 \text{C}_6\text{H}_5\text{OH} \rightleftharpoons \text{C}_{12}\text{H}_{10}\text{O}_2 \]
Let \( \alpha \) be the degree of association. The relationship between the Van't Hoff factor \( i \) and \( \alpha \) is given by:
\[ i = 1 - \frac{\alpha}{2} \]

### Step 6: Solve for \( \alpha \)
Substituting the value of \( i \):
\[ 0.636 = 1 - \frac{\alpha}{2} \]
Rearranging gives:
\[ \frac{\alpha}{2} = 1 - 0.636 = 0.364 \]
Thus,
\[ \alpha = 2 \times 0.364 = 0.728 \]

### Step 7: Conclusion
The degree of association \( \alpha \) of phenol in benzene is approximately 0.73.