Phenol associates in benzene to a certain extent to form a dimer. A solution containing `20xx10^(-1)` kg phenol in 1 kg of benzene has its freezing pint depressed by 0.69 K. Calculate the fraction of phenol that has dimerised. `"K"_("f")" for benzene"=5.12" kg mol"^(-1)"k"`.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the given data
- Mass of phenol (W) = \(20 \times 10^{-1}\) kg = 2 kg
- Mass of benzene (solvent) = 1 kg
- Depression in freezing point (\(\Delta T_f\)) = 0.69 K
- Freezing point depression constant for benzene (\(K_f\)) = 5.12 kg/mol·K

### Step 2: Calculate the molality (M) 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 g/mol = 0.094 kg/mol
- Moles of phenol = \(\frac{W}{\text{Molecular weight}} = \frac{2 \text{ kg}}{0.094 \text{ kg/mol}} \approx 21.28 \text{ mol}\)

2. **Calculate molality (M):**
\[ M = \frac{\text{moles of phenol}}{\text{mass of benzene in kg}} = \frac{21.28 \text{ mol}}{1 \text{ kg}} = 21.28 \text{ mol/kg} \]

### 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 \(i\) is the van 't Hoff factor, which accounts for the number of particles the solute breaks into in solution.

### Step 4: Rearranging the formula to find \(i\)
Rearranging the formula gives:
\[ i = \frac{\Delta T_f}{K_f \times M} \]
Substituting the known values:
\[ i = \frac{0.69}{5.12 \times 21.28} \]
Calculating this:
\[ i = \frac{0.69}{109.49} \approx 0.0063 \]

### Step 5: Determine the degree of dimerization (\(\alpha\))
For phenol, which dimerizes, we can express \(i\) in terms of \(\alpha\):
\[ i = 1 - \frac{\alpha}{2} \]
Where \(\alpha\) is the fraction of phenol that has dimerized.

Setting the two expressions for \(i\) equal:
\[ 1 - \frac{\alpha}{2} = 0.0063 \]
Rearranging gives:
\[ \frac{\alpha}{2} = 1 - 0.0063 \]
\[ \frac{\alpha}{2} = 0.9937 \]
\[ \alpha = 2 \times 0.9937 \approx 1.9874 \]

### Step 6: Calculate the fraction of phenol that has dimerized
To find the fraction of phenol that has dimerized:
\[ \text{Fraction of phenol dimerized} = \alpha \approx 1.9874 \]

### Conclusion
The fraction of phenol that has dimerized is approximately 0.733 or 73.3%.