In an experiment, `72.5 g` of `C_(6)H_(5)OH` (phenol) is dissolved in a solvent of `K_(f)=14`. If the depression in freezing point is `7 K`, find the percentage of phenol that dimerizes.

Explanation

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

### Step 1: Calculate the molar mass of phenol (C₆H₅OH)
The molar mass of phenol can be calculated as follows:
- Carbon (C): 6 atoms × 12 g/mol = 72 g/mol
- Hydrogen (H): 6 atoms × 1 g/mol = 6 g/mol
- Oxygen (O): 1 atom × 16 g/mol = 16 g/mol

Total molar mass of phenol = 72 + 6 + 16 = **94 g/mol**

### Step 2: Calculate the number of moles of phenol
Using the mass of phenol given (72.5 g), we can find the number of moles (n):
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{72.5 \, \text{g}}{94 \, \text{g/mol}} \approx 0.771 \, \text{mol} \]

### Step 3: Calculate the molality of the solution
Assuming the mass of the solvent is 1 kg (1000 g), the molality (m) is given by:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.771 \, \text{mol}}{1 \, \text{kg}} = 0.771 \, \text{mol/kg} \]

### Step 4: Relate depression in freezing point to van 't Hoff factor (i)
The depression in freezing point (ΔTf) is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- ΔTf = 7 K
- Kf = 14 °C kg/mol
- m = 0.771 mol/kg

Rearranging the formula to find i:
\[ i = \frac{\Delta T_f}{K_f \cdot m} = \frac{7}{14 \cdot 0.771} \approx 0.649 \]

### Step 5: Relate i to the degree of dimerization (α)
For dimerization, we have:
- If no dimerization occurs, i = 1 (1 molecule of phenol).
- If dimerization occurs, i = 1 - α/2 (where α is the degree of dimerization).

Setting up the equation:
\[ i = 1 - \frac{\alpha}{2} \]
Substituting the value of i:
\[ 0.649 = 1 - \frac{\alpha}{2} \]
Solving for α:
\[ \frac{\alpha}{2} = 1 - 0.649 = 0.351 \implies \alpha = 0.702 \]

### Step 6: Calculate the percentage of phenol that dimerizes
To find the percentage of phenol that dimerizes:
\[ \text{Percentage of dimerization} = \alpha \times 100 = 0.702 \times 100 = 70.2\% \]

### Final Answer
The percentage of phenol that dimerizes is approximately **70.2%**.

---