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

Answer

To solve the problem, we need to determine the percentage of phenol that dimerizes when 75.2 g of phenol is dissolved in a solvent with a freezing point depression of 7 K and a cryoscopic constant (Kf) of 14.

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of phenol (C₆H₅OH), \( W_b = 75.2 \, \text{g} \)
- Freezing point depression, \( \Delta T_f = 7 \, \text{K} \)
- Cryoscopic constant, \( K_f = 14 \, \text{K kg/mol} \)

2. **Use the formula for freezing point depression:**
\[ \Delta T_f = K_f \cdot m \]
where \( m \) is the molality of the solution.

3. **Rearranging the formula to find molality:**
\[ m = \frac{\Delta T_f}{K_f} = \frac{7}{14} = 0.5 \, \text{mol/kg} \]

4. **Calculate the number of moles of phenol:**
- Molar mass of phenol (C₆H₅OH):
\[ \text{C: } 6 \times 12 + \text{H: } 6 \times 1 + \text{O: } 16 = 72 + 6 + 16 = 94 \, \text{g/mol} \]
- Number of moles of phenol:
\[ n = \frac{W_b}{\text{Molar mass}} = \frac{75.2}{94} \approx 0.799 \, \text{mol} \]

5. **Assume the mass of the solvent (water) is 1 kg (1000 g):**
- Therefore, the molality \( m \) can also be expressed as:
\[ m = \frac{n}{\text{mass of solvent in kg}} = \frac{0.799}{1} = 0.799 \, \text{mol/kg} \]

6. **Relate the degree of dimerization to the van 't Hoff factor (i):**
- Let \( \alpha \) be the degree of dimerization. The dimerization of phenol can be represented as:
\[ 2 \text{C}_6\text{H}_5\text{OH} \rightleftharpoons \text{(C}_6\text{H}_5\text{OH)}_2 \]
- If 1 mole of phenol dimerizes, then at equilibrium:
- Moles of un-dimerized phenol = \( 1 - \alpha \)
- Moles of dimerized phenol = \( \frac{\alpha}{2} \)
- Thus, the total moles at equilibrium:
\[ n_{\text{total}} = (1 - \alpha) + \frac{\alpha}{2} = 1 - \frac{\alpha}{2} \]

7. **Calculate the van 't Hoff factor (i):**
\[ i = \frac{n_{\text{total}}}{n_{\text{initial}}} = \frac{1 - \frac{\alpha}{2}}{1} \]
\[ i = 1 - \frac{\alpha}{2} \]

8. **Relate the van 't Hoff factor to the abnormal molecular mass:**
- The abnormal molecular mass (M_ab) can be calculated using the formula:
\[ M_{\text{ab}} = \frac{\text{Molar mass}}{i} \]
- We know that the molar mass of phenol is 94 g/mol, and we have calculated the abnormal molecular mass to be:
\[ M_{\text{ab}} = \frac{94}{0.625} \approx 150.4 \, \text{g/mol} \]

9. **Setting up the equation:**
\[ 1 - \frac{\alpha}{2} = \frac{94}{150.4} \]
- Solving for \( \alpha \):
\[ 1 - \frac{\alpha}{2} = 0.625 \implies \frac{\alpha}{2} = 0.375 \implies \alpha = 0.75 \]

10. **Calculate the percentage of phenol that dimerizes:**
\[ \text{Percentage of dimerization} = \alpha \times 100 = 0.75 \times 100 = 75\% \]

### Final Answer:
The percentage of phenol that dimerizes is **75%**.