`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.

Options

A. `75%`
B. `80%`
C. `70%`
D. `100%`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Mass of phenol (C₆H₅OH) = 75.2 g
- Cryoscopic constant (Kf) = 14 K kg/mol
- Depression in freezing point (ΔTf) = 7 K

### Step 2: Calculate the number of moles of phenol
The molar mass of phenol (C₆H₅OH) can be calculated as follows:
- C: 12.01 g/mol × 6 = 72.06 g/mol
- H: 1.008 g/mol × 6 = 6.048 g/mol
- O: 16.00 g/mol × 1 = 16.00 g/mol

Total molar mass of phenol = 72.06 + 6.048 + 16.00 = 94.108 g/mol (approximately 94 g/mol)

Now, calculate the number of moles of phenol:
\[ \text{Number of moles of phenol} = \frac{\text{mass}}{\text{molar mass}} = \frac{75.2 \, \text{g}}{94 \, \text{g/mol}} \approx 0.8 \, \text{mol} \]

### Step 3: Calculate the molality of the solution
Assuming we have 1 kg of solvent, the molality (m) can be calculated as:
\[ \text{Molality} (m) = \frac{\text{number of moles of solute}}{\text{mass of solvent in kg}} = \frac{0.8 \, \text{mol}}{1 \, \text{kg}} = 0.8 \, \text{mol/kg} \]

### Step 4: Use the depression in freezing point formula
The formula for depression in freezing point is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( \Delta T_f \) = 7 K (given)
- \( K_f \) = 14 K kg/mol (given)
- \( m \) = 0.8 mol/kg (calculated)

Substituting the values into the equation:
\[ 7 = i \cdot 14 \cdot 0.8 \]
Solving for \( i \):
\[ i = \frac{7}{14 \cdot 0.8} = \frac{7}{11.2} \approx 0.625 \]

### Step 5: Relate the Van’t Hoff factor to dimerization
For dimerization, the reaction can be represented as:
\[ 2 \text{C}_6\text{H}_5\text{OH} \rightleftharpoons \text{(C}_6\text{H}_5\text{OH})_2 \]
Let \( n \) be the initial number of moles of phenol, and \( \alpha \) be the degree of dimerization. The final number of moles after dimerization can be expressed as:
\[ \text{Final moles} = n - n\alpha + \frac{n\alpha}{2} = n(1 - \frac{\alpha}{2}) \]
Thus, the Van’t Hoff factor \( i \) can be expressed as:
\[ i = \frac{n(1 - \frac{\alpha}{2})}{n} = 1 - \frac{\alpha}{2} \]
From our previous calculation, we found \( i \approx 0.625 \). Therefore:
\[ 1 - \frac{\alpha}{2} = 0.625 \]
Solving for \( \alpha \):
\[ \frac{\alpha}{2} = 1 - 0.625 = 0.375 \implies \alpha = 0.75 \]

### Step 6: Calculate the percentage of phenol that dimerizes
The percentage of phenol that dimerizes is given by:
\[ \text{Percentage} = \alpha \times 100 = 0.75 \times 100 = 75\% \]

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

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