Phenol associates in benzene to from dimer `(C_(6)H_(5)OH)_(2)` . The freezing point of a solution containing 5g of phenol in 250 g of benzene is lowered by `0.70^(@)C`. Calculate the degree of association of phenol in benzene .(`k_(f)` for benzene `= 5.12 Km^(-1)`)

Explanation

To solve the problem of calculating the degree of association of phenol in benzene, we will follow these steps:

### Step 1: Gather the Given Data
- Mass of phenol (solute), \( W_b = 5 \, \text{g} \)
- Mass of benzene (solvent), \( W_a = 250 \, \text{g} \)
- Depression in freezing point, \( \Delta T_f = 0.70 \, ^\circ C \)
- Freezing point depression constant for benzene, \( K_f = 5.12 \, \text{K kg}^{-1} \)

### Step 2: Use the Freezing Point Depression Formula
The formula for freezing point depression is given by:
\[ \Delta T_f = K_f \cdot \frac{W_b}{M_b \cdot W_a} \cdot 1000 \]
Where:
- \( M_b \) is the molar mass of the solute (phenol in this case).

### Step 3: Rearrange the Formula to Find Molar Mass
Rearranging the formula to solve for \( M_b \):
\[ M_b = K_f \cdot \frac{W_b}{\Delta T_f \cdot W_a} \cdot 1000 \]

### Step 4: Substitute the Values
Substituting the known values into the rearranged formula:
\[ M_b = 5.12 \cdot \frac{5}{0.70 \cdot 250} \cdot 1000 \]

### Step 5: Calculate Molar Mass
Calculating \( M_b \):
\[ M_b = 5.12 \cdot \frac{5}{175} \cdot 1000 = 5.12 \cdot 0.02857 \cdot 1000 \approx 146.3 \, \text{g/mol} \]

### Step 6: Compare with Actual Molar Mass of Phenol
The actual molar mass of phenol \( C_6H_5OH \) is approximately \( 94 \, \text{g/mol} \). Since the calculated molar mass \( (146.3 \, \text{g/mol}) \) is greater than the actual molar mass, it indicates that phenol is associating in benzene.

### Step 7: Calculate Van't Hoff Factor (i)
The van't Hoff factor \( i \) is given by:
\[ i = \frac{\text{actual molar mass}}{\text{observed molar mass}} = \frac{94}{146.3} \approx 0.64 \]

### Step 8: Set Up the Association Equation
For the association of phenol into dimers:
\[ 2 \, C_6H_5OH \rightleftharpoons (C_6H_5OH)_2 \]
Let \( \alpha \) be the degree of association. Initially, we have 1 mole of phenol and 0 moles of dimer.

### Step 9: Express Moles After Association
After association:
- Moles of phenol = \( 1 - \alpha \)
- Moles of dimer = \( \frac{\alpha}{2} \)

### Step 10: Write the Expression for i
The total number of moles after association is:
\[ i = \frac{(1 - \alpha) + \frac{\alpha}{2}}{1} = 1 - \frac{\alpha}{2} \]
Setting this equal to the calculated \( i \):
\[ 0.64 = 1 - \frac{\alpha}{2} \]

### Step 11: Solve for α
Rearranging gives:
\[ \frac{\alpha}{2} = 1 - 0.64 = 0.36 \implies \alpha = 0.72 \]

### Step 12: Calculate Percentage of Association
The percentage of association is given by:
\[ \text{Percentage of association} = \alpha \times 100 = 72\% \]

### Final Answer
The degree of association of phenol in benzene is **72%**.

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