Phenol dimerises in benzene having van't Hoff factor 0.54. what is the degree of association?

Options

A. 0.46
B. 0.54
C. 0.27
D. 0.92

Explanation

To solve the problem of determining the degree of association of phenol dimerizing in benzene with a van't Hoff factor of 0.54, we can follow these steps:

### Step 1: Understand the Reaction
Phenol (C₆H₅OH) dimerizes in benzene to form a dimer. The reaction can be represented as:
\[ 2 \text{C}_6\text{H}_5\text{OH} \rightleftharpoons \text{(C}_6\text{H}_5\text{OH)}_2 \]

### Step 2: Define the Degree of Association
Let α be the degree of association. This represents the fraction of phenol molecules that dimerize. If we start with 1 mole of phenol, then:
- At equilibrium, α moles of phenol will dimerize to form \( \frac{\alpha}{2} \) moles of dimer.
- The remaining phenol will be \( 1 - \alpha \) moles.

### Step 3: Calculate the Total Moles at Equilibrium
The total number of moles at equilibrium can be expressed as:
\[ \text{Total moles} = (1 - \alpha) + \frac{\alpha}{2} \]
This simplifies to:
\[ \text{Total moles} = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2} \]

### Step 4: Relate to the van't Hoff Factor
The van't Hoff factor (i) is given as 0.54. The van't Hoff factor is related to the total number of moles at equilibrium compared to the initial moles:
\[ i = \frac{\text{Total moles at equilibrium}}{\text{Initial moles}} \]
Substituting the values, we have:
\[ 0.54 = \frac{1 - \frac{\alpha}{2}}{1} \]
Thus, we can equate:
\[ 1 - \frac{\alpha}{2} = 0.54 \]

### Step 5: Solve for α
Rearranging the equation gives:
\[ \frac{\alpha}{2} = 1 - 0.54 \]
\[ \frac{\alpha}{2} = 0.46 \]
Multiplying both sides by 2:
\[ \alpha = 0.92 \]

### Conclusion
The degree of association (α) of phenol in benzene is 0.92.

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