Acetic acid exists in benzene solution in the dimeric form. In an actual experiment, the van't Hoff factor was found to be 0.52. Then, the degree of dissociation of acetic acid is

Options

A. 0.48
B. 0.88
C. 0.96
D. 0.52

Explanation

To solve the problem, we need to determine the degree of dissociation (α) of acetic acid in a benzene solution where it exists in dimeric form. Given that the van't Hoff factor (i) is 0.52, we can follow these steps:

### Step 1: Understanding the dimeric form of acetic acid
Acetic acid (CH₃COOH) can form dimers in solution. The dimer can be represented as (CH₃COOH)₂. Initially, we have 1 mole of acetic acid that can potentially form dimers.

### Step 2: Setting up the dissociation
Let α be the degree of dissociation of acetic acid. Initially, we have:
- 1 mole of acetic acid (CH₃COOH)
- 0 moles of dimer (CH₃COOH)₂

After dissociation:
- The amount of acetic acid that dissociates is α moles.
- The remaining acetic acid is (1 - α) moles.
- The dimer formed is α/2 moles (since 2 moles of acetic acid form 1 dimer).

### Step 3: Total moles after dissociation
The total number of moles after dissociation can be expressed as:
\[ \text{Total moles} = (1 - α) + \frac{α}{2} \]
This simplifies to:
\[ \text{Total moles} = 1 - α + \frac{α}{2} = 1 - \frac{α}{2} \]

### Step 4: Calculating the van't Hoff factor (i)
The van't Hoff factor (i) is defined as the ratio of the total number of moles after dissociation to the initial number of moles. Thus:
\[ i = \frac{\text{Total moles after dissociation}}{\text{Initial moles}} = \frac{1 - \frac{α}{2}}{1} \]
This simplifies to:
\[ i = 1 - \frac{α}{2} \]

### Step 5: Using the given van't Hoff factor
We know from the problem that the van't Hoff factor (i) is 0.52. Therefore, we can set up the equation:
\[ 0.52 = 1 - \frac{α}{2} \]

### Step 6: Solving for α
Rearranging the equation gives:
\[ \frac{α}{2} = 1 - 0.52 = 0.48 \]
Multiplying both sides by 2:
\[ α = 0.96 \]

### Conclusion
The degree of dissociation (α) of acetic acid in the benzene solution is 0.96.