A compound X undergoes tetramerisation in a given organic solvent. The van't Hoff factor is

Options

A. 4
B. 0.25
C. 0.125
D. 2

Explanation

To solve the problem regarding the van't Hoff factor (i) for a compound X that undergoes tetramerization, we will follow these steps:

### Step 1: Understand Tetramerization
Tetramerization is the process where four molecules of a compound combine to form a single molecule. In this case, we can represent the compound as \( A \). Therefore, when tetramerization occurs, four moles of \( A \) combine to form one mole of \( A_4 \).

### Step 2: Define the Dissociation
Let’s assume that \( \alpha \) moles of \( A \) dissociate. This means that out of the initial moles of \( A \), \( \alpha \) moles are involved in forming the tetramer \( A_4 \).

### Step 3: Calculate Remaining Moles
If \( \alpha \) moles of \( A \) are used to form \( A_4 \), the remaining moles of \( A \) will be:
- Initial moles of \( A \): 1
- Moles of \( A \) that remain: \( 1 - \alpha \)
- Moles of \( A_4 \) formed: \( \frac{\alpha}{4} \)

### Step 4: Calculate the van't Hoff Factor (i)
The van't Hoff factor \( i \) is defined as the total number of particles in solution after dissociation. Therefore, we can express \( i \) as:
\[ i = \text{(remaining moles of A)} + \text{(moles of A}_4\text{)} \]
Substituting the values we found:
\[ i = (1 - \alpha) + \frac{\alpha}{4} \]

### Step 5: Simplify the Expression
Now, we can simplify the expression for \( i \):
\[ i = 1 - \alpha + \frac{\alpha}{4} \]
To combine the terms, we can express \( -\alpha \) as \( -\frac{4\alpha}{4} \):
\[ i = 1 - \frac{4\alpha}{4} + \frac{\alpha}{4} = 1 - \frac{3\alpha}{4} \]

### Step 6: Consider Complete Dissociation
Assuming complete dissociation, we set \( \alpha = 1 \):
\[ i = 1 - \frac{3 \cdot 1}{4} = 1 - \frac{3}{4} = \frac{1}{4} = 0.25 \]

### Final Answer
Thus, the van't Hoff factor \( i \) for the compound X undergoing tetramerization is:
\[ \boxed{0.25} \]

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