The van't Hoff factor `i` for an electrolyte which undergoes dissociation in one solvent and association in other solvent respectively:

Options

A. greater than one and less then one
B. less then one and greater than one
C. less then one and less than one
D. greater then one and greater than one

Answer

To solve the question regarding the van't Hoff factor \( i \) for an electrolyte that undergoes dissociation in one solvent and association in another solvent, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Van't Hoff Factor**:
The van't Hoff factor \( i \) is defined as the ratio of the number of particles in solution after dissociation or association to the number of particles before dissociation or association. Mathematically, it can be expressed as:
\[ i = \frac{\text{Number of particles after}}{\text{Number of particles before}} \]

2. **Dissociation in One Solvent**:
When an electrolyte dissociates, it breaks down into its constituent ions. For example, if an electrolyte \( AB \) dissociates into \( A^+ \) and \( B^- \):
- Before dissociation: 1 particle (the compound \( AB \))
- After dissociation: 2 particles (\( A^+ \) and \( B^- \))

Therefore, the van't Hoff factor \( i \) for dissociation is:
\[ i = \frac{2}{1} = 2 \]
This indicates that \( i > 1 \) for dissociation.

3. **Association in Another Solvent**:
In contrast, when an electrolyte undergoes association, multiple particles combine to form a single particle. For example, if \( n \) particles of \( A \) associate to form \( A_n \):
- Before association: \( n \) particles
- After association: 1 particle (\( A_n \))

Therefore, the van't Hoff factor \( i \) for association is:
\[ i = \frac{1}{n} \]
Since \( n \) is greater than 1, this indicates that \( i < 1 \) for association.

4. **Conclusion**:
- For dissociation, \( i > 1 \)
- For association, \( i < 1 \)

Thus, the final answer is that the van't Hoff factor \( i \) is greater than 1 for dissociation and less than 1 for association.