Match the entries of column I with appropriate entries of column II and choose the correct option out of the four options given.
`{:("Column I (van't Hoff factor)","Column II (Behaviour)"),("(A) "i gt 1,"(p) There is association."),("(B) "i lt 1,"(q) There is dissociation."),("(C) "i=1,"(r) Impossible"),("(D) "i=0,"(s) No association or dissociation"):}`

Options

A. A-p, B-q, C-s, D-r
B. A-s, B-r, C-p, D-q
C. A-r, B-s, C-q, D-p
D. A-q, B-p, C-s, D-r

Answer

To solve the problem of matching the entries from Column I (van't Hoff factor) with the appropriate entries from Column II (Behavior), we need to understand the meaning of the van't Hoff factor (i) and how it relates to the behavior of solutes in solution.

### Step-by-Step Solution:

1. **Understanding the van't Hoff Factor (i)**:
- The van't Hoff factor (i) indicates the number of particles a solute breaks into when it dissolves in a solvent.
- For non-electrolytes that do not dissociate, i = 1.
- For electrolytes that dissociate, i can be greater than 1 (for complete dissociation) or less than 1 (for association).

2. **Analyzing each case in Column I**:
- **(A) i > 1**: This indicates that the solute dissociates into more particles than it originally was.
- **Match with (q)**: There is dissociation.
- **(B) i < 1**: This indicates that the solute associates to form fewer particles than it originally was.
- **Match with (p)**: There is association.
- **(C) i = 1**: This indicates that the solute does not dissociate or associate; it remains as one particle.
- **Match with (s)**: No association or dissociation.
- **(D) i = 0**: This is an impossible scenario for a solute in solution since it cannot have a van't Hoff factor of zero.
- **Match with (r)**: Impossible.

3. **Final Matches**:
- (A) i > 1 → (q) There is dissociation.
- (B) i < 1 → (p) There is association.
- (C) i = 1 → (s) No association or dissociation.
- (D) i = 0 → (r) Impossible.

### Summary of Matches:
- (A) → (q)
- (B) → (p)
- (C) → (s)
- (D) → (r)