Which of the following has highest value of Van't Hoff factor?

Options

A. 0.1 M `Al_(2) (SO_(4))_(3)`
B. 0.1 M `C_(6)H_(12)O_(6)`
C. 0.1 M `K_(2)SO_(4)`
D. `0.1 M NaCl`

Answer

To determine which of the given solutions has the highest Van't Hoff factor (i), we need to analyze each solute and how it dissociates in water. The Van't Hoff factor represents the number of particles the solute breaks into when dissolved.

### Step-by-Step Solution:

1. **Identify the Solutes**: The question provides four solutes. We need to analyze each one:
- Al2(SO4)3
- C6H12O6 (glucose)
- K2SO4
- NaCl

2. **Determine the Nature of Each Solute**:
- **Al2(SO4)3**: This is an electrolyte and will dissociate into ions.
- **C6H12O6**: This is a non-electrolyte and will not dissociate.
- **K2SO4**: This is an electrolyte and will dissociate into ions.
- **NaCl**: This is also an electrolyte and will dissociate into ions.

3. **Calculate the Van't Hoff Factor (i)**:
- **For Al2(SO4)3**:
- Dissociation: Al2(SO4)3 → 2Al^3+ + 3SO4^2-
- Total ions = 2 + 3 = 5
- Therefore, i = 5.

- **For C6H12O6**:
- Since it is a non-electrolyte, it does not dissociate.
- Therefore, i = 1.

- **For K2SO4**:
- Dissociation: K2SO4 → 2K^+ + SO4^2-
- Total ions = 2 + 1 = 3
- Therefore, i = 3.

- **For NaCl**:
- Dissociation: NaCl → Na^+ + Cl^-
- Total ions = 1 + 1 = 2
- Therefore, i = 2.

4. **Compare the Values of i**:
- Al2(SO4)3: i = 5
- C6H12O6: i = 1
- K2SO4: i = 3
- NaCl: i = 2

5. **Conclusion**: The solute with the highest Van't Hoff factor is Al2(SO4)3 with a value of i = 5.

### Final Answer:
The solution with the highest Van't Hoff factor is **Al2(SO4)3**.

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