Which of the following will have maximum depression in freezing point ?

Options

A. `0.5 M Li_(2)SO_(4)`
B. 1 M KCI
C. `0.5 M BaCI_(2)`
D. `1 M Al_(2)(SO_(4))_(3)`

Explanation

To determine which of the given solutions will have the maximum depression in freezing point, we need to analyze each option based on the number of particles produced when the solute dissociates in solution. The depression in freezing point is a colligative property, which depends on the number of solute particles in a solution.

### Step-by-Step Solution:

1. **Understand the Concept**:
Depression in freezing point (ΔTf) is given by the formula:
\[ \Delta T_f = K_f \times C \times i \]
where:
- \( K_f \) = cryoscopic constant (constant for a given solvent),
- \( C \) = molarity of the solution,
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into).

2. **Identify the Options**:
- 0.5 molar Li2SO4
- 1 molar KCl
- 0.5 molar BaCl2
- 1 molar Al2(SO4)3

3. **Calculate the Number of Particles for Each Option**:

- **For 0.5 molar Li2SO4**:
- Dissociation: \( \text{Li}_2\text{SO}_4 \rightarrow 2 \text{Li}^+ + \text{SO}_4^{2-} \)
- Number of ions produced: 3 (2 Li⁺ + 1 SO₄²⁻)
- \( i = 3 \)
- Total particles = \( 0.5 \times 3 = 1.5 \)

- **For 1 molar KCl**:
- Dissociation: \( \text{KCl} \rightarrow \text{K}^+ + \text{Cl}^- \)
- Number of ions produced: 2 (1 K⁺ + 1 Cl⁻)
- \( i = 2 \)
- Total particles = \( 1 \times 2 = 2 \)

- **For 0.5 molar BaCl2**:
- Dissociation: \( \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2 \text{Cl}^- \)
- Number of ions produced: 3 (1 Ba²⁺ + 2 Cl⁻)
- \( i = 3 \)
- Total particles = \( 0.5 \times 3 = 1.5 \)

- **For 1 molar Al2(SO4)3**:
- Dissociation: \( \text{Al}_2(\text{SO}_4)_3 \rightarrow 2 \text{Al}^{3+} + 3 \text{SO}_4^{2-} \)
- Number of ions produced: 5 (2 Al³⁺ + 3 SO₄²⁻)
- \( i = 5 \)
- Total particles = \( 1 \times 5 = 5 \)

4. **Compare the Total Particles**:
- Li2SO4: 1.5 particles
- KCl: 2 particles
- BaCl2: 1.5 particles
- Al2(SO4)3: 5 particles

5. **Conclusion**:
The solution with the maximum depression in freezing point is the one with the highest number of particles, which is **1 molar Al2(SO4)3** (5 particles). Therefore, the answer is:
**Option D: 1 molar Al2(SO4)3.**