Which of the following solutions shows maximum depression in freezing point?

Options

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

Answer

To determine which of the given solutions shows the maximum depression in freezing point, we will use the formula for depression in freezing point, which is:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \(\Delta T_f\) = depression in freezing point
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(K_f\) = freezing point depression constant (a constant for the solvent)
- \(m\) = molality of the solution

### Step-by-Step Solution:

1. **Identify the van 't Hoff factor (i)** for each solute:
- **Li2SO4**: Dissociates into 2 Li\(^+\) and 1 SO4\(^{2-}\), so \(i = 3\).
- **NaCl**: Dissociates into 1 Na\(^+\) and 1 Cl\(^-\), so \(i = 2\).
- **Al2(SO4)3**: Dissociates into 2 Al\(^{3+}\) and 3 SO4\(^{2-}\), so \(i = 5\).
- **BaCl2**: Dissociates into 1 Ba\(^{2+}\) and 2 Cl\(^-\), so \(i = 3\).

2. **Determine the molality (m)** for each solution:
- **Li2SO4**: \(m = 0.5\)
- **NaCl**: \(m = 1\)
- **Al2(SO4)3**: \(m = 0.5\)
- **BaCl2**: \(m = 0.5\)

3. **Calculate the depression in freezing point (\(\Delta T_f\)) for each solution**:
- **For Li2SO4**:
\[ \Delta T_f = i \cdot m = 3 \cdot 0.5 = 1.5 \]
- **For NaCl**:
\[ \Delta T_f = i \cdot m = 2 \cdot 1 = 2 \]
- **For Al2(SO4)3**:
\[ \Delta T_f = i \cdot m = 5 \cdot 0.5 = 2.5 \]
- **For BaCl2**:
\[ \Delta T_f = i \cdot m = 3 \cdot 0.5 = 1.5 \]

4. **Compare the values of \(\Delta T_f\)**:
- Li2SO4: 1.5
- NaCl: 2
- Al2(SO4)3: 2.5
- BaCl2: 1.5

5. **Conclusion**:
The solution with the maximum depression in freezing point is **Al2(SO4)3** with a \(\Delta T_f\) of 2.5.

### Final Answer:
**Al2(SO4)3 shows the maximum depression in freezing point.**