Which of the following has minimum freezing point

Options

A. 0.1 M `K_(2)Cr_(2)O_(7)`
B. 0.1 M `NH_(4)CI`
C. 0.1 M `BaSO_(4)`
D. 0.1 M `Al_(2)(SO_(4))_(3)`

Explanation

To determine which of the given solutions has the minimum freezing point, we will use the concept of colligative properties, specifically the depression in freezing point. The depression in freezing point (\( \Delta T_f \)) can be calculated using the formula:

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

Where:
- \( \Delta T_f \) = depression in freezing point
- \( i \) = van 't Hoff factor (the number of particles the solute dissociates into)
- \( k_f \) = cryoscopic constant of the solvent (for water, \( k_f = 1 \))
- \( m \) = molality of the solution

Since all the solutions have the same molality, the depression in freezing point will depend primarily on the value of \( i \).

### Step-by-Step Solution:

1. **Identify the Solutions**: We have four solutions to consider:
- 0.1 M \( K_2Cr_2O_7 \)
- 0.1 M \( NH_4Cl \)
- 0.1 M \( Na_2SO_4 \)
- 0.1 M \( Al_2(SO_4)_3 \)

2. **Calculate the van 't Hoff Factor (\( i \)) for Each Solution**:
- For \( K_2Cr_2O_7 \):
- Dissociation: \( K_2Cr_2O_7 \rightarrow 2K^+ + Cr_2O_7^{2-} \)
- \( i = 2 + 1 = 3 \)

- For \( NH_4Cl \):
- Dissociation: \( NH_4Cl \rightarrow NH_4^+ + Cl^- \)
- \( i = 1 + 1 = 2 \)

- For \( Na_2SO_4 \):
- Dissociation: \( Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-} \)
- \( i = 2 + 1 = 3 \)

- For \( Al_2(SO_4)_3 \):
- Dissociation: \( Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-} \)
- \( i = 2 + 3 = 5 \)

3. **Compare the Values of \( i \)**:
- \( K_2Cr_2O_7 \): \( i = 3 \)
- \( NH_4Cl \): \( i = 2 \)
- \( Na_2SO_4 \): \( i = 3 \)
- \( Al_2(SO_4)_3 \): \( i = 5 \)

4. **Determine the Depression in Freezing Point**:
- Since \( \Delta T_f \) is directly proportional to \( i \), the solution with the highest \( i \) will have the greatest depression in freezing point.
- The highest \( i \) value is for \( Al_2(SO_4)_3 \) with \( i = 5 \).

5. **Conclusion**:
- The solution with the minimum freezing point is \( Al_2(SO_4)_3 \) because it has the highest van 't Hoff factor, leading to the greatest depression in freezing point.

### Final Answer:
The solution with the minimum freezing point is \( Al_2(SO_4)_3 \).