Consider `1M` solutions of the following salts. State which solution will have the lowest freezing point.

Options

A. `Na_(2)SO_(4)`
B. `BaCI_(2)`
C. `NaCI`
D. `AI_(2)(SO_(4))_(3)`

Explanation

To determine which 1M solution of the given salts has the lowest freezing point, we will use the formula for freezing point depression:

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

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 property of the solvent, which is constant for water)
- \(m\) = molality of the solution (in this case, it is 1M, which we can consider as 1 for simplicity)

### Step-by-Step Solution:

1. **Identify the van't Hoff factor (i) for each salt:**
- **Na2SO4**: Dissociates into 2 Na\(^+\) and 1 SO4\(^{2-}\), giving a total of 3 ions. Thus, \(i = 3\).
- **BaCl2**: Dissociates into 1 Ba\(^{2+}\) and 2 Cl\(^-\), giving a total of 3 ions. Thus, \(i = 3\).
- **NaCl**: Dissociates into 1 Na\(^+\) and 1 Cl\(^-\), giving a total of 2 ions. Thus, \(i = 2\).
- **Al2(SO4)3**: Dissociates into 2 Al\(^{3+}\) and 3 SO4\(^{2-}\), giving a total of 5 ions. Thus, \(i = 5\).

2. **Compare the van't Hoff factors (i):**
- Na2SO4: \(i = 3\)
- BaCl2: \(i = 3\)
- NaCl: \(i = 2\)
- Al2(SO4)3: \(i = 5\)

3. **Determine which salt has the highest van't Hoff factor:**
- The highest \(i\) value is for Al2(SO4)3, which has \(i = 5\).

4. **Conclusion:**
- Since the freezing point depression (\(\Delta T_f\)) is directly proportional to the van't Hoff factor \(i\), the solution with the highest \(i\) will have the lowest freezing point.
- Therefore, **Al2(SO4)3** will have the lowest freezing point among the given solutions.

### Final Answer:
The solution with the lowest freezing point is **Al2(SO4)3**.