Among the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?

Options

A. KCl
B. `C_(6)H_(12)O_(6)`
C. `Al(2)(SO_(4))_(3)`
D. `K_(2)SO_(4)`

Explanation

To solve the problem of which 0.10 m aqueous solution exhibits the largest freezing point depression, we will follow these steps:

### Step 1: Understand the Freezing Point Depression Formula
The freezing point depression (\( \Delta T_f \)) can be calculated using the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( i \) = Van't Hoff factor (the number of particles the solute dissociates into)
- \( K_f \) = freezing point depression constant (which is constant for a given solvent)
- \( m \) = molality of the solution

### Step 2: Identify the Given Solutions
The solutions provided are:
1. KCl (Potassium Chloride)
2. C₆H₁₂O₆ (Glucose)
3. Al₂(SO₄)₃ (Aluminum Sulfate)
4. K₂SO₄ (Potassium Sulfate)

### Step 3: Determine the Van't Hoff Factor (\( i \)) for Each Solution
- **KCl**: Dissociates into 2 ions (K⁺ and Cl⁻). Thus, \( i = 2 \).
- **C₆H₁₂O₆**: Does not dissociate in solution. Thus, \( i = 1 \).
- **Al₂(SO₄)₃**: Dissociates into 5 ions (2 Al³⁺ and 3 SO₄²⁻). Thus, \( i = 5 \).
- **K₂SO₄**: Dissociates into 3 ions (2 K⁺ and 1 SO₄²⁻). Thus, \( i = 3 \).

### Step 4: Compare the Values of \( i \)
Now we can summarize the Van't Hoff factors:
- KCl: \( i = 2 \)
- C₆H₁₂O₆: \( i = 1 \)
- Al₂(SO₄)₃: \( i = 5 \)
- K₂SO₄: \( i = 3 \)

### Step 5: Identify the Solution with the Largest Freezing Point Depression
Since \( \Delta T_f \) is directly proportional to \( i \) (as \( K_f \) and \( m \) are constants for this comparison), the solution with the highest \( i \) will have the largest freezing point depression.

From our calculations, Al₂(SO₄)₃ has the highest \( i \) value of 5.

### Conclusion
The solution that will exhibit the largest freezing point depression among the given options is:
**Al₂(SO₄)₃ (Aluminum Sulfate)**.

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