Which solution will show maximum elevation in boiling point?

Options

A. `0.1 M KCl`
B. `0.1 M BaCl_(2)`
C. `0.1 M FeCl_(3)`
D. `0.1 M Fe_(2)(SO_(4))_(3)`

Explanation

To determine which solution will show the maximum elevation in boiling point, we can follow these steps:

### Step 1: Understand the Concept of Elevation in Boiling Point
The elevation in boiling point (\( \Delta T_b \)) occurs when a non-volatile solute is added to a solvent. The formula to calculate the elevation in boiling point is given by:
\[ \Delta T_b = K_b \cdot m \cdot i \]
where:
- \( K_b \) = ebullioscopic constant of the solvent (a constant for a given solvent)
- \( m \) = molality of the solution
- \( i \) = van 't Hoff factor (the number of particles the solute dissociates into)

### Step 2: Analyze the Given Solutions
We have four solutions to consider, all at a concentration of 0.1 M:
1. Potassium Chloride (KCl)
2. Barium Chloride (BaCl₂)
3. Ferric Chloride (FeCl₃)
4. Ferrous Sulfate (Fe₂(SO₄)₃)

### Step 3: Determine the van 't Hoff Factor (\( i \)) for Each Solution
- **KCl**: Dissociates into \( K^+ \) and \( Cl^- \) → \( i = 1 + 1 = 2 \)
- **BaCl₂**: Dissociates into \( Ba^{2+} \) and \( 2 Cl^- \) → \( i = 1 + 2 = 3 \)
- **FeCl₃**: Dissociates into \( Fe^{3+} \) and \( 3 Cl^- \) → \( i = 1 + 3 = 4 \)
- **Fe₂(SO₄)₃**: Dissociates into \( 2 Fe^{3+} \) and \( 3 SO₄^{2-} \) → \( i = 2 + 3 = 5 \)

### Step 4: Compare the van 't Hoff Factors
Now we can summarize the \( i \) values:
- KCl: \( i = 2 \)
- BaCl₂: \( i = 3 \)
- FeCl₃: \( i = 4 \)
- Fe₂(SO₄)₃: \( i = 5 \)

### Step 5: Determine the Maximum Elevation in Boiling Point
Since all solutions have the same concentration, the solution with the highest \( i \) value will show the maximum elevation in boiling point. From our calculations, \( Fe₂(SO₄)₃ \) has the highest \( i \) value of 5.

### Conclusion
Therefore, the solution that will show the maximum elevation in boiling point is:
**0.1 M Ferrous Sulfate (Fe₂(SO₄)₃)**.

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