Which of the following solutions will have the highest boiling point ?

Options

A. `0.1 " M " FeCl_3`
B. `0.1 " M " BaCl_2`
C. `0.1 " M " NaCl_3`
D. `0.1` M Urea

Answer

To determine which solution has the highest boiling point, we can use the concept of boiling point elevation, which is a colligative property. The boiling point elevation (\( \Delta T_b \)) can be calculated using the formula:

\[ \Delta T_b = i \cdot K_b \cdot m \]

Where:
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( K_b \) = ebullioscopic constant of the solvent (which is constant for a given solvent)
- \( m \) = molality of the solution

### Step 1: Identify the Solutions
Let's consider the following solutions:
1. \( \text{FeCl}_3 \)
2. \( \text{BaCl}_2 \)
3. \( \text{NaCl} \)
4. Urea

### Step 2: Determine the van 't Hoff Factor (\( i \))
- For \( \text{FeCl}_3 \):
- It dissociates into 1 \( \text{Fe}^{3+} \) and 3 \( \text{Cl}^- \) ions.
- \( i = 1 + 3 = 4 \)

- For \( \text{BaCl}_2 \):
- It dissociates into 1 \( \text{Ba}^{2+} \) and 2 \( \text{Cl}^- \) ions.
- \( i = 1 + 2 = 3 \)

- For \( \text{NaCl} \):
- It dissociates into 1 \( \text{Na}^+ \) and 1 \( \text{Cl}^- \) ion.
- \( i = 1 + 1 = 2 \)

- For Urea:
- Urea is a non-electrolyte and does not dissociate.
- \( i = 1 \)

### Step 3: Compare the van 't Hoff Factors
Now we can summarize the van 't Hoff factors:
- \( \text{FeCl}_3 \): \( i = 4 \)
- \( \text{BaCl}_2 \): \( i = 3 \)
- \( \text{NaCl} \): \( i = 2 \)
- Urea: \( i = 1 \)

### Step 4: Determine the Highest Boiling Point
Since the boiling point elevation (\( \Delta T_b \)) is directly proportional to \( i \) (assuming equal molality), the solution with the highest \( i \) will have the highest boiling point.

From our calculations:
- The highest \( i \) is for \( \text{FeCl}_3 \) with \( i = 4 \).

### Conclusion
Thus, the solution that will have the highest boiling point is the solution of \( \text{FeCl}_3 \).

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