Which one of following has maximum boiling point ?

Options

A. 0.2 M NaOH
B. 0.2 M `Na_2CO_3`
C. 0.1 M `AgNO_3`
D. 0.1 M `(NH_4)_2SO_4.FeSO_4.6H_2O`

Answer

To determine which substance has the maximum boiling point among the given options, we can use the concept of colligative properties, specifically the elevation of boiling point. Here’s a step-by-step solution:

### Step 1: Understand the Concept of Colligative Properties
Colligative properties depend on the number of solute particles in a solution rather than the identity of the solute. The elevation of boiling point (ΔTb) can be calculated using the formula:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \( \Delta T_b \) = elevation in boiling point
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)
- \( K_b \) = ebullioscopic constant of the solvent (a constant for a given solvent)
- \( m \) = molality of the solution

### Step 2: Identify the Van 't Hoff Factors
For each solute, we need to determine the van 't Hoff factor \( i \):
- For NaOH: When NaOH dissociates, it produces Na⁺ and OH⁻, so \( i = 2 \).
- For Na2CO3: When Na2CO3 dissociates, it produces 2 Na⁺ and CO3²⁻, so \( i = 3 \).

### Step 3: Compare the Concentrations
We are given concentrations:
- NaOH: 0.2 M
- Na2CO3: 0.2 M

Since both solutions have the same molality (0.2 M), we can focus on the van 't Hoff factor \( i \) to determine which solution will have a greater elevation in boiling point.

### Step 4: Calculate the Elevation of Boiling Point
Using the formula:
- For NaOH:
\[ \Delta T_b = i \cdot K_b \cdot m = 2 \cdot K_b \cdot 0.2 \]
- For Na2CO3:
\[ \Delta T_b = i \cdot K_b \cdot m = 3 \cdot K_b \cdot 0.2 \]

### Step 5: Conclusion
Since \( \Delta T_b \) for Na2CO3 is greater than that for NaOH due to the larger value of \( i \) (3 vs. 2), Na2CO3 will have a greater elevation of boiling point, thus it will have the maximum boiling point among the options provided.

### Final Answer
**Na2CO3 has the maximum boiling point.**