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 compound has the maximum boiling point among the given options, we can use the concept of boiling point elevation. The boiling point elevation 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 (which is constant for a given solvent)
- \(m\) = molality of the solution (which can be assumed to be equal to molarity for dilute solutions)

### Step-by-Step Solution:

1. **Identify the Compounds and Their Concentrations**:
We need to look at the compounds provided in the options and their respective concentrations.

2. **Calculate the van 't Hoff Factor (\(i\))**:
- For NaOH: It dissociates into Na⁺ and OH⁻, so \(i = 2\).
- For Na₂CO₃: It dissociates into 2 Na⁺ and CO₃²⁻, so \(i = 3\).
- For other compounds (not specified in the transcript), we would need to determine their dissociation and calculate \(i\) accordingly.

3. **Determine the Molality (or Molarity)**:
- The molality is given as 0.2 M for both NaOH and Na₂CO₃ in the example.

4. **Calculate \(\Delta T_b\) for Each Compound**:
- For NaOH:
\[ \Delta T_b = i \cdot K_b \cdot m = 2 \cdot K_b \cdot 0.2 \]
- For Na₂CO₃:
\[ \Delta T_b = i \cdot K_b \cdot m = 3 \cdot K_b \cdot 0.2 \]

5. **Compare \(\Delta T_b\)**:
- Since \(K_b\) is constant, we can compare the factors:
- For NaOH, \(\Delta T_b = 0.4 K_b\)
- For Na₂CO₃, \(\Delta T_b = 0.6 K_b\)

6. **Determine the Maximum Boiling Point**:
- The compound with the higher \(\Delta T_b\) will have a higher boiling point.
- Since \(0.6 K_b > 0.4 K_b\), Na₂CO₃ has the maximum boiling point.

### Conclusion:
The correct answer is that 0.2 M Na₂CO₃ has the maximum boiling point among the given options.