Which of the following aqueous solution has the highest boiling point

Options

A. `0.1MKNO_3`
B. `0.1MNa_3PO_4`
C. `0.1MBaCl_2`
D. `00.1MK_2SO_4`

Answer

To determine which of the following aqueous solutions has the highest boiling point, we will use the formula for boiling point elevation:

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

Where:
- \(\Delta T_b\) = elevation in boiling point
- \(i\) = van 't Hoff factor (number of particles the solute dissociates into)
- \(K_b\) = ebullioscopic constant (which is constant for a given solvent, in this case, water)
- \(m\) = molality of the solution

Since the molality (\(m\)) and the ebullioscopic constant (\(K_b\)) are the same for all solutions, the boiling point elevation (\(\Delta T_b\)) will depend solely on the van 't Hoff factor (\(i\)).

Let's analyze each option:

1. **1m KNO3**:
- KNO3 dissociates into 2 ions: \(K^+\) and \(NO_3^-\)
- Therefore, \(i = 2\)

2. **1m Na3PO4**:
- Na3PO4 dissociates into 4 ions: 3 \(Na^+\) and 1 \(PO_4^{3-}\)
- Therefore, \(i = 4\)

3. **1m BaCl2**:
- BaCl2 dissociates into 3 ions: 1 \(Ba^{2+}\) and 2 \(Cl^-\)
- Therefore, \(i = 3\)

4. **1m K2SO4**:
- K2SO4 dissociates into 3 ions: 2 \(K^+\) and 1 \(SO_4^{2-}\)
- Therefore, \(i = 3\)

Now, we summarize the van 't Hoff factors:
- KNO3: \(i = 2\)
- Na3PO4: \(i = 4\)
- BaCl2: \(i = 3\)
- K2SO4: \(i = 3\)

Since the boiling point elevation (\(\Delta T_b\)) is directly proportional to the van 't Hoff factor (\(i\)), the solution with the highest \(i\) will have the highest boiling point.

**Conclusion**:
The solution with the highest boiling point is **1m Na3PO4** (Option B) with \(i = 4\).

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