Which of the following aqueous solution will exhibit highest boiling point?

The correct option is 0.01M Na₂SO₄.

To determine which aqueous solution will exhibit the highest boiling point, we need to understand the colligative property known as elevation in boiling point. The elevation in boiling point ($\Delta T_b$) is directly proportional to the effective concentration of solute particles in the solution. This relationship is mathematically expressed as:

$\Delta T_b=i\cdot K_b\cdot m$

Where:
- $i$ is the van 't Hoff factor, representing the number of particles the solute dissociates into in solution.
- $K_b$ is the molal boiling point elevation constant of the solvent (water in this case, which is constant for all options).
- $m$ is the molality of the solution (which can be approximated by molarity, $M$, for dilute aqueous solutions).

Since $K_b$ is constant, the elevation in boiling point, and therefore the boiling point of the solution, is directly proportional to the product of the van 't Hoff factor and the molarity ($i\cdot M$). The solution with the highest value of $i\cdot M$ will have the highest boiling point elevation and thus the highest boiling point.

Let's calculate the value of $i\cdot M$ for each of the given solutions:

1. 0.01M KNO₃:
Potassium nitrate (${\mathrm{KNO}}_3$) is a strong electrolyte and dissociates completely in water as follows:
${\mathrm{KNO}}_3\left(\mathrm{aq}\right)\to K^{+}\left(\mathrm{aq}\right)+{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)$
Here, one formula unit produces 2 ions, so $i=2$.
$i\cdot M=2\cdot 0.01=0.02M$

2. 0.01M Na₂SO₄:
Sodium sulfate (${\mathrm{Na}}_2{\mathrm{SO}}_4$) dissociates completely in water as follows:
${\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to 2{\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)$
Here, one formula unit produces 3 ions (two sodium ions and one sulfate ion), so $i=3$.
$i\cdot M=3\cdot 0.01=0.03M$

3. 0.015M C₆H₁₂O₆ (Glucose):
Glucose is a non-electrolyte and does not dissociate in water, so $i=1$.
$i\cdot M=1\cdot 0.015=0.015M$

4. 0.01M Urea:
Urea is a non-electrolyte and does not dissociate in water, so $i=1$.
$i\cdot M=1\cdot 0.01=0.01M$

Comparing the effective concentrations ($i\cdot M$):
- KNO₃: 0.02 M
- Na₂SO₄: 0.03 M
- Glucose: 0.015 M
- Urea: 0.01 M

Since 0.01M Na₂SO₄ has the highest value of effective particle concentration ($0.03M$), it will experience the greatest elevation in boiling point and thus exhibit the highest boiling point.