Which of the following aqueous solutions should have the highest boiling point ?

Question

Accepted Answer

To determine which aqueous solution has the highest boiling point, we need to consider the concept of boiling point elevation, which is a colligative property. Colligative properties depend on the number of solute particles in a solution, not on the identity of the solute. The boiling point elevation can be calculated using the formula:

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

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

Let's analyze each option:

1. **1.0 M NaOH**: Sodium hydroxide (NaOH) dissociates into two ions: Na$^+$ and OH$^-$. Therefore, the van't Hoff factor $i$ is 2.

2. **1.0 M $Na_2SO_4$**: Sodium sulfate ($Na_2SO_4$) dissociates into three ions: 2 Na$^+$ and $SO_4^{2-}$. Therefore, the van't Hoff factor $i$ is 3. This means that $Na_2SO_4$ produces more particles in solution compared to the other options, leading to a higher boiling point elevation.

3. **1.0 M $NH_4NO_3$**: Ammonium nitrate ($NH_4NO_3$) dissociates into two ions: $NH_4^+$ and $NO_3^-$. Therefore, the van't Hoff factor $i$ is 2.

4. **1.0 M $KNO_3$**: Potassium nitrate ($KNO_3$) dissociates into two ions: K$^+$ and $NO_3^-$. Therefore, the van't Hoff factor $i$ is 2.

Given these dissociations, $Na_2SO_4$ has the highest van't Hoff factor of 3, meaning it produces the most particles in solution. This results in the greatest boiling point elevation among the options provided.

Therefore, Option 2 ($1.0~M~Na_2SO_4$) is correct because it has the highest boiling point due to the highest number of dissociated particles in solution. The other options have lower van't Hoff factors and thus result in lower boiling point elevations.