Which of these solutes raises the boiling point of water the most?

To determine which solute raises the boiling point of water the most, we can use the concept of boiling point elevation, which is dependent on the number of particles (ions or molecules) that a solute dissociates into in solution. The formula for boiling point elevation is:

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

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.

Now let's look at the solutes:

A. **Molecular sucrose (C12H22O11)** - Does not dissociate; \(i = 1\).

B. **Ionic sodium chloride (NaCl)** - Dissociates into 2 ions (Na⁺ and Cl⁻); \(i = 2\).

C. **Ionic potassium bromide (KBr)** - Dissociates into 2 ions (K⁺ and Br⁻); \(i = 2\).

D. **Ionic calcium chloride (CaCl2)** - Dissociates into 3 ions (Ca²⁺ and 2 Cl⁻); \(i = 3\).

E. **Ionic aluminum fluoride (AlF3)** - Dissociates into 4 ions (Al³⁺ and 3 F⁻); \(i = 4\).

Based on the van't Hoff factors:
- Sucrose: \(i = 1\)
- NaCl: \(i = 2\)
- KBr: \(i = 2\)
- CaCl2: \(i = 3\)
- AlF3: \(i = 4\)

The solute with the highest van't Hoff factor will produce the most significant boiling point elevation. Therefore, **ionic aluminum fluoride (AlF3)**, with \(i = 4\), will raise the boiling point of water the most.

**Answer: E. ionic aluminum fluoride (AlF3)**