Final answer:
The solution with the highest vapor pressure would be 0.10 m KClO4, because it has both a low molality and a low van 't Hoff factor, resulting in a lesser extent of vapor pressure lowering compared to the other solutions given.
Step-by-step explanation:
To determine which solution has the highest vapor pressure, we need to consider both the concentration of the solution and the number of particles it produces when it dissolves, because vapor pressure lowering is a colligative property.
Colligative properties depend on the number of solute particles in solution rather than the type of particles. For ionic compounds, the van 't Hoff factor (i) represents the number of particles the compound dissociates into when in solution. Strong electrolytes, which dissociate completely, have an i value greater than 1 (corresponding to the number of ions produced), while non-electrolytes have an i of 1 as they do not dissociate into ions.
- For Al(CIO₄)₃ (a strong electrolyte), which has a formula mass of Al3+ and 3CIO4-, the i value would be 4 (1 Al3+ ion + 3 CIO4- ions).
- For Ca(CIO₄)₂, the i value is 3 (1 Ca2+ ion + 2 CIO4- ions).
- For NaCl, the i value is 2 (1 Na+ ion + 1 Cl- ion).
- For C₂H₅OH (ethanol), it does not ionize, so the i value is 1.
- For KCIO₄, the i value is 2 (1 K+ ion + 1 CIO4- ion).
Higher i values lead to greater vapor pressure lowering because there are more particles to interact with the solvent and thus fewer solvent molecules available to escape into the vapor phase. However, solution B has the highest molality (0.50 m) and a relatively high i value (3), while the remaining solutions have lower molality values. Solution E (0.10 m KCIO₄) has both a low molality and an i value (2), which would tend to result in a lesser extent of vapor pressure lowering, therefore, 0.10 m KCIO₄ is the expected answer.