Consider the following solutions: \(0.010 \mathrm{~m} \mathrm{Na}_{3} \mathrm{PO}_{4}\) in water \(0.020 \mathrm{~m} \mathrm{CaBr}_{2}\) in water \(0.020 \mathrm{~m} \mathrm{KCl}\) in water \(0.020 \mathrm{~m} \mathrm{HF}\) in water \((\mathrm{HF}\) is a weak acid. \()\) a. Assuming complete dissociation of the soluble salts, which solution(s) would have the same boiling point as \(0.040 \mathrm{~m}\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in water? \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is a nonelectrolyte. b. Which solution would have the highest vapor pressure at \(28^{\circ} \mathrm{C}\) ? c. Which solution would have the largest freezing-point depression?

Colligative properties are properties that depend on the number of solute particles in the solution, not the nature of the solute particles. The colligative properties in question are boiling point elevation and freezing point depression. Molality (m) is the number of moles of solute divided by the mass of the solvent (in kg). Since boiling point elevation and freezing point depression are both proportional to the molality of the solution, we can simply compare the molality values of each solution.

For complete dissociation, the effective molality of ions in solution is given by the number of ions multiplied by the molarity of the solution. For Na3PO4: The molality is 0.010 m and it dissociates into 4 ions (3 Na+ and 1 PO43-). Effective molality: 0.010 m * 4 = 0.040 m For CaBr2: The molality is 0.020 m and it dissociates into 3 ions (1 Ca2+ and 2 Br-). Effective molality: 0.020 m * 3 = 0.060 m For KCl: The molality is 0.020 m and it dissociates into 2 ions (1 K+ and 1 Cl-). Effective molality: 0.020 m * 2 = 0.040 m For HF: The molality is 0.020 m, but since it's a weak acid, we will treat it as a non-electrolyte. The effective molality remains 0.020 m For C6H12O6: The molality is 0.040 m and as a non-electrolyte, its effective molality remains the same, 0.040 m

To find which solution would have the same boiling point as 0.040 m C6H12O6, we need to identify which solutions have the same effective molality. Comparing the effective molalities, we find that the 0.010 m Na3PO4 solution and the 0.020 m KCl solution both have an effective molality of 0.040 m. Therefore, these two solutions will have the same boiling point elevation as 0.040 m C6H12O6.

The solution with the highest vapor pressure will be the one with the lowest molality (since colligative properties are inversely proportional to vapor pressure). Among the given solutions, the 0.020 m HF solution has the lowest molality. Thus, the 0.020 m HF solution would have the highest vapor pressure at 28°C.

The freezing point depression is proportional to the effective molality of the solution. Among the given solutions, the 0.020 m CaBr2 solution has the highest effective molality (0.060 m). Therefore, the 0.020 m CaBr2 solution will have the largest freezing-point depression.