Consider the following solutions: 0.010\(m \mathrm{Na}_{3} \mathrm{PO}_{4}\) in water 0.020 \(m \mathrm{CaBr}_{2}\) in water 0.020 \(m \mathrm{KCl}\) in water 0.020 \(m \mathrm{HF}\) in water \((\mathrm{HF} \text { is a weak acid.) }\) 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?

In order to determine the total concentrations of solute particles in each of the given solutions, we need to consider how many ions will be generated upon dissociation of each solute. Here's the breakdown: - 0.010 m Na3PO4: \[\mathrm{Na}_{3} \mathrm{PO}_{4}(s) \rightarrow 3\mathrm{Na}^{+}(aq) + \mathrm{PO}_{4}^{3-}(aq)\] So, the concentration of solute particles in the solution is \(0.010 \times (3 + 1) = 0.040 \: m\). - 0.020 m CaBr2: \[\mathrm{CaBr}_{2}(s) \rightarrow \mathrm{Ca}^{2+}(aq) + 2\mathrm{Br}^{-}(aq)\] So, the concentration of solute particles in the solution is \(0.020 \times (1 + 2) = 0.060 \: m\). - 0.020 m KCl: \[\mathrm{KCl}(s) \rightarrow \mathrm{K}^{+}(aq) + \mathrm{Cl}^{-}(aq)\] So, the concentration of solute particles in the solution is \(0.020 \times (1 + 1) = 0.040 \: m\). - 0.020 m HF: Since HF is a weak acid, it does not completely dissociate. Therefore, we can consider its concentration to be 0.020 m. - 0.040 m C6H12O6: Since C6H12O6 is a nonelectrolyte, its concentration remains 0.040 m.

The boiling point elevation of a solution is directly proportional to the concentration of solute particles. In comparing the boiling points, we find: - 0.040 m C6H12O6 - 0.010 m Na3PO4 (0.040 m solute particles) - 0.020 m KCl (0.040 m solute particles) These three solutions will have the same boiling point due to their equal concentrations of solute particles. So, the answer to question (a) is: 0.010 m Na3PO4 and 0.020 m KCl will have the same boiling point as 0.040 m C6H12O6.

The vapor pressure lowering of a solution is directly proportional to the concentration of solute particles. The solution with the highest vapor pressure at \(28^{\circ} \mathrm{C}\) will be the one with the lowest concentration of solute particles: - 0.020 m HF Thus, the answer to question (b) is: 0.020 m HF has the highest vapor pressure at \(28^{\circ} \mathrm{C}\).

The freezing-point depression of a solution is directly proportional to the concentration of solute particles. The solution with the largest freezing-point depression will be the one with the highest concentration of solute particles: - 0.020 m CaBr2 (0.060 m solute particles) Thus, the answer to question (c) is: 0.020 m CaBr2 would have the largest freezing-point depression.