What is the boiling point of a solution that contains 3 moles of KBr in 2000 g of water? (K = 0.512C/m; molar mass of water = 18 g)

Question

BlackTom AI · Accepted Answer

To tackle boiling point elevation, start by recalling the formula: ΔTb = i · Kb · m, where i is the van't Hoff factor, Kb is the ebullioscopic constant for the solvent (water, here), and m is the molality.

First, determine the molality. You have 3 moles of KBr dissolved in 2000 g of water. The solvent mass is 2.0 kg, so m = moles solute / kg solvent = 3.0 mol / 2.0 kg = 1.5 mol/kg.

Next, account for the dissociation of KBr in solution. KBr dissociates into K+ and Br-, so ideally i ≈ 2.

Now compute ΔTb: ΔTb = i · Kb · m = 2 · 0.512 °C/m · 1.5 m = 0.512 · 3 = 1.536 °C (approximately 1.54 °C).

The boiling point of pure water is 100.0 °C. Adding the elevation gives Tb(solution) ≈ 100.0 °C + 1.54 °C ≈ 101.54 °C.

Compare with the answer choices:
- 97 °C: This is well below the boiling point and would imply a negative or negligible elevation, which contradicts the calculation.
- 99.7 °C: This is closer to no substantial boiling point rise and would correspond to a much smaller ΔTb than calculated; it misreads the effect of solute on boiling point.
- 101.4 °C: This matches the computed elevation (approximately 101.5 °C when rounded) and is the correct choice.
- 103 °C: This overestimates the boiling point elevation given the same data; it would require a larger ΔTb than computed.

Thus, the reasoning leads to a boiling point around 101.5 °C, making 101.4 °C the closest correct option.

Chemistry (H) - Varnold - Preparatory Academy - 2nd Semester - 2024/2025 2025 States of Matter and Solutions Practice Test

What is the boiling point of a solution that contains 3 moles of KBr in 2000 g of water? (K = 0.512C/m; molar mass of water = 18 g)

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