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 determine the boiling point elevation, we start with the formula ΔTb = i · Kb · m, where i is the van't Hoff factor, Kb is the ebullioscopic constant of the solvent, and m is the molality of the solution.

First, identify the solvent and its properties: water has a normal boiling point of 100.0°C and Kb = 0.512 °C/m.

Next, compute the molality. The solution contains 3 moles of KBr dissolved in 2000 g (2.0 kg) of water, so m = moles solute / kg solvent = 3 / 2.0 = 1.5 m.

Determine the van't Hoff factor i for KBr. Since KBr dissociates into two ions (K+ and Br−) in solution, i = 2.

Now plug the values into ΔTb: ΔTb = i · Kb · m = 2 · 0.512 °C/m · 1.5 m = 1.536 °C (roughly 1.54 °C).

Add the boiling point elevation to the normal boiling point of the solvent: Tb ≈ 100.0 °C + 1.54 °C ≈ 101.54 °C.

When choosing from the given options, 101.4°C is the closest to the calculated value (accounting for rounding in typical problem settings). The other options are not consistent with the calculated elevation:
- 97°C would imply a much larger decrease, which does not occur in boiling point elevation scenarios.
- 99.7°C is a smaller elevation than expected from the computed ΔTb.
- 103°C would correspond to a larger elevation than calculated.

Therefore, the provided correct-looking choice among the options is 101.4°C, aligning with the calculated boiling point elevation within rounding conventions.

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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