What is the boiling point of a solution of 0.1 mole of glucose in 200 mL of water? (K = 0.512C/m)

Question

BlackTom AI · Accepted Answer

To approach boiling-point elevation, I start by identifying the key relation: ΔTb = i · Kb · m. Here, the solute is glucose, which does not ionize in solution, so i = 1. Next, I need the molality m, which is moles of solute per kilogram of solvent. We have 0.1 moles of glucose in 200 mL of water. Assuming the density of water is close to 1 g/mL, 200 mL corresponds to about 200 g, i.e., 0.200 kg of solvent. Therefore, m = 0.1 mol / 0.200 kg = 0.5 m.

Now compute the boiling-point elevation: ΔTb = (1) · (0.512 °C·m) · (0.5 m) = 0.256 °C. The normal boiling point of water is 100.0 °C, so the new boiling point is 100.0 °C + 0.256 °C ≈ 100.256 °C. Depending on rounding, this is commonly reported as 100.26 °C.

Evaluating the answer choices:
- 100.06 °C: This would correspond to a ΔTb of about 0.06 °C, which is far smaller than the calculated 0.256 °C, implying a much smaller molality or an incorrect Kb value. This does not fit the given data.
- 100.13 °C: This implies a ΔTb around 0.13 °C, still significantly less than the expected 0.256 °C; the molality or Kb used would have to be roughly half of the actual calculated values, which is inconsistent with the provided Kb and solute amount.
- 100.26 °C: This matches the computed Boiling point, given the calculated ΔTb of about 0.256 °C and the standard water bp of 100.0 °C. This is the correct scale for the precision used here.
- 100.5 °C: A ΔTb of 0.5 °C would require either a larger molality or a larger Kb than given; with m = 0.5 m and Kb = 0.512 °C/m, the elevation is about 0.256 °C, not 0.5 °C, so this option is inconsistent with the data.

In summary, by applying the colligative property formula and plugging in the provided values, the computed boiling point shift aligns with the 100.26 °C option, while the other values do not correspond to the calculated ΔTb using i = 1, Kb = 0.512 °C/m, and m = 0.5.

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