A graph of vapour pressure (y-axis) against temperature (x-axis) shows two rising curves: the upper one for pure water and a lower one for a solution of 0.50 mol of glucose in 1.00 kg of water. A horizontal line at 1.013 bar cuts the water curve at 373.15 K and the solution curve at a higher temperature Tb. Kb for water is 0.52 K kg mol⁻¹. What is Tb?
Show the answer
The molality is 0.50 mol kg⁻¹, so ΔTb = Kb m = 0.52 × 0.50 = 0.26 K. The solution boils at 373.15 + 0.26 = 373.41 K, which is why its curve meets the 1.013 bar line further right.
- A.
This adds Kb itself (0.52 K) instead of Kb × m = 0.26 K.
- B.
This uses 1.86 K kg mol⁻¹, the freezing point depression constant of water, not Kb.
- D.
A non-volatile solute raises the boiling point; subtracting ΔTb is wrong.
NCERT: Class 12 Chemistry, Chapter 1