CBSE Class 12 Chemistry 2025: question 22
Set in the CBSE Class 12 board exam of 2025, paper 56/1/1, section C, question 22, for 3 marks. Try it first; the hints and our step-by-step solution are below.
Q22 · 3 marks · Short answerSolutions
A solution of glucose (molar mass = 180 g mol) in water has a boiling point of 100.20 °C. Calculate the freezing point of the same solution. Molal constants for water and are 1.86 K kg mol and 0.512 K kg mol respectively.
Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 56/1/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
Find the molality first, from the boiling point elevation.
Hint 2
The same molality applies to freezing point depression; the molar mass is not needed.
Hint 3
Use ΔTf = Kf·m, then subtract ΔTf from 0 °C.
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- ½ mark
Elevation in boiling point: °C, i.e. 0.20 K.
- ½ mark
Using :
- ½ mark
mol kg
- 1 mark
Depression in freezing point: K
- ½ mark
Freezing point of the solution: °C (about °C, or 272.42 K)
Answer: Freezing point ≈ −0.726 °C (≈ 272.42 K)
Where the marks are
- ΔTb = 0.20 K (½ mark)
- Uses ΔTb = Kb·m (½ mark)
- Molality = 0.390 mol/kg (½ mark)
- ΔTf = Kf·m = 0.726 K (1 mark)
- Tf ≈ −0.726 °C (½ mark)
Revise the chapter
- Chapter: Solutions (notes, formulas, where marks are lost and every past question)
- Topic: Elevation of Boiling Point and Depression of Freezing Point
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