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ChemistryCBSE Class 12 · Board exam 2025, set 56/1/1 · 3 marks · Short answer

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.

  1. Elevation in boiling point: °C, i.e. 0.20 K.

    ½ mark
  2. Using :

    ½ mark
  3. mol kg

    ½ mark
  4. Depression in freezing point: K

    1 mark
  5. Freezing point of the solution: °C (about °C, or 272.42 K)

    ½ mark

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

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.