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ChemistryCBSE Class 12 · Sample paper, session 2025-26 · 3 marks · Short answer

CBSE Class 12 Chemistry sample paper 2025-26: question 22

From the official CBSE sample question paper for 2025-26, 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

What will be the vapour pressure of a 1 molal aqueous solution of , assuming dissociation of to be 70 mole percent? (Vapor pressure of pure water at 25 C is 23.8 mmHg)

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

Open one hint at a time and try again before the next.

Hint 1

Find i first, using α = (i − 1)/(n − 1) with the number of ions formed by MgCl₂.

Hint 2

Then use relative lowering of vapour pressure = i × mole fraction of solute, with 1000 g water = 55.5 mol.

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. 1 molal means 1 mol MgCl in 1000 g water: mol and mol.

    ½ mark
  2. Dissociation: , so ions and .

    ½ mark
  3. ½ mark
  4. Raoult's law with van't Hoff factor:

    ½ mark
  5. ½ mark
  6. mm Hg

    ½ mark

Answer: Vapour pressure of the solution ≈ 22.8 mm Hg

Where the marks are

  • Moles of solute (1) and water (55.5) (½ mark)
  • Dissociation equation with n = 3 (½ mark)
  • Calculates i = 2.4 (½ mark)
  • Writes relative lowering formula including i (½ mark)
  • Correct substitution (½ mark)
  • Final Ps ≈ 22.8 mm Hg (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.