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.
- ½ mark
1 molal means 1 mol MgCl in 1000 g water: mol and mol.
- ½ mark
Dissociation: , so ions and .
- ½ mark
- ½ mark
Raoult's law with van't Hoff factor:
- ½ mark
- ½ mark
mm Hg
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
- Chapter: Solutions (notes, formulas, where marks are lost and every past question)
- Topic: Abnormal Molar Mass and van't Hoff Factor
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