CBSE Class 12 Chemistry sample paper 2026-27: question 28
From the official CBSE sample question paper for 2026-27, section C, question 28, for 3 marks. Try it first; the hints and our step-by-step solution are below.
Q28 · 3 marks · Short answerSolutions
(I) On adding 100 g of non-volatile solute A in 1000 g of water, the vapour gets reduced by 50 percent. Calculate the molar mass of solute A. (2)
(II) It has been observed that nitric acid solution containing 32% of water by mass, cannot be separated into its components through fractional distillation. Why? (1)
Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.
Hints
Open one hint at a time and try again before the next.
Hint 1
Relative lowering of vapour pressure equals the mole fraction of the solute.
Hint 2
A 50% drop is large, so do not use the dilute approximation; use n_B/(n_A+n_B).
Hint 3
For (II), think of mixtures that boil with constant composition.
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
(I) Relative lowering of vapour pressure:
- ½ mark
mol. A 50% reduction means .
- ½ mark
, so and mol.
- ½ mark
g mol⁻¹
- 1 mark
(II) Nitric acid with 32% water (68% HNO₃) forms a maximum-boiling azeotrope (negative deviation). Its liquid and vapour have the same composition, so it boils at constant temperature and cannot be separated by fractional distillation.
Answer: (I) g mol⁻¹; (II) it is an azeotrope (same composition in liquid and vapour).
Where the marks are
- Raoult's law expression for relative lowering (½ mark)
- Moles of water and value 0.5 (½ mark)
- Solving gives n_B = 55.56 mol (½ mark)
- Molar mass 1.8 g mol⁻¹ (½ mark)
- Azeotrope with identical liquid/vapour composition (1 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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