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(I) Aman prepared a… (Class 12 Chemistry SQP 2026-27, Q25)

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ChemistryCBSE Class 12 · Sample paper, session 2026-27 · 3 marks · Short answer

CBSE Class 12 Chemistry sample paper 2026-27: question 25

From the official CBSE sample question paper for 2026-27, section C, question 25, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q25 · 3 marks · Short answerSolutions

(I) Aman prepared a solution by dissolving 10 ml of Acetone and 10 ml of Carbon disulphide. The net volume of resulting solution was not found to be 20 mL.
a) What would be the observation in this case? Comment.
b) In case carbon disulphide is replaced by chloroform in the above case, what would be the expected changes in volume of the resulting solution? Support your answer with appropriate reason. (1+1)

(II) Calculate the concentration of oxygen (in ppm) present in 1 litre of sea water. Given that a litre of sea water weighs 1030 g and contains about g of dissolved oxygen. (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

Compare the solute–solvent attractions with those in the pure liquids.

Hint 2

Stronger new attractions shrink the volume; weaker ones expand it.

Hint 3

ppm = (mass of component / total mass) × 10⁶.

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. (I)(a) The volume of the solution is slightly more than 20 mL ().

    ½ mark

  2. The acetone–CS₂ attractions are weaker than the acetone–acetone and CS₂–CS₂ attractions. Molecules move apart, so volume increases (positive deviation from Raoult's law).

    ½ mark

  3. (I)(b) With chloroform, the volume is slightly less than 20 mL ().

    ½ mark

  4. Chloroform forms hydrogen bonds with acetone (). The stronger A–B attractions pull molecules closer, so volume decreases (negative deviation).

    ½ mark

  5. (II) ppm

    ½ mark

  6. ppm

    ½ mark

Answer: (I)(a) Volume > 20 mL (weaker interactions, positive deviation); (b) volume < 20 mL (H-bonding, negative deviation); (II) ≈ 5.8 ppm

Where the marks are

  • Volume more than 20 mL (½ mark)
  • Reason: weaker acetone–CS₂ interactions (½ mark)
  • Volume less than 20 mL with chloroform (½ mark)
  • Reason: H-bonding between chloroform and acetone (½ mark)
  • ppm formula with substitution (½ mark)
  • 5.8 ppm (½ mark)

Revise the chapter

  • Chapter: Solutions (notes, formulas, where marks are lost and every past question)
Practise this chapterBoard-style questions from Solutions, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2026-27: every question, by section.

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.