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A. An experiment… (Class 12 Chemistry SQP 2024-25, Q33a)

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ChemistryCBSE Class 12 · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 12 Chemistry sample paper 2024-25: question 33(a)

From the official CBSE sample question paper for 2024-25, section E, question 33(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(a) · 5 marks · Long answerSolutions

a. An experiment was carried out in the laboratory, to study depression in freezing point. 1M aqueous solution of and 1 M aqueous solution of glucose were taken. From the given figure identify solution 1 and solution 2. Give a plausible reason for your answer. (2)

b. The osmotic pressure of a solution of cane sugar was found to be 2.46 atm at 300 K. If the solution was diluted five times, calculate the osmotic pressure at the same temperature.
How can the osmotic pressure of the given cane sugar solution be decreased without changing its volume? Give a reason for your answer. (3)

Figure in the paper: Graph of vapour pressure (y-axis, with marks p, p1, p2) versus temperature (x-axis, with marks T2, T1, Tf). Curves for Solvent (through points A, B, C), Solution 1 (through D, F) and Solution 2 (through E, C) lie below the solvent curve; Solution 2 lies lowest, showing greater lowering of vapour pressure and larger freezing-point depression. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

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

Hint 1

The curve that meets the solid-solvent line at the lowest temperature has the largest ΔTf. Which solute gives more particles per formula unit?

Hint 2

In π = (n/V)RT, what happens to π when V becomes five times larger while n and T stay the same?

Hint 3

For the last part, V and n are fixed. Which remaining variable in π = CRT can you change?

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. a. In the graph, Solution 2 has the lowest vapour-pressure curve. It therefore meets the solid-solvent curve at the lowest temperature, giving the larger depression in freezing point. So **Solution 2 = 1 M and Solution 1 = 1 M glucose**.

    1 mark

  2. Reason: depression in freezing point is a colligative property, .

    • , so for complete dissociation.
    • Glucose does not dissociate, so .

    At the same molality, gives more solute particles and hence the larger .

    1 mark

  3. b. Use . At constant and fixed , , so .

    ½ mark

  4. Diluting five times gives . Then

    ½ mark

  5. 1 mark

  6. To lower the osmotic pressure without changing the volume, decrease the temperature. Since , at fixed concentration is directly proportional to the absolute temperature .

    1 mark

Answer: a. Solution 1 = glucose, Solution 2 = Al(NO₃)₃ (larger i, larger ΔTf). b. π = 0.492 atm; lower it by decreasing the temperature, since π ∝ T.

Where the marks are

  • Identifies Solution 2 as Al(NO₃)₃ and Solution 1 as glucose from the larger freezing-point depression (1 mark)
  • Reason based on ΔTf = iKf m and the higher van't Hoff factor of Al(NO₃)₃ (1 mark)
  • Uses π = (n/V)RT, i.e. π ∝ 1/V at constant T (½ mark)
  • Substitutes V₂ = 5V₁ (½ mark)
  • Gets π₂ = 0.492 atm (1 mark)
  • States that lowering the temperature lowers π because π ∝ T (1 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 2024-25: 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.