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A. Nitrogen gas is… (Class 12 Chemistry SQP 2024-25, Q17a)

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ChemistryCBSE Class 12 · Sample paper, session 2024-25 · 2 marks · Very short answer

CBSE Class 12 Chemistry sample paper 2024-25: question 17(a)

From the official CBSE sample question paper for 2024-25, section B, question 17(a), for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q17(a) · 2 marks · Very short answerCompetency-basedSolutions

a. Nitrogen gas is soluble in water. At temperature 293 K, the value of is 76.48 kbar. How would the solubility of nitrogen vary (increase, decrease or remain the same) at a temperature above 293 K, if the value of rises to 88.8 kbar. (1)
b. Chloroform (b.p. 61.2C) and acetone (b.p. 56C) are mixed to form an azeotrope. The mole fraction of acetone in this mixture is 0.339. Predict whether the boiling point of the azeotrope formed will be (i) 60C (ii) 64.5C or (iii) 54C. Defend your answer with reason. (1)

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

Write Henry's law and rearrange it for the mole fraction of the gas.

Hint 2

Does chloroform + acetone show positive or negative deviation? Think of the H-bond between them.

Hint 3

A negative-deviation azeotrope boils above or below both pure liquids?

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) By Henry's law, , so the mole fraction (solubility) of the gas is . Solubility is therefore inversely proportional to .

    ½ mark

  2. Above 293 K, rises from 76.48 kbar to 88.8 kbar. At the same partial pressure, falls, so the solubility of nitrogen decreases.

    ½ mark

  3. (b) The boiling point of the azeotrope will be (ii) 64.5 °C.

    ½ mark

  4. Reason: H-bonding between chloroform and acetone () is stronger than the attractions in the pure liquids. The mixture shows negative deviation from Raoult's law and forms a maximum-boiling azeotrope. Its boiling point is higher than both components (61.2 °C and 56 °C), so only 64.5 °C fits.

    ½ mark

Answer: (a) Solubility of N₂ decreases; (b) (ii) 64.5 °C (maximum-boiling azeotrope, negative deviation)

Where the marks are

  • Relates solubility inversely to K_H (x = p/K_H) (½ mark)
  • Concludes solubility of N₂ decreases as K_H increases (½ mark)
  • Chooses 64.5 °C (½ mark)
  • Reason: negative deviation gives a maximum-boiling azeotrope, so b.p. is above both components (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 2024-25: every question, by section.

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