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Answer the following. (Science 2025 31/1/1 Q34b)

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ScienceCBSE Class 10 · Board exam 2025, set 31/1/1 · 5 marks · Long answer

CBSE Class 10 Science 2025: question 34(b)

Set in the CBSE Class 10 board exam of 2025, paper 31/1/1, section D, question 34(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(b) · 5 marks · Long answerCarbon and its Compounds

Answer the following.

  1. (i)

    A carbon compound X is a good solvent. On reaction with sodium, X forms two products Y and Z. Z is used to convert vegetable oil into vegetable ghee. Identify and name X, Y and Z. Also write the equation of reaction of X with sodium to justify your answer.

    0 marks
  2. (ii)

    Write chemical equation to show what happens when ethanol: (I) burns in oxygen/air; (II) is heated at 443 K in excess conc. ; (III) reacts with acidified potassium dichromate.

    0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 31/1/1.

Hints

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

Hint 1

Which gas is used to hydrogenate oils into ghee? That gives you Z.

Hint 2

A carbon compound that is a good solvent and releases hydrogen with sodium is a common alcohol.

Hint 3

For (ii), think combustion, dehydration and oxidation of ethanol in turn.

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) X is a good solvent and reacts with sodium, so it must be an alcohol. **X = Ethanol (ethyl alcohol), **.

    ½ mark

  2. When sodium reacts with ethanol it forms the salt **Y = Sodium ethoxide, **.

    ½ mark

  3. The gas given off is **Z = Hydrogen, **. Hydrogen is used, with a nickel catalyst, to hydrogenate vegetable oil into vegetable ghee. This confirms Z.

    ½ mark

  4. Equation to justify: (X → Y + Z).

    ½ mark

  5. (ii)(I) Combustion: ethanol burns in oxygen with a clean blue flame. + heat + light.

    1 mark

  6. (ii)(II) Dehydration: hot concentrated sulphuric acid at 443 K removes water from ethanol and gives ethene. .

    1 mark

  7. (ii)(III) Oxidation: acidified potassium dichromate oxidises ethanol to ethanoic acid. . The orange dichromate turns green.

    1 mark

Answer: (i) X = ethanol (C₂H₅OH), Y = sodium ethoxide (C₂H₅ONa), Z = hydrogen (H₂); 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂. (ii)(I) C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O; (II) C₂H₅OH → C₂H₄ + H₂O (hot conc. H₂SO₄, 443 K); (III) C₂H₅OH + 2[O] → CH₃COOH + H₂O (acidified K₂Cr₂O₇)

Where the marks are

  • Identifies X as ethanol (½ mark)
  • Identifies Y as sodium ethoxide (½ mark)
  • Identifies Z as hydrogen gas (½ mark)
  • Writes equation of ethanol with sodium (½ mark)
  • Equation for combustion of ethanol giving CO₂ and H₂O (1 mark)
  • Equation for dehydration of ethanol to ethene with conc. H₂SO₄ at 443 K (1 mark)
  • Equation for oxidation of ethanol to ethanoic acid by acidified K₂Cr₂O₇ (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from Carbon and its Compounds, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 31/1/1: 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.