Class 10 Science Chapter 4: Carbon and its Compounds case study question with answers
One case-based question of the kind the board asks in Section E: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
| Compound | Formula | Molecular mass (u) | Boiling point (K) |
|---|---|---|---|
| Methane | CH₄ | 16 | 112 |
| Ethane | C₂H₆ | 30 | 185 |
| Propane | C₃H₈ | 44 | 231 |
| Butane | C₄H₁₀ | 58 | 273 |
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Answer: (a) C₅H₁₂, 72 u
The series is CₙH₂ₙ₊₂; n = 5 gives C₅H₁₂, and adding –CH₂– (14 u) to 58 u gives 72 u. C₆H₁₄ is two steps on.
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Answer: (d) butane
Only butane has a boiling point (273 K) above 268 K, so only butane condenses; the other three boil far below 268 K and stay gases.
Answer and marking scheme
The four form a homologous series: each differs from the next by –CH₂– (14 u). As molecular mass increases along the series, the boiling point increases. Their chemical properties are similar because all have the same kind of structure, saturated chains with only single bonds, and in a homologous series chemical properties stay similar while physical ones change gradually.
- boiling point rises with molecular mass along the homologous series (14 u steps) (1)
- same kind of compound (saturated, same series), so chemical properties stay similar (1)
Answer and marking scheme
(i) CH₄ + 2O₂ → CO₂ + 2H₂O + heat and light. (ii) CH₄ + Cl₂ → CH₃Cl + HCl (in sunlight).
- (i) CH₄ + 2O₂ → CO₂ + 2H₂O, balanced (1)
- (ii) CH₄ + Cl₂ → CH₃Cl + HCl, with sunlight (1)
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