Case Study
Passage with linked questions
Case Set 1
Case AnalysisPassage
A chemistry teacher demonstrates purification techniques to Class 11 students. She takes a mixture of naphthalene and sand and explains that naphthalene, when heated, directly converts from solid to vapour without passing through the liquid state. This property is called sublimation. She sets up a china dish containing the mixture, places a punctured filter paper over it, and puts an inverted funnel on top. On gentle heating, white crystals of pure naphthalene collect on the inner surface of the funnel, while sand remains in the dish. She then tells the class that the purity of the isolated naphthalene can be checked by determining its melting point, which should be sharp and match the known value for pure naphthalene.
Question 1: What is sublimation and which property of naphthalene makes it suitable for purification by this technique?
- Sublimation is the process by which certain solid substances directly convert from the solid state to the vapour state on heating, without passing through the liquid state.
- Naphthalene is sublimable, meaning it has sufficient vapour pressure below its melting point to pass directly to the vapour phase; this property separates it from non-sublimable sand.
- On cooling, the vapours of naphthalene condense back to the solid state (desublimation) on the cooler surface of the funnel, yielding pure naphthalene crystals.
Question 2: How does determining the melting point help confirm the purity of the isolated naphthalene?
- Pure organic compounds have sharp, well-defined melting points; the temperature remains constant throughout the melting process when the compound is pure.
- Impurities lower the melting point and broaden the melting range (the compound begins to melt at a lower temperature and melts over a range rather than at a single temperature).
- If the isolated naphthalene melts sharply at the known value for pure naphthalene (80 degrees C), purity is confirmed; a depressed or broad melting point indicates residual impurities.
Question 3: The teacher now has a mixture of two sublimable organic compounds P and Q with different vapour pressures. Explain why simple sublimation would not effectively separate them and suggest a better technique with justification.
- Simple sublimation relies on one component being sublimable and the other not; if both P and Q are sublimable, both will convert to vapour on heating and condense together, resulting in no separation.
- The mixture would need a technique that separates components based on a different physical property; chromatography (column chromatography or TLC) would be appropriate, exploiting the difference in adsorption affinity of P and Q on an adsorbent like silica gel.
- Alternatively, if P and Q have sufficiently different vapour pressures, sublimation at carefully controlled low temperatures might selectively sublime the more volatile component first, but this requires precise temperature control.