Assertion (A): Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97°C) and pr

NCERT Exemplar · Q55

Q. Assertion (A): Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97°C) and propanone (boiling point 56°C). Reason (R): Liquids with a difference of more than 20°C in their boiling points can be separated by simple distillation.

(i) Both A and R are correct and R is the correct explanation of A.

(ii) Both A and R are correct but R is not the correct explanation of A.

(iii) Both A and R are not correct.

(iv) A is not correct but R is correct.

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Concept understanding — Separation Techniques

Separation Techniques: The Intuition

Imagine you're making chai. You boil tea leaves in water, then pour the liquid through a strainer. The strainer catches the leaves; the tea flows through. You just performed a separation technique — you took a mixture (tea leaves + water) and isolated one component (the liquid tea) from the other (the solid leaves).

Every separation technique answers one question: How do I pull apart things that are mixed together, using their differences?

The key insight: You cannot separate things that are identical. Separation works only because the components of a mixture differ in at least one physical or chemical property — size, density, boiling point, solubility, magnetic behaviour, or something else.

Note

Separation techniques are physical processes. They do not change the chemical identity of the substances. The tea leaves remain tea leaves; the water remains water. No chemical reaction occurs.

The Precise Statement

Separation techniques are methods used to isolate individual components from a mixture by exploiting differences in their physical or chemical properties. The goal is to obtain one or more pure substances from a mixture, without altering the chemical nature of the components.

The Core Idea in One Sentence

Important

Every separation technique works because the components of the mixture differ in at least one measurable property. The method you choose depends entirely on which property differs.

A Quick Map of Common Separation Techniques

Here is how the choice of technique follows from the property difference:

Property Difference	Technique	Example
Particle size	Sieving, Filtration	Separating stones from sand; tea leaves from water
Density	Centrifugation, Decantation, Sedimentation	Separating cream from milk; mud from water
Boiling point	Distillation	Separating alcohol from water; crude oil into fractions
Solubility	Evaporation, Crystallisation	Obtaining salt from seawater
Magnetic property	Magnetic separation	Separating iron filings from sulphur powder
Volatility	Sublimation	Separating camphor from sand
Attraction to a stationary phase	Chromatography	Separating pigments in ink or plant leaves

How to Think About Any Separation Problem

When you see a mixture and need to decide how to separate it, ask yourself:

What are the components? (List them.)
What property is different between them? (Size? Density? Boiling point? Solubility? Magnetism?)
Which technique exploits that difference? (Match the property to the method above.)

That three-step chain — mixture → property difference → technique — is the entire logic of separation.

A Concrete Walkthrough

Mixture: Sand and iron filings.

Components: Sand (silicon dioxide) and iron (metal).
Property difference: Iron is magnetic; sand is not.
Technique: Magnetic separation. Pass a magnet over the mixture. Iron filings stick to the magnet; sand is left behind.

Mixture: Salt dissolved in water.

Components: Salt (sodium chloride) and water.
Property difference: Water boils at $100^\circ\text{C}$ and evaporates; salt does not boil or evaporate — it remains as a solid.
Technique: Evaporation (or distillation, if you also want to collect the water). Heat the solution. Water turns to vapour and leaves; salt crystals remain in the dish.

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Simple distillation works when the boiling-point difference is large enough — here the difference is $41^\circ\text{C}$ , so both Assertion and Reason are correct, and the Reason correctly explains the Assertion.

The key idea is that simple distillation separates liquids based on boiling-point differences. When one component vaporises much more readily than the other, the vapour can be condensed and collected as a pure distillate. The rule of thumb — a difference of more than $20^\circ\text{C}$ — ensures negligible overlap in boiling ranges, so the separation is clean.

Let’s check the numbers. Propanone boils at $56^\circ\text{C}$ and propan-1-ol at $97^\circ\text{C}$ . The difference is $97 - 56 = 41^\circ\text{C}$ , which is well above $20^\circ\text{C}$ . So the mixture satisfies the condition for simple distillation.

Why the Reason is correct.

The statement in R is a standard guideline: if two liquids differ by more than $20^\circ\text{C}$ in boiling point, simple distillation is sufficient. This is because the more volatile component (lower boiling point) will vaporise almost completely before the other begins to boil significantly. No fractional distillation column is needed.
Why the Assertion is correct.

Propanone (bp $56^\circ\text{C}$ ) is far more volatile than propan-1-ol (bp $97^\circ\text{C}$ ). On heating the mixture, propanone vaporises first and can be condensed and collected, leaving propan-1-ol behind. The $41^\circ\text{C}$ gap makes this a straightforward separation.
Connecting A and R. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
$Q. Which of the following is not related to purification of colloidal solutions? (A) Electrophoresis (B) Dialysis (C) Ultrafiltration (D) Electro dialysis$
›Reveal solutionSolution
This tests recognising which colloid technique is NOT a purification method; the answer is Electrophoresis.

Concept and Intuition

Colloidal sols are purified by removing dissolved electrolyte impurities that would otherwise destabilise them. Dialysis (using a semi-permeable membrane to let ions diffuse out), ultrafiltration (using a specially prepared filter paper to remove ions along with solvent, then re-dispersing), and electrodialysis (dialysis sped up with an applied electric field) are all purification methods. Electrophoresis, in contrast, is about the MOTION of colloidal particles themselves under an electric field — used to study/determine the charge on colloidal particles, not to purify them.

Step-by-Step Solution

Recall the standard purification methods for colloids taught in NCERT: dialysis, electrodialysis, ultrafiltration. …
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AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$Q. A mixture (X) contains two liquids with large difference in their boiling points. Another mixture (Y) contains two liquids with not much difference in their boiling points. Mixtures X and Y can be separated respectively by the methods (A) Simple distillation, Steam distillation (B) Simple distillation, Fractional distillation (C) Fractional distillation, Simple distillation (D) Steam distillation, Fractional distillation$
›Reveal solutionSolution
The choice of distillation method depends on how close the boiling points of the two liquids are: a large gap needs only simple distillation, while a small gap needs fractional distillation's repeated vaporisation-condensation cycles for good separation.

Concept and Intuition

Distillation separates miscible liquids based on differences in volatility (boiling point).

When boiling points are far apart, a single vaporisation–condensation step is enough: heat the mixture, the lower-boiling liquid vaporises and distils over first while the higher-boiling one is left behind largely undisturbed. This is simple distillation — used for mixture X.

When boiling points are close together, a single vaporisation step gives a vapour that's still a substantial mixture of both liquids (their vapour pressures at any given temperature are too similar). To get a clean separation, the vapour must be repeatedly re-condensed and re-vaporised as it rises through a fractionating column packed with surfaces that provide many theoretical plates — each equivalent to a fresh simple distillation. This is fractional distillation — used for mixture Y.

(Steam distillation, by contrast, is used for separating high-boiling substances that would decompose at their own boiling point, by co-distilling with steam at a lower effective temperature — it isn't the general tool for "large boiling point gap" mixtures as such.)

Step-by-Step Solution

Mixture X: large $\Delta T_b$ between the two liquids → simple distillation suffices. …
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AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$Q. Which method of purification is used for refining titanium? (A) Zone refining (B) Van Arkel method (C) Mond process (D) Liquation$
›Reveal solutionSolution
Titanium is purified by the Van Arkel–de Boer (iodide) method, which converts the crude metal to a volatile iodide and decomposes it on a hot filament.

Concept and Intuition

Different refining methods exploit different physical/chemical properties: zone refining exploits differing solubility in melt vs solid (for very high-purity semiconductors), Mond process exploits volatile carbonyl formation (for nickel), liquation exploits low melting point (for tin), and the Van Arkel method exploits the volatility of a metal halide (for reactive high-melting metals like Ti and Zr).

Step-by-Step Solution

Crude/impure titanium is heated with iodine to form volatile titanium(IV) iodide, $TiI_4$ .

The vapour of $TiI_4$ is passed over a red-hot tungsten filament (about 1700 K). …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$Q. Match the following List - I (Metal refining method) / List - II (Principle) A. Zone refining / I. Formation and decomposition of volatile compound B. Liquation / II. Difference in boiling points of the metal and impurities C. Vapour phase refining / III. Difference in solubilities of impurities in the molten and solid state of metal D. Distillation / IV. Difference between the melting point of the metal and the impurities The correct answer is (A) A-III, B-II, C-I, D-IV (B) A-III, B-IV, C-I, D-II (C) A-I, B-IV, C-II, D-III (D) A-II, B-III, C-I, D-IV$
›Reveal solutionSolution
Matching each refining method to its underlying physical principle: zone refining→solubility difference (III), liquation→melting-point difference (IV), vapour-phase refining→volatile-compound formation/decomposition (I), distillation→boiling-point difference (II).

Concept and Intuition

Each classical metal-refining method in metallurgy exploits one distinct physical/chemical property difference between the pure metal and its impurities. Recognizing which property each named method relies on is the whole question.

Step-by-Step Solution

Zone refining: a molten zone is moved along a rod of impure metal; impurities are more soluble in the melt than in the solid, so they get swept along with the moving molten zone, leaving purer solid behind. Principle = difference in solubilities of impurities in molten and solid states → III.

Liquation: the impure metal is heated on a sloping hearth just above its own melting point but below that of the impurities; the (lower-melting) metal melts and flows away, leaving the (higher-melting) impurities behind. Principle = difference in melting points of metal and impurities → IV.

Vapour-phase refining: the impure metal is converted to a volatile compound (e.g. Ni + CO → Ni(CO)₄ in the Mond process), which is then decomposed elsewhere to deposit the pure metal. Principle = formation and decomposition of a volatile compound → I. …
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AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
$\mathrm{Zr(Impure) + 2I_2 \rightarrow ZrI_4}$
›Reveal solutionSolution
The two-step iodide-formation-then-thermal-decomposition sequence shown is the textbook van Arkel method for refining metals like zirconium and titanium. The answer is (C).

Concept and Intuition

Different refining methods exploit different chemistry for different impurity/metal combinations. The van Arkel (iodide) method is specifically used for metals that form a volatile halide which can later be thermally decomposed to release the pure metal, leaving impurities behind (since impurities typically don't form the same volatile halide or decompose differently). This is distinct from zone refining (uses differential solubility of impurities in molten vs solid metal), the Mond process (uses volatile nickel carbonyl, specific to nickel), and electrolytic refining (uses an electrochemical cell with the impure metal as anode).

Step-by-Step Solution

Reaction I: impure Zr reacts with iodine vapour at a moderately low temperature to form volatile zirconium tetraiodide, $Zr(\text{impure}) + 2I_2 \rightarrow ZrI_4$ . Only zirconium (not its impurities) forms this volatile iodide, so this step purifies by selective volatilisation.

Reaction II: the $ZrI_4$ vapour is passed over a hot tungsten filament at about 1800 K, where it thermally decomposes back to pure zirconium metal (deposited on the filament) and iodine gas, which is recycled: $ZrI_4 \xrightarrow{1800\ K} Zr(\text{pure}) + 2I_2$ . …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$Q. What is the principle involved in froth floatation process? (A) Ore particles are heavier than gangue (B) Gangue particles are adsorbed by pine oil (C) Ore particles are preferentially wetted by oil and gangue by water (D) Ore particles are preferentially wetted by water and gangue by oil$
›Reveal solutionSolution
Froth flotation separates sulphide ore from gangue based on differential wetting by oil versus water.

Concept and Intuition

In froth flotation, a collector (like pine oil) is added to a water suspension (pulp) of the powdered ore, and air is blown through it. The sulphide ore particles are preferentially wetted by the oil, making their surfaces hydrophobic so they attach to air bubbles and rise as froth; the gangue (unwanted rocky matter) remains wetted by water (hydrophilic) and sinks.

Step-by-Step Solution

Identify the process: froth flotation, used mainly for concentrating sulphide ores.

The collector (oil, e.g. pine oil, xanthates) preferentially coats/wets the ore particles, making them water-repellent (hydrophobic).

Gangue particles remain wetted by water (hydrophilic), so they do not attach to air bubbles.

Air bubbles blown through the pulp attach to the oil-wetted (hydrophobic) ore particles, carrying them up as froth, which is skimmed off; gangue sinks to the bottom. …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$Q. Match the following List-I (Compound) A) Kieselghur B) Silica gel C) ZSM-5 D) Hydrated zeolites List-II (Use) I) Chromotographic material II) Softening of hard Water III) Filtration plants IV) To convert alcohol directly into gasoline The correct answer is (A) A-IV, B-III, C-II, D-I (B) A-IV, B-I, C-II, D-III (C) A-III B-IV, C-I, D-II (D) A-III, B-I, C-IV, D-II$
›Reveal solutionSolution
This tests recall of specific industrial/lab uses of silica-based materials: Kieselghur → filtration, silica gel → chromatography, ZSM-5 → alcohol-to-gasoline conversion, hydrated zeolites → water softening. Answer: (D).

Concept and Intuition

Several materials based on silica or aluminosilicates look similar chemically but are chosen industrially for very different physical/chemical properties — porosity for filtration, surface adsorption for chromatography, pore geometry for shape-selective catalysis, and ion-exchange capacity for softening water. Recognising which property is being exploited pins down the match.

Step-by-Step Solution

Kieselghur is a soft, porous, sedimentary rock made of fossilised diatom (silica) shells. Its high porosity and inertness make it the classic filter-aid used in filtration plants → matches III.

Silica gel is amorphous, highly porous SiO₂ with a huge surface area — the standard stationary-phase adsorbent used in column/TLC chromatography (also a desiccant) → matches I.

ZSM-5 is a synthetic zeolite with a specific pore size that is shape-selective; it is industrially used to dehydrate/convert alcohols directly into a mixture of hydrocarbons in the gasoline range → matches IV. …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$Q. Which of the following sets are correctly matched? Metal Refining process I) Hg distillation II) Cu poling III) B zone refining IV) Ti liquation (A) I, III & IV only (B) I, II & III only (C) II, III & IV only (D) I, II, III & IV$
›Reveal solutionSolution
Checking each metal–refining-process pair against standard extraction chemistry: Hg-distillation, Cu-poling, and B-zone refining are all correct, but Ti is refined by the van Arkel method, not liquation. Answer: (B).

Concept and Intuition

Each refining technique exploits a specific physical or chemical property of the impure metal: volatility (distillation), selective oxidation of an impurity (poling), differential solubility of impurity in a melt front (zone refining), or a low melting point relative to impurities (liquation). Matching the technique to the metal requires knowing which property is actually exploited for that metal.

Step-by-Step Solution

Hg – distillation (I): mercury has a very low boiling point and is quite volatile, so it is purified by simple distillation. Correct.

Cu – poling (II): impure (blister) copper containing Cu₂O is melted and stirred with green wood poles; the hydrocarbon gases released reduce Cu₂O back to Cu, removing oxygen impurity. Correct.

B (boron) – zone refining (III): boron (like germanium and silicon) is refined to ultra-high purity for semiconductor-grade use via zone refining, where a molten zone is passed along a rod and impurities concentrate in the melt. Correct. …
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AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$Q. Which method is used to purify liquids having very high boiling points and liquids which decompose at or below their boiling point? (A) Distillation (B) Fractional distillation (C) Distillation under reduced pressure (D) Steam distillation$
›Reveal solutionSolution
Reducing the external pressure lowers a liquid's boiling point, letting high-boiling or thermally unstable liquids vaporise before they decompose. The answer is (C) Distillation under reduced pressure.

Concept and Intuition

A liquid boils when its vapour pressure equals the surrounding (external) pressure. Boiling point is therefore not a fixed property of the substance alone — it depends on pressure. If we mechanically reduce the pressure above the liquid (using a vacuum pump), the liquid's vapour pressure reaches that lower external pressure at a much lower temperature, so it boils and distills off at a temperature well below its normal (1 atm) boiling point. This is exactly what is needed for:

Liquids whose normal boiling point is very high (so reaching it would waste energy or be impractical), and

Liquids that chemically decompose (oxidise, polymerise, char) at or before their normal boiling point is reached.

Step-by-Step Solution

Recall that ordinary (simple) distillation requires heating the liquid up to its normal boiling point at atmospheric pressure — unsuitable if that temperature causes decomposition.

Fractional distillation separates liquids of close boiling points using a fractionating column — it does not address the decomposition-at-high-temperature problem.

Steam distillation is used for substances that are steam-volatile and immiscible with water (lowers effective boiling temperature via partial pressure with steam) — it is a different technique, mainly for water-insoluble, steam-volatile organics, not generally for very-high-boiling pure liquids. …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$Q. Which of the following methods is useful for producing semiconductor grade metals of high purity ? (A) Liquation (B) Vapour phase refining (C) Electrolytic refining (D) Zone refining$
›Reveal solutionSolution
Zone refining exploits impurities being more soluble in the liquid phase than the solid phase of a metal, letting a travelling molten zone sweep impurities to one end — the method used industrially to get ultra-pure, semiconductor-grade silicon and germanium.

Concept and Intuition

Semiconductor applications need metals of extraordinarily high purity (parts-per-billion level), far beyond what electrolytic or chemical refining alone can achieve. Zone refining works because when a narrow molten band moves along a rod of the impure metal, impurities preferentially stay dissolved in the liquid rather than the freshly re-solidifying solid behind it, so repeated passes of the moving heater concentrate impurities at one end, leaving the rest ultra-pure.

Step-by-Step Solution

Liquation: separates a low-melting metal from higher-melting impurities by simply melting it off a sloped hearth — gives moderate purity, not semiconductor grade.

Vapour phase refining (e.g. Mond process, van Arkel method): converts the metal to a volatile compound and decomposes it back to pure metal — used for Ni, Ti, Zr, but not the standard route for Si/Ge.

Electrolytic refining: purifies metals like Cu via electrodeposition — good for 99.9%+ purity, not the extreme purity semiconductors need. …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$Q. From the following identify pairs in which metal is correctly matched with its refining process A. Zn - Distillation B. Sn - Liquation C. Ga - Zone refining D. Zr - Vapour phase refining (A) A, B, C only (B) B, D only (C) A, C, D only (D) A, B, C, D$
›Reveal solutionSolution
This tests memorized metal-to-refining-method pairings from the extraction-of-metals chapter.

Concept and Intuition

Different physical/chemical properties of metals call for different purification methods: volatility for distillation, low melting point for liquation, extreme purity requirements for zone refining, and formation of a volatile compound for vapour-phase refining.

Step-by-Step Solution

Zn — Distillation: zinc has a comparatively low boiling point (~907 °C), so impure zinc can simply be distilled away from higher-boiling impurities. Correct.

Sn — Liquation: tin has a low melting point; heating impure tin on a sloping hearth lets pure tin melt and flow away, leaving high-melting impurities behind. Correct.

Ga — Zone refining: gallium, like germanium, silicon, boron and indium, needs extremely high purity (semiconductor grade) and is refined by passing a moving molten zone along a rod so impurities concentrate in the melt and get swept to one end. Correct. …
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AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
$R_f$
›Reveal solutionSolution
The retardation factor $R_f$ is the ratio of the distance travelled by a spot to the distance travelled by the solvent front; here spot B travels 8 cm while the solvent front travels 12 cm, giving $R_f=2/3$ .

Concept and Intuition

In paper chromatography, each component of a mixture travels a characteristic fraction of the distance the solvent front travels, called the retardation factor $R_f = \dfrac{\text{distance moved by substance}}{\text{distance moved by solvent}}$ . It is always between 0 and 1 and is used to identify/compare components.

Step-by-Step Solution

Solvent front travels from the baseline (0 cm) to 12 cm.

Spot B travels from the baseline to 8 cm. …
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Q.Assertion (A): Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97°C) and propanone (boiling point 56°C). Reason (R): Liquids with a difference of more than 20°C in their boiling points can be separated by simple distillation.

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Q.Assertion (A): Simple distillation can help in separating a mixture of propan-1-ol (boiling point 97°C) and propanone (boiling point 56°C).
Reason (R): Liquids with a difference of more than 20°C in their boiling points can be separated by simple distillation.

Unlock everything free for 14 days