NCERT Solutions for Class 9th Science Chapter 5 End-of-chapter questions — Revise, Reflect, Refine

Book page 90 – 93 Updated on2026-09-08

Q1.

Which of the following mixtures are correctly classified as homogeneous (Hm) and heterogeneous (Ht)? Choose the correct option. (i) Air — Hm, Milk — Ht, Sugar solution — Hm, Smoke — Hm (ii) Brass — Ht, Fog — Ht, Vinegar — Ht, Muddy water — Hm (iii) Copper sulfate solution — Hm, Salt solution — Hm, Milk — Hm, Bronze — Hm (iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm

Answer

The correct option is (iv) Muddy water — Ht, Milk — Ht, Blood — Ht, Brass — Hm.

Option	The mistake in it
(i)	Smoke is called Hm. Smoke is solid carbon particles dispersed in air — a heterogeneous colloid.
(ii)	Three mistakes: brass is an alloy and therefore Hm; vinegar is acetic acid dissolved in water and therefore Hm; muddy water is a suspension and therefore Ht.
(iii)	Milk is called Hm. Milk is a colloid — fat droplets dispersed in water — and so is heterogeneous.
(iv)	All four are right: muddy water (suspension) Ht, milk (colloid) Ht, blood (colloid) Ht, brass (alloy) Hm.

Why milk and blood count as heterogeneous: They look uniform to the eye, but they are two-phase mixtures — droplets or cells of one substance dispersed through another, with a real boundary around every particle. Only a mixture that is uniform right down to the level of individual particles, such as a salt solution or a molten-and-cooled alloy, is homogeneous. The Tyndall effect is the practical test: milk and blood scatter a beam of light, a sugar solution does not.

Q2.

Choose the correct options, and explain the reason for the correct and incorrect options. Which among the following mixtures show the Tyndall Effect? A mixture of: (a) air and dust particles (b) copper sulfate and water (c) starch and water (d) acetone and water — (i) a and b (ii) b and d (iii) a and c (iv) c and d

Answer

The correct option is (iii) a and c.

Mixture	Type	Particle size	Tyndall effect?
(a) Air and dust particles	Suspension / aerosol	> 1000 nm	Yes — shows it
(b) Copper sulfate and water	True solution	< 1 nm	No
(c) Starch and water	Colloid	1 – 1000 nm	Yes — shows it
(d) Acetone and water	Solution of two miscible liquids	< 1 nm	No

Why (a) and (c) show it: both contain particles large enough to turn part of a light beam sideways, so the path of the beam becomes visible from the side.

Why (b) and (d) do not: in both, the substances are dissolved to the level of individual ions or molecules, far smaller than the wavelength of light (about 400 – 700 nm). Such particles barely disturb the light wave, so nothing is scattered and the beam passes through unseen. That copper sulfate solution is blue makes no difference — colour comes from absorption, scattering is a different phenomenon altogether.

The rule to remember: a mixture shows the Tyndall effect only if its dispersed particles are at least about as large as the wavelength of light. That is true for every colloid and every suspension, and false for every true solution.

Q3.

A mixture can be categorised as a solution, a suspension, or a colloid, each possessing distinct properties. Utilise the words or phrases provided in the box to fill in the Table 5.2. Words and phrases may be used more than once. [Large-sized particles; Particles remain evenly distributed; Small-sized particles (less than 1 nm diameter); Moderate-sized particles (1 – 1000 nm); Settles down when left undisturbed (more than 1000 nm in diameter); Does not settle down; Scatters light; Separates by filtration; Transparent; Salt solution; Milk; Sand in water; Smoke; Heterogeneous mixture; Cannot be separated by filtration; Mud; Butter; Brass.]

Answer

Table 5.2 completed

	Solution	Suspension	Colloid
Properties	Small-sized particles (less than 1 nm diameter); Transparent; Particles remain evenly distributed; Does not settle down; Cannot be separated by filtration	Large-sized particles; Settles down when left undisturbed (more than 1000 nm in diameter); Separates by filtration; Scatters light; Heterogeneous mixture	Moderate-sized particles (1 – 1000 nm); Particles remain evenly distributed; Does not settle down; Cannot be separated by filtration; Scatters light; Heterogeneous mixture
Examples	Salt solution; Brass	Sand in water; Mud	Milk; Smoke; Butter

How to place each phrase: work from particle size. Small particles cannot settle, cannot be trapped by filter paper and cannot scatter light — that gives the whole Solution column, and salt solution and brass are the two examples in the box that are uniform at particle level. Large particles do all three, giving the Suspension column with sand in water and mud. The Colloid column then takes one property from each side: it does not settle and cannot be filtered (like a solution), but it scatters light and is heterogeneous (like a suspension). Milk, smoke and butter all fit that description.

Tip: “Scatters light” and “Heterogeneous mixture” are used twice each — the question warns you that some phrases repeat, and these are the ones.

Q4.

Solve the following problems: (i) A cake recipe uses dry ingredients, namely 75 g of sugar for 420 g of all-purpose flour and 5 g of sodium hydrogencarbonate. Express the concentration of each component in the mixture using an appropriate method. (ii) A brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.

Answer

(i) All three ingredients are solids, so the appropriate method is the mass by mass percentage (% m/m).

Total mass of the mixture = 75 g + 420 g + 5 g = 500 g

% m/m of sugar = (75 g ÷ 500 g) × 100 = 15 % m/m
% m/m of flour = (420 g ÷ 500 g) × 100 = 84 % m/m
% m/m of sodium hydrogencarbonate = (5 g ÷ 500 g) × 100 = 1 % m/m

Check: 15 % + 84 % + 1 % = 100 % ✓

(ii) “70 % copper by mass” means 70 g of copper in every 100 g of brass.

Mass of copper = (70 ÷ 100) × 120 g = 84 g
Mass of zinc = 120 g − 84 g = 36 g

Check: % m/m of zinc = (36 g ÷ 120 g) × 100 = 30 %, and 70 % + 30 % = 100 % ✓

Why % m/m and not % m/v here: mass by volume percentage needs a volume of solution, and powders and solid alloys have no well-defined solution volume — a jar of flour also holds air between the grains, so its volume is not a reliable measure of how much flour there is. Mass does not suffer from that problem, so a solid mixture is always described by mass. The chapter uses the same convention for milk powder and spice mixtures.

Tip: The percentages of all the components of a mixture must add up to 100. Use that as a quick check on every concentration calculation.

Q5.

The label on a cooking oil pack says one litre (910 g). If this oil is mixed with water, will it form a separate layer? If so, which substance will be on top? How will you separate the two layers? Also, draw the diagram of the apparatus used.

Answer

Yes, it forms a separate layer, and the oil floats on top. The two are separated with a separating funnel.

Step 1 — find the density of the oil

Density = mass ÷ volume
Volume of oil = 1 L = 1000 mL = 1000 cm³, mass = 910 g
Density of oil = 910 g ÷ 1000 cm³ = 0.91 g cm⁻³
Density of water = 1.00 g cm⁻³

Step 2 — compare. 0.91 g cm⁻³ < 1.00 g cm⁻³, so the oil is lighter than an equal volume of water. Oil and water are also immiscible. Hence they form two layers, with the oil as the upper layer and water below.

Separating funnel. The less dense oil floats as the upper layer; opening the stopcock lets the denser water run out first.

Step 3 — how to separate them

Pour the mixture into a separating funnel mounted on a laboratory stand and close it with the glass stopper.
Let it stand undisturbed until two clear layers form with a sharp boundary.
Remove the stopper and open the stopcock slowly. The lower layer (water) runs out into a conical flask.
Close the stopcock the moment the boundary reaches the tap. Collect the small mixed portion separately and discard it.
Open the stopcock again and collect the upper layer (oil) in a fresh, clean container.

Why this works: A separating funnel exploits exactly two facts — that the liquids do not mix, and that they have different densities. Gravity does the sorting on its own, arranging the denser liquid at the bottom; the tap at the very bottom then lets you draw off that liquid alone. No heating, no chemicals and no filter paper are needed.

Q6.

Assertion (A): Solutions do not exhibit the Tyndall effect. Reason (R): The particles in solutions are larger than 100 nm, so they cannot scatter light. Choose the correct option: (i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.

Answer

The correct option is (iii) A is true, but R is false.

Assertion — true. A true solution does not show the Tyndall effect. Shine a laser through a salt or copper sulfate solution and the path of the beam cannot be seen from the side.

Reason — false, on two counts.

Stated in R: particle size of a solution > 100 nm ✗
Correct value: particle size of a solution < 1 nm ✓

The size is wrong. Solute particles in a solution are smaller than 1 nm, not larger than 100 nm.
The logic is wrong too. Being larger would make a particle scatter light more, not less. It is precisely because the particles are so small that they fail to scatter.

The physics behind it: Visible light has wavelengths of about 400 – 700 nm. A particle much smaller than that wavelength hardly disturbs the passing light wave, so almost nothing is scattered sideways and the beam stays invisible. Once particles reach roughly the size of the wavelength — as in a colloid (1 – 1000 nm) or a suspension (> 1000 nm) — they scatter strongly and the beam lights up.

Tip: In assertion–reason questions, always check the reason on its own before asking whether it explains the assertion. Here the assertion is a correct fact but the reason contains a factual error, which forces option (iii).

Q7.

How would you separate the mixtures given in Table 5.3? Mention the reason for choosing your method. If a mixture cannot be separated, explain why. [Mud from muddy water; Plasma from other components in the blood sample; Naphthalene and sand; Chalk powder and common salt; Common salt and water; Oil from water; Pigments of the flower]

Answer

Table 5.3 completed

Mixture	Method of separation	Reason for selection
Mud from muddy water	Sedimentation and decantation, then filtration; add alum (coagulation) or centrifuge if it stays cloudy	Mud is an insoluble suspension with particles larger than 1000 nm, so it settles and is held back by filter paper. Very fine particles need to be clumped by a coagulant or thrown down by centrifugation.
Plasma from other components in the blood sample	Centrifugation	Blood is a colloid; its cells do not settle under gravity and pass through filter paper. Spinning produces a much stronger outward force, and the denser cells collect at the bottom leaving the lighter plasma on top.
Naphthalene and sand	Sublimation	Naphthalene changes directly from solid to vapour below its melting point and is recovered by deposition on a cool surface; sand does not sublime and is left behind.
Chalk powder and common salt	Add water, filter, then evaporate or crystallise the filtrate	Salt is soluble in water and chalk is not. Filtration takes out the chalk as residue; evaporating the filtrate returns the salt.
Common salt and water	Evaporation (to recover the salt) or distillation (to recover both)	Salt is a non-volatile solid dissolved in the water, so it stays behind when the water is driven off. Distillation also condenses and collects the water.
Oil from water	Separating funnel	The two are immiscible and have different densities, so they form two layers and the lower one can be run off through the stopcock.
Pigments of the flower	Paper chromatography	The pigments differ in how strongly they are held by the paper and how readily they dissolve in the solvent, so they travel different distances up the strip.

The common thread: every method here picks out one difference between the components — particle size (filtration, centrifugation), the ability to sublime (sublimation), solubility (dissolve-and-filter), volatility (evaporation, distillation), density (separating funnel) or rate of movement on paper (chromatography). Choosing a technique is really a matter of asking: in what single property do these two components differ most?

Q8.

Two miscible liquids, A and B, are present in a mixture. The boiling point of A is 60 °C and the boiling point of B is 90 °C. Suggest a method to separate them. Also, draw a labelled diagram of the method suggested.

Answer

Separate them by simple distillation.

Boiling point of A = 60 °C
Boiling point of B = 90 °C
Difference = 90 °C − 60 °C = 30 °C
30 °C is more than the required 25 °C → simple distillation will work

Distillation set-up. The lower-boiling liquid vaporises first, the condenser turns the vapour back into liquid, and the distillate collects in the conical flask.

Procedure

Take the mixture in a distillation flask fitted with a thermometer, the bulb of the thermometer level with the mouth of the side arm. Connect the side arm to a water condenser and place a conical flask as the receiver.
Heat the flask gently over a wire gauze. The thermometer rises and then steadies at about 60 °C, which shows that liquid A alone is boiling.
The vapour of A passes into the condenser, where circulating cold water removes its heat and turns it back into a liquid. Pure A drips into the conical flask.
When the thermometer reading starts climbing above 60 °C, stop and change the receiver — liquid A has all distilled over.
Liquid B, with the higher boiling point, is left behind in the distillation flask; it can be collected there or distilled over separately at about 90 °C.

Why the boiling-point gap decides the method: Some of the higher-boiling liquid always evaporates too, so the vapour is never quite pure. A gap of about 25 °C or more means that while A is boiling, B contributes so little vapour that the distillate is effectively pure A. Here the gap is 30 °C, which is comfortably enough. Had the gap been smaller than 25 °C — as for acetone (56 °C) and alcohol (78 °C) — simple distillation would fail and fractional distillation would be needed instead.

Q9.

Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?

Answer

Point of comparison	Evaporation	Crystallization	Distillation
What is done	The solvent is driven off into the air	A hot saturated solution is cooled slowly	The liquid is boiled off, then condensed and collected
What you obtain	The solute only; the solvent is lost	Pure crystals of the solute; the liquid left over is discarded	Both components, each in pure form
Purity of the solid obtained	Low — every dissolved impurity is left behind with it	High — impurities stay in the solution	The liquid collected is pure
Apparatus	An open dish; often just sun and wind	Beaker, funnel, watch glass — simple	Distillation flask, thermometer, condenser, receiver
Cost and effort	Lowest	Moderate	Highest — needs continuous heating and cooling water

When to prefer each

Evaporation — when the solute is a non-volatile solid, you do not need the solvent back, and purity does not matter much. Obtaining common salt from seawater in salt pans is the standard example; the sun does the work free of charge.
Crystallization — when you need the solid pure and its solubility changes sharply with temperature. Purifying copper sulfate, or separating a compound from the impurities formed along with it, is done this way, and it gives well-shaped crystals as a bonus.
Distillation — when the liquid is what you want, or when you must recover both components. Separating acetone from water, obtaining drinking water from salty water, and extracting the fragrance of flowers in the Deg-Bhapka method of Kannauj all need distillation.

Why crystallization is preferred to evaporation for purification: Evaporation removes the solvent completely, so whatever was dissolved in it — the wanted solute and the unwanted impurities alike — is dumped together in the dish. Crystallization removes only the excess solute. Because the impurities are present in much smaller amounts, they are nowhere near their own saturation level and simply stay in the liquid, which is then poured away. That is the whole reason crystallization purifies and evaporation does not.

Q10.

Blood is an example of a colloidal mixture. (i) What would happen if blood behaved like a true suspension inside the body? (ii) In a blood sample, identify the dispersed phase and the dispersion medium.

Answer

(i) If blood behaved like a true suspension, its cells would be large enough for gravity to pull them down — and that would be fatal.

The cells would settle. Wherever the blood slowed down, red cells would sink to the lower side of the vessel, leaving almost cell-free plasma above. The blood would no longer be uniform.
Oxygen supply would fail. Red blood cells carry oxygen to every tissue. If they settled out, the blood reaching the brain and the muscles would carry very little oxygen.
Capillaries would block. Settled and clumped cells would jam the finest capillaries, cutting off the supply of nutrients and the removal of wastes from those tissues.
The heart would be strained as it tried to push a mixture that keeps separating out, and clots could form.

(ii) In a blood sample:

Component of the colloid	What it is in blood
Dispersed phase	The blood cells and proteins — red blood cells, white blood cells, platelets and plasma proteins
Dispersion medium	Plasma, which is mostly water with dissolved salts, glucose and hormones

Why being a colloid is exactly what the body needs: Colloidal particles are small enough that the constant molecular bombardment of the medium keeps them evenly dispersed, so blood stays uniform however long you stand still, and every drop that reaches a tissue carries its share of cells. Yet the particles are still large enough to be pushed aside by a strong enough force — which is precisely why a blood bank can spin donated blood in a centrifuge and split it into plasma, platelets, white cells and red cells for four different patients.

Q11.

You are given a mixture of sand, common salt and naphthalene (Fig. 5.25a). The Fig. 5.25b depicts various steps used to separate the components of this mixture. Identify and write down the correct sequence of separation techniques.

Answer

The correct sequence is 1 → 3 → 2, that is sublimation → filtration → evaporation (crystallization).

The order in which the three techniques must be applied to a mixture of sand, common salt and naphthalene.

Order	Step in Fig. 5.25b	Technique	What is obtained
First	Step 1 — inverted funnel over a china dish on a tripod, heated	Sublimation	Naphthalene collects on the cool funnel wall; sand and salt remain in the dish
Second	Step 3 — the mixture stirred with water and poured through a filter funnel	Filtration	Sand stays on the filter paper; salt solution passes through as the filtrate
Third	Step 2 — the filtrate heated in a china dish	Evaporation / crystallization	Water boils away, leaving pure common salt

Why the order cannot be changed: Sublimation must come first, while everything is dry. If you added water at the start, the salt would dissolve and the naphthalene would be wet, and you could no longer heat the mixture gently to sublime it. Filtration must come before evaporation, because once the water is boiled off the salt and sand are back together as a dry mixture and nothing has been achieved. So each step uses a different property in turn — first the ability to sublime, then solubility in water, then volatility of the solvent.

Tip: Naphthalene is insoluble in water, so an alternative order — filter first to remove salt, then sublime the naphthalene from the sand — would also work chemically. But the steps shown in Fig. 5.25b are set up for 1 → 3 → 2, which is also the more efficient route because the mixture is handled dry to begin with.

Q12.

Why is distillation an effective method for separating a mixture of water and acetone?

Answer

Because water and acetone are miscible — so no separating funnel can work — but their boiling points are far apart, which is exactly what distillation needs.

Boiling point of acetone = 56 °C
Boiling point of water = 100 °C
Difference = 100 °C − 56 °C = 44 °C
44 °C is well above the minimum of about 25 °C → distillation is effective

On heating, acetone reaches its boiling point long before the water does. The thermometer holds steady near 56 °C while acetone vapour passes into the condenser, is cooled by the circulating water and collects in the receiver as pure acetone. The water, which needs 100 °C, stays behind in the distillation flask. Both liquids are recovered.

Why a large gap matters so much: Every liquid gives off some vapour below its boiling point, so the vapour above a mixture is never made of one substance only. What decides the purity of the distillate is the proportion. At 56 °C, acetone is boiling vigorously while water contributes only a trace of vapour, so the distillate is essentially pure acetone. If the two boiling points were within 25 °C of each other, both liquids would vaporise together in comparable amounts and simple distillation would fail — that mixture would need fractional distillation.

Q13.

Answer the following questions with the help of the data given in Table 5.4. (i) What mass of potassium nitrate would be needed to prepare its saturated solution in 50 g of water at 40 °C? (ii) A student makes a saturated solution of potassium chloride in water at 80 °C and leaves the solution to cool at room temperature (25 °C). What would she observe as the solution cools? Explain. (iii) What is the effect of a change in temperature on the solubility of salts? Also, compare the changes in the solubility of the four given salts with increasing temperature from 10 °C to 80 °C.

Answer

(i) 31 g of potassium nitrate.

From Table 5.4, solubility of KNO₃ at 40 °C = 62 g per 100 g of water
Mass needed for 50 g of water = (62 g ÷ 100 g) × 50 g
= 31 g of potassium nitrate

(ii) She would see crystals of potassium chloride separating out as the solution cools, so that the liquid becomes cloudy and then a solid layer collects at the bottom of the beaker.

Solubility of KCl at 80 °C = 54 g per 100 g of water
Solubility of KCl at 20 °C = 35 g, at 30 °C = 37.4 g per 100 g of water
So at 25 °C the solubility is about (35 + 37.4) ÷ 2 ≈ 36 g per 100 g of water
Mass that must crystallise out ≈ 54 g − 36 g = about 18 g per 100 g of water

Why the crystals appear: At 80 °C the 100 g of water is holding 54 g of KCl, the most it can hold at that temperature. As the temperature falls, so does the maximum the water can hold. By 25 °C it can keep only about 36 g in solution, so the extra 18 g has nowhere to stay and separates out as solid crystals. This is crystallization — and if she lets it cool slowly and undisturbed, the crystals will be large and well formed.

(iii) For all four salts the solubility increases as the temperature increases — but by very different amounts.

Salt	Solubility at 10 °C	Solubility at 80 °C	Increase	How many times
Potassium nitrate	21 g	167 g	146 g	≈ 8.0 times
Ammonium chloride	24 g	66 g	42 g	≈ 2.8 times
Potassium chloride	35 g	54 g	19 g	≈ 1.5 times
Sodium chloride	36 g	37 g	1 g	≈ 1.03 times

All values are in grams per 100 g of water. The order of increase is potassium nitrate > ammonium chloride > potassium chloride > sodium chloride.

Why this matters in practice: How steeply the solubility rises decides which separation method to use. Potassium nitrate is ideal for crystallization — cool a hot saturated solution and a huge mass of it comes out. Sodium chloride is almost unaffected by temperature, so cooling gives virtually nothing; that is why common salt is obtained by evaporating seawater instead of by cooling it.

Q14.

Three students, A, B and C, are preparing sugar solutions for an experiment: Student A dissolves 20 g of sugar in 80 g of water. Student B dissolves 20 g of sugar in 100 g of water. Student C dissolves 30 g of sugar in 80 g of water. (i) Calculate the mass percentage (% m/m) concentration of sugar in each student’s solution. (ii) Whose solution is the most concentrated? Explain why.

Answer

(i) Remember that the denominator is the mass of the solution, that is sugar + water — not the mass of water alone.

% m/m = (mass of solute ÷ mass of solution) × 100

Student A: mass of solution = 20 g + 80 g = 100 g
% m/m = (20 g ÷ 100 g) × 100 = 20 % m/m

Student B: mass of solution = 20 g + 100 g = 120 g
% m/m = (20 g ÷ 120 g) × 100 = 16.7 % m/m (to 3 significant figures)

Student C: mass of solution = 30 g + 80 g = 110 g
% m/m = (30 g ÷ 110 g) × 100 = 27.3 % m/m (to 3 significant figures)

(ii) Student C’s solution is the most concentrated, at about 27.3 % m/m.

Student	Sugar	Water	Solution	% m/m
A	20 g	80 g	100 g	20.0 %
B	20 g	100 g	120 g	16.7 %
C	30 g	80 g	110 g	27.3 %

Why C wins: Compare the students in pairs and the reason becomes clear. A and B use the same 20 g of sugar, but B spreads it through more water, so B’s solution is the most dilute — more solvent for the same solute means lower concentration. A and C use the same 80 g of water, but C adds more sugar, so C is more concentrated than A — more solute in the same solvent means higher concentration. C therefore beats both. Concentration is a ratio: it depends on how much solute there is compared with the solution, never on the amount of solute alone.

Tip: A very common error is to divide by the mass of water (20 ÷ 80 = 25 % for student A). Always add the solute to the solvent first to get the mass of the solution.

Q15.

Examine Fig. 5.26. (i) Identify the separation technique marked as ‘S’. (ii) Label the apparatus A, B and C. (iii) Which of the following mixtures can be separated by the technique identified above? Use the data given in Table 5.5. Mixtures: (a) water — acetone (b) water — salt (c) acetone — alcohol (d) sand — salt (e) alcohol — chloroform (f) alcohol — benzene

Answer

(i) The technique marked S is distillation (simple distillation).

(ii) Labels

Label in Fig. 5.26	Apparatus	Its job
A	Distillation flask	Holds the mixture and is heated so that the lower-boiling liquid vaporises
B	Water condenser	Cold water circulating in the outer jacket cools the vapour and turns it back into liquid
C	Conical flask (receiver)	Collects the pure distillate as it drips out of the condenser

A thermometer is fixed at the mouth of the flask so that its bulb is level with the side arm, to show the temperature of the vapour actually passing over.

(iii) Distillation will separate (a) water — acetone and (b) water — salt.

Mixture	Boiling points from Table 5.5	Difference	Can distillation separate it?
(a) water — acetone	100 °C and 56 °C	44 °C	Yes — gap is more than 25 °C
(b) water — salt	Salt is a non-volatile solid	—	Yes — the water distils over and the salt is left in the flask
(c) acetone — alcohol	56 °C and 78 °C	22 °C	No — gap is less than 25 °C; needs fractional distillation
(d) sand — salt	Both are solids	—	No — nothing vaporises; dissolve in water and filter instead
(e) alcohol — chloroform	78 °C and 61 °C	17 °C	No — gap is less than 25 °C
(f) alcohol — benzene	78 °C and 80 °C	2 °C	No — the boiling points are almost the same

The rule being applied: simple distillation separates two miscible liquids only when their boiling points differ by at least about 25 °C, and it separates a liquid from a dissolved solid whenever the solid is non-volatile. Where the gap is smaller — (c), (e) and (f) — both liquids vaporise together in comparable amounts, the distillate is a mixture, and fractional distillation must be used instead. Where nothing can vaporise at all, as in (d), distillation has nothing to work with.

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