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NCERT Solutions for Class 9 Science Chapter 5 Exploring Mixtures and Their Separation (PDF) – 2027

NCERT Solutions for Class 9 Science Chapter 5 Exploring Mixtures and Their Separation – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.

Sugar from sugarcane, salt from seawater, plasma from a drop of blood — every one of them is a mixture being taken apart. This chapter sorts mixtures into solutions, suspensions and colloids, learns to state how much solute a solution holds, and then builds the separation toolkit: crystallization, distillation, chromatography, the separating funnel, sublimation, centrifugation and coagulation.

NCERT Solutions for Class 9 Science Chapter 5 Exploring Mixtures and Their Separation PDF

All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.

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NCERT Solutions for Class 9 Science Chapter 5 Exploring Mixtures and Their Separation – Questions and Answers

All 81 questions of the chapter (textbook pages 72 – 93) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.

Think It Over — Page 72Chapter opener

Think It Over
Q1Why do suspended particles settle in muddy water over time but not in milk?
Answer

Because the two mixtures hold particles of very different size. Muddy water is a suspension; milk is a colloid.

Mud particles: diameter > 1000 nm
Milk (fat and protein) particles: diameter 1 – 1000 nm
Q2How is evaporation different from boiling?
Answer

Evaporation is a surface change that goes on at every temperature; boiling is a bulk change that happens only at one fixed temperature.

Point of comparisonEvaporationBoiling
Where it happensOnly at the free surface of the liquidThroughout the liquid — bubbles form inside it
TemperatureAt all temperatures below the boiling pointOnly at the boiling point (100 °C for water at normal pressure)
SpeedSlow and quietFast and vigorous
Energy sourceDraws heat from the liquid itself and the surroundings, so the liquid coolsHeat must be supplied continuously from outside
Everyday exampleWet clothes drying, sweat cooling youWater bubbling in a kettle
Q3Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree?
Answer

Because the air under the tree carries dust, smoke and tiny water droplets, and these particles scatter the sunlight sideways into your eye. This is the Tyndall effect.

In-text Questions — Page 735.1 How Can We Classify Mixtures?

Q1Is the mixture of oil and water homogeneous or heterogeneous?
Answer

Heterogeneous. Oil and water are immiscible — they refuse to mix, and however hard you stir they settle back into two separate layers with a visible boundary between them.

Q2Can you think of some other heterogeneous mixtures?
Answer

Yes — any mixture in which you can point to more than one part is heterogeneous.

  • Solid in liquid: sand in water, chalk powder in water, muddy water, tea leaves in water.
  • Liquid in liquid: mustard oil and water, kerosene and water.
  • Solid in solid: iron filings and sulfur, sand and common salt, a handful of mixed dal.
  • Solid or liquid in gas: smoke (solid particles in air), fog and clouds (water droplets in air), dust in air.

Activity 5.1: Let us experiment — Group activity — Page 735.1 How Can We Classify Mixtures?

Activity
Q1Are the particles visible in each mixture? Record your observations.
Answer

Only in beaker B. The three mixtures are salt in water (A), chalk powder in water (B) and a few drops of milk in water (C).

BeakerMixtureAre particles visible?Appearance
ACommon salt + 50 mL waterNoClear and transparent — you can read print through it
BChalk powder + 50 mL waterYesCloudy; white specks clearly seen swirling in the water
CA few drops of milk + 50 mL waterNoUniformly milky-white, but no separate specks are seen
Q2Direct the light from a laser pointer through the beakers containing the mixtures (Fig. 5.3) and observe it from the side of the beaker in a direction perpendicular to the laser beam. Record your observations.
A(a) salt and waterB(b) chalk powder and waterC(c) milk and waterlaser pointer
Fig. 5.3, page 73 — redrawn sketch of the set-up: a laser beam sent through each of the three beakers A, B and C and viewed from the side.
Answer

The path of the beam is invisible in A, brightly visible in B and visible in C.

BeakerPath of the laser beam seen from the sideWhat it tells you
A — salt and waterNot visibleA true solution; particles are far too small to scatter light
B — chalk and waterBright, sharply visibleA suspension; large particles scatter light strongly
C — milk and waterVisibleA colloid; the mixture looks uniform yet still scatters light
Safety first: Never look straight into the laser beam. Observe only from the side, at right angles to the beam.
Q3Predict what you would observe in each of the beakers if you leave them undisturbed for a few minutes.
Answer
  • Beaker A (salt and water): no change at all. The solution stays clear and uniformly salty from top to bottom.
  • Beaker B (chalk and water): the chalk settles as a white layer at the bottom, leaving clearer water above. This settling is called sedimentation.
  • Beaker C (milk and water): no change. It stays evenly milky; nothing settles even after hours.
Q4Set up a filtration apparatus and filter each mixture separately. Is there any residue left on the filter paper?
Answer

Only beaker B leaves a residue.

BeakerResidue on the filter paperFiltrate
A — salt and waterNoneClear salt solution (still salty — the salt passes through)
B — chalk and waterWhite chalk residueClear water
C — milk and waterNoneMilky liquid — unchanged
Q5Based on your observations, do you think these are the same types of mixtures or are they different?
Answer

They are three different types of mixture. Put the four tests side by side and each beaker gives its own pattern of answers.

TestA — salt + waterB — chalk + waterC — milk + water
Particles visible?NoYesNo
Tyndall effect?NoYesYes
Settles on standing?NoYesNo
Residue on filtering?NoYesNo
Type of mixtureSolution (homogeneous)Suspension (heterogeneous)Colloid (heterogeneous)

In-text Questions — Page 745.2 Solutions; 5.2.1 Concentration of a solution

Q1In what proportion are a solute and a solvent present in a solution? Can these be expressed quantitatively?
Answer

There is no single fixed proportion — a solution can be dilute or concentrated. But the proportion actually used can be stated exactly, and that number is called the concentration of the solution.

Concentration = amount of solute present in a given amount of solvent or of solution

At Class 9 level it is written as a percentage in one of three ways:

% m/m = (mass of solute ÷ mass of solution) × 100
% m/v = (mass of solute ÷ volume of solution) × 100
% v/v = (volume of solute ÷ volume of solution) × 100
Q2When farmers spray pesticides on their crops, they must mix the right amount of pesticide with a fixed amount of water to prepare a solution. If they do not do so, what is likely to happen?
Answer

The spray will fail in one of two ways.

  • Too little pesticide: the concentration is below what the pest can withstand, so the insects survive and the crop is not protected. Repeated weak doses can even help resistant pests to multiply.
  • Too much pesticide: the excess scorches the crop, builds up in the soil, is washed into ponds and rivers, and leaves residues on the grain that reach the people who eat it.
Q3Can you think of other such examples?
Answer

Yes — almost every prepared solution has a proportion that must be respected.

SituationWhy the exact proportion matters
ORS for a child with diarrhoeaThe salt and sugar must match the concentration of body fluids, or water will not be absorbed properly
Saline drip in a hospital0.9 % m/v NaCl matches blood; stronger or weaker damages the blood cells
Bleaching solution for cleaning a floorToo weak does not disinfect; too strong corrodes the surface and irritates the skin
Sugar syrup for jam or a cold drinkDecides taste and how long the food can be preserved
Fluoride in toothpaste, chlorine in drinking waterA useful amount protects; more than that is harmful
Sugar and milk in a cup of teaSame ingredients, different proportion, completely different taste

Pause and Ponder — Page 765.2.2 How do we express concentration?

Pause and Ponder
Q1A common talcum powder contains 4 % m/m zinc oxide, which acts as an antiseptic. How much zinc oxide is present in 300 g of the talcum powder?
Answer

12 g of zinc oxide.

% m/m = (mass of solute ÷ mass of mixture) × 100
4 = (mass of ZnO ÷ 300 g) × 100
mass of ZnO = (4 × 300 g) ÷ 100
mass of ZnO = 12 g
Q2Your mother gives you a bottle of orange juice concentrate to mix with water and serve it to your visiting friends. She asks you to mix two tablespoons of the concentrate with water in a glass tumbler. If each tablespoon measures 15 mL and you make 150 mL of juice per person, what is the % v/v of orange juice concentrate in the mixture you prepared?
Answer

20 % v/v.

Volume of concentrate (solute) = 2 × 15 mL = 30 mL
Volume of juice (solution) = 150 mL
% v/v = (volume of solute ÷ volume of solution) × 100
% v/v = (30 mL ÷ 150 mL) × 100
% v/v = 20 % v/v
Q3Vinegar, used as a food preservative and additive, contains 5 % v/v acetic acid. Glacial acetic acid is a liquid, i.e., 100% acetic acid. If you want to make vinegar from glacial acetic acid, how would you proceed?
Answer

Measure 5 mL of glacial acetic acid and add water to it until the total volume of the solution is 100 mL. That gives 5 % v/v vinegar.

Required: % v/v = 5, volume of solution = 100 mL
5 = (volume of acetic acid ÷ 100 mL) × 100
volume of acetic acid = (5 × 100 mL) ÷ 100 = 5 mL
Water to be added ≈ 100 mL − 5 mL = 95 mL

Safety first: add the acid slowly to the water (never water to the acid), under an adult’s supervision.

Activity 5.2: Let us represent solubility graphically — Page 775.2.3 Solubility of substances

Activity
Q1Based on the information from the above graph, predict which of the two compounds, ‘A’ or ‘B’, will dissolve more in a given amount of water at a given temperature?
0501001502002503003504001020304050607080241287Compound ‘B’Compound ‘A’Temperature (°C)Solubility (g per 100 g of water)
Fig. 5.6, page 77 — solubility curves of compounds ‘A’ and ‘B’ in water.
Answer

Compound ‘B’ — at every temperature on the graph its curve lies far above that of compound ‘A’.

At 20 °C: B ≈ 205 g per 100 g water, A ≈ 37 g per 100 g water
At 60 °C: B = 287 g per 100 g water, A ≈ 56 g per 100 g water
At 80 °C: B ≈ 360 g per 100 g water, A ≈ 66 g per 100 g water
Q2(i) The solubility of compound ‘A’ in water at 20 °C is ______________ (less than/more than/similar to) its solubility at 60 °C.
Answer

Less than.

Solubility of A at 20 °C ≈ 37 g per 100 g of water
Solubility of A at 60 °C ≈ 56 g per 100 g of water
Increase ≈ 56 g − 37 g = 19 g per 100 g of water
Q3(ii) The solubility of compound ‘B’ at 20 °C is ____________ (less than/more than/similar to) its solubility at 60 °C.
Answer

Less than.

Solubility of B at 20 °C ≈ 205 g per 100 g of water
Solubility of B at 60 °C = 287 g per 100 g of water
Increase ≈ 287 g − 205 g = 82 g per 100 g of water
Q4(iii) The solubility of _____________ increases more than that of ______________ with an increase in the temperature.
Answer

The solubility of compound ‘B’ increases more than that of compound ‘A’ with an increase in the temperature.

From 20 °C to 60 °C:
Compound B: 287 g − 205 g ≈ 82 g per 100 g of water
Compound A: 56 g − 37 g ≈ 19 g per 100 g of water
Compound B gains about 82 ÷ 19 ≈ 4 times as much

In-text Questions — Page 775.2.3 Solubility of substances

Q1What do you think will happen if you make a saturated solution at a higher temperature and cool it slowly?
Answer

Part of the dissolved solute will come out of the solution as pure solid crystals. This is the process of crystallization.

Saturated solution of compound B at 60 °C: 287 g in 100 g of water
Solubility of B at 40 °C (from Fig. 5.6): 241 g per 100 g of water
Mass that must separate out = 287 g − 241 g = 46 g

Activity 5.3: Let us prepare — Page 785.3.1 Crystallization

Activity
Q1You may also try:
1. Place 1 – 2 mL of the saturated copper sulfate solution on a small glass plate or a lamination sheet.
2. Leave it for some time. What do you observe?
Answer

The water evaporates from the drop of saturated copper sulfate solution on the glass plate, and small blue crystals of copper sulfate are left behind, spread over the plate.

Q2Did you get crystals? If yes, is this a good way to experiment? Explain.
Answer

Yes, small crystals form as the water evaporates — but this is not a good way to experiment. The thin film dries too fast, so the crystals are tiny and badly shaped; all the impurities are left mixed with them; and the result depends on the room temperature and breeze, so it cannot be controlled or repeated. A better method is to filter the hot saturated solution and let it cool slowly and undisturbed (steps 1 – 6).

Think as a Scientist — Page 795.3.1 Crystallization

Think as a Scientist
Q1If a hot, saturated solution of copper sulfate is cooled rapidly in ice-cold water, smaller and less well-formed crystals will form than if it is cooled slowly at room temperature. How would you design and perform an experiment to test this hypothesis?
Answer

A controlled experiment in which only the rate of cooling is changed:

  1. Prepare a hot saturated copper sulfate solution and filter it while hot.
  2. Divide it into two equal parts in two identical beakers.
  3. Cool beaker 1 quickly by standing it in ice-cold water; let beaker 2 cool slowly at room temperature. Cover both and do not disturb them.
  4. After the same time, filter out the crystals from each beaker, rinse them with a little cold water and dry them.
  5. Compare the crystals under a magnifying glass — their size and how well-shaped they are.

Expected result: beaker 1 gives many small, poorly formed crystals; beaker 2 gives fewer, larger, well-shaped crystals. This supports the hypothesis.

Pause and Ponder — Page 795.3.1 Crystallization

Pause and Ponder
Q4Refer to the solubility curves given in Activity 5.2. If equal masses of hot, saturated solutions of compounds ‘A’ and ‘B’ are cooled from 80 °C to 60 °C, which solution is likely to deposit more solid?
Answer

The solution of compound ‘B’ will deposit far more solid.

Step 1 — read the solubilities off Fig. 5.6 (in g per 100 g of water):

Compound B: at 80 °C ≈ 360 g   at 60 °C = 287 g
Compound A: at 80 °C ≈ 66 g   at 60 °C ≈ 56 g

Step 2 — solid deposited per 100 g of water

Compound B: 360 g − 287 g ≈ 73 g
Compound A: 66 g − 56 g ≈ 10 g
Q5Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain.
Answer

Yes. Slow evaporation gives large crystals; fast evaporation gives many small ones.

Rate of evaporationWhat happens in the solutionCrystals obtained
Slow (shallow pan, mild sun, still air)The solution creeps past saturation gently, so only a few nuclei form and each keeps growingFew, large, well-shaped crystals
Fast (strong heating, dry wind, thin film)The solution shoots far past saturation, so a huge number of nuclei appear togetherMany tiny, poorly formed crystals

Activity 5.4: Let us describe a process — Page 795.3.1 Crystallization

Activity
Q1Observe Fig. 5.9, it shows how salt crystals are obtained from seawater. Can you describe the process in your own words?
SeawaterSaturated solutionSalt crystals
Fig. 5.9, page 79 — steps followed in the salt manufacturing process.
Answer

Salt is obtained from seawater by evaporating the water, which leads to crystallization:

  1. Seawater, a dilute solution of salt, is let into wide, shallow pans near the coast.
  2. The sun and wind evaporate the water. The salt stays behind, so the solution becomes more and more concentrated until it is a saturated solution.
  3. As more water evaporates, the water left can no longer hold all the salt, so the extra salt separates out as salt crystals. These are collected, washed and dried.

Ready to Go Beyond — Page 795.3.1 Crystallization

Ready to Go Beyond
Q1Have you visited a place where large crystal deposits can be observed in nature?
Answer

Natural crystals grow wherever a mineral-rich solution is left undisturbed for a very long time — in caves, in mines and inside the Earth’s crust.

Sample answer: On a school trip to Sohra (Cherrapunji) in Meghalaya we went into the Mawsmai Cave (Fig. 5.10a). Water dripping through the limestone roof carries dissolved calcium compounds. As each drop hangs at the roof and slowly loses water and carbon dioxide, the solution becomes saturated and a little solid is deposited. Over thousands of years these deposits build up into stalactites hanging down and stalagmites rising from the floor. Quartz (Fig. 5.10b), with its clear six-sided pointed crystals, is another beautiful example found in mines and rock cavities.

In-text Questions — Page 805.3.2 Distillation

Q1When we allow a solvent to evaporate to get a solute, such as salt from a salt solution, the solvent disappears. However, we may also want to recover the solvent. What should we do then?
Answer

Use distillation instead of plain evaporation.

Evaporation: solution → solute recovered, solvent lost to the air
Distillation: solution → solute left in the flask and solvent condensed and collected
Q2While experimenting in the laboratory, a student accidentally mixed acetone and water, two miscible liquids. How can we separate the two miscible liquids?
Answer

By distillation, provided their boiling points differ by at least about 25 °C.

Boiling point of acetone = 56 °C
Boiling point of water = 100 °C
Difference = 100 °C − 56 °C = 44 °C, which is more than 25 °C → distillation will work

Heat the mixture in a distillation flask. Acetone, the lower-boiling liquid, boils first at about 56 °C; the thermometer bulb at the mouth of the side arm holds steady near 56 °C while it distils over. Its vapour passes through the water condenser, is cooled and collected in the conical flask as pure acetone. Water, needing 100 °C, stays behind in the distillation flask.

Q3Is it possible to separate the mixture of two miscible liquids by evaporation and obtain both the liquids?
Answer

No. Evaporation can give you back only one of the two — and even that one is not obtained pure.

  • The liquid that evaporates escapes into the air and is lost; there is no arrangement to catch it. So one component is gone for good.
  • Both liquids evaporate at the same time, because evaporation goes on at all temperatures. The liquid left behind therefore still contains some of the other liquid — it is not pure either.

Activity 5.5: Let us investigate — Page 825.3.3 Paper Chromatography

Activity
Q1A pencil line is drawn 2 cm from the bottom of a strip of chromatographic paper, a black sketch-pen spot is marked at its centre, and the strip is stood in a thin layer of water in a gas jar, with the water level below the spot.
Observe the paper as the water rises through the paper. What do you notice?
Answer

The water creeps steadily upwards through the paper, well above the level in the jar, and when the wet front reaches the ink spot it begins to drag the ink up with it. The single black spot stretches into a streak, and colours start to appear at the edges.

Q2As the water rises, the ink starts to separate into different colour spots. What can you infer from this?
Answer

That the “black” ink is not a single substance at all — it is a mixture of several coloured dyes, and each dye travels up the paper at its own speed.

  • Because separate coloured spots appear, at least that many different components are present.
  • The number of spots tells you how many components the ink contains.
  • The spot that has travelled farthest is the component least held by the paper and most attracted to the water.

In-text Questions — Page 825.3.3 Paper Chromatography

Q1Will water work as a solvent in every case?
Answer

No. Water works only when the components of the mixture actually dissolve in water. In many cases a different solvent, or a mixture of solvents, must be used.

Mixture to be separatedSuitable mobile phaseReason
Black or coloured sketch-pen inkWater, or 2 % m/v salt solutionThe dyes are water-soluble
Green pigments of spinach leavesAlcohol, or a solvent mixtureChlorophyll is insoluble in water but dissolves in alcohol
Coloured pigments of flower petalsAlcohol or a water–alcohol mixtureDepends on which pigment the petal contains

Pause and Ponder — Page 825.3.3 Paper Chromatography

Pause and Ponder
Q6State whether the following statements are True or False. Also, correct the False statements.
(i) Salt can be separated from a salt solution by evaporation or distillation.
(ii) Distillation can be used for separation of two liquids even when these have the same boiling point.
(iii) In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment.
(iv) Evaporation and crystallization are the same processes.
Answer
StatementTrue / FalseCorrected statement, where needed
(i) Salt can be separated from a salt solution by evaporation or distillation.True—
(ii) Distillation can be used for separation of two liquids even when these have the same boiling point.FalseDistillation can separate two miscible liquids only when their boiling points differ by at least about 25 °C.
(iii) In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment.FalseThe solvent level should be below the sample spot at the beginning of the experiment.
(iv) Evaporation and crystallization are the same processes.FalseThey are different: evaporation removes the solvent to leave the solute behind, while crystallization deposits a pure solid as regular crystals from a saturated solution.

Activity 5.6: Let us separate — Page 835.4.1 Separation of two immiscible liquids

Activity
Q1A mixture of 5 mL mustard oil and 20 mL water is poured into a 50 mL separating funnel. Let it stand undisturbed. What do you observe?
Glass stopperSeparating funnelMustard oilWaterStopcockConical flaskLaboratory stand
Fig. 5.16, page 83 — redrawn sketch of the separating-funnel set-up used in Activity 5.6.
Answer

The mixture, cloudy at first from being poured, soon clears into two sharply separated layers — a yellow layer of mustard oil (5 mL) floating on a colourless layer of water (20 mL), with a flat, clearly visible boundary between them.

Q2The yellow-coloured mustard oil forms the upper layer and water forms the lower layer. Can you explain why?
Answer

Because mustard oil is less dense than water, so an equal volume of it weighs less and it floats.

Density of water = 1.00 g/cm³
Density of mustard oil ≈ 0.91 g/cm³
Since 0.91 g/cm³ < 1.00 g/cm³, the oil rises to the top

What if … — Page 835.4.1 Separation of two immiscible liquids

What if …
Q1two immiscible liquids of the same density are mixed in a separating funnel, how will the layers form?
Answer

They will not form one layer above the other. With equal densities neither liquid floats or sinks, so one liquid stays as drops or blobs spread anywhere inside the other (the drops may join into larger blobs). There is no flat boundary between them, so a separating funnel cannot separate them.

In-text Questions — Page 835.4.1 Separation of two immiscible liquids

Q1Can you think of any heterogeneous mixtures with a gas as one of the components?
Answer

Yes. Gases mix freely with one another, so a mixture of gases is almost always homogeneous — but if the other component is a solid or a liquid spread through the gas, the mixture is heterogeneous.

MixtureWhat is dispersed in the gasType
SmokeFine solid carbon and ash particles in airHeterogeneous
Fog, mist, cloudsTiny liquid water droplets in airHeterogeneous
Dust in the airSolid dust particles in airHeterogeneous
Air freshener spray, an aerosolLiquid droplets in airHeterogeneous
Hydrogen and oxygen (rocket fuel)Gas mixed with gasHomogeneous

Activity 5.7: Let us explore — Page 845.4.2 Sublimation

Activity
Q1Observe the inner wall of the funnel carefully. Do you notice any solid deposits?
Answer

Yes — a layer of white solid camphor collects on the cool inner wall of the inverted funnel, while the sand stays behind in the china dish.

Camphor (solid) —heat→ camphor vapour —cooling on the funnel wall→ camphor (solid)
Solid → vapour directly = sublimation
Vapour → solid directly = deposition

Pause and Ponder — Page 845.4.2 Sublimation

Pause and Ponder
Q7Why do immiscible liquids form two separate layers in a separating funnel?
Answer

Because they do not dissolve in each other, and they have different densities.

Denser liquid → sinks → lower layer
Less dense liquid → floats → upper layer
Example: water (1.00 g/cm³) below, mustard oil (≈ 0.91 g/cm³) above
Q8Is sublimation different from evaporation? Justify.
Answer

Yes, they are different. Sublimation is solid → vapour directly; evaporation is liquid → vapour from the surface. The key difference is the state you start from and whether a liquid stage appears at all.

Point of comparisonSublimationEvaporation
Change of stateSolid → vapour, with no liquid stageLiquid → vapour
Starting materialA solidA liquid
Where it occursAt the surface of the solidAt the surface of the liquid
Which substancesOnly a few — camphor, naphthalene, ammonium chloride, dry iceEvery liquid, at all temperatures
Reverse processDeposition (vapour → solid)Condensation (vapour → liquid)
Used to separateA sublimable solid from a non-sublimable one (camphor from sand)A non-volatile solute from its solvent (salt from salt solution)

In-text Questions — Page 845.4.2 Sublimation

Q1Can you think of any other mixtures where sublimation can be used to separate the components?
Answer

Yes — any mixture in which exactly one component sublimes on gentle heating.

MixtureComponent that sublimesComponent left behind
Naphthalene and sandNaphthaleneSand
Ammonium chloride and common saltAmmonium chlorideCommon salt
Camphor and common saltCamphorCommon salt
Iodine and sandIodineSand
Dry ice mixed with an ordinary solidSolid carbon dioxideThe other solid
Q2Is it possible to dissolve one metal in another?
Answer

Not at room temperature — but yes, if the metals are melted first. The homogeneous solid formed when such a molten mixture solidifies is called an alloy.

AlloyApproximate compositionWhy it is made
BrassAbout 80 % copper and 20 % zincHarder than copper, easy to shape, does not corrode readily
BronzeAbout 80 % copper and 20 % tinStrong and hard-wearing; used for statues and bells
Stainless steelIron with carbon 0.03 – 0.8 %, chromium 16 – 18 %, nickel 10.0 – 14.0 %, molybdenum 2.0 – 3.0 %Strong and rust-resistant; used for utensils and instruments

In-text Questions — Page 855.4.3 Suspensions

Q1How can we separate mud from water?
Answer

Start with the simplest method and add another only if the water is still not clear.

  1. Sedimentation and decantation. Let the muddy water stand. The heavier mud particles settle at the bottom; pour off the clearer water carefully without disturbing the sediment.
  2. Filtration. Pass it through filter paper or a clean cotton cloth. Particles larger than the pores are held back as residue.
  3. Coagulation, then filtration. Add powdered alum (fitkari). It makes the fine particles clump into larger flocs that settle by gravity and can then be removed by decantation or filtration.
  4. Centrifugation. Spin a sample at high speed; the centrifugal force drives even very fine particles to the bottom of the tube.
Q2If the muddy water is still not clear even after keeping for some time, how can it be cleaned?
Answer

By centrifugation and/or coagulation — the two techniques the chapter introduces for exactly this situation.

Coagulation. Add a pinch of powdered alum, a white crystalline substance used in water purification. Alum makes the fine suspended particles stick together in clumps. These larger, heavier clumps settle rapidly under gravity — the process called sedimentation — and the clear water above can then be decanted or filtered off.

Centrifugation. Take the cloudy water in a tube and spin it at high speed. The tubes swing out to the horizontal, and the centrifugal force acting outwards on each particle is many times greater than its weight, so even the finest particles are driven to the bottom of the tube while clear liquid remains on top.

Threads of Curiosity — Page 855.4.3 A. Centrifugation

Threads of Curiosity
Q1The spinning game is a folk dance called phugadi in Marathi and kikli in Punjabi. What is this called in your local language?
Answer

The same two-person spinning game is played all over India and has a different name in each language.

Language / regionName of the game or dance
Marathi (Maharashtra)Phugadi (फुगडी)
Punjabi (Punjab)Kikli (ਕਿੱਕਲੀ)
Konkani (Goa)Fugdi

Sample answer: “In my language we call it ______. We hold each other’s crossed hands and spin, and the faster we spin the harder we feel ourselves being pulled outwards.” Ask your grandparents — they will almost certainly know a local name for it.

Bridging Science and Society — Page 86The Paperfuge

Bridging Science and Society
Q1Find out more about such simple technological inventions that improve access to medical care in remote places.
Answer

The paperfuge is one of a whole family of frugal medical devices: they use everyday materials and no electricity, cost very little, and can be carried to a village that has neither a laboratory nor a reliable power supply.

DeviceHow it worksWhat it makes possible
PaperfugeA cardboard disc on twisted strings, like a button whirligig; spinning it at very high speed separates blood by centrifugationDetecting malaria and anaemia without electricity
FoldscopeA microscope folded from a paper sheet with a tiny glass lensSeeing microbes and blood cells in the field, at very low cost
Paper-strip diagnostic testsA drop of blood or urine travels along treated paper by capillary action and changes colourInstant tests for pregnancy, malaria, diabetes and infections
Solar-powered vaccine carrierAn insulated box cooled by a small solar panelVaccines kept cold on the way to villages with no power
Jaipur footA rubber and wood artificial limb made by hand from local materialsAffordable mobility for people who have lost a leg

Activity 5.8: Let us make a model — Page 865.4.3 A. Centrifugation

Activity
Q1Make your own centrifuge with a cardboard disc and thick thread. You will be able to see how the heavier particles move outwards. It is a fun and hands-on way to understand the science behind separation. Which mixture would you like to separate using this mini centrifuge?
Answer

Mixture chosen: muddy water.

Pass a thick thread through two holes near the centre of a stiff cardboard disc and tie the ends into a loop. Tape two small, tightly capped tubes of muddy water on opposite sides of the disc, near its rim. Spin the disc by pulling the strings apart and letting them come together, again and again, for 2 – 3 minutes.

Result: the mud collects at the outer end of each tube, with clearer water above it — the heavier particles have moved outwards.

In-text Questions — Page 865.4.3 B. Coagulation; 5.4.4 Colloids

Q1Can you think of any other coagulation processes used in everyday life?
Answer

Yes — coagulation is at work in the kitchen, at the water works, and even in your own body.

Everyday processCoagulant usedWhat clumps together
Making paneer or chhena from milkLemon juice or vinegar (an acid)Milk proteins clump and separate from the whey
Setting curd from milkLactic acid made by the bacteria in the starterMilk proteins form a soft solid network
Purifying drinking water at a water worksAlum (fitkari)Fine clay and silt particles form settleable flocs
Sewage treatmentAlum or other coagulantsSuspended solids settle out before filtration
Blood clotting over a cutThe body’s own clotting proteinsBlood cells are trapped in a fibrous mesh, sealing the wound
Setting an egg while cookingHeatEgg proteins clump into a solid white
Q2If the components of blood can be separated by centrifugation and the blood coagulates, is it a suspension? However, we cannot see blood cells with the naked eye. Is it a solution?
Answer

Neither — blood is a colloid. It is not a solution, because its particles (blood cells and large protein particles) are far bigger than dissolved particles: they scatter light and can be separated by centrifugation. It is not like an ordinary suspension such as muddy water either: blood looks uniform, its cells cannot be seen with the naked eye, and its components do not settle out quickly as mud does.

Bridging Science and Society — Page 87Donate Blood

Bridging Science and Society
Q1Do you know your blood group? Find out!
Answer

Your blood group is decided by which marker proteins sit on the surface of your red blood cells. In the common ABO and Rh system there are eight groups: A+, A−, B+, B−, AB+, AB−, O+ and O−.

How to find out: a blood group test needs only a drop or two of blood. It is done free at almost every blood donation camp, and at any hospital or pathology laboratory. Some school health check-ups also record it. Write it down and keep it in your school diary — in an emergency it saves valuable minutes.

In-text Questions — Page 875.4.4 Colloids

Q1Let us try to understand what colloids are. How are they different from solutions and suspensions?
Answer

A colloid is a mixture in which particles of one substance, sized between 1 nm and 1000 nm, stay evenly dispersed through another substance without dissolving and without settling.

PropertySolutionColloidSuspension
NatureHomogeneousHeterogeneous (looks homogeneous)Heterogeneous
Particle sizeLess than 1 nm1 – 1000 nmMore than 1000 nm
Particles visible?NoNoYes, to the naked eye
Separated by filtration?NoNoYes
Settles on standing?NoNoYes
Tyndall effectNot shownShownShown
Q2Can you think of some other substances that could be colloids?
Answer

Colloids are everywhere in daily life — in the kitchen, in the bathroom shelf, in the sky and in your own body.

ColloidDispersed phaseDispersion medium
MilkLiquid fat dropletsWater (an oil-in-water emulsion)
ButterWater dropletsFat (a water-in-oil emulsion)
BloodBlood cells and proteinsPlasma
Ice cream, tomato sauceFat and solid particlesWater
Cloud, fog, mistWater dropletsAir
SmokeSolid carbon particlesAir
Cold cream, body lotion, vanishing creamWater or oil dropletsOil or water

In-text Questions — Page 885.5 Tyndall Effect

Q1Have you ever observed light scattering by particles in your surroundings? Think of some more examples!
Answer

Yes — wherever a beam of light passes through air or a liquid that carries suspended particles, its path becomes visible. That is the Tyndall effect.

  • A shaft of sunlight entering a dark room through a small hole or a gap in the curtain, picked out by the dust in the air.
  • Sunlight streaming through the gaps between the leaves of a dense tree, or through mist in an early-morning forest.
  • The beams of floodlights in a sports stadium (Fig. 5.24), and the cone of light from a cinema projector.
  • Car headlights on a foggy night; the beam is visible far ahead because water droplets scatter the light.
  • A laser or a torch shone through milk diluted with water, or through a beaker of dilute soap solution.

Pause and Ponder — Page 885.5 Tyndall Effect

Pause and Ponder
Q9Clouds are made up of tiny water droplets or ice crystals floating in the air. Based on what you know about solutions, suspensions and colloids, what type of mixture do you think clouds are and why?
Answer

A cloud is a colloid — an aerosol, with tiny liquid water droplets or ice crystals as the dispersed phase and air as the dispersion medium.

TestWhat a cloud doesConclusion
Are the droplets dissolved in the air?No — they stay as separate droplets of liquid waterNot a solution
Do they settle out?No — a cloud floats for hours or daysNot a suspension
Does it scatter light?Yes — that is why clouds look white and cast shadowsShows the Tyndall effect
Do you see separate droplets?No — the cloud looks like one smooth massParticles are of colloidal size → a colloid
Q10Why do cities with a lot of smoke and dust in the air often look hazy?
Answer

Because the smoke and dust particles suspended in the city air scatter sunlight in all directions — the Tyndall effect on the scale of a whole city.

Activity 5.9 — Page 885.5 Tyndall Effect

Activity
Q1Complete Table 5.1 and review what you have learnt about solutions, suspensions and colloids.
S. No.PropertySolutionSuspensionColloid
1.Nature (homogeneous/heterogeneous)   
2.Particle size   
3.Visibility   
4.Separation by filtration   
5.Settling   
6.Tyndall effect   
Table 5.1, page 88 — Properties of different types of mixtures (as printed, with the cells left blank for you to fill).
Answer

Table 5.1: Properties of different types of mixtures

S. No.PropertySolutionSuspensionColloid
1.Nature (homogeneous / heterogeneous)HomogeneousHeterogeneousHeterogeneous, though it looks homogeneous
2.Particle sizeLess than 1 nmMore than 1000 nm1 – 1000 nm
3.VisibilityParticles not visible; the mixture is transparentParticles visible to the naked eyeParticles not visible; the mixture looks uniform
4.Separation by filtrationCannot be separatedCan be separated — leaves a residueCannot be separated
5.SettlingDoes not settleSettles on standing (sedimentation)Does not settle
6.Tyndall effectNot shownShownShown

Examples: solution — salt or sugar in water, vinegar, brass; suspension — chalk powder in water, muddy water, sand in water; colloid — milk, blood, smoke, ice cream, cloud.

In-text Questions — Page 89Separation in nature and in daily life

Q1Imagine trying to separate all the ingredients in a lemonade once they have been mixed. Can you do it?
Answer

Not completely — and certainly not by any one method. Lemonade (shikanji) is water with dissolved sugar, salt, lemon juice with its acids and pigments, and often bits of pulp and ice. You can get some components back, but not every one of them in pure form.

ComponentMethod that would workHow well
Pulp and seedsFiltrationEasily removed — they are a suspension
WaterDistillationRecovered pure
Sugar and salt togetherEvaporation or crystallizationRecovered, but mixed with each other
Coloured pigments of lemonPaper chromatographyOnly in traces, not in usable amounts
Sugar separated from saltRepeated fractional crystallizationVery difficult — both are soluble in water

Revise, Reflect, Refine — Pages 90 – 93End-of-chapter questions

Revise, Reflect, Refine
Q1Which 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.

OptionThe 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.
Q2Choose 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.

MixtureTypeParticle sizeTyndall effect?
(a) Air and dust particlesSuspension / aerosol> 1000 nmYes — shows it
(b) Copper sulfate and waterTrue solution< 1 nmNo
(c) Starch and waterColloid1 – 1000 nmYes — shows it
(d) Acetone and waterSolution of two miscible liquids< 1 nmNo

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: they are true solutions. The dissolved particles are smaller than 1 nm, far too small to scatter light, so the beam passes through unseen. (The blue colour of copper sulfate solution does not matter — colour is not scattering.)

Q3A 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.]
SolutionSuspensionColloid
Properties
______________
______________
Properties
______________
______________
Properties
______________
______________
Examples
______________
______________
Examples
______________
______________
Examples
______________
______________
Table 5.2, page 90 — as printed, with blanks for the properties and examples of each type of mixture.
Answer

Table 5.2 completed

SolutionSuspensionColloid
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
Q4Solve 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

(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
Q5The 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.

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/cm³
Density of water = 1.00 g/cm³

Compare: 0.91 g/cm³ < 1.00 g/cm³, and oil does not mix with water, so the oil forms the upper layer.

Glass stopperCooking oil, density0.91 g/cm³ (upper layer)Water, density1.00 g/cm³ (lower layer)StopcockConical flask —water drained firstStand
Separating funnel: oil forms the upper layer, water the lower layer.

To separate them:

  1. Pour the mixture into a separating funnel fixed on a stand and let it stand until two clear layers form.
  2. Open the stopcock slowly and run the lower layer (water) into a flask. Close it when the boundary reaches the tap, and discard the small mixed portion.
  3. Collect the upper layer (oil) in a separate, clean container.
Q6Assertion (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 ✓
Q7How 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.
MixtureMethod of separationReason for selection
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  
Table 5.3, page 91 — as printed, with the two right-hand columns blank.
Answer

Table 5.3 completed

MixtureMethod of separationReason for selection
Mud from muddy waterSedimentation and decantation, then filtration; add alum (coagulation) or centrifuge if it stays cloudyMud 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 sampleCentrifugationBlood cannot be separated by filtration, and its cells settle only very slowly on standing. Spinning produces a much stronger outward force, so the denser cells collect at the bottom, leaving the lighter plasma on top.
Naphthalene and sandSublimationNaphthalene 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 saltAdd water, filter, then evaporate or crystallise the filtrateSalt is soluble in water and chalk is not. Filtration takes out the chalk as residue; evaporating the filtrate returns the salt.
Common salt and waterEvaporation (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 waterSeparating funnelThe 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 flowerPaper chromatographyThe 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.
Q8Two 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
ThermometerWater outletWater inletWater condenserDistillationflaskMixture ofacetone andwaterConical flaskStandBurneracetone vapour risesvapour cools → distillate
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

  1. Put the mixture in a distillation flask fitted with a thermometer and joined to a water condenser; place a conical flask as the receiver.
  2. Heat gently. The thermometer steadies at about 60 °C while liquid A boils; its vapour is cooled in the condenser and collects as pure A in the conical flask.
  3. When the temperature starts rising above 60 °C, change the receiver. Liquid B (boiling point 90 °C) remains in the distillation flask.
Q9Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?
Answer
Point of comparisonEvaporationCrystallizationDistillation
What is doneThe solvent is driven off into the airA hot saturated solution is cooled slowlyThe liquid is boiled off, then condensed and collected
What you obtainThe solute only; the solvent is lostPure crystals of the solute; the liquid left over is discardedBoth components, each in pure form
Purity of the solid obtainedLow — every dissolved impurity is left behind with itHigh — impurities stay in the solutionThe liquid collected is pure
ApparatusAn open dish; often just sun and windBeaker, funnel, watch glass — simpleDistillation flask, thermometer, condenser, receiver
Cost and effortLowestModerateHighest — 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.
Q10Blood 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 colloidWhat it is in blood
Dispersed phaseThe blood cells and proteins — red blood cells, white blood cells, platelets and plasma proteins
Dispersion mediumPlasma, which is mostly water with dissolved salts, glucose and hormones
Q11You 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.
Fig. 5.25: (a)123Fig. 5.25: (b)
Fig. 5.25, page 92 — redrawn sketch. (a) the mixture as it is given to you; (b) the three set-ups, numbered 1, 2 and 3, shown in the book.
Answer

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

OrderStep in Fig. 5.25bTechniqueWhat is obtained
FirstStep 1 — inverted funnel over a china dish on a tripod, heatedSublimationNaphthalene collects on the cool funnel wall; sand and salt remain in the dish
SecondStep 3 — the mixture stirred with water and poured through a filter funnelFiltrationSand stays on the filter paper; salt solution passes through as the filtrate
ThirdStep 2 — the filtrate heated in a china dishEvaporation / crystallizationWater boils away, leaving pure common salt
Q12Why 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.

Q13Answer 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.
Salts10 °C20 °C30 °C40 °C60 °C80 °C
Potassium nitrate21324562106167
Sodium chloride363636.336.53737
Potassium chloride353537.4404654
Ammonium chloride243741415566
Table 5.4, page 92 — Solubility of various salts (in g per 100 g of water) at different temperatures.
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.

Solubility of KCl: 54 g at 80 °C; about 36 g at 25 °C (between 35 g at 20 °C and 37.4 g at 30 °C), per 100 g of water
Mass that crystallises out ≈ 54 g − 36 g = about 18 g per 100 g of water

(iii) For all four salts the solubility increases as the temperature increases — but by very different amounts (for sodium chloride, only very slightly).

SaltSolubility at 10 °CSolubility at 80 °CIncreaseHow many times
Potassium nitrate21 g167 g146 g≈ 8.0 times
Ammonium chloride24 g66 g42 g≈ 2.8 times
Potassium chloride35 g54 g19 g≈ 1.5 times
Sodium chloride36 g37 g1 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.

Q14Three 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.

StudentSugarWaterSolution% m/m
A20 g80 g100 g20.0 %
B20 g100 g120 g16.7 %
C30 g80 g110 g27.3 %
Q15Examine 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
ABCS ……………………………………
Fig. 5.26, page 93 — redrawn sketch of the apparatus, with the parts marked A, B and C and the technique marked S, exactly as printed.
SolventWaterAcetoneAlcoholChloroformBenzene
Temperature (°C)100 °C56 °C78 °C61 °C80 °C
Table 5.5, page 93 — Boiling points of some compounds.
Answer

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

(ii) Labels

Label in Fig. 5.26ApparatusIts job
ADistillation flaskHolds the mixture and is heated so that the lower-boiling liquid vaporises
BWater condenserCold water circulating in the outer jacket cools the vapour and turns it back into liquid
CConical flask (receiver)Collects the pure distillate as it drips out of the condenser

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

MixtureBoiling points from Table 5.5DifferenceCan distillation separate it?
(a) water — acetone100 °C and 56 °C44 °CYes — gap is more than 25 °C
(b) water — saltSalt is a non-volatile solid—Yes — the water distils over and the salt is left in the flask
(c) acetone — alcohol56 °C and 78 °C22 °CNo — gap is less than 25 °C; needs fractional distillation
(d) sand — saltBoth are solids—No — nothing vaporises; dissolve in water and filter instead
(e) alcohol — chloroform78 °C and 61 °C17 °CNo — gap is less than 25 °C
(f) alcohol — benzene78 °C and 80 °C2 °CNo — the boiling points are almost the same

The Journey Beyond — Page 93Projects and investigations

The Journey Beyond
Q1Demonstrate the Tyndall effect using different colloids. Create a series of experiments showcasing how light scatters in colloids making the beam visible. Use laser pointers (Safety first: Use it under the supervision of an adult), flashlights, or other light sources for your demonstrations.
Answer

What a good demonstration must contain: at least one true solution as a control, several colloids, and the beam observed from the side at right angles each time — so that the difference is a comparison and not just a single observation.

Sample plan

  1. Take five identical glass tumblers. Fill them with: (1) plain water, (2) a clear sugar or salt solution, (3) water with two drops of milk, (4) water with a few drops of soap solution shaken with a drop of oil, (5) very dilute starch paste in water.
  2. Darken the room. Shine a laser pointer or a narrow torch beam through each tumbler in turn, and look at the beam from the side, at right angles to it.
  3. Record for each: is the path of the beam visible? Is it bright or faint?
SampleType of mixtureExpected observation
Water; sugar solutionSolutionBeam path invisible
Diluted milkColloid (emulsion)Beam path clearly visible
Soap and oil in waterColloid (emulsion)Beam path visible
Dilute starch pasteColloidBeam path visible
Q2Make crystals of different compounds (common salt, epsom salt, sugar, borax, nickel sulfate, etc.). Observe them under a magnifying glass or a microscope. Note down their colours and shapes.
Answer

Method. For each compound, dissolve it in warm water until no more will dissolve — a saturated solution. Filter it while warm, pour it into a clean, undisturbed container, cover it, and let it cool slowly. Leave it for a day or two, then lift out the crystals, rinse them with a little cold water and dry them on a watch glass. Examine each under a magnifying glass and sketch what you see.

CompoundColourTypical crystal shape
Common salt (sodium chloride)Colourless / whiteCubes
SugarColourlessLong six-sided blocks (as in mishri)
Epsom salt (magnesium sulfate)ColourlessLong needles
BoraxWhiteShort prisms
Copper sulfate / nickel sulfateBlue / greenSlanted blocky crystals
Q3Do red leaves also contain green pigments? Investigate it using paper chromatography.
Answer

Expected answer: yes, most red leaves also contain chlorophyll — the red pigment simply masks it. Paper chromatography can reveal the green.

What you should see: the single spot separates into several bands — usually a red or purple band, a green band of chlorophyll and often a yellow-orange band of carotenoid. The green band proves that the red leaf does contain chlorophyll.

Q4You can try chromatography to find the number of components present in a food colour (green, orange, yellow, etc.) or in coloured mouth fresheners (fennel seeds).
Answer

Method. Dissolve a little of the food colour, or soak the coloured coating off a few sugar-coated fennel seeds, in a few drops of water. Spot the coloured liquid on the pencil line of a chromatography strip, dry it, and stand the strip in water (or in 2 % m/v salt solution) with the solvent level below the spot. Let the solvent rise, then dry the strip.

SampleTypical resultConclusion
Green food colourTwo spots — blue and yellowA mixture of two dyes
Orange food colourOften a red and a yellow spotA mixture
Coloured fennel-seed coatingOne or more spots depending on the brandShows which dyes have been used
Q5Design an educational game where players identify and apply separation techniques to different mixtures through interactive challenges and hands-on activities.
Answer

Sample answer — “Separation Snap” (a card game):

  • Mixture cards: sand + salt, oil + water, camphor + sand, muddy water, blood, black ink, seawater, acetone + water, brass.
  • Technique cards: filtration, sublimation, separating funnel, centrifugation, coagulation, chromatography, distillation, evaporation, crystallization, “cannot be separated by physical methods”.
  • Rules: a player wins a mixture card by laying the correct technique card and naming the property used (particle size, solubility, density, boiling point, sublimation). Some mixtures need techniques in the right order; brass is a trap card. Every fifth turn, the group actually performs a separation at the bench.
Q6If you are camping outdoors and running short on clean water, you can obtain clean water by distillation. Can you think of a set-up with the items available to you?
Answer

Sample answer — a pot-and-lid still: Half fill a large pot with the dirty water and stand a small empty cup inside it on a stone, above the water level. Cover the pot with its lid turned upside down, so that the lowest point of the lid is right above the cup, and keep the lid cool with a wet cloth or cold stones. Heat the pot on the camp fire. Steam rises, condenses on the cool lid, runs to its lowest point and drips into the cup as clean water, while the mud and salt stay in the pot.

The Quest Continues … — Page 93An open question for you to carry forward

The Quest Continues …
Q1Can we create artificial blood that works just as real blood for all patients?
Answer

Not yet — but scientists are working on it, and the difficulty is exactly the one this chapter is about: blood is a colloid with several different jobs, not a single substance you can bottle.

Chapter 5 of 12 – Science

Also for This Chapter

NCERT Solutions Class 9 Science Chapter 5 Exploring Mixtures and Their Separation – Overview

ClassClass 9
SubjectScience
Chapter5 – Exploring Mixtures and Their Separation
Textbook pages72 – 93
Questions solved81
Session2027
FormatFull answers on this page + PDF

NCERT Solutions Class 9 Science Chapter 5 – FAQs

Where can I download NCERT Solutions Class 9 Science Chapter 5 Exploring Mixtures and Their Separation PDF?

Read every answer on this page, or download the free PDF from the table at the top.

Is this for the 2027 session?

Yes, it follows the latest NCERT Class 9 Science book for 2027.

How many questions are solved in NCERT Class 9 Science Chapter 5?

All 81 questions of the chapter (textbook pages 72 – 93) – the in-text questions, activities and the end-of-chapter exercises.

How should I use these solutions?

Read the chapter in the NCERT book first, attempt the questions yourself, then check your answers here.

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