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.
| Chapter | Download | |
|---|---|---|
| Solutions PDF | Exploring Mixtures and Their Separation | View PDF |
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
Because the two mixtures hold particles of very different size. Muddy water is a suspension; milk is a colloid.
Milk (fat and protein) particles: diameter 1 – 1000 nm
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 comparison | Evaporation | Boiling |
|---|---|---|
| Where it happens | Only at the free surface of the liquid | Throughout the liquid — bubbles form inside it |
| Temperature | At all temperatures below the boiling point | Only at the boiling point (100 °C for water at normal pressure) |
| Speed | Slow and quiet | Fast and vigorous |
| Energy source | Draws heat from the liquid itself and the surroundings, so the liquid cools | Heat must be supplied continuously from outside |
| Everyday example | Wet clothes drying, sweat cooling you | Water bubbling in a kettle |
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?
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.
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?
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).
| Beaker | Mixture | Are particles visible? | Appearance |
|---|---|---|---|
| A | Common salt + 50 mL water | No | Clear and transparent — you can read print through it |
| B | Chalk powder + 50 mL water | Yes | Cloudy; white specks clearly seen swirling in the water |
| C | A few drops of milk + 50 mL water | No | Uniformly milky-white, but no separate specks are seen |
The path of the beam is invisible in A, brightly visible in B and visible in C.
| Beaker | Path of the laser beam seen from the side | What it tells you |
|---|---|---|
| A — salt and water | Not visible | A true solution; particles are far too small to scatter light |
| B — chalk and water | Bright, sharply visible | A suspension; large particles scatter light strongly |
| C — milk and water | Visible | A colloid; the mixture looks uniform yet still scatters light |
- 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.
Only beaker B leaves a residue.
| Beaker | Residue on the filter paper | Filtrate |
|---|---|---|
| A — salt and water | None | Clear salt solution (still salty — the salt passes through) |
| B — chalk and water | White chalk residue | Clear water |
| C — milk and water | None | Milky liquid — unchanged |
They are three different types of mixture. Put the four tests side by side and each beaker gives its own pattern of answers.
| Test | A — salt + water | B — chalk + water | C — milk + water |
|---|---|---|---|
| Particles visible? | No | Yes | No |
| Tyndall effect? | No | Yes | Yes |
| Settles on standing? | No | Yes | No |
| Residue on filtering? | No | Yes | No |
| Type of mixture | Solution (homogeneous) | Suspension (heterogeneous) | Colloid (heterogeneous) |
In-text Questions — Page 745.2 Solutions; 5.2.1 Concentration of a solution
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.
At Class 9 level it is written as a percentage in one of three ways:
% m/v = (mass of solute ÷ volume of solution) × 100
% v/v = (volume of solute ÷ volume of solution) × 100
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.
Yes — almost every prepared solution has a proportion that must be respected.
| Situation | Why the exact proportion matters |
|---|---|
| ORS for a child with diarrhoea | The salt and sugar must match the concentration of body fluids, or water will not be absorbed properly |
| Saline drip in a hospital | 0.9 % m/v NaCl matches blood; stronger or weaker damages the blood cells |
| Bleaching solution for cleaning a floor | Too weak does not disinfect; too strong corrodes the surface and irritates the skin |
| Sugar syrup for jam or a cold drink | Decides taste and how long the food can be preserved |
| Fluoride in toothpaste, chlorine in drinking water | A useful amount protects; more than that is harmful |
| Sugar and milk in a cup of tea | Same ingredients, different proportion, completely different taste |
Pause and Ponder — Page 765.2.2 How do we express concentration?
12 g of zinc oxide.
4 = (mass of ZnO ÷ 300 g) × 100
mass of ZnO = (4 × 300 g) ÷ 100
mass of ZnO = 12 g
20 % v/v.
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
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.
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
Compound ‘B’ — at every temperature on the graph its curve lies far above that of compound ‘A’.
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
Less than.
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
Less than.
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
The solubility of compound ‘B’ increases more than that of compound ‘A’ with an increase in the temperature.
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
Part of the dissolved solute will come out of the solution as pure solid crystals. This is the process of crystallization.
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
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?
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.
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
A controlled experiment in which only the rate of cooling is changed:
- Prepare a hot saturated copper sulfate solution and filter it while hot.
- Divide it into two equal parts in two identical beakers.
- 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.
- After the same time, filter out the crystals from each beaker, rinse them with a little cold water and dry them.
- 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
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 A: at 80 °C ≈ 66 g at 60 °C ≈ 56 g
Step 2 — solid deposited per 100 g of water
Compound A: 66 g − 56 g ≈ 10 g
Yes. Slow evaporation gives large crystals; fast evaporation gives many small ones.
| Rate of evaporation | What happens in the solution | Crystals obtained |
|---|---|---|
| Slow (shallow pan, mild sun, still air) | The solution creeps past saturation gently, so only a few nuclei form and each keeps growing | Few, large, well-shaped crystals |
| Fast (strong heating, dry wind, thin film) | The solution shoots far past saturation, so a huge number of nuclei appear together | Many tiny, poorly formed crystals |
Activity 5.4: Let us describe a process — Page 795.3.1 Crystallization
Salt is obtained from seawater by evaporating the water, which leads to crystallization:
- Seawater, a dilute solution of salt, is let into wide, shallow pans near the coast.
- 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.
- 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
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
Use distillation instead of plain evaporation.
Distillation: solution → solute left in the flask and solvent condensed and collected
By distillation, provided their boiling points differ by at least about 25 °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.
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
Observe the paper as the water rises through the paper. What do you notice?
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.
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
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 separated | Suitable mobile phase | Reason |
|---|---|---|
| Black or coloured sketch-pen ink | Water, or 2 % m/v salt solution | The dyes are water-soluble |
| Green pigments of spinach leaves | Alcohol, or a solvent mixture | Chlorophyll is insoluble in water but dissolves in alcohol |
| Coloured pigments of flower petals | Alcohol or a water–alcohol mixture | Depends on which pigment the petal contains |
Pause and Ponder — Page 825.3.3 Paper Chromatography
(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.
| Statement | True / False | Corrected 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. | False | Distillation 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. | False | The solvent level should be below the sample spot at the beginning of the experiment. |
| (iv) Evaporation and crystallization are the same processes. | False | They 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
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.
Because mustard oil is less dense than water, so an equal volume of it weighs less and it floats.
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
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
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.
| Mixture | What is dispersed in the gas | Type |
|---|---|---|
| Smoke | Fine solid carbon and ash particles in air | Heterogeneous |
| Fog, mist, clouds | Tiny liquid water droplets in air | Heterogeneous |
| Dust in the air | Solid dust particles in air | Heterogeneous |
| Air freshener spray, an aerosol | Liquid droplets in air | Heterogeneous |
| Hydrogen and oxygen (rocket fuel) | Gas mixed with gas | Homogeneous |
Activity 5.7: Let us explore — Page 845.4.2 Sublimation
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.
Solid → vapour directly = sublimation
Vapour → solid directly = deposition
Pause and Ponder — Page 845.4.2 Sublimation
Because they do not dissolve in each other, and they have different densities.
Less dense liquid → floats → upper layer
Example: water (1.00 g/cm³) below, mustard oil (≈ 0.91 g/cm³) above
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 comparison | Sublimation | Evaporation |
|---|---|---|
| Change of state | Solid → vapour, with no liquid stage | Liquid → vapour |
| Starting material | A solid | A liquid |
| Where it occurs | At the surface of the solid | At the surface of the liquid |
| Which substances | Only a few — camphor, naphthalene, ammonium chloride, dry ice | Every liquid, at all temperatures |
| Reverse process | Deposition (vapour → solid) | Condensation (vapour → liquid) |
| Used to separate | A 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
Yes — any mixture in which exactly one component sublimes on gentle heating.
| Mixture | Component that sublimes | Component left behind |
|---|---|---|
| Naphthalene and sand | Naphthalene | Sand |
| Ammonium chloride and common salt | Ammonium chloride | Common salt |
| Camphor and common salt | Camphor | Common salt |
| Iodine and sand | Iodine | Sand |
| Dry ice mixed with an ordinary solid | Solid carbon dioxide | The other solid |
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.
| Alloy | Approximate composition | Why it is made |
|---|---|---|
| Brass | About 80 % copper and 20 % zinc | Harder than copper, easy to shape, does not corrode readily |
| Bronze | About 80 % copper and 20 % tin | Strong and hard-wearing; used for statues and bells |
| Stainless steel | Iron 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
Start with the simplest method and add another only if the water is still not clear.
- 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.
- Filtration. Pass it through filter paper or a clean cotton cloth. Particles larger than the pores are held back as residue.
- 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.
- Centrifugation. Spin a sample at high speed; the centrifugal force drives even very fine particles to the bottom of the tube.
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
The same two-person spinning game is played all over India and has a different name in each language.
| Language / region | Name 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
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.
| Device | How it works | What it makes possible |
|---|---|---|
| Paperfuge | A cardboard disc on twisted strings, like a button whirligig; spinning it at very high speed separates blood by centrifugation | Detecting malaria and anaemia without electricity |
| Foldscope | A microscope folded from a paper sheet with a tiny glass lens | Seeing microbes and blood cells in the field, at very low cost |
| Paper-strip diagnostic tests | A drop of blood or urine travels along treated paper by capillary action and changes colour | Instant tests for pregnancy, malaria, diabetes and infections |
| Solar-powered vaccine carrier | An insulated box cooled by a small solar panel | Vaccines kept cold on the way to villages with no power |
| Jaipur foot | A rubber and wood artificial limb made by hand from local materials | Affordable mobility for people who have lost a leg |
Activity 5.8: Let us make a model — Page 865.4.3 A. Centrifugation
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
Yes — coagulation is at work in the kitchen, at the water works, and even in your own body.
| Everyday process | Coagulant used | What clumps together |
|---|---|---|
| Making paneer or chhena from milk | Lemon juice or vinegar (an acid) | Milk proteins clump and separate from the whey |
| Setting curd from milk | Lactic acid made by the bacteria in the starter | Milk proteins form a soft solid network |
| Purifying drinking water at a water works | Alum (fitkari) | Fine clay and silt particles form settleable flocs |
| Sewage treatment | Alum or other coagulants | Suspended solids settle out before filtration |
| Blood clotting over a cut | The body’s own clotting proteins | Blood cells are trapped in a fibrous mesh, sealing the wound |
| Setting an egg while cooking | Heat | Egg proteins clump into a solid white |
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
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
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.
| Property | Solution | Colloid | Suspension |
|---|---|---|---|
| Nature | Homogeneous | Heterogeneous (looks homogeneous) | Heterogeneous |
| Particle size | Less than 1 nm | 1 – 1000 nm | More than 1000 nm |
| Particles visible? | No | No | Yes, to the naked eye |
| Separated by filtration? | No | No | Yes |
| Settles on standing? | No | No | Yes |
| Tyndall effect | Not shown | Shown | Shown |
Colloids are everywhere in daily life — in the kitchen, in the bathroom shelf, in the sky and in your own body.
| Colloid | Dispersed phase | Dispersion medium |
|---|---|---|
| Milk | Liquid fat droplets | Water (an oil-in-water emulsion) |
| Butter | Water droplets | Fat (a water-in-oil emulsion) |
| Blood | Blood cells and proteins | Plasma |
| Ice cream, tomato sauce | Fat and solid particles | Water |
| Cloud, fog, mist | Water droplets | Air |
| Smoke | Solid carbon particles | Air |
| Cold cream, body lotion, vanishing cream | Water or oil droplets | Oil or water |
In-text Questions — Page 885.5 Tyndall Effect
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
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.
| Test | What a cloud does | Conclusion |
|---|---|---|
| Are the droplets dissolved in the air? | No — they stay as separate droplets of liquid water | Not a solution |
| Do they settle out? | No — a cloud floats for hours or days | Not a suspension |
| Does it scatter light? | Yes — that is why clouds look white and cast shadows | Shows the Tyndall effect |
| Do you see separate droplets? | No — the cloud looks like one smooth mass | Particles are of colloidal size → a colloid |
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
| S. No. | Property | Solution | Suspension | Colloid |
|---|---|---|---|---|
| 1. | Nature (homogeneous/heterogeneous) | |||
| 2. | Particle size | |||
| 3. | Visibility | |||
| 4. | Separation by filtration | |||
| 5. | Settling | |||
| 6. | Tyndall effect |
Table 5.1: Properties of different types of mixtures
| S. No. | Property | Solution | Suspension | Colloid |
|---|---|---|---|---|
| 1. | Nature (homogeneous / heterogeneous) | Homogeneous | Heterogeneous | Heterogeneous, though it looks homogeneous |
| 2. | Particle size | Less than 1 nm | More than 1000 nm | 1 – 1000 nm |
| 3. | Visibility | Particles not visible; the mixture is transparent | Particles visible to the naked eye | Particles not visible; the mixture looks uniform |
| 4. | Separation by filtration | Cannot be separated | Can be separated — leaves a residue | Cannot be separated |
| 5. | Settling | Does not settle | Settles on standing (sedimentation) | Does not settle |
| 6. | Tyndall effect | Not shown | Shown | Shown |
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
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.
| Component | Method that would work | How well |
|---|---|---|
| Pulp and seeds | Filtration | Easily removed — they are a suspension |
| Water | Distillation | Recovered pure |
| Sugar and salt together | Evaporation or crystallization | Recovered, but mixed with each other |
| Coloured pigments of lemon | Paper chromatography | Only in traces, not in usable amounts |
| Sugar separated from salt | Repeated fractional crystallization | Very difficult — both are soluble in water |
Revise, Reflect, Refine — Pages 90 – 93End-of-chapter questions
(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
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. |
(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
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: 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.)
| Solution | Suspension | Colloid |
|---|---|---|
| Properties ______________ ______________ | Properties ______________ ______________ | Properties ______________ ______________ |
| Examples ______________ ______________ | Examples ______________ ______________ | Examples ______________ ______________ |
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 |
(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.
(i) All three ingredients are solids, so the appropriate method is the mass by mass percentage (% m/m).
% 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 zinc = 120 g − 84 g = 36 g
Yes, it forms a separate layer, and the oil floats on top. The two are separated with a separating funnel.
Density of the oil:
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.
To separate them:
- Pour the mixture into a separating funnel fixed on a stand and let it stand until two clear layers form.
- 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.
- Collect the upper layer (oil) in a separate, clean container.
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.
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.
Correct value: particle size of a solution < 1 nm ✓
| Mixture | Method of separation | Reason 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 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 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 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. |
Separate them by simple distillation.
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
Procedure
- Put the mixture in a distillation flask fitted with a thermometer and joined to a water condenser; place a conical flask as the receiver.
- 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.
- When the temperature starts rising above 60 °C, change the receiver. Liquid B (boiling point 90 °C) remains in the distillation flask.
| 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.
(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 |
The correct sequence is 1 → 3 → 2, that is sublimation → filtration → evaporation (crystallization).
| 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 |
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 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.
(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.
| Salts | 10 °C | 20 °C | 30 °C | 40 °C | 60 °C | 80 °C |
|---|---|---|---|---|---|---|
| Potassium nitrate | 21 | 32 | 45 | 62 | 106 | 167 |
| Sodium chloride | 36 | 36 | 36.3 | 36.5 | 37 | 37 |
| Potassium chloride | 35 | 35 | 37.4 | 40 | 46 | 54 |
| Ammonium chloride | 24 | 37 | 41 | 41 | 55 | 66 |
(i) 31 g of potassium nitrate.
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.
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).
| 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.
• 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.
(i) Remember that the denominator is the mass of the solution, that is sugar + water — not the mass of water alone.
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 % |
(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
| Solvent | Water | Acetone | Alcohol | Chloroform | Benzene |
|---|---|---|---|---|---|
| Temperature (°C) | 100 °C | 56 °C | 78 °C | 61 °C | 80 °C |
(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 |
(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 Journey Beyond — Page 93Projects and investigations
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
- 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.
- 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.
- Record for each: is the path of the beam visible? Is it bright or faint?
| Sample | Type of mixture | Expected observation |
|---|---|---|
| Water; sugar solution | Solution | Beam path invisible |
| Diluted milk | Colloid (emulsion) | Beam path clearly visible |
| Soap and oil in water | Colloid (emulsion) | Beam path visible |
| Dilute starch paste | Colloid | Beam path visible |
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.
| Compound | Colour | Typical crystal shape |
|---|---|---|
| Common salt (sodium chloride) | Colourless / white | Cubes |
| Sugar | Colourless | Long six-sided blocks (as in mishri) |
| Epsom salt (magnesium sulfate) | Colourless | Long needles |
| Borax | White | Short prisms |
| Copper sulfate / nickel sulfate | Blue / green | Slanted blocky crystals |
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.
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.
| Sample | Typical result | Conclusion |
|---|---|---|
| Green food colour | Two spots — blue and yellow | A mixture of two dyes |
| Orange food colour | Often a red and a yellow spot | A mixture |
| Coloured fennel-seed coating | One or more spots depending on the brand | Shows which dyes have been used |
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.
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
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
- Previous: Chapter 4 Describing Motion Around Us
- Next: Chapter 6 How Forces Affect Motion
- All chapters: NCERT Solutions Class 9 Science
Also for This Chapter
NCERT Solutions Class 9 Science Chapter 5 Exploring Mixtures and Their Separation – Overview
| Class | Class 9 |
| Subject | Science |
| Chapter | 5 – Exploring Mixtures and Their Separation |
| Textbook pages | 72 – 93 |
| Questions solved | 81 |
| Session | 2027 |
| Format | Full 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?
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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.