Matter, Mixtures and Separation

Start with two questions

A clear liquid may be pure water, a sugar solution or another substance. Looking clear cannot answer the useful chemistry questions: What is it made of? What property could separate its components?

By the end of this lesson you will classify matter in two different ways, explain a physical change with a particle model, and choose a separation from its purpose and the available property differences. You need ordinary reading and simple counting. The optional percentage box is not required for the five checks or the module review.

We will reason from descriptions and diagrams. Do not taste, directly smell, heat or mix unknown substances, build apparatus or perform the chemical processes described here. A familiar name does not make an unknown sample safe.

1. State and composition are different questions

Matter has mass and occupies space. In a simple particle model, matter consists of very small particles with spaces and interactions between them. The useful particle may be an atom, molecule or ion; those words are explained more fully in the next lesson. A dust speck is not a single atom.

At stated conditions, a solid usually has a definite shape and volume; a liquid has a definite volume but takes the container's shape; a gas spreads to fill the available container. Solids and liquids are much less compressible than gases, rather than absolutely impossible to compress. Particles in a solid vibrate about positions. In a liquid they can move past one another; in a gas they move through relatively large spaces. Changing arrangement and motion does not require each particle to swell.

The other question is composition:

• A pure substance is one chemical substance. It may be an element or a compound. Here, 'pure' does not mean healthy, natural or safe.
• An element contains only one kind of element's atoms. Oxygen is an element even when its atoms are joined in pairs, written O₂.
• A compound contains different elements chemically combined in a fixed composition. Water is a compound: H₂O indicates two hydrogen atoms for each oxygen atom in a water molecule.
• A mixture contains two or more substances together. Its composition can vary. Mixing does not, by itself, give all the components a new common chemical identity.

A compound is not a mixture merely because its formula contains different letters. A mixture may contain compounds as well as elements. Some compounds form extended ionic structures rather than separate molecules; the next lesson develops that distinction.

Worked case CF01-A: classify along both axes

The descriptions deliberately give the information needed about purity.

• Pure ice: solid; pure compound. It is still water.
• Pure oxygen gas: gas; element. Two oxygen atoms in O₂ are still atoms of the same element.
• A clear sugar solution: liquid; mixture. Sugar and water are present together, although individual components are not visible.
• Sand dispersed in water: a heterogeneous mixture containing solid and liquid components. Do not force every mixture into a single physical-state label.

Figure CF1-V01. Two questions, two labels

Read the state column separately from composition. The particle symbols in a model need a key; colour alone cannot identify an element or prove purity.

2. A state change can leave the substance unchanged

A physical change changes such features as state, size, shape or arrangement while the chemical substance remains the same. A chemical change produces different substances. Whether a change is easy to reverse is not the definition: cutting a sheet into tiny pieces may be hard to undo but does not, by itself, create a new substance. Reaction evidence is developed in lesson 4.

Worked case CF01-E: ice becomes water

Figure CF1-V02. Representative particle model, not to scale

Ice: ↔ marks vibration about positions. Each [H₂O] is one unchanged water group.

[H₂O]↔   [H₂O]↔   [H₂O]↔
[H₂O]↔   [H₂O]↔   [H₂O]↔

Liquid water: arrows mark possible movement; the six groups are less ordered.

       [H₂O]→        [H₂O]↗
[H₂O]↙       [H₂O]↓
      [H₂O]→   [H₂O]↗

Key: H is hydrogen and O is oxygen; each labelled group contains two H atoms and one O atom. Both panels represent the same six water groups, not all the molecules in a real container. In ice they vibrate about positions; in liquid they can move past one another.

Pause before reading on: Should the second panel contain larger H₂O groups, separate H and O atoms, or the same H₂O groups in a different arrangement?

Answer: The same H₂O groups. The substance remains water, so this is a physical change and the sample remains a compound. Replacing H₂O groups by separate H and O atoms would represent a different process, not melting. The picture is not to scale and represents a tiny portion of the material. It does not imply that solid particles are motionless or that all substances expand on melting; water is an important reason not to use that blanket claim.

3. Uniform appearance is useful evidence, not proof of purity

A homogeneous mixture has uniform composition throughout the region being considered. A salt or sugar solution is a familiar example. A heterogeneous mixture has nonuniform composition at the relevant scale, such as a visible sand-water mixture.

In a solution, the solute is dissolved in the solvent. In a simple sugar-water solution, sugar is the solute and water the solvent. The dissolved particles do not settle out on ordinary standing and pass ordinary filter paper. Dissolving does not mean the solute has vanished. Solubility is the maximum amount of a specified solute that can dissolve in a stated amount of solvent under stated conditions, including temperature and pressure when relevant. It is not the speed at which a spoonful disappears. A more concentrated solution contains more solute relative to the stated amount of solution or solvent; always specify the comparison basis.

A suspension contains dispersed particles large enough to make the mixture heterogeneous. In the simple examples here, they settle on standing and can be retained by ordinary filter paper. Very fine suspended particles may need longer to settle or a different separation, so the observation conditions matter.

A colloid also has dispersed material in a medium. It may look uniform to the eye but is heterogeneous at the smaller particle scale. Its dispersed particles are intermediate in size between those in a true solution and the coarse particles in our suspension examples. They commonly remain dispersed on ordinary standing, pass ordinary filter paper and scatter enough light to make a beam path visible. This visible scattering is called the Tyndall effect in the colloid context. Suspensions can scatter light too. Passing a paper filter does not make a sample a solution or a pure substance.

Worked case CF01-C: combine the observations

S, T and U are known mixtures, compared within the idealised school model. The standing interval is ten minutes and the filter is ordinary filter paper.

• S: no settling; passes the paper; no visible light path in the stated beam test
• T: visible sediment within ten minutes; retained by the paper
• U: remains dispersed during the interval; passes the paper; visible light path

S fits a solution, T a suspension and U a colloid within this supplied comparison. S's observations without the known-mixture premise would also be compatible with a pure liquid. Scattering alone would not separate a colloid from a suspension. Ten minutes without settling is not a universal proof of colloid identity. If a real sample remains ambiguous, more evidence, such as a suitable particle-size measurement, is needed.

Figure CF1-V03. Known mixtures: combine the evidence

The known-mixture premise and all supplied observations matter. Scattering by itself is not unique to a colloid.

4. Choose a separation by property and purpose

Ask three questions in order: What components are present? Which property differs? Which component must be recovered? The method must use that difference without assuming a chemical reaction will separate the mixture.

Learn the following core decisions:

• Filtration / निस्यंदन: a barrier retains suitable insoluble particles while liquid passes. The retained material is the residue; the liquid passing through is the filtrate. Ordinary filter paper does not remove dissolved salt or sugar.
• Evaporation / वाष्पीकरण: solvent enters the vapour phase and can leave a nonvolatile solute behind. Ordinary open evaporation does not collect the solvent. Evaporation can happen without boiling.
• Crystallisation / क्रिस्टलीकरण: under a supplied change in conditions, solubility changes and the desired solid forms crystals from solution. This can help separate it from impurities with different behaviour. It does not guarantee perfect purity, and no crystallisation calculation is needed here.
• Distillation / आसवन: a suitable component enters the vapour phase, is condensed and is collected. For water with a nonvolatile dissolved solute, this provides a way to collect water separately. The suitability of a liquid mixture depends on its properties; do not assume all liquids are separated equally easily.
• Separating-funnel principle / पृथक्करण कीप का सिद्धांत: immiscible liquids form separate layers under stated conditions. A described apparatus can separate those layers. Density helps determine which layer is lower; density difference alone does not create layers in two mutually miscible liquids.

Worked case CF01-B: the recovery goal changes the answer

A described laboratory sample has coarse insoluble sand, dissolved sodium chloride and water. The sand is retained by ordinary filter paper.

Stage 1: Filtration separates sand as residue. Dissolved salt stays with water in the filtrate. A claim that the filter also catches dissolved salt fails at this first check.

Goal A, recover salt: With the salt stated to be nonvolatile and stable under the conditions, evaporation can leave it behind. A suitable crystallisation design can be chosen when the supplied solubility behaviour supports crystal recovery and purification.

Goal B, collect water: A distillation design includes condensation and collection of water vapour. Ordinary evaporation alone loses the water to the surroundings.

Figure CF1-V04. Branch by the recovery goal

Sand + dissolved salt + water → ordinary filtration

• Residue: coarse sand
• Filtrate: dissolved salt + water

From that filtrate:

This is a choice between specified processes, not a claim that one method is universally best. Nothing here proves that the collected water is safe to drink.

5. Transfer the same reasoning to other methods

The following contrast panel supplies each principle. You do not need ten memorised material-method pairs. In any question using an unfamiliar method from this panel, the relevant principle will be given.

Worked contrast panel CF01-F

Gravity or spinning: If insoluble particles settle under gravity, this is sedimentation. Separating the upper liquid from the settled material is decantation. If the supplied fact says spinning makes finer dispersed particles settle more quickly, the matching method is centrifugation. The difference is the stated settling behaviour, not merely whether the sample looks cloudy.

Magnetic or nonmagnetic: If one solid is attracted by a magnet and the other is not, magnetic separation uses that difference. It offers no separating advantage when neither component responds in the described situation.

Dyes that travel differently: Dissolved dyes can all pass ordinary filter paper. If they interact differently with a stationary paper phase and a moving solvent, they can move at different rates and separate by paper chromatography. Different colours help us see bands; colour difference itself is not the separating force.

Direct vapour or crystals from solution: When one supplied solid changes directly into vapour and later deposits as solid while the other does not under the same conditions, the useful behaviour is sublimation followed by deposition. Crystallisation instead uses the stated behaviour of a solution. Do not confuse the two because both may produce a solid at the end.

Separate liquid layers or one uniform liquid phase: Immiscible layers support the separating-funnel principle. Putting a homogeneous liquid mixture into the same apparatus does not create separate layers. A suitable volatility-based method would need its own supplied conditions.

These are explanations of described systems, not instructions to operate equipment. The sorting-station analogy connects these choices: a method is useful only when it can act on a relevant property difference.

Optional bridge: concentration by mass, CF01-D

This box is optional and is not tested in this lesson's compulsory checks or the module review.

A final solution contains 9 g of solute and 141 g of solvent, with no material lost. Solution mass = 9 + 141 = 150 g. Mass percentage = (9/150) × 100 = 6%. Dividing by 141 g would use solvent mass instead of the required solution mass. State the percentage basis rather than treating every concentration percentage as the same quantity.

Check your reasoning

Try the five questions before reading the explanations. This is free, untimed learning practice with no negative marking. It is not an official RRB paper or a topic-weightage prediction.

1. A sample contains only carbon dioxide, whose carbon and oxygen are chemically combined in a fixed composition. Another sample is ordinary dry air containing several gases. Which classification is correct?

A. Carbon dioxide: compound; air: mixture B. Both are compounds C. Carbon dioxide: mixture; air: compound D. Both are elements

2. Pure ice melts in a closed container without any chemical reaction. Which particle-model explanation fits?

A. Each water particle grows into a new substance B. Particles were completely motionless in the ice C. Water remains H₂O; arrangement and freedom of movement change D. The H₂O groups split into separate hydrogen and oxygen atoms

3. A described sample contains coarse sand, fully dissolved sugar and water. Sand is insoluble and is retained by ordinary filter paper. What does one filtration stage achieve?

A. Both sand and dissolved sugar remain on the paper B. Only sugar remains on the paper C. Nothing remains on the paper D. Sand remains on the paper; dissolved sugar passes through with water

4. The goal is to collect water separately from a salt solution. The salt is nonvolatile under the stated conditions. Which conceptual method includes a stage that collects the water again after it becomes vapour?

A. Ordinary filtration alone B. Distillation with condensation and collection C. Evaporation into the open air alone D. Magnetic separation

5. Three samples are already known to be mixtures. In an idealised school comparison, V passes ordinary filter paper, does not settle during 15 minutes and gives no visible light path. W forms visible sediment during the same interval and is retained by the paper. X stays dispersed during the interval, passes the paper and shows a light path. Which statement respects this evidence and its limits?

A. V fits a solution and W a suspension; X fits a colloid in this model, but scattering alone would not prove it B. V must be a pure liquid because its light path is invisible C. X must be a true solution because it passes filter paper D. Every light-scattering mixture must be a colloid

Answers and option diagnoses

1. A. Carbon dioxide: compound; air: mixture

A is correct. The first sample is one chemical substance made from different elements in a fixed composition. Air contains several substances mixed together. B wrongly gives the mixture a single compound identity. C reverses the two categories. D ignores both the different elements in carbon dioxide and the several substances in air. Being gaseous does not decide composition.

2. C. Water remains H₂O; arrangement and freedom of movement change

C preserves substance identity while explaining the change of state. In ice, particles vibrate about positions; in liquid water they can move past one another. A invents a new substance and particle growth. B treats a solid as having no particle motion. D describes breaking the water groups, which is not melting. The drawing need not claim that all materials expand on melting.

3. D. Sand remains on the paper; dissolved sugar passes through with water

D is correct. The stage separates the stated coarse insoluble particles from the liquid. Dissolved sugar stays in the filtrate with water. A wrongly treats dissolved sugar as coarse particles. B selects the component that passes through and misses the sand. C contradicts the stated retention of sand. Filtration has not recovered dry sugar or proved the liquid drinkable.

4. B. Distillation with condensation and collection

B is correct: water vapour is condensed and collected in the described distillation design. A does not separate dissolved salt from water. C may leave salt behind but loses the water to the surroundings instead of collecting it. D has no relevant magnetic-property difference in this problem. Collected water is not automatically certified safe for drinking; that is a different question.

5. A. V fits a solution and W a suspension; X fits a colloid in this model, but scattering alone would not prove it

A uses the combined observations and keeps the school-model limit. B contradicts the given fact that V is a mixture; an invisible path does not establish purity anyway. C ignores X’s visible scattering and treats passing ordinary paper as sufficient to identify a solution. D is false because suspensions can also scatter light. A short no-settling interval alone is not a universal identification test; an ambiguous real sample needs further evidence.

Where this lesson fits and sources

The RRB CEN 09/2025 official notice, §14.1, printed p.28, places General Science at Class 10 CBSE level. It does not name this lesson as a separate chapter. These foundations help you understand later chemistry; the RRB topic list is illustrative, not exhaustive.

Concept references: NCERT Exploration, Grade 9, Chapter 5, first edition April 2026, §§5.1–5.5, printed pp.73–88; NIOS Secondary Science and Technology 212, Lesson 2, §§2.1–2.9, printed pp.22–44, edition not established. Current NCERT leads the mixture treatment; numerical and scientific shortcuts in older materials have not been imported. All teaching prose, case explanations, visual designs and questions here are original. This lesson is part of a developing course, not the whole General Science syllabus.

Analogy

Imagine a sorting station with different gates. A size-based gate, a magnet and a settling area help only when the relevant properties differ. First identify the difference; then decide what you need to keep. This models property-based separation, but real mixtures need not have visibly separate pieces, and the analogy does not explain chemical bonds or guarantee complete separation.

Quick reference

• State and composition are separate classifications
• Pure substance: one chemical substance; element or compound
• Mixture: multiple substances; uniform appearance does not prove purity
• Melting water preserves H₂O identity; particles change arrangement/movement
• Solution, suspension and colloid: use combined observations and stated conditions
• Scattering alone is not unique to colloids; ordinary filtration is not a purity test
• Separation: components → property difference → recovery goal → method limit
• Ordinary filtration retains suitable insoluble particles, not dissolved salt
• Evaporation may recover nonvolatile solute; distillation can collect suitable solvent
• Immiscible layers support layer separation; density difference alone is insufficient
• Other-method principles are supplied for transfer, not a recall catalogue
• Clear or separated water is not automatically safe to drink
• Optional mass % = solute mass / solution mass ×100; not in compulsory review