This MCQ module is based on: Exploring Mixtures and their Separation — NCERT Exercises
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Exploring Mixtures and their Separation — NCERT Exercises
🎓 Class 9
Science
CBSE
Theory
Ch 5 — Exploring Mixtures and their Separation
⏱ ~15 min
🌐 Language: [gtranslate]
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This assessment will be based on: Exploring Mixtures and their Separation — NCERT Exercises
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Chapter Summary — The Big Ideas
Pure substance
One kind of particle, fixed composition, sharp m.p./b.p.
Mixture
Two or more substances, any ratio, components keep their identity.
Homogeneous
Uniform throughout (salt water, brass, air).
Heterogeneous
Visible boundaries (sand+iron, oil+water).
Solution
Particles < 1 nm, no Tyndall effect, cannot be filtered.
Suspension
Particles > 100 nm, settle on standing, can be filtered.
Colloid
Particles 1–100 nm, do not settle, show Tyndall effect.
Mass %
\((\text{solute mass}/\text{solution mass})\times 100\).
Mass/Volume %
\((\text{solute g}/\text{solution mL})\times 100\).
Solubility
g of solute that saturates 100 g of solvent at given T.
Henry’s Law
Gas solubility ∝ pressure of gas above the liquid.
Element vs Compound
Element: one type of atom. Compound: fixed-ratio bonded atoms.
Keyword Grid
Pure SubstanceMade of one kind of particle.
MixtureTwo or more substances together.
HomogeneousSame throughout, no boundary.
HeterogeneousDifferent parts visible.
SolutionHomogeneous mix; particles < 1 nm.
SoluteSubstance dissolved in solvent.
SolventThe dissolving medium.
SuspensionHeterogeneous; particles > 100 nm.
ColloidParticles 1–100 nm; show Tyndall.
Tyndall EffectScattering of light by colloidal particles.
Saturated SolutionCannot dissolve more solute at given T.
Solubilityg of solute per 100 g solvent (saturated).
Henry’s LawGas solubility ∝ pressure.
EvaporationBoil off solvent; recover solute.
CentrifugationSpin to separate by density.
Separating FunnelDrains immiscible liquid layers.
ChromatographyDifferent speeds across paper.
DistillationSeparate by boiling point (>25 °C).
Fractional DistillationColumn for close boiling points.
CrystallisationCool saturated solution; pure crystals form.
ElementOne kind of atom.
CompoundFixed-ratio chemical combination.
NCERT Exercises — Solved
Q1
Which separation techniques will you apply for the separation of the following?
(a) Sodium chloride from its solution in water
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride
(c) Small pieces of metal in the engine oil of a car
(d) Different pigments from an extract of flower petals
(e) Butter from curd
(f) Oil from water
(g) Tea leaves from tea
(h) Iron pins from sand
(i) Wheat grains from husk
(j) Fine mud particles suspended in water
(a) Evaporation or crystallisation. (b) Sublimation — warming the mixture sublimes ammonium chloride leaving sodium chloride. (c) Filtration. (d) Paper chromatography. (e) Centrifugation (the churning of curd is essentially centrifugation by hand). (f) Separating funnel. (g) Filtration through a strainer. (h) Magnetic separation. (i) Winnowing. (j) Sedimentation followed by decantation, or centrifugation for finer particles.
Q2
Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
1. Take water (the solvent) in a kettle and heat it. 2. Add tea leaves — they are insoluble, so they remain as separate solids. 3. Add sugar (the solute); it dissolves because it is soluble in water. The water + sugar form a true solution. 4. Add milk and boil. 5. Pour through a strainer. The clear liquid that passes through is the filtrate (the tea you drink). The wet tea leaves left in the strainer are the residue.
Q3
Pragya tested the solubility of three different substances at different temperatures and collected the data: at 20°C, KNO₃ = 32 g, NaCl = 36 g, KCl = 35 g (per 100 g water). At 40°C, KNO₃ = 62 g, NaCl = 36 g, KCl = 40 g.
(a) What mass of KNO₃ will be needed to make a saturated solution in 50 g of water at 40°C?
(b) Pragya makes a saturated solution of KCl in water at 60°C and on cooling she observes some solid coming out. Explain.
(c) Find the solubility of each salt at 20°C. Which has the highest solubility?
(d) What is the effect of change of temperature on the solubility of a salt?
(a) Solubility of KNO₃ at 40°C = 62 g per 100 g water. For 50 g water, mass needed = \(\dfrac{62}{2} = 31\) g.
(b) On cooling, the solubility of KCl decreases, so the extra dissolved salt that the cooler water can no longer hold separates out as solid (crystallisation).
(c) Solubilities at 20°C: KNO₃ = 32 g, NaCl = 36 g, KCl = 35 g per 100 g water. NaCl has the highest solubility at 20°C.
(d) For most solid solutes, solubility increases as temperature rises. The increase is sharp for salts like KNO₃ and very small for salts like NaCl.
(b) On cooling, the solubility of KCl decreases, so the extra dissolved salt that the cooler water can no longer hold separates out as solid (crystallisation).
(c) Solubilities at 20°C: KNO₃ = 32 g, NaCl = 36 g, KCl = 35 g per 100 g water. NaCl has the highest solubility at 20°C.
(d) For most solid solutes, solubility increases as temperature rises. The increase is sharp for salts like KNO₃ and very small for salts like NaCl.
Q4
Explain the following giving examples: (a) Saturated solution (b) Pure substance (c) Colloid (d) Suspension.
(a) Saturated solution — one in which no more solute can be dissolved at the given temperature. Example: a salt solution made by stirring NaCl into water until grains begin to sit at the bottom.
(b) Pure substance — a single kind of particle, fixed composition and properties. Example: distilled water, copper, sodium chloride.
(c) Colloid — heterogeneous mixture with particles 1–100 nm that do not settle and show the Tyndall effect. Example: milk, fog, smoke.
(d) Suspension — heterogeneous mixture with particles > 100 nm that settle on standing. Example: chalk in water, muddy river water.
(b) Pure substance — a single kind of particle, fixed composition and properties. Example: distilled water, copper, sodium chloride.
(c) Colloid — heterogeneous mixture with particles 1–100 nm that do not settle and show the Tyndall effect. Example: milk, fog, smoke.
(d) Suspension — heterogeneous mixture with particles > 100 nm that settle on standing. Example: chalk in water, muddy river water.
Q5
Classify each of the following as a homogeneous or heterogeneous mixture: soda water, wood, air, soil, vinegar, filtered tea.
Homogeneous: soda water, air (in clean conditions), vinegar, filtered tea.
Heterogeneous: wood, soil.
Heterogeneous: wood, soil.
Q6
How would you confirm that a colourless liquid given to you is pure water?
Boil it in a clean flask. If it boils at exactly 100°C at standard atmospheric pressure and freezes at exactly 0°C, and if it leaves no residue on complete evaporation, the liquid is pure water. A pure substance has a sharp, fixed boiling point and freezing point.
Q7
Which of the following materials fall in the category of a ‘pure substance’? (a) Ice (b) Milk (c) Iron (d) Hydrochloric acid (e) Calcium oxide (f) Mercury (g) Brick (h) Wood (i) Air.
Pure substances: (a) Ice, (c) Iron, (e) Calcium oxide, (f) Mercury. The rest are mixtures.
Q8
Identify the solutions among the following mixtures: (a) Soil (b) Sea water (c) Air (d) Coal (e) Soda water.
Solutions are: (b) Sea water, (c) Air, (e) Soda water — all are homogeneous. Soil and coal are heterogeneous.
Q9
Which of the following will show the Tyndall effect? (a) Salt solution (b) Milk (c) Copper sulphate solution (d) Starch solution.
(b) Milk and (d) Starch solution — both are colloids and scatter light. Salt solution and copper sulphate solution are true solutions and do not show the Tyndall effect.
Q10
Classify each into element, compound or mixture: sodium, soil, sugar solution, silver, calcium carbonate, tin, silicon, coal, air, soap, methane, carbon dioxide, blood.
Elements: sodium, silver, tin, silicon.
Compounds: calcium carbonate, methane, carbon dioxide.
Mixtures: soil, sugar solution, coal (impure carbon), air, soap, blood.
Compounds: calcium carbonate, methane, carbon dioxide.
Mixtures: soil, sugar solution, coal (impure carbon), air, soap, blood.
Q11
Which of the following are chemical changes? (a) Growth of a plant (b) Rusting of iron (c) Mixing of iron filings and sand (d) Cooking of food (e) Digestion of food (f) Freezing of water (g) Burning of a candle.
Chemical changes (new substance with new properties is formed): (a) Growth of a plant, (b) Rusting of iron, (d) Cooking of food, (e) Digestion of food, (g) Burning of a candle (the wax that burns). Mixing iron and sand is physical (separable by magnet); freezing of water is physical (only state changes).
Q12 (Numerical)
Calculate the mass percent of a solution containing 20 g of common salt dissolved in 180 g of water.
Mass of solute = 20 g; Mass of solvent = 180 g; Mass of solution = 200 g.
\(\text{Mass \%} = \dfrac{20}{200}\times 100 = 10\%\).
\(\text{Mass \%} = \dfrac{20}{200}\times 100 = 10\%\).
Q13 (Numerical)
A solution contains 50 mL of alcohol mixed with 150 mL of water. Calculate the volume percent of alcohol.
Total volume of solution = 50 + 150 = 200 mL.
\(\text{Volume \%} = \dfrac{50}{200}\times 100 = 25\%\) alcohol by volume.
\(\text{Volume \%} = \dfrac{50}{200}\times 100 = 25\%\) alcohol by volume.
Q14 (Numerical)
The solubility of potassium chloride at 30°C is 37 g per 100 g of water. If a student dissolves 18.5 g of KCl in 50 g of water at 30°C, will the solution be saturated, unsaturated or supersaturated? Justify with a calculation.
For 50 g water, the saturating mass = \(\dfrac{37}{2} = 18.5\) g. The student has dissolved exactly 18.5 g, so the solution is just saturated. Adding even one more crystal would not dissolve at 30°C.
Q15
Name the technique used to separate (a) butter from curd, (b) salt from sea water, (c) camphor from salt.
(a) Centrifugation (churning). (b) Evaporation in salt pans, followed by crystallisation for purer salt. (c) Sublimation — camphor sublimes on warming, leaving salt behind.
Q16
What is the difference between physical and chemical changes? Give one example of each.
A physical change alters only the appearance, state or shape; no new substance is formed and the change is usually reversible. Example: melting of ice into water.
A chemical change produces one or more new substances with different properties; energy is usually released or absorbed and the change is generally not reversible. Example: rusting of iron forms reddish-brown iron oxide.
A chemical change produces one or more new substances with different properties; energy is usually released or absorbed and the change is generally not reversible. Example: rusting of iron forms reddish-brown iron oxide.
Q17 (Application)
Why does a soda bottle fizz only when its cap is opened? Explain using Henry’s Law.
Inside a sealed bottle, carbon dioxide is dissolved in water under high pressure. By Henry’s Law, the mass of gas dissolved is proportional to the pressure of the gas above the liquid. When the cap is opened, the pressure suddenly drops to atmospheric pressure, the solubility of CO₂ falls and the excess gas escapes rapidly as bubbles — producing the fizz.
Frequently Asked Questions — NCERT Exercises & Intext Questions
How do I solve NCERT Class 9 Science Chapter 5 (Exploring Mixtures and their Separation) exercise questions for the CBSE board exam?
Solve NCERT Chapter 5 — Exploring Mixtures and their Separation — exercise questions by first reading the question carefully, writing down the given data, recalling the relevant concepts like pure substance, mixture, solution, and applying them step by step. This Part 4 covers every intext and end-of-chapter exercise from the NCERT textbook. Write balanced equations, label diagrams clearly and show each step — CBSE Class 9 examiners award step marks even if the final answer has a small slip. Practising these solutions strengthens conceptual clarity and builds speed for both the school exam and the upcoming Class 10 board exam.
Are the NCERT intext questions from Exploring Mixtures and their Separation important for the Class 9 Science exam?
Yes, NCERT intext questions for Chapter 5 Exploring Mixtures and their Separation are highly important for the CBSE Class 9 Science exam. Many questions in school and competitive papers are directly lifted or only slightly modified from these intext questions, and they test the foundational concepts — pure substance, mixture, solution — that chapter-end questions and the Class 10 board build on. Attempt every intext question first, then move on to the exercises. This practice ensures complete NCERT coverage, which is the CBSE syllabus's primary source.
What types of questions from Exploring Mixtures and their Separation are asked in the Class 9 Science exam?
The Class 9 Science paper (CBSE pattern) asks a mix of question types from Exploring Mixtures and their Separation: 1-mark MCQ and assertion-reason, 2-mark short answers, 3-mark explanations, 5-mark long answers with diagrams or derivations, and 4-mark competency-based / case-study questions. These test understanding of pure substance, mixture, solution, solubility. Practising every NCERT exercise and intext question prepares you to answer all of these formats with confidence.
How many marks does Chapter 5 — Exploring Mixtures and their Separation — typically carry in the Class 9 Science paper?
Chapter 5 — Exploring Mixtures and their Separation — is part of the CBSE Class 9 Science syllabus and typically contributes 5–9 marks in the annual paper, depending on the year's weightage. Questions are drawn from definitions, reasoning, numerical/descriptive problems and diagrams on topics like pure substance, mixture, solution. Solving the NCERT exercises in this part is essential because CBSE directly references the NCERT Exploration textbook for question design.
Where can I find step-by-step NCERT solutions for Chapter 5 Exploring Mixtures and their Separation Class 9 Science?
You can find complete, step-by-step NCERT solutions for Chapter 5 Exploring Mixtures and their Separation Class 9 Science on MyAiSchool. Every intext and end-of-chapter exercise question is solved with full working, labelled diagrams and CBSE-aligned mark distribution. Solutions highlight key points about pure substance, mixture, solution that examiners look for. This makes revision quick and exam-focused for Class 9 CBSE students.
What is the best way to revise Exploring Mixtures and their Separation for the Class 9 Science exam?
The best way to revise Exploring Mixtures and their Separation for the CBSE Class 9 Science exam is a three-pass approach. First pass: skim the chapter and note down key terms like pure substance, mixture, solution in a one-page mind map. Second pass: solve every NCERT intext and exercise question without looking at the solution, then self-check. Third pass: attempt sample papers and competency-based questions under timed conditions. This structured revision secures full marks for this chapter.
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