Class 9 SEBA Science Chapter 2 Solutions – Is Matter Around Us Pure? (2026–27) | Assam Eduverse
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SEBA Class 9 Science Chapter 2 Is Matter Around Us Pure? solutions are prepared by Assam Eduverse strictly according to the latest SEBA / ASSEB syllabus 2026–27. These SEBA Class 9 Science Chapter 2 Is Matter Around Us Pure? solutions are designed specifically for students searching for SEBA Class 9 Science Chapter 2 Is Matter Around Us Pure? solutions that are accurate, updated, and exam-oriented. This page offers complete SEBA Class 9 Science Chapter 2 Is Matter Around Us Pure? solutions, making it a reliable and trusted source for SEBA Class 9 Science Chapter 2 Is Matter Around Us Pure? solutions based entirely on the official SEBA Class 9 Science textbook.
The SEBA Class 9 Science Chapter 2 Is Matter Around Us Pure? solutions explain all concepts covered in SEBA Class 9 Science Chapter 2 Is Matter Around Us Pure? solutions, including pure substances, mixtures, types of mixtures, solutions, suspensions, colloids, physical and chemical changes, separation techniques, and concentration of solutions. These SEBA Class 9 Science Chapter 2 Is Matter Around Us Pure? solutions help students understand theory, definitions, numericals, and reasoning questions using SEBA Class 9 Science Chapter 2 Is Matter Around Us Pure? solutions written in simple, clear, and exam-friendly language. Every SEBA Class 9 Science Chapter 2 Is Matter Around Us Pure? solution strictly follows the official ASSEB Class 9 Science Chapter 2 solutions format and marking scheme.
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SEBA / ASSEB Class 9 Science Chapter 2 – Is Matter Around Us Pure? Intext Questions & Answers (Latest Syllabus 2026–27)
📝 Page 15
Q1: What is meant by a substance?
Answer: A substance is a pure single form of matter that consists of only one type of particles. All the constituent particles of a pure substance have the same chemical nature. For example, pure water, pure salt, or pure sugar are substances because they contain only one kind of matter with uniform composition throughout.
Q2: List the points of differences between homogeneous and heterogeneous mixtures.
Answer: The main differences between homogeneous and heterogeneous mixtures are:
Homogeneous mixtures have uniform composition throughout and their components cannot be seen separately with naked eyes. Examples include salt dissolved in water, sugar in water, and air.
Heterogeneous mixtures have non-uniform composition with physically distinct parts that can be seen separately. Examples include sand and water, oil and water, and mixtures of salt and iron filings.
In homogeneous mixtures, the properties are uniform throughout, while in heterogeneous mixtures, different parts may show different properties.
📝 Page 18
Q1: Differentiate between homogeneous and heterogeneous mixtures with examples.
Answer: Homogeneous mixtures are those where the composition is uniform throughout and components are not visible separately. The mixing is so complete that it appears as a single substance. Examples include sugar solution, salt solution, air, and brass. Heterogeneous mixtures have non-uniform composition where different components can be clearly distinguished and seen separately. Examples include sand and iron filings, oil and water, granite rock, and soil.
Q2: How are sol, solution and suspension different from each other?
Answer: A solution is a homogeneous mixture where particles are extremely small (less than 1 nm), cannot be seen by naked eyes, do not scatter light, and remain stable.
A suspension is a heterogeneous mixture where particles are large enough to be seen, scatter light, and settle down when left undisturbed.
A sol (colloidal solution) is between solution and suspension, where particles are too small to be seen but large enough to scatter light and show Tyndall effect, and they remain stable without settling.
Q3: To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.
Answer: Given data:
Mass of solute (sodium chloride) = 36 g
Mass of solvent (water) = 100 g
Mass of solution = Mass of solute + Mass of solvent = 36 g + 100 g = 136 g
Concentration by mass percentage = (Mass of solute/Mass of solution) × 100
= (36/136) × 100 = 26.47%
📝 Page 24
Q1: How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25°C), which are miscible with each other?
Answer: Since kerosene and petrol are miscible liquids with a significant difference in boiling points (more than 25°C), they can be separated by simple distillation. In this process, the mixture is heated in a distillation flask. The component with lower boiling point (petrol) will vaporize first, travel through the condenser where it gets cooled and condensed, and can be collected separately. The component with higher boiling point (kerosene) remains in the distillation flask.
Q2: Name the technique to separate:
(i) butter from curd
(ii) salt from sea-water
(iii) camphor from salt
Answer:
(i) Centrifugation – Butter from curd is separated using centrifugation because butter particles are lighter and separate when spun rapidly.
(ii) Evaporation – Salt from sea-water is separated by evaporation where water evaporates leaving behind salt crystals.
(iii) Sublimation – Camphor from salt is separated by sublimation since camphor directly changes from solid to gas state on heating, leaving salt behind.
Q3: What type of mixtures are separated by the technique of crystallisation?
Answer: Crystallisation technique is used to separate mixtures where we need to obtain a pure solid from its solution. This method is particularly useful for separating soluble solids from their solutions and for purifying impure solid samples. It is better than simple evaporation when the solid may decompose on heating or when impurities might remain dissolved. Examples include purifying salt from sea water and separating alum crystals from impure samples.
📝 Page 24
Q1: Classify the following as chemical or physical changes:
- cutting of trees
- melting of butter in a pan
- rusting of almirah
- boiling of water to form steam
- passing of electric current through water and the water breaking down into hydrogen and oxygen gases
- dissolving common salt in water
- making a fruit salad with raw fruits
- burning of paper and wood
Answer:
Physical changes: cutting of trees, melting of butter in a pan, boiling of water to form steam, dissolving common salt in water, making a fruit salad with raw fruits.
Chemical changes: rusting of almirah, passing of electric current through water and breaking it down into hydrogen and oxygen gases, burning of paper and wood.
Physical changes involve change in physical state or appearance without forming new substances, while chemical changes result in formation of new substances with different chemical properties.
Q2: Try segregating the things around you as pure substances or mixtures.
Answer:
Pure substances: Distilled water, pure salt, pure sugar, copper wire, aluminum foil, pure gold ornaments, pure iron nails.
Mixtures: Air we breathe, tap water, milk, tea, coffee, soil, wood, paper, food items like bread, fruits, vegetables, gasoline, and most everyday materials we use.
Most naturally occurring materials are mixtures because they contain more than one type of substance mixed together.
SEBA Class 9 Science Chapter 2 – Is Matter Around Us Pure? Textbook Exercise Questions & Solutions | 2026–27
📝 Page 28-29
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
Answer:
(a) Evaporation – Heat the solution to evaporate water, leaving sodium chloride crystals behind
(b) Sublimation – Heat the mixture; ammonium chloride will sublime and can be collected separately
(c) Filtration – Use filter paper to separate metal pieces from oil
(d) Chromatography – Different pigments will separate based on their solubility
(e) Centrifugation – Spin rapidly to separate lighter butter from heavier curd
(f) Separating funnel – Oil and water form separate layers due to different densities
(g) Filtration – Use a strainer or filter to separate tea leaves from liquid tea
(h) Magnetic separation – Use a magnet to attract iron pins, leaving sand behind
(i) Winnowing – Use wind to blow away lighter husk, leaving heavier grains
(j) Filtration or Sedimentation and decantation – Allow mud to settle and carefully pour clear water
Q2: Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Answer: To make tea, first I would boil water which acts as the solvent. Then I add tea leaves to the boiling water. The tea leaves contain soluble compounds that dissolve in water, forming a solution of tea. When sugar is added, it being soluble, dissolves completely in the solution. The tea leaves themselves are insoluble in water and remain suspended. Finally, I strain the mixture using a sieve where the clear tea becomes the filtrate and the used tea leaves form the residue that remains in the strainer.
Q3: Pragya tested the solubility of three different substances at different temperatures and collected data. Answer the following based on the solubility data:
(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?
Answer: From the table, at 313 K, the solubility of potassium nitrate is 62 g per 100 g of water. For 50 g of water:
Required mass = (62 × 50)/100 = 31 g of potassium nitrate
(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
Answer: As the solution cools from 353 K to room temperature, she would observe crystals of potassium chloride precipitating out of the solution. This happens because the solubility of potassium chloride decreases with decrease in temperature (from 54 g at 353 K to 35 g at 293 K per 100 g water). The excess salt that cannot remain dissolved at lower temperature crystallizes out.
(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?
Answer: At 293 K, the solubility values are:
Potassium nitrate: 32 g per 100 g water
Sodium chloride: 36 g per 100 g water
Potassium chloride: 35 g per 100 g water
Ammonium chloride: 37 g per 100 g water
Ammonium chloride has the highest solubility at 293 K.
(d) What is the effect of change of temperature on the solubility of a salt?
Answer: For most salts like potassium nitrate, potassium chloride, and ammonium chloride, solubility increases with increase in temperature. However, sodium chloride shows very little change in solubility with temperature change. The effect varies for different salts – some show dramatic increase (potassium nitrate), while others remain almost constant (sodium chloride).
Q4: Explain the following giving examples:
(a) saturated solution (b) pure substance (c) colloid (d) suspension
Answer:
(a) Saturated solution: A solution in which no more solute can be dissolved at a given temperature. For example, when no more sugar can be dissolved in water at room temperature, the solution becomes saturated.
(b) Pure substance: A substance that consists of only one type of particles with same chemical nature throughout. Examples include pure water, pure salt (sodium chloride), pure gold, and pure oxygen gas.
(c) Colloid: A heterogeneous mixture where particle size is between solution and suspension. Particles are too small to be seen but large enough to scatter light (Tyndall effect). Examples include milk, fog, smoke, and gelatin.
(d) Suspension: A heterogeneous mixture where particles are large enough to be seen with naked eyes and settle down when left undisturbed. Examples include chalk powder in water, muddy water, and sand in water.
Q5: Classify each of the following as a homogeneous or heterogeneous mixture: soda water, wood, air, soil, vinegar, filtered tea.
Answer:
Homogeneous mixtures: Soda water, air, vinegar, filtered tea
Heterogeneous mixtures: Wood, soil
Homogeneous mixtures have uniform composition throughout, while heterogeneous mixtures have visible different components or phases.
Q6: How would you confirm that a colourless liquid given to you is pure water?
Answer: To confirm if a colourless liquid is pure water, I would check its physical properties:
- Boiling point: Pure water boils at exactly 100°C at standard atmospheric pressure
- Freezing point: Pure water freezes at exactly 0°C
- Electrical conductivity: Pure water does not conduct electricity
- Evaporation test: On complete evaporation, pure water should leave no residue
- Density: Pure water has a density of 1 g/cm³ at 4°C
Any deviation from these standard values indicates the presence of impurities.
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
Answer: Pure substances from the list are: (a) Ice, (c) Iron, (d) Hydrochloric acid, (e) Calcium oxide, (f) Mercury.
These consist of only one type of substance with uniform chemical composition. Milk, brick, wood, and air are mixtures containing multiple substances.
Q8: Identify the solutions among the following mixtures:
(a) Soil (b) Sea water (c) Air (d) Coal (e) Soda water
Answer: Solutions from the given mixtures are: (b) Sea water, (c) Air, (e) Soda water.
These are homogeneous mixtures where components are completely dissolved and uniformly distributed. Soil and coal are heterogeneous mixtures.
Q9: Which of the following will show “Tyndall effect”?
(a) Salt solution (b) Milk (c) Copper sulphate solution (d) Starch solution
Answer: (b) Milk and (d) Starch solution will show Tyndall effect.
These are colloidal solutions where particles are large enough to scatter light but small enough to remain suspended, making the path of light beam visible.
Q10: Classify the following into elements, compounds and mixtures:
(a) Sodium (b) Soil (c) Sugar solution (d) Silver (e) Calcium carbonate (f) Tin (g) Silicon (h) Coal (i) Air (j) Soap (k) Methane (l) Carbon dioxide (m) Blood
Answer:
Elements: (a) Sodium, (d) Silver, (f) Tin, (g) Silicon
Compounds: (e) Calcium carbonate, (k) Methane, (l) Carbon dioxide
Mixtures: (b) Soil, (c) Sugar solution, (h) Coal, (i) Air, (j) Soap, (m) Blood
Elements cannot be broken down further, compounds have fixed composition of different elements, and mixtures contain multiple substances in variable proportions.
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
Answer: Chemical changes are: (a) Growth of a plant, (b) Rusting of iron, (d) Cooking of food, (e) Digestion of food, (g) Burning of a candle.
These processes involve formation of new substances with different chemical properties. Mixing of iron filings and sand, and freezing of water are physical changes as no new substances are formed.
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