GSEB Class 9 Science Solutions Chapter 2 Is Matter Around Us Pure

 

Question 1. What is meant by a pure substance?
Answer: A pure substance is one that consists of a single kind of particle, meaning all component particles of the substance possess the same chemical nature. Pure substances can be categorized as elements or compounds.
In simple words: A pure substance contains only one type of particle, and all its parts have the same chemical properties.

Exam Tip: Remember to mention both 'single type of particle' and 'same chemical nature' for a complete definition of a pure substance.

 

Question 2. List the points of difference between homogeneous and heterogeneous mixtures.
Answer: Homogeneous mixtures have a uniform composition throughout and show no visible boundaries between their parts. Heterogeneous mixtures, however, possess a non-uniform composition and have distinct boundaries of separation between their constituents.
In simple words: Homogeneous mixtures look the same everywhere, while heterogeneous mixtures have different parts that you can see.

Exam Tip: When differentiating, always provide a clear point for each type of mixture regarding composition and visibility of boundaries.

 

Question 3. Differentiate between homogeneous and heterogeneous mixtures with examples.
Answer: A homogeneous mixture is a mixture that has a uniform composition throughout its entire volume. For instance, mixtures of salt in water, sugar in water, copper sulphate in water, iodine in alcohol, alloys, and air all have uniform composition across the mixture. On the other hand, a heterogeneous mixture is a mixture that has a non-uniform composition throughout its entire volume. For example, mixtures of sodium chloride and iron filings, salt and sulphur, oil and water, chalk powder in water, wheat flour in water, and milk and water are not uniform throughout the mixture.
In simple words: Homogeneous mixtures look the same all over, like sugar in water. Heterogeneous mixtures have different parts you can easily see, like oil and water.

Exam Tip: Provide clear examples for both types of mixtures to illustrate the difference effectively and ensure full marks.

 

Question 4. How are sol, solution and suspension different from each other?
Answer: Sol, solution, and suspension differ primarily in particle size, uniformity, and light scattering properties. Solutions have very small particles that are uniformly dispersed and don't scatter light. Suspensions have large, visible particles that settle over time and scatter light. Sols (colloids) have intermediate-sized particles that are stable but can scatter light (Tyndall effect).

In simple words: Solutions are uniform and clear, like sugar water. Sols are somewhat uniform but cloudy, like milk. Suspensions are cloudy, and particles eventually settle, like muddy water.

Exam Tip: For differentiation, using a table format helps in clearly presenting the differences across multiple characteristics like particle size, visibility, and stability.

 

Question 3. 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: Concentration refers to the amount of solute that is dissolved in a specific quantity of solvent. Here, 36 g of sodium chloride is dissolved in 100 g of water. Therefore, the concentration of the solution is 36 g per 100 g of water.
In simple words: To find the concentration, you see how much salt is put into 100 grams of water. Here, it's 36 grams of salt for every 100 grams of water.

Exam Tip: Concentration for saturated solutions is usually expressed as mass of solute per 100g of solvent at a specific temperature. Be clear about the units.

 

Question 4. 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: Kerosene and petrol are miscible liquids, and the difference between their boiling points is more than 25°C. Because of this, they can be separated easily by using the method of distillation. The setup for this method is shown in the figure above. In this technique, the mixture of kerosene and petrol is placed in a distillation flask with a thermometer. A beaker, a water condenser, and a Bunsen burner are also needed. The apparatus is arranged as depicted in the figure. The mixture is then slowly heated, with continuous monitoring of the thermometer. Kerosene will vaporize first and then condense in the water condenser. The condensed kerosene is collected from the condenser outlet, while petrol remains in the distillation flask.
In simple words: Since kerosene and petrol mix well and have different boiling points, we can separate them by heating the mixture (distillation). Kerosene turns into gas first, gets cooled, and collects as liquid, leaving petrol behind.

Exam Tip: Distillation is suitable for separating miscible liquids with a significant difference (more than 25°C) in their boiling points.

 

Question 5. Name the technique to separate
(i) butter from curd
(ii) salt from sea water
(iii) camphor from salt.
Answer:
(i) Butter can be separated from curd by centrifugation.
(ii) Salt can be separated from sea water by evaporation.
(iii) Camphor can be separated from salt by sublimation.
In simple words: To get butter from curd, use a centrifuge. To get salt from sea water, let the water evaporate. To separate camphor from salt, use sublimation.

Exam Tip: Match the separation technique to the physical properties of the substances being separated (e.g., density for centrifugation, boiling/evaporation points for evaporation, sublimation for solids that vaporize directly).

 

Question 6. What type of mixtures are separated by the technique of crystallisation?
Answer: Solids, where the impurities are either insoluble or are more soluble than the main solids in a specific solvent, can be readily separated using the technique of crystallisation. For example, impure copper sulphate, alum, nitre, and sea salt can all be easily purified by crystallisation.
In simple words: Crystallisation is used to clean up solid mixtures where the unwanted bits either don't dissolve or dissolve much better than the solid you want.

Exam Tip: Crystallisation is a purification technique for solids, especially when impurities have different solubilities at various temperatures.

 

Question 7. Classify the following as chemical or physical changes:
(i) cutting of trees,
(ii) melting of butter in a pan,
(iii) rusting of almirah,
(iv) boiling of water to form steam,
(v) passing of electric current through water and the water breaking down into hydrogen and oxygen gases,
(vi) dissolving common salt in water,
(vii) making a fruit salad with raw fruits, and
(viii) burning of paper and wood.
Answer:
(i) Physical change
(ii) Physical change
(iii) Chemical change
(iv) Physical change
(v) Chemical change
(vi) Physical change
(vii) Physical change
(viii) Chemical change.
In simple words: Physical changes alter how something looks but not what it is, like cutting wood or melting butter. Chemical changes create new substances, like rust forming or paper burning.

Exam Tip: A chemical change produces new substances, often irreversible, while a physical change alters state or appearance but not chemical identity.

 

Question 8. Try segregating the things around you as pure substances or mixtures.
(a) distilled water
(b) curd
(c) diamond
(d) ice-cream
(e) kerosene oil
(f) cooking oil
(g) steel
(h) graphite
(i) raw rubber
(j) vulcanized rubber
(k) solder wire.
Answer: Pure substances - Distilled water, diamond, graphite, raw rubber. Mixtures - Curd, ice-cream, kerosene oil, cooking oil, steel, vulcanized rubber, solder wire (alloy of lead and tin).
In simple words: Pure substances are single items like distilled water or diamond. Mixtures are combinations of different things, like curd or ice-cream.

Exam Tip: Remember that alloys (like steel and solder wire) are considered mixtures because their components can be present in varying proportions and retain some of their original properties.

 

Activity 2.1
Answer: Observation: Group A and B obtained a mixture that has a uniform composition throughout. These mixtures are known as homogeneous mixtures or solutions. Although both groups A and B obtained copper sulphate solution, the intensity of color of the solutions is different. Different color intensities demonstrate that a homogeneous mixture can have a variable composition. Groups C and D obtained mixtures that contain physically distinct parts and have non-uniform compositions. These types of mixtures are called heterogeneous mixtures. Conclusion: A mixture that has a uniform composition throughout is called a homogeneous mixture or solution. A mixture that has variable composition throughout is called a heterogeneous mixture or solution.
In simple words: Groups A and B made even mixtures (homogeneous solutions) with different color strengths. Groups C and D made uneven mixtures (heterogeneous mixtures) with visible separate parts.

Exam Tip: When describing observations, always connect them back to the definition of the concept being studied, such as homogeneous vs. heterogeneous mixtures.

 

Activity 2.2
Answer: Observation:

• The particles in group A and B will not be visible, whereas in group C and D, they would be visible.
• In group A and B, the particles will not show the path of a beam of light, whereas in group C and D, the particles will show the path of a beam of light from a torch.
• The mixture is stable only in group A and B, as it does not settle down when kept for some time.
• For D, it is not observed rapidly. When filtering the mixture, a residue is left on the filter paper in the case of group C.

Conclusion: Groups A and B have true solutions. Group C has a suspension. Group D has a colloidal solution.
In simple words: Groups A and B showed clear, stable solutions that didn't scatter light. Group C had a suspension where particles settled and were trapped by a filter. Group D had a colloid, which scatters light.

Exam Tip: When distinguishing between solutions, colloids, and suspensions, focus on particle visibility, stability (settling), and the Tyndall effect (light scattering).

 

Activity 2.3
Answer: Observation: When no more solute can be dissolved in a solution at a given temperature, it is called a saturated solution. So, at any particular temperature, a solution that has dissolved as much solute as it is capable of dissolving is said to be a saturated solution. No, the amount of salt and sugar or barium chloride that can be dissolved in water at a given temperature is not the same. By increasing the temperature, more solute can be added. Conclusion: The amount of solute present in the saturated solution at the given temperature is called its solubility. We can infer from the activity that different substances in a given solvent have different solubilities at particular temperatures.
In simple words: A saturated solution holds the maximum solute at a specific temperature. Different substances dissolve differently, and heating the solvent lets more solute dissolve. Solubility means how much solute dissolves at a certain temperature.

Exam Tip: Clearly define saturated solution and solubility, and note the effect of temperature on solubility, as it is a common point of confusion.

 

Activity 2.4
Answer: Observation: The liquid present in the watch glass evaporates when heated. A residue of ink will be left on the watch glass in a dry state. We can separate the volatile component (solvent) from its non-volatile solute using the method of evaporation. Ink is a mixture of a dye in water or a solvent. Conclusion: From this activity, we can conclude that ink is not a single substance, but is a mixture of dye and water/solvent.
In simple words: When ink is heated, the water evaporates, leaving the dry dye behind. This shows ink is a mixture, not a pure substance, and evaporation can separate its parts.

Exam Tip: Evaporation is a simple technique for separating a non-volatile solute from a volatile solvent. Always state the conclusion clearly based on the observation.

 

Activity 2.5
Answer: Observation: The soluble fats in the milk collide rapidly to form larger particles, creating cream. The cream formed separates, leaving behind fat-free milk. Centrifugation works on the idea that denser particles are forced to move to the bottom, while lighter particles stay at the top of the test tube when spun quickly. Conclusion: Cream can be separated from milk by the process of centrifugation.
In simple words: Spinning milk fast (centrifugation) pushes heavier fat particles (cream) to the bottom, separating them from the lighter fat-free milk.

Exam Tip: Centrifugation is ideal for separating fine suspended particles from a liquid, especially when density differences are subtle, like in milk.

 

Activity 2.6
Answer: Observation: Two distinct layers of kerosene oil and water are formed when a separating funnel is left undisturbed for some time. This process is based on the idea that immiscible liquids separate out into layers depending on their densities. The liquid with heavier density stays at the bottom as the lower layer, while the liquid with lighter density is present as the upper layer. Conclusion: A separating funnel helps in separating two immiscible liquids based on the difference in the densities of the liquids in the mixture.
In simple words: When you let oil and water sit, they form layers because they don't mix and have different densities. A separating funnel helps pour out the heavier bottom layer first, leaving the lighter top layer.

Exam Tip: Immiscible liquids with different densities can be efficiently separated using a separating funnel. Always identify which liquid forms the lower layer.

 

Activity 2.7
Answer: Observation: The ink we use has water as the solvent, and the dye is soluble in it. So, when the water rises on the filter paper, it carries the dye particles along with it. Yes, different colored spots of dyes are obtained because a dye is a mixture of two or more colors. The colored component is more soluble in water; therefore, it rises faster on the paper strip. Conclusion: Colored components of a mixture can be separated by using chromatography.
In simple words: When water moves up filter paper through ink, it carries different ink colors at different speeds, showing that ink is a mix of colors. This separation method is called chromatography.

Exam Tip: Chromatography separates components based on their different affinities for the stationary phase (paper) and the mobile phase (solvent), useful for separating colors or pigments.

 

Activity 2.8
Answer: Observation: The thermometer shows an increase in temperature. At 56.5°C, the thermometer reading becomes constant for some time. At this temperature, acetone vaporizes, condenses, and can be collected from the condenser outlet. Water is left behind in the distillation flask. Acetone boils at 56.5°C, while water boils at 100°C. The two components of the mixture separate because acetone is more volatile than water. Conclusion: The distillation process is used for the separation of components of a mixture containing two miscible liquids that boil without decomposition and have a sufficient difference in their boiling points.
In simple words: When heating a mixture of acetone and water, acetone boils and turns to gas at a lower temperature (56.5°C). It then cools and turns back into liquid, separating from the water. This is distillation, used for liquids that mix but have different boiling points.

Exam Tip: Distillation effectively separates miscible liquids with different boiling points, especially when one component is significantly more volatile. Remember to mention both vaporization and condensation.

 

Activity 2.9
Answer: Observation: The crystals of pure copper sulphate are observed in the china dish. The crystals are not all alike. The pure crystals can be separated from the liquid in the china dish by filtration. Conclusion: Pure crystals of copper sulphate can be obtained from an impure sample by the process of crystallisation.
In simple words: We saw copper sulphate crystals form from a liquid. These pure crystals can be taken out by filtering, meaning crystallisation helps get pure solids from impure mixtures.

Exam Tip: Crystallisation is a purification technique that often relies on cooling a hot saturated solution to form pure crystals, leaving impurities in the solution.

 

Activity 2.10
Answer: Observation: The material obtained by group 1 is a mixture of two substances: iron and sulphur. The material obtained by group 2 is a compound. The properties of the mixture are the same as those of its constituents. We can also observe that the texture and the color of the compound are the same throughout. A black substance is formed, and it no longer gets attracted to a magnet. Hence, group 1 obtained a material with magnetic properties. The gas obtained by group 1 is hydrogen gas. It is a colorless, odorless, and combustible gas. So, it is not advisable to do the combustion test for hydrogen in the class. The gas obtained by group 2 is hydrogen sulphide. It is a colorless gas with the smell of rotten eggs. Although the starting materials were the same, the product obtained by both groups showed different properties. Group 1 carried out an activity involving a physical change, whereas in group 2, a chemical change occurred. Conclusion: Heat is not needed for the formation of a mixture, but it is needed for the formation of a compound.
In simple words: Group 1 made a mixture (iron and sulfur) with original properties. Group 2 made a compound, which changed into a new black substance that wasn't magnetic and had different gases. This shows mixtures don't need heat to form, but compounds do, and they have new properties.

Exam Tip: This activity highlights key differences between mixtures and compounds: mixtures retain component properties and can be separated easily, while compounds have new properties and require chemical changes (often with energy input) to form or separate.

 

Question 1. 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) Crystallisation or Evaporation
(b) Sublimation
(c) Filtration
(d) Chromatography
(e) Centrifugation
(f) Separating funnel
(g) Filtration
(h) Magnetic separation
(i) Blowing air
(j) Using alum.
In simple words: Different methods are used to separate things. For salt from water, you evaporate the water. For ammonium chloride, you use sublimation. Filtration works for metal in oil or tea leaves from tea. To get butter from curd, use centrifugation. Oil from water uses a separating funnel. Iron pins from sand need a magnet. Wheat grains from husk use blowing air, and fine mud uses alum.

Exam Tip: When choosing separation techniques, consider the physical properties of the components, such as solubility, boiling point, density, magnetic properties, and particle size.

 

Question 2. Write the steps you would use for making tea. Use the words-solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Answer: Take the solvent, water, in a kettle and heat it. When the solvent boils, add the solute, milk. Milk and water form a solution. Then, pour some tea leaves over a sieve. Slowly pour the hot solution of milk over the tea leaves. The color of the tea leaves goes into solution as filtrate. The remaining tea leaves, being insoluble, remain as residue. Add the required sugar, which dissolves, and the tea is now ready.
In simple words: Heat water (solvent), add milk (solute) to make a solution. Pour this hot solution over tea leaves in a sieve. The colored liquid that passes through is the filtrate; the tea leaves left behind are the insoluble residue. Add sugar, which dissolves, and your tea is made.

Exam Tip: Ensure you correctly identify and use all the specified scientific terms (solution, solvent, solute, dissolve, soluble, insoluble, filtrate, residue) in their proper context when describing the process.

 

Question 3. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (result are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).

(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?
(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.
(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?
(d) What is the effect of change of temperature on the solubility of a salt?
Answer:
(a) At 313 K, potassium nitrate needed to produce a saturated solution in 100 grams of water = 62 g. Therefore, the amount of potassium nitrate to produce a saturated solution in 50 grams of water = 31 g.
(b) Some amount of dissolved potassium chloride will reappear as undissolved solid as the solubility of the solute decreases with the decrease in temperature. As the solution cools, the potassium chloride becomes less soluble and will start to crystallize out of the solution.
(c) Solubility of each salt at 293 K is as follows:

• Potassium nitrate - 32 g
• Sodium chloride - 36 g
• Potassium chloride - 35 g
• Ammonium chloride - 37 g

Ammonium chloride salt has the highest solubility at this temperature.
(d) The solubility of a salt generally increases with an increase in temperature.
In simple words: (a) You need 31g of potassium nitrate for 50g of water at 313K. (b) When potassium chloride solution cools, some solid will form again because it dissolves less in cold water. (c) At 293K, ammonium chloride dissolves the most. (d) Hotter water usually dissolves more salt.

Exam Tip: Carefully read the temperature and mass values from the table. For cooling saturated solutions, remember that most solids become less soluble as temperature drops, leading to crystallization.

 

Question 4. Explain the following giving examples.
(a) saturated solution
(b) pure substance
(c) colloid
(d) suspension
Answer:
(a) A solution in which no more solute can be dissolved at a particular temperature is known as a saturated solution. For example, in an aqueous solution of sugar, no more sugar can be dissolved at room temperature.
(b) A pure substance is a substance consisting of a single type of particles, meaning all constituent particles of the substance have the same chemical properties. For example, water, sugar, salt, etc.
(c) A colloid is a heterogeneous mixture whose particles are not as small as a solution's but they are so small that they cannot be seen with the naked eyes. When a beam of light is passed through a colloid, the path of the light becomes visible. For example, milk, smoke, etc.
(d) A suspension is a heterogeneous mixture in which solids are dispersed in liquids. The solute particles in suspension do not dissolve but remain suspended throughout the medium. For example, paints, muddy water, chalk water mixtures, etc.
In simple words: (a) A saturated solution holds the most solute at a given temperature, like sugar in water. (b) A pure substance has only one type of particle, like water or salt. (c) A colloid is a cloudy mixture where tiny particles scatter light, like milk. (d) A suspension is a cloudy mixture where solid particles float but eventually settle, like muddy water.

Exam Tip: For each term, provide a clear definition and a relevant example. Focus on particle size, homogeneity, and light scattering (Tyndall effect) to differentiate colloids, solutions, and suspensions.

 

Question 5. Classify each of the following as a homogeneous or heterogeneous mixture: soda water, wood, air, soil, vinegar, filtered tea.
Answer:
(a) Homogeneous mixtures - Soda water, vinegar, and filtered tea are homogeneous mixtures. Air is also a homogeneous mixture if dust particles and other suspended impurities are excluded.
(b) Heterogeneous mixtures - Wood and soil are heterogeneous mixtures. Air is also a heterogeneous mixture if dust particles and other suspended impurities are included.
In simple words: Soda water, vinegar, and filtered tea are homogeneous (even). Wood, soil, and air with dust are heterogeneous (uneven).

Exam Tip: Remember that "air" can be either homogeneous (pure, filtered air) or heterogeneous (with dust or smog), depending on the presence of impurities.

 

Question 6. How would you confirm that a colourless liquid given to you is pure water?
Answer: Every liquid has a characteristic boiling point. Pure water has a boiling point of 100°C (373 K) at 1 atmospheric pressure. If the given colorless liquid boils at even slightly above or below 100°C, then the given liquid is not pure water. Thus, by observing the boiling point, we can confirm whether a given colorless liquid is pure water or not.
In simple words: You can check if a clear liquid is pure water by seeing its boiling point. Pure water always boils at 100°C. If it boils at a different temperature, it's not pure.

Exam Tip: A unique boiling point (and freezing point) is a characteristic property of a pure substance. Impurities typically elevate the boiling point and depress the freezing point.

 

Question 7. Which of the following materials falls 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: The following materials fall in the category of a "pure substance”:
(a) Ice
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury
In simple words: Ice, iron, hydrochloric acid, calcium oxide, and mercury are pure substances because they are made of only one type of material or molecule.

Exam Tip: Pure substances can be elements (like iron, mercury) or compounds (like ice- \(H_2O\), hydrochloric acid, calcium oxide). Milk, brick, wood, and air are mixtures.

 

Question 8. Identify the solutions among the following mixtures:
(a) Soil
(b) Seawater
(c) Air
(d) Coal
(e) Soda water.
Answer: The following mixtures are solutions:
(b) Seawater
(c) Air
(e) Soda water.
In simple words: Seawater, air, and soda water are solutions because their parts are evenly mixed and appear as a single phase.

Exam Tip: Solutions are homogeneous mixtures where the solute is completely dissolved in the solvent, appearing uniform throughout.

 

Question 9. Which of the following will show "Tyndall effect"?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution
Answer: Tyndall effect is shown by colloidal solutions. Here, milk and starch solution are colloids; therefore, they will show the Tyndall effect.
In simple words: The Tyndall effect, where light scatters through a mixture, happens with colloids. Both milk and starch solution are colloids, so they will show this effect.

Exam Tip: The Tyndall effect is a key characteristic for distinguishing colloids from true solutions. True solutions (like salt or copper sulphate solutions) do not scatter light.

 

Question 10. 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
Answer: Elements: Sodium, silver, tin and silicon. Compounds: Calcium carbonate, methane and carbon dioxide. Mixtures: Soil, sugar solution, coal, air, soap and blood.
In simple words: Sodium, silver, tin, and silicon are elements. Calcium carbonate, methane, and carbon dioxide are compounds. Soil, sugar solution, coal, air, soap, and blood are mixtures.

Exam Tip: Elements are pure substances made of one type of atom. Compounds are pure substances made of two or more elements chemically combined. Mixtures are combinations of substances that retain their individual properties and are not chemically bonded.

 

Question 11. Which of the following are chemical changes?
(a) Growth of a plant
(b) Rusting of iron
(c) Mixing iron filings and sand
(d) Cooking of food
(e) Digestion of food
(f) Freezing of water
(g) Burning of a candle.
Answer: These specific alterations are considered chemical changes:
(a) Growth of a plant
(b) Rusting of iron
(d) Cooking of food
(e) Digestion of food
(g) Burning of a candle.
In simple words: Chemical changes create new substances. The plant growing, iron rusting, food cooking, digestion, and a candle burning all involve forming new materials or structures.

Exam Tip: Remember that physical changes alter form but not chemical identity, while chemical changes result in new substances. Growth, rusting, cooking, and burning are common examples of chemical transformations.