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Page No. 15

Q1. What is meant by a substance?
Ans: A pure substance is one that is made up of only one kind of particle, either atoms or molecules. It has definite composition and distinct properties.  Examples of Pure SubstancesA pure substance consists of only one type of particle, either atoms or molecules. It has a definite composition and distinct properties. Examples include:

• Oxygen
• Carbon

In contrast, a mixture contains two or more pure substances. For instance:

• Sea water is a mixture of salt and water.
• Soil contains various organic and inorganic materials.

Key points about mixtures:

• Mixtures can be separated into their components through physical processes.
• Each component retains its own properties.

Types of mixtures include:

• Homogeneous mixtures have a uniform composition (e.g., sugar in water).
• Heterogeneous mixtures have a non-uniform composition (e.g., sand and salt).

In summary, a pure substance has consistent properties, while a mixture contains multiple substances that can vary in composition.

Q2. List the points of differences between homogeneous and heterogeneous mixtures.

Ans:

Page No. 18

Q1. Differentiate between homogeneous and heterogeneous mixtures with examples.
Ans: The following are the differences between heterogeneous and homogenous mixtures.

Q2. How are sol, solution and suspension different from each other?
Ans:

Q3. To make a saturated solution, 36g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.

Ans: Mass of solute (NaCl): 36 g

Mass of solvent (H₂O): 100 g

Mass of solution: 136 g (NaCl + H₂O)

Concentration: Calculated as follows:

• Concentration = (Mass of solute / Mass of solution) × 100
• Concentration = (36 g / 136 g) × 100
• Concentration = 26.47%

Thus, the concentration of the solution is 26.47%.Page No. 19

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, and
• burning of paper and wood.

Ans: The following is the classification into physical and chemical change:

Q2. Try segregating the things around you as pure substances or mixtures.
Ans: Listed below are the classifications based on pure substances and mixtures:
 Page No. 22

Exercises

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.
Ans: (a) In water, sodium chloride in its solution can be separated through the process of evaporation (as well as crystallization). 
(b) The sublimation technique is appropriate as ammonium chloride supports sublimation. 
(c) Tiny metal pieces in the engine oil of a car can be filtered manually.
(d) Chromatography can be used to separate different pigments from an extract of flower petals.
(e) The technique of churning can be applied to separate butter from curd. It is based on the concept of difference in density. 
(f) To separate oil from water, which are two immiscible liquids which vary in their densities, using a funnel can be an effective method. 
(g) Tea leaves can be manually separated from tea using simple filtration methods. 
(h) Iron pins can be separated from sand either manually or with the use of magnets as the pins exhibit strong magnetic quality, which can be a key characteristic taken into consideration. 
(i) The differentiating property between husk and wheat is that there is a difference in their mass. If treated with a small amount of wind energy, a remarkable variation in the moving distance is noticed. Hence, to separate them, the sedimentation/winnowing procedure can be applied. 
(j) Due to the property of water, sand or fine mud particles tend to sink in the bottom as it is denser, provided they are undisturbed. Through the process of sedimentation/decantation, water can be separated from fine mud particles, as the technique is established on obtaining clear water by tilting it out.Page No. 23

Q2. Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Ans:  Steps for Making Tea

• Heat a cup of milk, which acts as the solvent.
• Add tea powder or leaves, the solute, to the boiling milk.
• Observe that the tea powder remains insoluble while boiling.
• Add sugar to the boiling solution and stir.
• Sugar, being a solute, is soluble in the milk.
• Continue stirring until the sugar completely dissolves, achieving saturation.
• Once the raw smell of tea leaves disappears, remove the solution from heat.
• Filter the mixture to separate the tea powder, which becomes the residue.
• The liquid that passes through is the filtrate, containing the dissolved sugar and milk.

Q3. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results 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?
Ans: (a) Given:
Mass of potassium nitrate required to produce a saturated solution in 100 g of water at 313 K = 62g
To find:
Mass of potassium nitrate required to produce a saturated solution in 50 g of water =?
Required amount = 62 x 50/100 = 31
Hence, 31 g of potassium nitrate is required.
(b) The solubility of potassium chloride in water is decreased when a saturated solution of potassium chloride loses heat at 353 K. Consequently, Pragya would observe crystals of potassium chloride, which would have surpassed its solubility at low temperatures.
(c) As per the given data, that is
Solubility of potassium nitrate at 293K = 32 g
Solubility of sodium chloride at 293K = 36 g
Solubility of potassium chloride at 293K = 35 g
Solubility of ammonium chloride at 293K = 37g
We can observe from this data that ammonium chloride has the highest solubility at 293K.
(d) Effect of change of temperature on the solubility of salts:
The table clearly depicts that the solubility of the salt is dependent upon the temperature and increases with an increase in temperature. With this, we can infer that when a salt arrives at its saturation point at a specific temperature, there is a propensity to dissolve more salt through an increase in the temperature of the solution.

Q4. Explain the following, giving examples. 
(a) Saturated solution 
(b) Pure substance 
(c) Colloid 
(d) Suspension
Ans: (a) Saturated solution: It is the state in a solution at a specific temperature when a solvent is no longer soluble without an increase in temperature. Example: Excess carbon leaves off as bubbles from a carbonated water solution saturated with carbon.
(b) Pure substance: A substance is said to be pure when it comprises only one kind of molecule, atom or compound without adulteration with any other substance or any divergence in the structural arrangement. Examples: Sulphur, diamonds etc.
(c) Colloid: A Colloid is an intermediate between solution and suspension. It has particles of various sizes that range between 2 to 1000 nanometers. Colloids can be distinguished from solutions using the Tyndall effect. Tyndall effect is defined as the scattering of light (light beam) through a colloidal solution. Examples: Milk and gelatin.
(d) Suspension: It is a heterogeneous mixture that comprises solute particles that are insoluble but are suspended in the medium. These particles that are suspended are not microscopic but visible to bare eyes and are large enough (usually larger than a micrometer) to undergo sedimentation.

Q5. Classify each of the following as a homogeneous or heterogeneous mixture. 
soda water, wood, air, soil, vinegar, filtered tea.
Ans: The following is the classification of the given substances into homogenous and heterogenous mixtures.

Q6. How would you confirm that a colorless liquid given to you is pure water?
Ans: We can confirm if a colorless liquid is pure by setting it to boil. If it boils at 100°C, it is said to be pure. But if there is a decrease or increase in the boiling point, we infer that water has added impurities and, hence not pure.Page No. 24

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
Ans: The following substances from the above-mentioned list are pure substances:

• Iron
• Ice
• Hydrochloric acid
• Calcium oxide
• Mercury

Q8. Identify the solutions among the following mixtures.
(a) Soil
(b) Seawater
(c) Air
(d) Coal
(e) Soda water
Ans: The following are the solutions from the above-mentioned list of mixtures:

• Sea water
• Air
• Soda water

Q9. Which of the following will show the “Tyndall effect”?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution
Ans:  Milk and starch solution demonstrate the Tyndall effect because they are colloidal solutions. In these solutions, light is scattered, making its path visible.

• The Tyndall effect occurs when light passes through a colloid.
• Colloidal solutions contain particles that are small enough to scatter light.
• Examples of colloids include milk and starch solutions.

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 
(i) Carbon dioxide 
(m) Blood
Ans:
 

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
Ans: Among the options listed, the following are considered chemical changes:

• Rusting of iron
• Cooking of food
• Digestion of food
• Burning of a candle

The growth of a plant is a complex process involving both chemical and physical changes, while mixing iron filings and sand and freezing water are not chemical changes.

Page No. 3

Q1. Which of the following are matter?
Chair, air, love, smell, hate, almonds, thought, cold, lemon water, the smell of perfume.
Ans: The following substances are matter: Chair, Air, Almonds, Lemon water, and the smell of perfume (The smell is caused by volatile substances which are matter, as they occupy space and have mass).

Matter around us

Q2. Give reasons for the following observation.
The smell of hot sizzling food reaches you several meters away, but to get the smell from cold food, you have to go close.
Ans: When the air is heated, the particles in it gain more kinetic energy and move faster. This is why the smell of hot food travels farther, allowing a person to sense it even from several meters away.

Q3. A diver is able to cut through water in a swimming pool. Which property of matter does this observation show?
Ans: The particles of every matter have a force of attraction between them. This force keeps the particles together in a matter. In the case of water, the force of attraction between particles is less in comparison to solids. Thus, water molecules flow easily, giving way to a diver.

Showing less intermolecular force between liquid molecules.

Q4. What are the characteristics of the particles of matter?
Ans:  The characteristics of particles of matter are:
(i) Presence of intermolecular spaces between particles.
(ii) Particles are in constant motion.
(iii) They attract each other.Page No. 6

Q1. The mass per unit volume of a substance is called density. 
(density=mass/volume). Arrange the following in the order of increasing density – air, exhaust from the chimneys, honey, water, chalk, cotton and iron.
Ans: The following substances are arranged in increasing density: Air < Exhaust from chimney <  Cotton < Water <  Honey < Chalk < Iron.

Q2. (a) Tabulate the differences in the characteristics of states of matter.
(b) Comment upon the following: rigidity, compressibility, fluidity, filling a gas container, shape, kinetic energy and density.
Ans:  (a) The differences in the characteristics of the three states of matter solid, liquid and gas are:-

(b)
(i) Rigidity: It is the property of matter that continues to remain in its shape when treated with an external force.
(ii) Compressibility: Particles have the ability to reduce their intermolecular space when an external force is applied, which increases their density. This characteristic is called compressibility.
(iii) Fluidity: It is the ability of a substance to flow or move about freely.
(iv) Filling the gas container: The particles in a container take their shape as they randomly vibrate in all possible directions.
(v) Shape: It is the definite structure of an object within an external boundary.
(vi) Kinetic energy: Motion allows particles to possess energy, which is referred to as kinetic energy. The increasing order of kinetic energy possessed by various states of matter is Solids < Liquids < Gases.
Mathematically, it can be expressed as K.E = 1/2 mv², where ‘m’ is the mass and ‘v’ is the velocity of the particle.
(vii) Density: It is the mass of a unit volume of a substance. It is expressed as d = M/V, where ‘d’ is the density, ‘M’ is the mass and ‘V’ is the volume of the substance

Q3. Give reasons 
(a) A gas fills completely the vessel in which it is kept. 
(b) A gas exerts pressure on the walls of the container.
(c) A wooden table should be called a solid. 
(d) We can easily move our hand in the air, but to do the same through a solid block of wood, we need a karate expert.
Ans: 
(a) There is a low force of attraction between gas particles. The particles in the filled vessel are free to move about.
(b) Gaseous particles have the weakest attraction force and move randomly in all directions. When a gas particle hits the walls of its container, it applies a force, creating pressure on the walls.
(c) The hardwood table has a clear shape and volume. The wood particles are tightly packed and do not change to fit the shape of a container. This gives the table its solid properties.
(d) The air particles are spread far apart with a lot of space between them, which is why we can move our hands freely through the air. However, in a solid block, the particles are held tightly together by a strong force of attraction, leaving little or no space between them. That’s why breaking a solid block would require the strength of a karate expert.

Q4. Liquids generally have lower density as compared to solids. But you must have observed that ice floats on water. Find out why?
Ans: 

• The mass per unit volume of a substance is called density (density = mass/volume). 
• As the volume of a substance increases, its density decreases. In general, solids have higher density than liquids. Water is also a liquid, so it should also have less density than that of a solid, which is ice.
• Though ice is solid, it has a cage-like structure; hence, there are a large number of empty spaces between its particles. These spaces are larger than the spaces between the particles of water. Thus, for a given mass of water, the volume of ice is greater than that of water.
• Hence, the density of ice is less than that of water. A substance with a lower density than water can float on water. Therefore, ice floats on water.

Page No. 9

Q1. Convert the following temperature to Celsius scale:
(a) 300 K 
(b) 573 K
Ans: To convert a temperature from the Kelvin scale to the Celsius scale, you simply subtract 273 from the given Kelvin temperature.

(a)  (300 – 273)°C = 27°C
(b) (573 – 273)°C = 300°C

Q2. What is the physical state of water at:
(a) 250°C 
(b) 100°C
Ans: 
(a) At 250°C, the physical state of water is gas as the temperature is beyond its boiling point.
(b) At 100°C, it is in the transition state (both liquid and gaseous states) as the water is at its boiling point. Hence, it would be present in both liquid and gaseous state.

Q3. For any substance, why does the temperature remain constant during the change of state?
Ans: It is due to the latent heat as the heat supplied to increase the temperature of the substance is used up to transform the state of matter of the substance; hence, the temperature stays constant. Latent heat is the heat energy needed to change a substance from one form to another without changing its temperature. For example, when ice melts into water, it absorbs heat (latent heat of fusion) without increasing in temperature. Similarly, when water evaporates into steam, it also absorbs heat (latent heat of vaporization) without a temperature change.

Q4. Suggest a method to liquefy atmospheric gases.
Ans: To transform a gas into a liquid, it is necessary to bring its constituent particles or molecules closer. This can be achieved with atmospheric gases by either increasing the pressure or lowering the temperature. Page No. 10

Q1. Why does a desert cooler cool better on a hot dry day?
Ans: A desert cooler works better on hot, dry days because the high temperature and low humidity increase the rate of evaporation. This greater evaporation results in more effective cooling.

Q2. How does the water kept in an earthen pot (matka) become cool during summer?
Ans: An earthen pot has tiny pores that allow water to seep through and evaporate from its surface. This evaporation requires energy, which is drawn from the water inside the pot, making the water cooler.

Q3. Why does our palm feel cold when we put some acetone or petrol or perfume on it?
Ans: Acetone, petrol, and perfume are highly volatile substances. When applied to our palm, they evaporate quickly, absorbing heat from the skin and making our palm feel cold.

Q4. Why are we able to sip hot tea or milk faster from a saucer rather than a cup?
Ans: A saucer has a larger surface area than a cup, which allows for faster evaporation. This rapid evaporation cools the tea or milk more quickly, enabling us to sip it faster.

Q5. What type of clothes should we wear in summer?
Ans: In summer, it is advisable to wear light-colored cotton clothes. Light colors reflect heat, and cotton is breathable, allowing sweat to evaporate and providing a cooling effect on the skin.

Exercises 

Q1. Convert the following temperature to Celsius scale.
(a) 293K 
(b) 470K
Ans: To convert a temperature from the Kelvin scale to the Celsius scale, you simply subtract 273 from the given Kelvin temperature because 0°C=273K.
(a) 293K= (293 – 273)°C = 20°C
(b) 470K= (470 – 273)°C = 197°C

Q2. Convert the following temperatures to the kelvin scale.
(a) 25°C 
(b) 373°C
Ans: 
0°C = 273K
(a) 25°C = (25+273)K = 298K
(b) 373°C = (373+273)K = 646K

Q3. Give reason for the following observations:
(a) Naphthalene balls disappear with time without leaving any solid.
(b) We can get the smell of perfume sitting several metres away.
Ans: 
(a) At room temperature, naphthalene balls undergo sublimation, which means they change directly from a solid to a gaseous state without undergoing the intermediate state, i.e., the liquid state.

Naphthalene Balls

(b) It is because perfumes contain a volatile organic solvent that can easily diffuse through air and, hence, carry the fragrance to people sitting several metres away.

Q4. Arrange the following substances in increasing order of forces of attraction between the particles— water, sugar, oxygen.
Ans: Oxygen (gas) < Water (liquid) < Sugar (solid)

Q5. What is the physical state of water at: 
(a) 25°C, (b) 0°C, (c) 100°C?
Ans: The physical state of water at different temperatures is as follows:
(a) At 25°C, water is in a liquid state (typical room temperature).
(b) At 0°C, water is at its freezing point, so both solid (ice) and liquid (water) phases can be observed.
(c) At 100°C, water is at its boiling point, resulting in the presence of both liquid water and gaseous water (water vapor).

Q6. Give two reasons to justify.
(a) Water at room temperature is a liquid.
(b) An iron almirah is a solid at room temperature.
Ans:

(a) Water remains in a liquid state at room temperature for two main reasons:

1. Its melting or freezing point is lower than room temperature, while its boiling point is higher (100°C).
2. Water has no fixed shape and flows to take the shape of its container, which indicates that it occupies a fixed volume but does not have a defined shape.

(b) An iron almirah is a solid at room temperature for the following reasons:

1. Both the melting and boiling points of iron are above room temperature, meaning it remains solid under these conditions.
2. An iron almirah is rigid and maintains a definite shape, and metals typically have a high density, further confirming that it is solid at room temperature.

Q7. Why is ice at 273 K more effective in cooling than water at the same temperature?
Ans: At 273 K, ice will absorb heat energy or latent heat from the medium during melting to transform into water. As a result, ice has a greater cooling impact than water at the same temperature since water does not absorb the excess heat from the medium.

Q8. What produces more severe burns, boiling water or steam?
Ans: Steam produces severe burns. It is because it is an exothermic reaction that releases a high amount of heat, which it consumes during vaporization.

Q9. Name A, B, C, D, E and F in the following diagram showing change in its state.

Ans: Interconversion of three states of matter: Using temperature or pressure, any state of matter can be turned into another.
(A) Solid to Liquid → Melting (or) fusion (or) liquefaction
(B) Liquid to Gas → Evaporation (or) vaporization
(C) Gas to liquid → Condensation
(D) Liquid to Solid → Solidification
(E) Solid to Gas → Sublimation
(F) Gas to Solid → Deposition