case study questions chemistry class 9 chapter 2

Class 9 Science Case Study Questions Chapter 2 Is Matter Around Us Pure

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Case study Questions in Class 9 Science Chapter 2 are very important to solve for your exam. Class 9 Science Chapter 2 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Class 9 Science Case Study Questions Chapter 2 Is Matter Around Us Pure?

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In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Is Matter Around Us Pure? Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Science Chapter 2 Is Matter Around Us Pure?

Case Study/Passage-Based Questions

Case Study 1: Akshita wants to separate the mixture of dyes constituting a sample of ink. She marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in the figure. The filter paper was removed when the water moved near the top of the filter paper.

case study questions chemistry class 9 chapter 2

(i) Identify the technique used by the Akshita. (a) Sedimentation (b) Filtration (c) Chromatography (d) Distillation

Answer: (c) Chromatography.

(ii) What would you expect to see, if the ink contains three different coloured components? (a) We will not see any band on the filter paper. (b) We would see three bands on the filter paper at various lengths. (c) We would see infinite bands on the filter paper. (d) We would see the single band on the filter paper.

Answer: (b) The components of the ink will travel with water and we would see three bands on the filter paper at various lengths.

(iii) An application where you can use this technique is: (a) To separate salt from sand (b) To separate the wheat from the husk (c) To separate oil from water (d) To separate drugs from the blood.

Answer: (d) To separate drugs from blood.

(iv) The above process is used for the separation of : (a) insoluble substances (b) single solute that dissolves in the soluble solvent. (c) solutes that dissolve in the same solvent. (d) solutes that dissolve in the different solvents.

Answer: (c) For the separation of those solutes that dissolve in the same solvent.

(v) What is chromatography? (a) It is an agricultural method to separate grains (b) A method to separate magnetic impurities from non-magnetic impurities

(c) The process of separating the suspended particles of an insoluble substance (d) Method of separating and identifying various components in a mixture, which are present in small trace quantities.

Answer: (d) Method of separating and identifying various components in a mixture, which are present in small trace quantities.

Case Study 2: A homogeneous mixture of two or more substances is called a true solution. it consists of solute and solvent. The particle size of the true solution is less than 1 nanometer. A suspension is a heterogeneous mixture in which the solute particle does not dissolve but remains suspended throughout the bulk of the medium. A colloid is a mixture that is actually heterogeneous but appears to be homogeneous as the particles are uniformly spread throughout the solution.

(i) which one of the following is most stable?

A)True solution

B)Suspensions

D) both A and B

Answer: A)True solution

ii) which type of mixture can be separated by filtration?

D)All of these

Answer: B)Suspensions

iii) which statement is incorrect about the Tyndall effect. *

A)True solution shows Tyndall effect

B)Suspensions show the Tyndall effect

C)Colloid show Tyndall effect

D)Both B and C show the Tyndall effect

Answer: A)True solution shows Tyndall effect

iv) Which is the correct order of stability of solution *

A) True < Colloid<Suspension

B)Colloid<Suspension<True

C)Colloid<True<Suspension

D)Suspension<Colloid<True

Answer: D)Suspension

Case Study 3:

Matter can be classified into two categories: pure substances and mixtures. Pure substances are made up of a single type of particle and cannot be separated into other substances by physical methods. They have definite and constant properties. On the other hand, mixtures are made up of two or more substances that are physically combined and can be separated into their individual components. Mixtures can be further classified into homogeneous and heterogeneous mixtures. Homogeneous mixtures are uniform in composition, meaning the components are evenly distributed throughout the mixture. Heterogeneous mixtures, on the other hand, have non-uniform composition with visible different parts. It is important to understand the nature of matter around us and differentiate between pure substances and mixtures to comprehend their properties and behavior.

What is the main characteristic of a pure substance? a) Made up of two or more substances b) Cannot be separated into other substances c) Has non-uniform composition d) Components are evenly distributed Answer: b) Cannot be separated into other substances

Which of the following is an example of a pure substance? a) Air b) Saltwater c) Gold d) Soil Answer: c) Gold

How are mixtures different from pure substances? a) Mixtures have definite and constant properties b) Mixtures are made up of a single type of particle c) Mixtures cannot be separated into other substances d) Mixtures are physically combined and can be separated Answer: d) Mixtures are physically combined and can be separated

Which type of mixture has a non-uniform composition? a) Homogeneous mixture b) Heterogeneous mixture Answer: b) Heterogeneous mixture

What is the primary reason for understanding the nature of matter around us? a) To separate mixtures into pure substances b) To comprehend the properties and behavior of matter c) To classify mixtures into homogeneous and heterogeneous d) To identify the components in pure substances Answer: b) To comprehend the properties and behavior of matter

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Case Study and Passage Based Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure?

Last modified on: 2 years ago
Reading Time: 4 Minutes

Case Study Questions:

Question 1:

Akshita wants to separate the mixture of dyes constituting a sample of ink. She marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in figure. The filter paper was removed when the water moved near the top of the filter paper.

(i) Identify the technique used by the Akshita. (a) Sedimentation (b) Filtration (c) Chromatography (d) Distillation

(iii) An application where you can use this technique is: (a) To separate salt from sand (b) To separate wheat from husk (c) To separate oil from water (d) To separate drugs from blood.

(iv) The above process is used for the separation of : (a) insoluble substances (b) single solute that dissolves in soluble solvent. (c) solutes that dissolve in the same solvent. (d) solutes that dissolve in the different solvents.

(v) What is chromatography ? (a) It is an agricultural method to separate grains (b) A method to separate magnetic impurities from non-magnetic impurities (c) The process of separating the suspended particles of an insoluble substance (d) Method of separating and identifying various components in a mixture, which are present in small trace quantities.

Case Study and Passage Based Questions for Class 9 Science Chapter 3 Atoms and Molecules

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Class 9 Science Case...

Class 9 Science Case Study Questions

Table of Contents

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

If you are wondering how to solve class 9 science case study questions, then myCBSEguide is the best platform to choose. With the help of our well-trained and experienced faculty, we provide solved examples and detailed explanations for the recently added Class 9 Science case study questions.

You can find a wide range of solved case studies on myCBSEguide, covering various topics and concepts. Class 9 Science case studies are designed to help you understand the application of various concepts in real-life situations.

The rationale behind Science

Science is crucial for Class 9 students’ cognitive, emotional, and psychomotor development. It encourages curiosity, inventiveness, objectivity, and aesthetic sense.

In the upper primary stage, students should be given a variety of opportunities to engage with scientific processes such as observing, recording observations, drawing, tabulating, plotting graphs, and so on, whereas in the secondary stage, abstraction and quantitative reasoning should take a more prominent role in science teaching and learning. As a result, the concept of atoms and molecules as matter’s building units, as well as Newton’s law of gravitation, emerges.

Science is important because it allows Class 9 Science students to understand the world around us. It helps to find out how things work and to find solutions to problems at the Class 9 Science level. Science is also a source of enjoyment for many people. It can be a hobby, a career, or a source of intellectual stimulation.

Case study questions in Class 9 Science

The inclusion of case study questions in Class 9 science CBSE is a great way to engage students in critical thinking and problem-solving. By working through real-world scenarios, Class 9 Science students will be better prepared to tackle challenges they may face in their future studies and careers. Class 9 Science Case study questions also promote higher-order thinking skills, such as analysis and synthesis. In addition, case study questions can help to foster creativity and innovation in students. As per the recent pattern of the Class 9 Science examination, a few questions based on case studies/passages will be included in the CBSE Class 9 Science Paper. There will be a paragraph presented, followed by questions based on it.

Examples of Class 9 science class case study questions

Class 9 science case study questions have been prepared by myCBSEguide’s qualified teachers. Class 9 case study questions are meant to evaluate students’ knowledge and comprehension of the material. They are not intended to be difficult, but they will require you to think critically about the material. We hope you find Class 9 science case study questions beneficial and that they assist you in your exam preparation.

The following are a few examples of Class 9 science case study questions.

Class 9 science case study question 1

due to its high compressibility
large volumes of a gas can be compressed into a small cylinder
transported easily
all of these
shape, volume
volume, shape
shape, size
size, shape
the presence of dissolved carbon dioxide in water
the presence of dissolved oxygen in the water
the presence of dissolved Nitrogen in the water
liquid particles move freely
liquid have greater space between each other
both (a) and (b)
none of these
Only gases behave like fluids
Gases and solids behave like fluids
Gases and liquids behave like fluids
Only liquids are fluids

Answer Key:

(d) all of these
(a) shape, volume
(b) the presence of dissolved oxygen in the water
(c) both (a) and (b)
(c) Gases and liquids behave like fluids

Class 9 science case study question 2

12/32 times
18 g of O 2
18 g of CO 2
18 g of CH 4
1 g of CO 2
1 g of CH 4 CH 4
2 moles of H2O
20 moles of water
6.022 × 1023 molecules of water
1.2044 × 1025 molecules of water
(I) and (IV)
(II) and (III)
(II) and (IV)
Sulphate molecule
Ozone molecule
Phosphorus molecule
Methane molecule
(c) 8/3 times
(d) 18g of CH 4
(c) 1g of H 2
(d) (II) and (IV)
(c) phosphorus molecule

Class 9 science case study question 3

collenchyma
chlorenchyma
It performs photosynthesis
It helps the aquatic plant to float
It provides mechanical support
Sclerenchyma
Collenchyma
Epithelial tissue
Parenchyma tissues have intercellular spaces.
Collenchymatous tissues are irregularly thickened at corners.
Apical and intercalary meristems are permanent tissues.
Meristematic tissues, in its early stage, lack vacuoles, muscles
(I) and (II)
(III) and (I)
Transpiration
Provides mechanical support
Provides strength to the plant parts
None of these
(a) Collenchyma
(b) help aquatic plant to float
(b) Sclerenchyma
(d) Only (III)
(c) provide strength to plant parts

Cracking Class 9 Science Case Study Questions

There is no one definitive answer to Class 9 Science case study questions. Every case study is unique and will necessitate a unique strategy. There are, nevertheless, certain general guidelines to follow while answering case study questions.

To begin, double-check that you understand the Class 9 science case study questions. Make sure you understand what is being asked by reading it carefully. If you’re unclear, seek clarification from your teacher or tutor.
It’s critical to read the Class 9 Science case study material thoroughly once you’ve grasped the question. This will provide you with a thorough understanding of the problem as well as the various potential solutions.
Brainstorming potential solutions with classmates or other students might also be beneficial. This might provide you with multiple viewpoints on the situation and assist you in determining the best solution.
Finally, make sure your answer is presented simply and concisely. Make sure you clarify your rationale and back up your claim with evidence.

A look at the Class 9 Science Syllabus

The CBSE class 9 science syllabus provides a strong foundation for students who want to pursue a career in science. The topics are chosen in such a way that they build on the concepts learned in the previous classes and provide a strong foundation for further studies in science. The table below lists the topics covered in the Class 9 Science syllabus of the Central Board of Secondary Education (CBSE). As can be seen, the Class 9 science syllabus is divided into three sections: Physics, Chemistry and Biology. Each section contains a number of topics that Class 9 science students must study during the course.

CBSE Class 9 Science (Code No. 086)

Theme: Materials Unit I: Matter-Nature and Behaviour Definition of matter; solid, liquid and gas; characteristics – shape, volume, density; change of state-melting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation. Nature of matter: Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Particle nature and their basic units: Atoms and molecules, Law of constant proportions, Atomic and molecular masses. Mole concept: Relationship of mole to mass of the particles and numbers. Structure of atoms: Electrons, protons and neutrons, valency, the chemical formula of common compounds. Isotopes and Isobars.

Theme: The World of the Living Unit II: Organization in the Living World Cell – Basic Unit of life: Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes – basic structure, number. Tissues, Organs, Organ System, Organism: Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).

Theme: Moving Things, People and Ideas Unit III: Motion, Force and Work Motion: Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, derivation of equations of motion by graphical method; elementary idea of uniform circular motion. Force and Newton’s laws: Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration. Elementary idea of conservation of Momentum. Gravitation: Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall. Floatation: Thrust and Pressure. Archimedes’ Principle; Buoyancy. Work, energy and power: Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy. Sound: Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Theme: Food Unit IV: Food Production Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.

PRESCRIBED BOOKS:

Science-Textbook for class IX-NCERT Publication
Assessment of Practical Skills in Science-Class IX – CBSE Publication
Laboratory Manual-Science-Class IX, NCERT Publication
Exemplar Problems Class IX – NCERT Publication

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myCBSEguide offers high-quality study materials that cover all of the topics in the Class 9 Science curriculum.
myCBSEguide provides practice questions and mock examinations to assist students in the best possible preparation for their exams.
On our myCBSEguide app, you’ll find a variety of solved Class 9 Science case study questions covering a variety of topics and concepts. These case studies are intended to help you understand how certain principles are applied in real-world settings
myCBSEguide is that the study material and practice problems are developed by a team of specialists who are always accessible to assist students with any questions they may have. As a result, students may be confident that they will receive the finest possible assistance and support when studying for their exams.

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Case Study Questions of Chapter 2 Is Matter Around Us Pure? PDF Download

Case study Questions on Class 9 Science Chapter 2 are very important to solve for your exam. Class 9 Science Chapter 2 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Science Chapter 2 Is Matter Around Us Pure?

Is Matter Around Us Pure? Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 9 Science Chapter 2 Is Matter Around Us Pure?

Case Study/Passage-Based Questions

Question 1:

Akshita wants to separate the mixture of dyes constituting a sample of ink. She marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in the figure. The filter paper was removed when the water moved near the top of the filter paper.

(i) Identify the technique used by the Akshita. (a) Sedimentation (b) Filtration (c) Chromatography (d) Distillation

Answer: (c) Chromatography.

Answer: (b) The components of the ink will travel with water and we would see three bands on the filter paper at various lengths.

(iii) An application where you can use this technique is: (a) To separate salt from sand (b) To separate the wheat from the husk (c) To separate oil from water (d) To separate drugs from the blood.

Answer: (d) To separate drugs from blood.

Answer: (c) For the separation of those solutes that dissolve in the same solvent.

(v) What is chromatography? (a) It is an agricultural method to separate grains (b) A method to separate magnetic impurities from non-magnetic impurities

(c) The process of separating the suspended particles of an insoluble substance (d) Method of separating and identifying various components in a mixture, which are present in small trace quantities.

Answer: (d) Method of separating and identifying various components in a mixture, which are present in small trace quantities.

Question 2:

A homogeneous mixture of two or more substances is called a true solution. it consists of solute and solvent. The particle size of the true solution is less than 1 nanometer. A suspension is a heterogeneous mixture in which the solute particle does not dissolve but remains suspended throughout the bulk of the medium. A colloid is a mixture that is actually heterogeneous but appears to be homogeneous as the particles are uniformly spread throughout the solution.

(i) which one of the following is most stable?

A)True solution

B)Suspensions

D) both A and B

Answer: A)True solution

ii) which type of mixture can be separated by filtration?

D)All of these

Answer: B)Suspensions

iii) which statement is incorrect about the Tyndall effect. *

A)True solution shows Tyndall effect

B)Suspensions show the Tyndall effect

C)Colloid show Tyndall effect

D)Both B and C show the Tyndall effect

Answer: A)True solution shows Tyndall effect

iv) Which is the correct order of stability of solution *

A) True < Colloid<Suspension

B)Colloid<Suspension<True

C)Colloid<True<Suspension

D)Suspension<Colloid<True

Answer: D)Suspension

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Science Chapter 2 Is Matter Around Us Pure? with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Science Is Matter Around Us Pure? Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible

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NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure

NCERT Solutions for Class 9 Science (chemistry) Chapter 2 Is Matter Around Us Pure are given below. In these solutions, we have answered all the intext and exercise questions provided in NCERT class 9 science textbook. Class 9 NCERT Solutions Science Chapter 2 provided in this article are strictly based on the CBSE syllabus and curriculum. Students can easily download these solutions in PDF format for free from our app.

Class 9 Science Chapter 2 Textbook Questions and Answers

Intext Questions Page No. 15

Questions 1: what do mean by a pure substance?

Answer: A pure substance is the one that consists of a single type of particles, i.e., all constituent particles of the substance have the same chemical nature. Pure substances can be Classified as elements or compounds.

Question 2: List the points of differences between homogeneous and heterogeneous mixtures.

Page No. 18

Question 1: Differentiate between homogeneous and heterogeneous mixtures with examples.

Answer: A homogeneous mixture is a mixture having a uniform composition throughout the mixture. For example, mixtures of salt in water, sugar in water, copper sulphate in water, iodine in alcohol, alloy, and air have uniform compositions throughout the mixtures.

On the other hand, a heterogeneous mixture is a mixture having a non-uniform composition throughout the mixture. For example, composition of mixtures of sodium chloride and iron fillings, salt and sulphur, oil and water, chalk powder in water, wheat flour in water, milk and water are not uniform throughout the mixtures.

Question 2: How are sol, solution and suspension different from each other?

Answer: Sol is a heterogeneous mixture. In this mixture, the solute particles are so small that they cannot be seen with the naked eye. Also, they seem to be spread uniformly throughout the mixture. The Tyndall effect is observed in this mixture. For example: milk of magnesia, mud

Solution is a homogeneous mixture. In this mixture, the solute particles dissolve and spread uniformly throughout the mixture. The Tyndall effect is not observed in this mixture.

For example: salt in water, sugar in water, iodine in alcohol, alloy

Suspensions are heterogeneous mixtures. In this mixture, the solute particles are visible to the naked eye, and remain suspended throughout the bulk of the medium. The Tyndall effect is observed in this mixture.

For example: chalk powder and water, wheat flour and water

NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure part 1

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.

NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure part 2

PAGE NO. 24 (I)

Question 1: 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: A mixture of two miscible liquids having a difference in their boiling points more than 25°C can be separated by the method of distillation. Thus, kerosene and petrol can be separated by distillation.

NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure part 3

In this method, the mixture of kerosene and petrol is taken in a distillation flask with a thermometer fitted in it. We also need a beaker, a water condenser, and a Bunsen burner. The apparatus is arranged as shown in the above figure. Then, the mixture is heated slowly. The thermometer should be watched simultaneously. Kerosene will vaporize and condense in the water condenser. The condensed kerosene is collected from the condenser outlet, whereas petrol is left behind in the distillation flask.

Question 2: 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.

Question 3: What type of mixtures is separated by the technique of crystallization?

Answer: By the technique of crystallization, pure solids are separated from impurities. For example, salt obtained from sea is separated from impurities; crystals of alum (Phitkari) are separated from impure samples.

PAGE NO 24(II)

Question 1: 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 water breaking into hydrogen and oxygen gases.
Dissolving common salt in water
Making a fruit salad with raw fruits, and
Burning of paper and wood

Answer: Cutting of trees → Physical change

Melting of butter in a pan → Physical change

Rusting of almirah → Chemical change

Boiling of water to form steam → Physical change

Passing of electric current through water, and water breaking down into hydrogen and oxygen gas → Chemical change

Dissolving common salt in water → Physical change

Making a fruit salad with raw fruits → Physical change

Burning of paper and wood → Chemical change

Question 2: Try segregating the things around you as pure substances or mixtures.

Answer: Listed below are the classifications based on pure substances and mixtures:

Question 1: Which separation techniques will 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) Sodium chloride from its solution in water → Evaporation

(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride → Sublimation

(d) Different pigments from an extract of flower petals → Chromatography

(e) Butter from curd → Centrifugation

(f) Oil from water → Using separating funnel

(g) Tea leaves from tea → Filtration

(h) Iron pins from sand → Magnetic separation

(i) Wheat grains from husk → Winnowing

(j) Fine mud particles suspended in water → Centrifugation

Question 2: Write the steps you would use for making tea. Use the words – solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.

Answer: 1. Take a cup of water in a container as solvent and heat it. 2. Add sugar in it which is solute. Heat it till all sugar dissolves. 3. You get a solution of water and sugar. 4. Sugar is soluble in water completely. 5. Add half a tea-spoon of tea-leaves, it is insoluble in water. 6. Boil the content, add milk which is also soluble in water, boil again. 7. Filter the tea with the help of strainer, the tea collected in cup is filtrate and the tea leaves collected on the strainer is residue.

NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure part 4

Question 3: 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.

(d) What is the effect of change of temperature on the solubility of a salt?

Answer: (a) Mass of potassium nitrate required to produce a saturated solution in 100 g of water at 313 K = 62g

∴ Mass of potassium nitrate required to produce a saturated solution in 50 g of water = (62 × 50)/100 = 31 Hence 31 g of potassium nitrate is required.

(b) Some amount of dissolved Potassium Chloride will reappear as undissolved solid as solubility of solute decreases with the decrease of temperature.

Potassium nitrate – 32 g
Sodium chloride – 36 g
Potassium chloride – 35 g
Ammonium chloride – 37 g
Ammonium chloride has the highest solubility at 293 K.

(d) Solubility of salt increases with the increase in temperature.

Question 4: 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 of the solid (solute) can be dissolved at a given temperature is called a saturated solution. Suppose 50 gm of a solute is the maximum amount that can be dissolved in 100 gm water at 298 K. Then 150 gm of solution so obtained is the saturated solution at 298 K.

A saturated solution is a solution in which the maximum amount of solute has been dissolved at a given temperature. The solution cannot dissolve beyond that amount of solute at that temperature. Any more solute added will settle down at the bottom of the container as a precipitate. Suppose 500 g of a solvent can dissolve a maximum of 150 g of a particular solute at 40°C. Then, the solution obtained by dissolving 150 g of that solute in 500 g of that solvent at 300 K is said to be a saturated solution at 300 K.

Pure Substance: A pure substance consists of a single of matter or particles and cannot be separated into other kind of matter by any physical process. Pure substances always have the same colour, taste and texture at a given temperature and pressure. For example, pure water is always colourless, odorless and tasteless and boils at 373 K at normal atmospheric pressure.

Colloid: Colloid A colloid is a heterogeneous mixture. The size of the solutes in this mixture is so small that they cannot be seen individually with naked eyes, and seems to be distributed uniformly throughout the mixture. The solute particles do not settle down when the mixture is left undisturbed. This means that colloids are quite stable. Colloids cannot be separated by the process of filtration. They can be separated by centrifugation. Colloids show the Tyndall effect. For example, milk, butter, foam,fog, smoke, clouds.

Suspension: Suspension Suspensions are heterogeneous mixtures. The solute particles in this mixture remain suspended throughout the bulk of the medium. The particles can be seen with naked eyes. Suspension shows the Tyndall effect. The solute particles settle down when the mixture is left undisturbed. This means that suspensions are unstable. Suspensions can be separated by the method of filtration. For example, mixtures of chalk powder and water, wheat flour and water.

Question 5. Classify each of the following as a homogeneous or heterogeneous mixture: soda water, wood, air. soil, vinegar, filtered tea.

Answer: Homogeneous: Soda water, vinegar, filtered tea. Heterogeneous: Wood, air, soil.

Question 6. How would, you confirm that a colourless liquid given to you is pure water?

Answer: We can confirm if a colourless 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 hence not pure.

Question 7. 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) Back (h) Wood (i) Air.

Answer: Following substances from the above-mentioned list are pure substances:

Hydrochloric acid
Calcium oxide

Question 8. Identify the solutions among the following mixtures. (a) Soil (b) Sea water (c) Air (d) Coal (e) Soda water.

Answer: The following are the solutions from the above-mentioned list of mixture:

Question 9. Which of the following will show “Tyndall effect”? (a) Salt solution (b) Milk (c) Copper sulphate solution (d) Starch solution.

Answer: Milk and starch solution will show the “Tyndall effect”.

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 (i) Air (j) Soap (k) Methane (l) Carbon dioxide (m) Blood

NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure part 5

Question 11. 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 candle

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

CBSE Class 9 Science NCERT Solutions Chapter 2 helps students to clear their doubts and to score good marks in the board exam. All the questions are solved by experts with a detailed explanation that will help students complete their assignments & homework. Having a good grasp over CBSE NCERT Solutions for Class 9 Science will further help the students in their preparation for board exams and other competitive exams such as NTSE, Olympiad, etc.

NCERT Solutions for Class 9 Science Chapter 2 PDF

Below we have listed the topics discussed in NCERT Solutions for Class 9 Science Chapter 2. The list gives you a quick look at the different topics and subtopics of this chapter.

NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure

23rd December 2023

NCERT Solutions for Class 9 Science (Chemistry) Chapter 2 Is Matter Around Us Pure provides detailed answers for all in-text and exercise Questions. These solutions contain an in-depth explanation of each topic involved in the chapter. Students studying in class 9 can access these solutions for free in PDF format.

All these solutions are prepared by expert teachers and updated for the current academic session. NCERT Solutions for Class 9 Science (Chemistry) Chapter 2 Is Matter Around Us Pure help students to understand the fundamental concepts given in class 9 Science textbook. We have prepared the answers to all the questions in an easy and well-structured manner. It helps students to grasp the chapter easily.

Class 9 Science Is Matter Around Us Pure Intext Questions and Answers

PAGE NO. 15

Questions 1: what do mean by a pure substance?

Answer: A pure substance is a material that has a constant composition and consistent properties throughout the sample. It can be either an element, which is made up of only one kind of atom, or a compound, which is made up of two or more types of atoms chemically bonded together in a fixed ratio.

Question 2: List the points of differences between homogeneous and heterogeneous mixtures.

PAGE NO. 18

Question 1: Differentiate between homogeneous and heterogeneous mixtures with examples.

On the other hand, a heterogeneous mixture is a mixture having a non-uniform composition throughout the mixture. For example: the composition of mixtures of sodium chloride and iron filings, salt and sulphur, oil and water, chalk powder in water, wheat flour in water, milk and water are not uniform throughout the mixtures.

Question 2: How are sol, solution and suspension different from each other?

Answer:

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: Mass of solute (sodium chloride) = 36 g (Given) Mass of solvent (water) = 100 g (Given) Then, mass of solution = Mass of solute + Mass of solvent = (36 + 100) g = 136 g Therefore, concentration (mass by mass percentage) of the solution

NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure image 1

PAGE NO. 19

1. Classify the following as physical or chemical 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 water breaking into hydrogen and oxygen gases.
Dissolving common salt in water
Making a fruit salad with raw fruits, and
Burning of paper and wood

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

Question 2: Try segregating the things around you as pure substances or mixtures.

Answer: Pure substance: Water, salt, sugar etc.

Mixture: Salt water, soil, wood, air, cold drink, rubber, sponge, fog, milk, butter, clothes, food.

Class 9 Science Matter in Our Surroundings Exercise Questions

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.

(a) Crystallization or Evaporation. (b) Sublimation. (c) Centrifugation or Sedimentation. (d) Chromatography. (e) Centrifugation. (f) Separating funnel. (g) Hand-picking. (h) Magnetic separation. (i) Winnowing. (j) Centrifugation.

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. Heat it. When the solvent boils, add the solute, milk. Milk and water forms a solution. Then pour some tea leaves over a sieve. Pour slowly hot solution of milk over tea leaves. Colour of tea leaves goes into solution as filtrate. The remaining tea leaves being insoluble remains as residue. Add requisite sugar which dissolves and the tea is ready.

Question 3. 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 a 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 for saturated solution of 100 grams of water = 62 g Potassium nitrate for saturated solution of 50 grams of water = 31 g

(b) Some amount of dissolved Potassium Chloride will reappear as undissolved solid as solubility of solute decreases with the decrease of temperature.

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) Solubility of salt increases with the increase in temperature.

Question 4: Explain the following giving examples. (a) Saturated solution, (b) Pure substance, (c) Colloid, (d) Suspension.

(b) Pure Substance: A pure substance consists of a single of matter or particles and cannot be separated into other kind of matter by any physical process. Pure substances always have the same colour, taste and texture at a given temperature and pressure. For example, pure water is always colourless, odourless and tasteless and boils at 373 K at normal atmospheric pressure.

(c) Colloid: Colloids are heterogeneous mixtures the particle size is too small to be seen with a naked eye, but it is big enough to scatter light. The particles are called the dispersed phase and the medium in which they are distributed is called the dispersion medium. Colloids are useful in industry and daily life. A colloid has the following characteristics:

It is a heterogeneous mixture.
The size of particles of a colloid lies between 1 – 100 nm and cannot be seen by naked eyes.
The particles of colloid can scatter a beam of light passing through it and make the path visible.
The particles of colloid cannot be separated from the mixture by filtration. The process of separation of colloidal particles is known as ‘centrifugation’.
They do not settle down when left undisturbed. In other words, colloids are quite stable e.g. smoke, milk, fog, cloud etc.

(d) Suspension: It is a heterogeneous mixture that comprises of 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.

Question 5. Classify each of the following as a homogeneous or heterogeneous mixture: soda water, wood, air. soil, vinegar, filtered tea.

Answer: Homogeneous: Soda water, vinegar, filtered tea. Heterogeneous: Wood, air, soil.

Question 6. How would, you confirm that a colourless liquid given to you is pure water?

Answer: By finding the boiling point of a given colourless liquid. If the liquid boils at 100°C at atmospheric pressure, then it is pure water. This is because pure substances have fixed melting and boiling point.

Question 7. 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) Back (h) Wood (i) Air.

Answer: Pure substances are: Ice, iron, hydrochloric acid, calcium oxide and mercury.

Question 8. Identify the solutions among the following mixtures. (a) Soil (b) Seawater (c) Air (d) Coal (e) Soda water.

Answer: Solutions are: Sea water soda water and air.

Question 9. Which of the following will show “Tyndall effect”? (a) Salt solution (b) Milk (c) Copper sulphate solution (d) Starch solution.

Answer: Milk and starch solution.

Answer:

Answer: Chemical changes are: (a) Growth of a plant (b) Rusting of iron (c) Cooking of food (d) Digestion of food (e) Burning of a candle

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CBSE Class 9 Science Important Case Study Questions with Answers for Term 2 Exam 2022 (PDF)

Check important case study questions of cbse class 9 science to prepare for the cbse term 2 exam 2022. all these questions have been put together by subject experts..

CBSE Class 9 Term 2 Exam 2022: Important case based questions for CBSE Class 9 Science are provided here students to prepare for the upcoming Term 2 Exam 2022. All the questions provided below are curated by the subject experts. These questions are really helpful to revise important concepts and prepare the case study questions for the exam. Answers to all questions have been provided for reference. So, students should practice the chapter-wise questions to clearly understand the right way to attempt the case based questions. Download the chapter-wise questions in PDF.

Check some of the important case study questions below:

Q. Read the following and answer the questions :

A student was asked by his teacher to verify the law of conservation of mass in the laboratory. He prepared 5% aqueous solutions of NaCl and Na 2 SO 4 . He mixed 10 mL of both these solutions in a conical flask. He weighed the flask on a balance. He then stirred the flask with a rod and weighed it after sometime. There was no change in mass.

Was the student able to verify the law of conservation of mass?
If not, what was the mistake committed by him?
In your opinion, what he should have done?
What is the molar mass of Na 2 SO 4 ?
No, he could not verify the law of conservation of mass in-spite of the fact that there was no change in mass.
No chemical reaction takes place between NaCl and Na 2 SO 4 . This means that no reaction actually took place in the flask.
He should have performed the experiment by using aqueous solutions of BaCl 2 and Na 2 SO 4 . A chemical reaction takes place in this case and a white precipitate of BaSO 4 is formed.
Will the weight of the precipitate be the same as that of the reactants before mixing?
If not, what she should have done?
Which law of chemical combination does this support?
State the law of conservation of mass.
No, it will not be the same.
She should have weighed the total contents of the beaker after the reaction and not the precipitate alone.
It supports the law of conservation of mass.
Mass can neither be created nor destroyed during a chemical reaction.

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Class 9 Science Chapter 2 Important Extra Questions

Class 9 Science Chapter 2 Important Questions of Is Matter Around Us Pure? Class 9 Science Chapter 2 Extra Questions contains all long and short answers questions with solutions and explanations. All the questions based on elements, compounds and mixtures are taken for the description of the chapter 2 of class ix science. These sets of questions include all the basic and important questions which are crucial for the school tests and final term exams. These revision questions of high school science ensure scoring good marks in exams and clearing all doubts in chapter 2 of 9th science. UP Board students also use the same NCERT Textbooks as the CBSE Students. So, these questions are important for UP Board students also.

Important Extra Questions for class 9 science chapter 2 Is Matter Around Us Pure is given below. Proper answers of each question is also given just below the questions. We have tried to cover entire chapter 2 of class 9 science NCERT at our level best. If someone want to add some more questions in this section, please send us the questions with answer. We will include the questions, if it is up to marks and useful for the students.

Class 9 Science Chapter 2 Important Extra Questions Set – 1

What is meant by a pure substance.

A pure substance is one which made up of only one kind of particle either atoms or molecules. For example, oxygen, carbon etc.

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

Mass of sodium chloride = 36g Mass of water = 100g Total mass of the solution = (36 + 100) g = 136g Concentration (mass percentage) of solution = (Mass of NaCl )/(Mass of solution ) × 100 = 36g/(136g ) × 100 = 26. 47g

Which separation technique will you apply to separate containing kerosene and petrol (difference in their boiling points is more than 250C) which are miscible with each other?

The miscible mixture of kerosene and petrol can be separated by fractional distillation. On heating the distillation flask, petrol having lower boiling points than kerosene, distils our first.

Name the technique to separate (i) butter and curd (ii) salt and water (iii) camphor from salt.

(i) Centrifugation (ii) Evaporation (iii) Sublimation

How can you change a saturated solution to an unsaturated solution without adding any more solvent to it?

By heating the saturated solution, it becomes unsaturated.

Class 9 Science Chapter 2 Important Extra Questions Set – 2

Sucrose crystals obtained from sugarcane and beetroot are mixed together. will it to be a passing pure substance or a mixture give reason for the same..

It is a pure substance because chemical composition of sugar crystals is same irrespective of its source.

Based on which factor a solution is said to diluted, concentration or saturated?

A solution is said to be diluted, concentrated or saturated on the basis of the amount of solute dissolved in the solution.

Identify solute and solvent in tincture of iodine.

Iodine is the solute and alcohol is the solvent.

What is mass per cent of a solution?

It is defined as the mass in grams of the solute present in one hundred grams of a solution.

What are the two components of a colloidal solution?

The two components of a colloidal solution are dispersed phase and dispersing medium.

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,
Dissolving common salt in water
Making a fruit salad with raw fruits and
Burning of paper and wood.
Physical change
Chemical change

Class 9 Science Chapter 2 Important Extra Questions Set – 3

In what respect does not true solution differ from a colloidal solution.

A true solution is homogeneous whereas a colloidal solution is heterogenous.

Two liquids A and B are miscible with each other at room temperature. Which separation technique will you apply to separate the mixture of A and B if the difference in their boiling points is 270C?

It is a process that separate a pure solid in the form of its crystals from a solution.

Define crystallisation.

Why is crystallisation technique considered better than simply evaporation to purify solids.

Crystallisation is considered better than simply evaporation because (i) some solids decompose or get charred on heating to dryness. (ii) some impurities may remain dissolved in the solution even after filtration. On evaporation these contaminate the solid.

Why is water called universal solvent?

Water is known as universal solvent because it has the ability to dissolve wide variety of substances.

Class 9 Science Chapter 2 Important Extra Questions Set – 4

Which of the are physical changes melting of iron metal, rusting of iron, bending of iron rod, drawing a wire of iron metal..

Melting of iron metal, bending of iron rod and drawing a wire of iron metal are physical changes because there is no change in the chemical composition of iron, only its form is changing.

Name two elements which exist in liquid state at room temperature.

Gallium and mercury exist in liquid state at room temperature.

An unknown substance A on thermal decomposition produces B and C What is an element, a compound or a mixture?

Unknown substance A should be a compound because elements and mixture do not decompose.

Identify the elements from the following substances sulphur, brine, hydrochloride acid, water, neon, paper, sugar.

The elements are sulphur and neon.

Sea water can be classified as homogenous as well as heterogeneous mixture; Comment.

Sea water is a mixture of salts and water which cannot be separated except by evaporation. Therefore, sea water is considered homogenous. Sea water also contains mud, decayed plants, etc., other than salts and water, so it is heterogeneous also.

Class 9 Science Chapter 2 Important Extra Questions Set – 5

Why is it not possible to distinguish particles of a solute from the solvent in solution.

A true solution is homogenous in nature. The solute and solvent particles are very small. They cannot be distinguishing even under a microscope.

Explain why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do.

Particle size in a suspension is large than those in a colloidal solution. Also molecular interaction in a suspension is not strong enough to keep the particles suspended and hence they settle down.

Identify colloids and true solution from the following: Pond water, frog, aluminium paints, vinegar and glucose solution.

True solution are vinegar and glucose solution. Colloids are frog, aluminium paint.

Give two examples each for (i) Aerosol, (ii) Emulsion.

(i) Aerosol: Clouds smoke (ii) Emulsion: Milk, face cream

Smoke and fog both are aerosols. In what way are they different?

Both fog and smoke have gas as the dispersion medium. The only difference is that the dispersed phase in fog is liquid and in smoke it is a solid.

Properties of Colloid

It is a heterogeneous mixture.
Size of particle is too small to be seen by naked eye.
They do not settle down when left undisturbed.
They scatter light passing through them making its path visible.
They cannot be separated by the process of filtration.

Class 9 Science Chapter 2 Important Extra Questions Set – 6

What do you understand by the term distillation give its one application..

Distillation is used to get back acetone. We know that boiling point of water is 1000C and since acetone is more volatile it will separate our first.

Define chromatography and give its one application.

Chromatography is a technique used for the separation of a mixture of solutes brought about by distribution of dissolved material between two immiscible phases, one of which is mobile phase and the other part is stationary phase. It is useful in forensic science to detect and identify trace amounts of substances in the contents of bladder or stomach.

Is water an element or a compound? Give one reason in support of your statement.

Water is a compound because of the following reasons: (i) It is compound of two different elements, hydrogen and oxygen which cannot be separated by physical methods. They can be separated only by electrolysis. (ii) The physical and chemical properties of hydrogen and oxygen are entirely different from the properties of water.

Suggest separation technique (s) one would need to employ to separate the following mixtures: (a) Mercury and water, (b) Potassium chloride and ammonium chloride, (c) Common salt water and sand, (d) Kerosene oil, water and salt

(a) Separation by using separating funnel. (b) Sublimation (c) Filtration to separate sand followed by evaporation/distillation. (d) Separation by using separating funnel to separate kerosene oil followed by evaporation or distillation.

What would you observe when? (a) a saturated solution of potassium chloride prepared at 600C is allowed to cool at room temperature? (b) an aqueous sugar solution is heated to dryness? (c) a mixture of iron filings and Sulphur powder is heated strongly?

(a) Solid potassium chloride will separate out. (b) Initially the water will evaporate ad then sugar will get charred. (c) Iron suplhide will be formed.

Name the process associated with the following

Dry ice is kept at room temperature and at one atmospheric pressure.
A potassium permanganate crystal is in a beaker and water is poured into a beaker wit stirring.
An acetone bottle is left open and the bottle becomes empty.
Milk is churned to separate cream from it.
Setting of sand when a mixture of sand and water is left undisturbed for some time.
Fine beam of light entering through a small hole in a dark room, illuminates the particles its paths.
Sublimation
Dissolution/ diffusion
Evaporation/ diffusion
Centrifugation
Sedimentation
Scattering of light.

Class 9 Science Chapter 2 Important Extra Questions Set – 7

On heating, calcium carbonate gets converted into calcium oxide and carbon dioxide. (a) is this a physical or a chemical change (b) can you prepare one acidic and one basic solution by using the products formed in above process if so, write the chemical equation involved..

(a) Chemical change. (b) Acidic and basic solution can be prepared by dissolving the products of the above process water. CaO + H₂O → Ca(OH)₂ (basic solution) CO₂ + H₂O → H₂CO₃ (acidic solution)

Classify the following into metals, non-metals and metalloids: (i) Germanium (ii) Boron (iii) Diamond (iv) Iodine (v) Copper (vi) Helium

Metal – Copper Non Metals – Diamond, iodine and helium Metalloids – Germanium, boron.

Distinguish between physical change and chemical change.

Physical change: (i) In a physical change, one physical properties such as colour, physical state, density, volume, etc. change; chemical properties remain unchanged. (ii) No new substance is formed in a physical change. (iii) Very little or no energy in the form of heat, light or sound is usually absorbed or given out in a physical change. (iv) A physical change is a temporary change. (v) The original form of a substance can be regained by simply physical methods. (vi) A physical change is reversible. Chemical change: (i) In a chemical change, the chemical composition and chemical properties undergo a changes. (ii) A new substance is formed in a chemical change. (iii) A chemical change is always accompanied by absorption or evolution of energy. (iv) A chemical change is a permanent change. (v) Original substance cannot be obtained by simply physical methods. (vi) A chemical change is irreversible.

Distinguish between metals and non-metals.

(a) Metals: (i) They have lustre (sheen). (ii) They are malleable and ductile. (iii) They have high density and high melting and boiling points. (iv) Except mercury and gallium all other metals are solid at room temperature. (v) They are sonorous. (vi) They are good conductors of heat and electricity. (vii) They are generally hard (except sodium or potassium). (viii) They have high tensile strength. (b) Non-metal: (i) They are non-lustrous. (ii)The are neither malleable nor ductile. (iii) They have low density and low melting and boiling points. (iv) Non- metals may exist in solid, liquid or gaseous states at room temperature. (v) They are not sonorous. (vi) They are poor conductors of heat and electricity. (vii) Non-metals are generally soft. (except diamond) (viii) They have low tensile strength.

Distinguish between compounds and mixtures.

(a) Compound: (i) Compounds are formed as a result of chemical reactions between two or more elements or compounds. (ii) The components of a compound are always present in a definite ratio by the mass. (iii) The properties of a compound are entirely different from its constituents. (iv) Compounds are always homogenous in nature. (v) Compound formation is accompanied by absorption or evolution of light, heat or electrical energy. (vii)Melting and boiling points of a compound are usually sharp are fixed. (viii) The constituents of a compound cannot be separated by physical by physical or mechanical means. They can however be separated by chemical methods. (b) Mixtures: (i) Mixtures are formed by simply mixing two or more constituents. There are no chemical reactions between the constituents. (ii) The properties of a mixture are same as those of its constituents. (iii) The components of a mixture may be present in any ratio. (iv) Mixture are usually heterogeneous (except in solution) (v) Heat, light or electrical energy may not be evolved or absorbed during the formation of a mixture. (vii) The components of a mixture can be easily separated by physical methods.

Class 9 Science Chapter 2 Important Extra Questions Set – 8

Define element.

Robert Boyle discovered element in 1661 an element is a basic form of matter that cannot be broken down into simpler substance by chemical reaction.

What is Tyndall effect?

The scattering of a beam of light is called Tyndall effect this happens due to the scattering of light by the particles of dust and seen when the light passes through a dense forest.

Write the principal of evaporation method?

The principle of evaporation method is that a volatile component is separated by its non-volatile component.

What is fractional distillation method or which method is used to separate gases from air?

The method which is used to separate the component of a mixture in which the components have difference less than 25K in their boiling points is called fractional distillation method. It is used to separate different gases from air.

What is a concentration of a solution?

The concentration of a solution is the amount of a solute that present in the given amount of a solution. Concentration of solution = (amount of solute )/(amount of solution)

Classify the following Physical and Chemical Changes with reason

Drying of a shirt in the sun.
Rising of hot air over a radiator.
Burning of kerosene in a lantern.
Change in the colour of black tea on adding lemon juice to it.
Churning of milk cream to get butter.
It is a physical change because water is converted from its liquid state to gaseous state because the sun’s heat.
It is a physical change because water in the radiator is converted from a liquid state to gaseous state.
It is a chemical change because combustion of kerosene occurs and new products are formed.
It is a chemical change because there is a reaction between the citric acid in the lemon and the compounds of tea resulting in formation of new product.
It is a physical change because the cream suspended in the milk is separated by churning.

One Word Questions

Non- metals are usually poor conductors of heat and electricity. They are non-lustrous, non-sonorous, non-malleable and are coloured.

Name a lustrous non-metal.
Name a non-metal which exists as a liquid at room temperature.
The allotropic form of a non-metal is a good conductor of electricity. Name the allotrop.
Name a non-metal which is known a to form the largest number of compound.
Name a non-metal other than carbon which shows allotropy.
Name a nonmetal which is required for combustion.

Answers of One Word Questions

You are provided with a mixture containing sand, iron filings, ammonium chloride and sodium chloride. Describe the procedure you would use o separate these constituent from the mixture.

(i) Removing iron filings from the mixture by magnetic separation. Take the mixture in a Petri dish and roll a bar magnet over it. Iron filings will get attach to the magnet and thus separate from the mixture. (ii) Removing ammonium chloride by sublimation Transfer the remaining mixture into china dish and heat it. On heating, ammonium chloride sublime and solidifies on condensation. The mixture containing sand and sodium chloride left behind in the china dish. (iii) Removing sand by filtration Make a solution of sand and sodium chloride in water. Filter the solution. Sodium chloride will dissolve in water and sand is left as residue on the filter paper. (iv) Evaporate the filtrate to dryness to get sodium chloride or by crystallisation.

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NCERT Exemplar for Class 9 Science - Is Matter Around Us Pure - Free PDF Download
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NCERT Exemplar Solutions for Class 9 Science (Chemistry) Chapter 2 - Is Matter around us Pure

Multiple choice questions.

1. Which of the following statements are true for pure substances?

i. Pure substances contain only one kind of particle.

ii. Pure substances may be compounds or mixtures.

iii. Pure substances have the same composition throughout.

iv. Pure substances can be exemplified by all elements other than nickel.

(i) and (ii)

(i) and (iii)

(iii) and (iv)

(ii) and (iii)

Ans: (b) (i) and (iii)

The pure substances include only one kind of particle and have the same composition throughout. Hence the correct answer is option (b).

2. Rusting of an article made up of iron is called:-

a. corrosion and it is a physical as well as chemical change.

b. dissolution and it is a physical change.

c. corrosion and it is a chemical change.

d. dissolution and it is a chemical change.

Ans : (c) corrosion and it is a chemical change.

The rusting of iron is a chemical change. It results in the corrosion of iron.

Hence the correct answer is option (c).

3. A mixture of sulphur and carbon disulphide is:-

heterogeneous and shows Tyndall effect.

homogeneous and shows Tyndall effect.

heterogeneous and does not show Tyndall effect.

homogeneous and does not show Tyndall effect.

Ans : (d) homogeneous and does not show Tyndall effect.

The given mixture of the sulphur and the carbon disulphide is a homogeneous mixture. The carbon disulphide is used as a solvent for the sulphur. The Tyndall effect is caused by the scattering of light by insoluble particles present in a solution. The colloids and suspensions, which are heterogeneous mixtures, contain insoluble particles, and thus show Tyndall effect. Hence the correct answer is option (d).

4. Tincture of iodine has antiseptic properties. This solution is made by dissolving:-

iodine in potassium iodide.

iodine in vaseline.

iodine in water.

iodine in alcohol.

Ans : (d) iodine in alcohol

A solution of "Tincture of Iodine" is made by dissolving 2 - 7% iodine into the alcohol.The tincture solutions are characterised by the presence of alcohol because the alcohol is a good solvent. The iodine does not dissolve in water. The tincture of Iodine is also used as an antiseptic. Hence the correct answer is option (d).

5. Which of the following are homogeneous in nature?

(i) ice (ii) wood (iii) soil (iv) air

(ii) and (iv)

(i) and (iv)

Ans : (c) (i) and (iv)

The ice and air are homogeneous mixtures. The soil is a heterogeneous mixture and the wood also has a heterogeneous composition. Hence the correct answer is option (c).

6. Which of the following are physical changes?

i. Melting of iron metal

ii. Rusting of iron

iii. Bending of an iron rod

iv. Drawing a wire of iron metal

(i), (ii) and (iii)

(i), (ii) and (iv)

(i), (iii) and (iv)

(ii), (iii) and (iv)

Ans : (c) (i), (iii) and (iv)

The rusting of iron is a chemical change. A new substance like iron oxide is formed in case rusting of iron takes place. In the other given options, no new substance is formed, and they are physical changes. Hence the correct answer is option (c).

7. Which of the following are chemical changes?

i. Decaying of wood

ii. Burning of wood

iii. Sawing of wood

iv. Hammering of a nail into a piece of wood

(ii) and (iii)

(i) and (iv)

Ans: (a) (i) and (ii)

The burning and decay result in the formation of new substances and they cannot be reversed. Therefore, they are under chemical changes. The sawing of wood and hammering of a nail into a piece of wood are under physical changes. Hence the correct answer is option (a).

8. Two substances, A and B, were made to react to form a third substance A2B according to the following reaction: 2A + B → A2B. Which of the following statements concerning this reaction are incorrect?

i. The product A2B shows the properties of substances A and B.

ii. The product will always have a fixed composition.

iii. The product so formed cannot be classified as a compound.

iv. The product so formed is an element.

(i), (ii) and (iii),

(ii), (iii) and (iv)

Ans: (c) (i), (iii) and (iv)

The properties of the compound A2B that is the product formed as a result of the reaction are different from the properties of its constituent elements - "A" and "B". The A2B will always have "A" and "B" in the fixed composition by the mass. Hence the correct answer is option (c).

9. Two chemical species X and Y combine together to form a product P which contains both X and Y. X + Y → P

X and Y cannot be broken down into simpler substances by simple chemical reactions.

Which of the following concerning the species X, Y and P are correct?

i. P is a compound.

ii. X and Y are compounds.

iii. X and Y are elements.

iv. P has a fixed composition.

(i), (iii) and (iv)

Ans : (d) (i), (iii) and (iv)

Since "X" and "Y" cannot be broken down further by the simple chemical reactions are not compounds, they are elements. Therefore (iii) is correct and (ii) is incorrect. The product "P" is a compound of elements "X" and "Y". Therefore (i) is correct. The "P" compound will always have a fixed composition of the constituent elements "X" and "Y". Therefore (iv) is correct. Hence the correct answer is option (d).

Short Answer Questions

10.Suggest separation technique(s) one would need to employ to separate the following mixtures:

a. Mercury and water

Ans : By separating the funnel. The mercury and water are immiscible liquids. The mercury is more dense than the water and can be separated by using a separating funnel.

b. Potassium chloride and ammonium chloride

Ans : By sublimation method. The ammonium chloride is a sublimating compound and can be separated from the mixture through the sublimation method

c. Common salt, water and sand

Ans : By filtration of sand followed by the evaporation of water OR by centrifugation to separate sand followed by the evaporation/distillation of the water.

d. Kerosene oil, water and salt

Ans : By separating funnel to separate kerosene oil followed by the evaporation / distillation of the water.

11. Which of the tubes in Fig. 2.1 (a) and (b) will be more effective as a condenser in the distillation apparatus?

Ans : The tube (a) will be more effective than tube (b) as a condenser in the distillation apparatus. The presence of marbles increases the surface area that comes in contact with the vapours. This allows more time for the condensation of vapours and therefore the first column would be more effective than the second column without the marbles.

12. Salt can be recovered from its solution by evaporation. Suggest some other technique for the same?

Ans : The salt can also be recovered from its solution by the crystallization technique.

13. The ‘sea-water’ can be classified as a homogeneous as well as heterogeneous mixture. Comment.

Ans : The sample of sea-water collected from the surface of sea: The sea water can be considered as a homogeneous mixture when it has salts and water only.

The sample of the sea-water collected from the deep sea: The sea water will be a heterogeneous mixture when the sample is collected from the deeper layers and it contains salts, water, mud, decayed plants, etc.

14. While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56 °C). What technique can be employed to get back the acetone? Justify your choice.

Ans : The acetone can be separated from the mixture of the salt solution and water by the distillation method. There is a considerable difference in the boiling point of the acetone which is 56 °C and the boiling point of the water which is 100 °C. Therefore when the solution is heated, the acetone will evaporate before the water does and it can be collected after the condensation.

15. What would you observe when:

a. a saturated solution of potassium chloride prepared at 60 °C is allowed to cool to room temperature.

Ans : The solid potassium chloride will separate out from the saturated solution when the temperature of the solution reduces from the 60°C to the room temperature. A change in the temperature affects the solubility of the solid solute.

b. an aqueous sugar solution is heated to dryness.

Ans : When the aqueous sugar solution is heated, the water will initially evaporate. When the solution is heated to the dryness, the sugar will get charred.

c. a mixture of iron filings and sulphur powder is heated strongly.

Ans : The iron will combine chemically with the sulphur when heated strongly and iron sulphide (FeS) will be formed.

16. Explain why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do.

Ans : In a suspension, the size of the particles is relatively larger than that of a colloidal solution. Also, the molecular interaction in a suspension is not strong enough to keep the particles suspended. Therefore the particles settle down when the suspension is left undisturbed for some time. On the other hand, the molecular interaction in a colloidal solution like milk is strong enough and does not allow the particles to settle down.

17. Smoke and fog both are aerosols. In what way are they different?

Ans : Both the - fog and smoke - have gas as the dispersion medium that is the continuous phase. The difference lies in the dispersed phase that is the suspended phase. The dispersed phase in the fog is liquid like water droplets whereas the dispersed phase in the smoke is solid like the particulate matter.

18. Classify the following as physical or chemical properties:

a. The composition of a sample of steel is: 98% iron, 1.5% carbon and 0.5% other elements.

Ans : The physical property of iron: The steel is an alloy of iron with about one percent carbon.

b. Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.

Ans : The chemical property of zinc: The zinc is a reactive metal. It displaces hydrogen from the hydrochloric acid and the formation of zinc chloride takes place.

c. Metallic sodium is soft enough to be cut with a knife.

Ans : The physical property of sodium: The sodium is a soft metal.

d. Most metal oxides form alkalis on interacting with water.

Ans : The chemical property of metallic oxides: The metallic oxides react with water and form the alkalies.

19. The teacher instructed three students ‘A’, ‘B’ and ‘C’ respectively to prepare a 50% (mass by volume) solution of sodium hydroxide (NaOH).

‘A’ dissolved 50 g of NaOH in 100 ml of water, ‘B’ dissolved 50 g of NaOH in 100 g of water while ‘C’ dissolved 50 g of NaOH in water to make 100 ml of solution. Which one of them has made the desired solution and why?

Ans : Student ' C ' has made the desired solution.

Mass by volume % = (Mass of solute / Volume of solution) X Then, 100 =(50g / 100ml)×100

mass by volume

20. Name the process associated with the following:

a. Dry ice is kept at room temperature and at one atmospheric pressure.

Ans: By the sublimation process.(The dry ice or the solid carbon dioxide sublimes at lower temperatures into the large volumes of CO 2 gas and is dangerous to handle. It can also cause burns by freezing).

b. A drop of ink placed on the surface of water contained in a glass spreads throughout the water.

Ans : By the diffusion of ink in water.

c. A potassium permanganate crystal is in a beaker and water is poured into the beaker with stirring.

Ans : By the dissolution/diffusion of the potassium permanganate.

d. A acetone bottle is left open and the bottle becomes empty.

Ans : By the evaporation and diffusion of the acetone.

e. Milk is churned to separate cream from it.

Ans : By the centrifugation of the milk.

f. Settling of sand when a mixture of sand and water is left undisturbed for some time.

Ans : By the sedimentation of sand.

g. Fine beam of light entering through a small hole in a dark room, illuminates the particles in its paths.

Ans : By the scattering of light (Tyndall effect).

21. You are given two samples of water labelled as ‘A’ and ‘B’. Sample ‘A’ boils at 100°C and sample ‘B’ boils at 102 °C. Which sample of water will not freeze at 0°C? Comment.

Ans : The sample ‘B’ will not freeze at 0°C because it may contain some impurities. At 1 atm, the boiling point of the pure water is 100°C and the freezing point of the pure water is 0°C.

22. What are the favourable qualities given to gold when it is alloyed with copper or silver for the purpose of making ornaments?

Ans : The pure gold that is 24-carat gold is very soft. It is alloyed with the silver or copper to impart strength while making the ornaments. An alloy that contains 20 parts of the gold and 4 parts of the silver is called a 20-carat gold.

23. An element is sonorous and highly ductile. Under which category would you classify this element? What other characteristics do you expect the element to possess?

Ans : The metals are sonorous and highly ductile; therefore this element can be classified as a metal. The other characteristics expected to be possessed by this element are – lustre, malleability, heat and the electrical conductivity.

24. Give an example each for the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures:

a. A volatile and a non-volatile component.

Ans : The ammonium chloride and the sodium chloride. The mixture can be separated by the sublimation process of the ammonium chloride.

b. Two volatile components with appreciable difference in boiling points.

Ans : The acetone and water. The mixture can be separated by the evaporation process or by the distillation process. The boiling point of the acetone is 56 °C. The boiling point of water is 100 °C. The difference in the boiling points is 44 °C.

c. Two immiscible liquids.

Ans : The kerosene and water. The mixture can be separated with the help of a separating funnel.

d. One of the components changes directly from solid to gaseous state.

Ans: The potassium chloride and the ammonium chloride. The mixture can be separated by the sublimation process of the ammonium chloride.

e. Two or more coloured constituents are soluble in some solvent.

Ans: The plant pigments. The mixture of the coloured constituents can be separated by the chromatography method.

25. Fill in the blanks:

a. A colloid is a mixture and its components can be separated by the technique known as .

Ans : heterogeneous, centrifugation.

b. Ice, water and water vapour look different and display different properties but they are the same.

Ans : physical, chemically.

c. A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for some time. The upper layer in the separating funnel will be of and the lower layer will be that of .

Ans : Water, chloroform. (The density of water is less than that of chloroform.)

d. A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25 K can be separated by the process called .

Ans : fractional distillation.

e. When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the of light by milk and the phenomenon is called . This indicates that milk is a solution.

Ans : scattering, Tyndall effect, colloidal

26. Sucrose (sugar) crystals obtained from sugarcane and beetroot are mixed together.

Will it be a pure substance or a mixture? Give reasons for the same.

Ans : It will be a pure substance compound since the chemical composition of sugar crystals will be the same whether obtained from the sugarcane or from beetroot.

27. Give some examples of Tyndall effects observed in your surroundings?

Ans : Following are some of the examples of the Tyndall effect:-

The sunlight entering a room through a ventilation near the ceiling.

The beam of light coming inside a forest through a canopy of trees.

28. Can we separate alcohol dissolved in water by using a separating funnel? If yes, then describe the procedure. If not, explain.

Ans : The water that is a universal solvent and the alcohol which is an organic solvent, are miscible liquids. Therefore, a separating funnel cannot be used to separate them from their mixture.

29. On heating calcium carbonate gets converted into calcium oxide and carbon dioxide.

a. Is this a physical or a chemical change?

Ans : The chemical change that is the decomposition of the calcium carbonate.

CaCO 3 → CaO + CO 2

b. Can you prepare one acidic and one basic solution by using the products formed in the above process? If so, write the chemical equation involved.

Ans : The acidic and the basic solutions can be prepared by dissolving the products of the above process in water.

The calcium oxide will form a basic solution.

CaO + H 2 O → Ca(OH) 2

(Basic solution)

The carbon dioxide will form an acidic solution.

CO 2 + H 2 O → H 2 CO 3

(Acidic solution)

30. Non-metals are usually poor conductors of heat and electricity. They are non lustrous, non-sonorous, non-malleable and are coloured.

a. Name a lustrous non-metal.

Ans : Iodine

b. Name a non-metal which exists as a liquid at room temperature.

Ans : Bromine

c. The allotropic form of a non-metal is a good conductor of electricity. Name the allotrope.

Ans : Graphite

d. Name a non-metal which is known to form the largest number of compounds.

Ans : Carbon

e. Name a non-metal other than carbon which shows allotropy.

Ans : Sulphur, phosphorus

f. Name a non-metal which is required for combustion.

Ans : Oxygen

31. Classify the substances given in Fig. 2.2 into elements and compounds:

Ans : Elements:

H 2 O, CaCO 3 .

Cu, Zn,O 2 , F 2 , Hg , Diamond Compounds: NaCl(aq) , Wood, Sand,

32. Which of the following are not compounds?

(a) Chlorine gas (b) Potassium chloride (c) Iron (d) Iron sulphide

(e) Aluminium (f) Iodine (g) Carbon (h) Carbon monoxide

(i) Sulphur powder

Ans: The following are not compounds: Chlorine gas, Iron, Aluminium, Iodine, Carbon, Sulphur powder.

Long Answer Questions

33. Fractional distillation is suitable for separation of miscible liquids with a boiling point difference of about 25 K or less. What part of fractional distillation apparatus makes it efficient and possess an advantage over a simple distillation process? Explain using a diagram.

Ans : Fractional distillation of miscible liquids

A fractionating column makes the process of the fractional distillation more efficient than the process of the simple distillation by providing increased possibilities for the condensation of the liquid. A fractionating column packed with glass beads or glass helices provides a large surface area for the vapours to collide and lose energy so that they can be condensed and can be distilled quickly. Increasing the length of the fractionating column would also increase the efficiency of the fractional distillation.

34. Answer the following:-

a. Under which category of mixtures will you classify alloys and why?

Ans : The alloys are homogeneous mixtures because they have a uniform composition throughout.

b. A solution is always a liquid. Comment.

Ans : No, a solution is not always a liquid. The solid solutions and the gaseous solutions are also possible. For example the brass is a solid solution and the air is a gaseous solution.

c. Can a solution be heterogeneous?

Ans : No, a solution cannot be heterogeneous. A solution is a homogenous mixture of two or more than two substances.

35. Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘B’. Part ‘A’ was heated strongly while Part ‘B’ was not heated. Dilute hydrochloric acid was added to both the Parts and evolution of gas was seen in both the cases. How will you identify the gases that evolved?

Ans : The following reaction will take place when part A (Mixture of Iron filings and sulphur) is heated:

Fe ( s ) + S ( s ) → Fe ( s )

When the dilute hydrochloric acid is added to the mixture, the following reaction takes place and the hydrogen sulphide gas is evolved.

FeS ( s ) + 2HCl (aq ) → FeCl 2 (aq) + H 2 S (g)

The hydrogen sulphide is a foul smelling gas and smells like rotten eggs.

When the dil. hydrochloric acid is added to the mixture of iron and sulphur (Part B), the following reaction takes place and the hydrogen gas is evolved. In this case, the sulphur does not participate in the reaction.

Fe (s) + S (s) + 2HCl (aq) → FeCl 2 (aq) + H 2 (g) + S (s)

When a burning matchstick is brought near the evolved gas, the matchstick burns with a pop sound. This confirms the evolution of the hydrogen gas.

36. A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in Fig. 2.3. The filter paper was removed when the water moved near the top of the filter paper.

i. What would you expect to see, if the ink contains three different coloured components?

Ans : If the ink contains three different coloured components, three different bands will be seen on the filter paper.

ii. Name the technique used by the child.

Ans : The child used the technique of Paper Chromatography.

iii. Suggest one more application of this technique.

Ans : The paper Chromatography can be used to separate the pigments present in the chlorophyll.

37. A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/tumbler in the box as shown in Fig.2.4. They were amazed to see that milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it?

a. Explain why the milk sample was illuminated. Name the phenomenon involved.

Ans : The milk is a colloid. The particulate matter present inside milk scatters the light passing through the milk and shows Tyndall effect.

b. Same results were not observed with a salt solution. Explain.

Ans : The salt solution is a homogeneous solution. The small particles present in a salt solution do not scatter light and therefore a salt solution does not exhibit the Tyndall effect.

c. Can you suggest two more solutions which would show the same effect as shown by the milk solution?

Ans : The detergent solution and the sulphur solution will also show the Tyndall effect.

38. Classify each of the following, as a physical or a chemical change. Give reasons.

a. Drying a shirt in the sun.

Ans : The drying of a shirt in the sun: The physical change. There is no chemical reaction that happens during the drying of a shirt under the sun.

b. Rising of hot air over a radiator.

Ans : The rising of hot air over a radiator: The physical change. The water in a radiator converts to vapours. The hot air becomes lighter and rises.

c. Burning of kerosene in a lantern.

Ans : The burning of kerosene in a lantern: The chemical change. The kerosene burns and combines with the atmospheric oxygen to form new chemical products.

d. Change the colour of black tea by adding lemon juice to it.

Ans : Change in the colour of black tea by adding lemon juice to it: The chemical change.

The lemon is a source of citric acid, ascorbic acid and malic acid. The black tea contains theaflavin antioxidants.

e. Churning of milk cream to get butter.

Ans : The churning of milk cream to get butter: The physical change. There is no chemical reaction that happens during the centrifugation of milk cream.

39. During an experiment the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution.

a. Are the two solutions of the same concentration?

Ans : No, the two solutions will not have the same concentration.

Mass % = \[\mathrm{\frac{mass\: of\: solute}{\textrm{[mass of solute + mass of solvent]}}\times 100}\]

b. Compare the mass % of the two solutions.

Ans : Mass % of solution made by Rames

= \[\mathrm{\frac{10}{110}\times 100}\]

Mass % of solution made by Sarika

= \[\mathrm{\frac{mass\: of\: solute}{\textrm{[mass of solute + mass of solvent]}}\times 100}\]

\[\mathrm{\frac{10}{110}\times 100}\]

Therefore the solution prepared by Sarika has a higher mass % than that prepared by Ramesh.

40. You are provided with a mixture containing sand, iron filings, ammonium chloride and sodium chloride. Describe the procedure you would use to separate these constituents from the mixture?

Ans : The components of the given mixture can be separated by the following methods:-

i . By using a magnet : By moving a magn.et over the mixture will result in the iron fillings getting stuck to the magnet. Hence the iron will be separated from the mixture.

ii . By the sublimation : The remaining mixture is heated in a china dish. The ammonium chloride is a sublimating substance and therefore it will evaporate without passing through the liquid phase. The crust of ammonium chloride can be collected by placing an inverted funnel on top of the china dish.

iii. By the sedimentation, decantation and filtration : The remaining mixture is dissolved in the water and allowed to settle for some time. The sand, being insoluble in water, settles at the bottom. The liquid is decanted in the other beaker. The liquid is then filtered to remove any traces of sand.

iv. By evaporation : The liquid is now a solution of salt in water. This is heated in a beaker so that the water evaporates. Once all the water evaporates, salt remains in the beaker.

41. Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represent the composition of the solution?

1.00 g of NaCl + 100 g of water

0.11 g of NaCl + 100 g of water

0.01 g of NaCl + 99.99 g of water

0.10 g of NaCl + 99.90 g of water

Ans : (c) 0.01g of NaCl + 99.99 g of water is the correct composition of the solution.

Mass % in (C) = \[\mathrm{\frac{mass\: of\: solute}{\textrm{[mass of solute + mass of solvent]}}\times 100}\]%

= \[\mathrm{\left ( \frac{.01}{.01+99.99} \right )\times 100}\]

42. Calculate the mass of sodium sulphate required to prepare its 20% (mass percent) solution in 100 g of water?

Ans : Let the mass of the sodium sulphate required to prepare the solution be "x" grams.

The mass of the solvent (water) is given as 100 g .

The mass of the solution would be (x +100)g

x gof solute (sodium sulphate) is dissolved in (x+100) g of solution

20% = x / [x +100]×100 Or, 20x + 2000 =100x Or, 80x = 2000

Or, x = 2000 / 80

Or, x = 25g

Hence 25g of sodium sulphate will be required to prepare its 20% solution in 100 g

Introduction to Chapter 2

Chapter 2 of class 9 science begins with an introduction of what matter is. Pure and impure substances have then been stressed upon in detail. Then, students learn about what metals and nonmetals are. A comparative table of their properties has then been explained. Some elements have properties of both metals as well as non-metals and are known as metalloids. That too has been explained. We then learn about the types of the mixture and the distinction between a compound and a mixture. Illustrations have been used for a clearer understanding. Solutions, suspensions, and colloids have then been explained. The chapter ends after the Separation of Mixtures has been dealt with.

FAQs on NCERT Exemplar for Class 9 Science - Is Matter Around Us Pure - Free PDF Download

1. What are the kinds of questions that have been asked in NCERT Exemplar for Class 9 science- Is matter around us pure?

There are multiple-choice questions, short answer questions, long answer questions in NCERT Exemplar for Class 9 science- Is matter around us pure.

These questions have been designed by the various scientific experts and science teachers as per the NCERT guidelines. They help students prepare for all sorts of examinations including competitive examinations. The book is available on Vedantu and is completely free of cost. All the study material on Vedantu is free of cost so that the students do not hesitate before getting their hands on quality academic material.

2. How are metals and non-metals different from each other?

Metals and nonmetals are both different from each other as they have different properties. Some examples of metals are iron, copper, and aluminium and some non-metals would be wood, rubber, glass.. The distinction between them and their full-fledged explanations are provided on Vedantu.com. The site has the complete study material that’s needed by students of Class 9 for science. The explanations are well and to the point. This facilitates smoother learning in students who wish to score higher grades.

3. How can one understand the concepts of NCERT science in Class 9 apart from the main textbooks?

One can check out NCERT Exemplar for Class 9 science on Vedantu.com as the revision notes and the solved questions are very effective when it comes to appearing for examinations. Students always need a revision book of some sort to go through right before their tests for a comprehensive understanding of the chapters that are there. Unless a particular concept is clear, it will become very tough for students to answer challenging questions that are posed to them. The books present on Vedantu have just the right amount of information that’s needed and have been made as per the NCERT guidelines.

4. Is common salt an organic compound?

Organic compounds are derived from living organism parts whereas inorganic compounds are derived from nonliving materials such as rocks and minerals. Common salt is an inorganic compound as it comes from sodium chloride mineral halites. The explanations for these have been summed up in NCERT Exemplar for Class 9 Science Chapter 2 - Is Matter Around Us Pure. The concepts have been simplified down to the students’ level of comprehension and are pretty interesting to go through. You can find the book online on Vedantu.

5. How do I make revision notes for NCERT Class 9 Science Chapter 2?

You can make revision notes by logging into Vedantu.com where you will find NCERT Exemplar for Class 9 Science Chapter 2 - Is Matter Around Us Pure. The chapter has revision notes as well as solved questions in all formats for you to learn. The language used is simple and scientific and the solved questions will prepare you for all those tests that you might have to take. You must go through each and every solved question to understand the chapter completely and make relevant notes for revision.

NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around us Pure?

Chapter 2 Is Matter Around us Pure NCERT Solutions for Class 9 Science

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Case Study Questions Class 9 Science Matter in our Surroundings

Case study questions class 9 science chapter 1 matter in our surroundings.

CBSE Class 9 Case Study Questions Science Matter in our Surroundings. Important Case Study Questions for Class 9 Exam. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Matter in our Surroundings.

At Case Study Questions there will given a Paragraph. In where some Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks or 4 marks.

CBSE Case Study Questions Class 9 Science – Matter in our Surroundings

Case study 1:.

1.) A matter is anything that has mass and occupies space. Pen, paper, clips, sand, air, ice, etc. are different forms of matter. Every matter is made up of small particles. These particles are so tiny that they can’t be seen with naked eyes. Let’s see about the different characteristics of particles of matter.

All matter is made up of very small particles.
.Particles of matter has spaces between them.
Particles of matter are continuously moving.
Particles of matter attract each other.

Answer the following questions by referring above paragraph.

i.) Which of following is not matter?

c.) smell of perfume

d.) None of these

ii.) Thoughts coming in our mind are example of matter. True or false

c.) None of these

iii.) Which of the following is true about particles of matter?

a.) Particles of matter has spaces between them

b.) Particles of matter are continuously moving

c.) Particles of matter attract each other

d.) All of these

iv.) Give 5 examples of matter in our surroundings

v.) Enlist all properties of particles of matter

Answer key-1

iv.) pen, pencil, notebook, ice and water

v.) Different characteristics of particles of matter are

Case Study 2:

2.) There are three states of matter – solid, liquid and gas.

Solids have a definite shape, distinct boundaries and fixed volumes, that is, have negligible compressibility. Solids have a tendency to maintain their shape when subjected to outside force. Solids may break under force but it is difficult to change their shape, so they are rigid.

Liquids have no fixed shape but have a fixed volume. They take up the shape of the container in which they are kept. Liquids flow and change shape, so they are not rigid but can be called fluid.

Gas as has indefinite shape, no fixed volume. Gas gets the shape and volume of container.

Gas has very low density hence are light. Gas can flow easily and hence are called fluid.

i.) Which of the following state of matter takes shape of container in which it is filled?

d.) Both b and c

ii.) Distance between particles of matter least in

iii.) Compressibility is least in case of

iv.) Give properties of solids.

v.) Give properties of Gases.

Answer key-2

iv.) properties of solid are given below

Solid has fixed volume.
Solid has fixed shape.
Solid has high density.
Solids are heavy.
Solid does not flow.

v.) Properties of gases are

Gas has indefinite shape
Gas has no fixed volume.
Gas gets the shape and volume of container.
Gas fills the container completely.
Gas has very low density.
Because of low density gas are light.
Gas can flow easily and hence are called fluid.

Case Study 3:

3.) What happens inside the matter during change of state? On increasing the temperature of solids, the kinetic energy of the particles increases. Due to the increase in kinetic energy, the

Particles start vibrating with greater speed. The energy supplied by heat overcomes the forces of attraction between the particles. The particles leave their fixed positions and start moving more freely. A stage is reached when the solid melts and is converted to a liquid. The minimum temperature at which a solid melts to become a liquid at the atmospheric pressure is called its melting point.

The temperature of the system does not change after the melting point is reached, till all the ice melts. This happens even though we continue to heat the beaker, that is, we continue to supply heat. This heat gets used up in changing the state by overcoming the forces of attraction between the particles. The amount of heat energy that is required to change 1 kg of a solid into liquid at atmospheric pressure at its melting point is known as the latent heat of fusion. So, particles in water at 0 0 C (273 K) have more energy as compared to particles in ice at the same temperature.

The temperature at which a liquid starts boiling at the atmospheric pressure is known as its boiling point. Boiling is a bulk phenomenon. Particles from the bulk of the liquid gain enough energy to change into the vapour state. A change of state directly from solid to gas without changing into liquid state is called sublimation and the direct change of gas to solid without changing into liquid is called deposition.

i.) A change of state directly from solid to gas without changing into liquid state is called

a.) Sublimation

b.) Deposition

c.) Boiling point

ii.) The direct change of gas to solid without changing into liquid is called

iii.) The energy supplied by heat to solid is used to overcome the forces of attraction between the particles. True or false

iv.) Define melting point and boiling point

v.) Define latent heat of fusion

Answer key-3

iv.) The minimum temperature at which a solid melts to become a liquid at the atmospheric pressure is called its melting point.

The temperature at which a liquid starts boiling at the atmospheric pressure is known as its boiling point.

v.) The amount of heat energy that is required to change 1 kg of a solid into liquid at atmospheric pressure at its melting point is known as the latent heat of fusion.

Case Study 4:

4 .) Do we always need to heat or change pressure for changing the state of matter? Can you quote some examples from everyday life where change of state from liquid to vapour takes place without the liquid reaching the boiling point? In the case of liquids, a small fraction of particles at the surface, having higher kinetic energy, is able to break away from the forces of attraction of other particles and gets converted into vapour. This phenomenon of change of a liquid into vapors at any temperature below its boiling point is called evaporation.

i.) Evaporation of liquid takes place at

a.) Boiling point

b.) Above boiling point

c.) Below boiling point

ii.) Evaporation takes place at surface of liquid because

a.) They are heavy as compare to other particles

b.) They have sufficient kinetic energy to break the force

c.) They are light weight as compare to other particles

iii.) During evaporation particles of liquid change into vapour

a.) From the surface

b.) From the bottom

c.) From all over the liquid

iv.) Define evaporation.

v.) Explain process of evaporation

Answer key-4

iv.) The phenomenon of change of a liquid into vapors at any temperature below its boiling point is called evaporation.

v.) In the case of liquids, a small fraction of particles at the surface, having higher kinetic energy, is able to break away from the forces of attraction of other particles and gets converted into vapour. This phenomenon of change of a liquid into vapors at any temperature below its boiling point is called evaporation.

Case Study 5:

5.) You must have observed that the rate of evaporation increases with–

an increase of surface area:
We know that evaporation is a surface phenomenon. If the surface area is increased, the rate of evaporation increases. For example, while putting clothes for drying up we spread them out.
an increase of temperature:

With the increase of temperature, more number of particles get enough kinetic energy to go into the vapour state.

In an open vessel, the liquid keeps on evaporating. The particles of liquid absorb energy from the surrounding to regain the energy lost during evaporation. This absorption of energy from the surroundings makes the surroundings cold. What happens when you pour some acetone (nail polish remover) on your palm? The particles gain energy from your palm or surroundings and evaporate causing the palm to feel cool. After a hot sunny day, people sprinkle water on the roof or open ground because the large latent heat of vaporization of water helps to cool the hot surface.

i.) Evaporation is surface phenomenon. True or false

ii.) As temperature increases the rate of evaporation is

a.) increases

b.) decreases

c.) remains constant

iii.) The rate of evaporation increases with

a.) Increase in wind speed

b.) Decrease in wind speed

c.) Does not have any effect from wind speed

iv.) What happens when you pour some acetone (nail polish remover) on your palm?

v.) We are able to sip hot tea from saucer than from cup. Why?

Answer key-5

iv.) The particles gain energy from your palm or surroundings and evaporate causing the palm to feel cool.

v.) We are able to sip hot tea from saucer than from cup. This is because saucer has large surface area, due to large surface area as compare to cut area tea evaporates at faster rate.

Thank you It helped me a lot

Why smell of Perfume is not a matter?

Because there is no particle

Because their are perfume particles suspended in air

These all case study questions are really helpful . Thanks

This is my first I was so nervous but these questions help me alot thank you

Smell of perfume is a matter because it have gas particles means perfume particles

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Class 9 Chemistry Mcqs
Class 9 Chemistry Mcqs Chapter 2 is Matter Around us Pure

Class 9 Chemistry Chapter 2 Is Matter Around Us Pure? MCQs

Class 9 chemistry MCQs with answers are provided here for chapter 2 Is Matter Around Us Pure?. These MCQs are based on the CBSE board curriculum and correspond to the most recent Class 9 chemistry syllabus. By practising these Class 9 Multiple choice questions, students will be able to quickly review all of the ideas covered in the chapter and prepare for the Class 9 Annual examinations as well as other entrance exams such as CTET and KVS.

Download Chapter 2 Is Matter Around Us Pure? MCQs PDF by clicking on the button below. Download PDF

Class 9 Is Matter Around Us Pure? MCQs

1.What is true about homogeneous mixture?

(a) Homogeneous mixture is the mixture of two or more than two components.

(b) In homogeneous mixture the composition and properties are uniform throughout the mixture

(d) none of the above

Solution: In the homogeneous mixture two or more than two components have same composition and properties throughout the mixture

2. Which of the following properties does not describe a compound?

(a) It is composed of two or more elements

(b) It is a pure substance.

(d) It is mixed in any proportion by mass

Solution: A compound is composed of two or more elements with a fixed proportion by mass. It is pure substance and cannot be separated into constituents by physical means.

3. In the tincture of iodine, find the solute and solvent?

(a) alcohol is the solute and iodine is the solvent

(b) iodine is the solute and alcohol is the solvent

(d) tincture of iodine is not a solution

Solution: A solution of iodine in alcohol known as ‘tincture of iodine’ here iodine is the solute and alcohol is the solvent

4. Which of the following is not a homogeneous mixture?

(a) Air (b) Tincture of iodine

Solution: Milk is a heterogeneous mixture. It is a colloidal solution of water and fat.

5. What is the statement?

“10 percent glucose in water by mass” signifies.

(a) 10 gram of glucose dissolved in 100 gram of water.

(b) 10 gram of glucose dissolved in 90 gram of water.

(d) 20 gram of glucose dissolved in 90 gram of water.

Solution: “10 percent glucose in water by mass” signifies that 10 gram of glucose dissolved in 90 gram of water.

6. Sol and gel are examples of ——————

(a) Solid-solid colloids

(b) Sol is a solid-liquid colloid and gel is liquid-solid colloid

(d) Sol is a liquid-solid colloid and gel is a solid-liquid colloid

Solution: Sol is a solid-liquid colloid and gel is liquid-solid colloid.

7. Solid solution in which the solute is gas ———–

(a) Copper dissolved in gold

(b) Camphor in nitrogen gas

(d) All of the above

Solution: Solution of hydrogen in palladium, here solute is hydrogen (gas) and the solvent is palladium.

8. An example of liquid metal and liquid non-metal is

(a) Gallium, mercury

(b) Mercury, chlorine

(d) Bromine, sulphur

Solution: An example of liquid metal and liquid non-metal is Mercury and Bromine respectively.

9. Which method is used to separate cream from milk?

(a) Centrifugation (b) Adsorption

Solution: By the process of centrifugation cream is separated from the milk.

10. Which of the statements is incorrect about the physical change?

(a) There is no gain or loss of energy.

(b) It is permanent and Irreversible

(d) No new substance is formed.

Solution: It is permanent and Irreversible is the incorrect statement about physical change.

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Class 9 Science Case Study Questions Chapter 2 Is Matter ...
Case Study/Passage-Based Questions. Case Study 1: Akshita wants to separate the mixture of dyes constituting a sample of ink. She marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in the figure. The filter paper was removed when the water moved near the top of the filter paper.
Case Study and Passage Based Questions for Class 9 Science Chapter 2 Is
Case Study Questions: Question 1: Akshita wants to separate the mixture of dyes constituting a sample of ink. She marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in figure. The filter paper was removed when the water moved near the top … Continue reading Case Study and Passage Based Questions for Class 9 Science Chapter 2 Is ...
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By practising these Class 9 important questions, students will be able to quickly review all of the ideas covered in the chapter and prepare for the Class 9 Annual examinations. Download Class 9 Chemistry Chapter 2 Is Matter Around Us Pure Important Questions with Answers PDF by clicking on the button below. Download PDF. Recommended Videos
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Question 2: A homogeneous mixture of two or more substances is called a true solution. it consists of solute and solvent. The particle size of the true solution is less than 1 nanometer. A suspension is a heterogeneous mixture in which the solute particle does not dissolve but remains suspended throughout the bulk of the medium.
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Important Extra Questions for class 9 science chapter 2 Is Matter Around Us Pure is given below. Proper answers of each question is also given just below the questions. We have tried to cover entire chapter 2 of class 9 science NCERT at our level best. If someone want to add some more questions in this section, please send us the questions with ...
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Lakhmir Singh Solutions for Class 9 Chemistry Chapter 2 Is Matter Around Us Pure are provided here in PDF. Matter is anything that has mass and occupies space. It may be solid, liquid, or gas. All the matter around us is not pure. Matter exists in two types: pure substances and mixtures.
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Answer: In terms of science, a pure substance may be defined as a single substance or matter which cannot be separated into other kinds of matter by any physical process. All pure elements and compounds are pure substances. Question 2. List points of differences between homogeneous and heterogeneous mixtures. Answer:
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Chemistry Worksheets Class 9 on Chapter 2 Is Matter Around Us Pure with Answers - Set 2. If we look around, everything we see is matter from books, pens, and pencils to we humans ourselves all matter. The matter is anything that has mass and occupies some space. We can further divide matter into pure and impure substances.
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2. Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate, and residue. Solution: (a) Into a vessel, add a cup of milk, which is the solvent, and supply it with heat. (b) Add tea powder or tea leaves to the boiling milk, which acts as a solute.
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At Case Study Questions there will given a Paragraph. In where some Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks or 4 marks. CBSE Case Study Questions Class 9 Science - Matter in our Surroundings Case Study 1: 1.) A matter is anything that has mass and occupies ...
Class 9 Chemistry Chapter 2 Is Matter Around Us Pure? MCQs
These MCQs are based on the CBSE board curriculum and correspond to the most recent Class 9 chemistry syllabus. By practising these Class 9 Multiple choice questions, students will be able to quickly review all of the ideas covered in the chapter and prepare for the Class 9 Annual examinations as well as other entrance exams such as CTET and KVS.

Class 9 Science Case Study Questions Chapter 2 Is Matter Around Us Pure

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Class 9 Science Chapter 2 Important Extra Questions

Class 9 Science Chapter 2 Important Extra Questions Set – 1

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

Which separation technique will you apply to separate containing kerosene and petrol (difference in their boiling points is more than 250C) which are miscible with each other?

Name the technique to separate (i) butter and curd (ii) salt and water (iii) camphor from salt.

How can you change a saturated solution to an unsaturated solution without adding any more solvent to it?

Class 9 Science Chapter 2 Important Extra Questions Set – 2

Based on which factor a solution is said to diluted, concentration or saturated?

Identify solute and solvent in tincture of iodine.

What is mass per cent of a solution?

What are the two components of a colloidal solution?

Classify the following as chemical or physical changes

Class 9 Science Chapter 2 Important Extra Questions Set – 3

Two liquids A and B are miscible with each other at room temperature. Which separation technique will you apply to separate the mixture of A and B if the difference in their boiling points is 270C?

Define crystallisation.

Why is water called universal solvent?

Class 9 Science Chapter 2 Important Extra Questions Set – 4

Name two elements which exist in liquid state at room temperature.

An unknown substance A on thermal decomposition produces B and C What is an element, a compound or a mixture?

Identify the elements from the following substances sulphur, brine, hydrochloride acid, water, neon, paper, sugar.

Sea water can be classified as homogenous as well as heterogeneous mixture; Comment.

Class 9 Science Chapter 2 Important Extra Questions Set – 5

Explain why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do.

Identify colloids and true solution from the following: Pond water, frog, aluminium paints, vinegar and glucose solution.

Give two examples each for (i) Aerosol, (ii) Emulsion.

Smoke and fog both are aerosols. In what way are they different?

Properties of Colloid

Class 9 Science Chapter 2 Important Extra Questions Set – 6

Define chromatography and give its one application.

Is water an element or a compound? Give one reason in support of your statement.

Suggest separation technique (s) one would need to employ to separate the following mixtures: (a) Mercury and water, (b) Potassium chloride and ammonium chloride, (c) Common salt water and sand, (d) Kerosene oil, water and salt

What would you observe when? (a) a saturated solution of potassium chloride prepared at 600C is allowed to cool at room temperature? (b) an aqueous sugar solution is heated to dryness? (c) a mixture of iron filings and Sulphur powder is heated strongly?

Name the process associated with the following

Class 9 Science Chapter 2 Important Extra Questions Set – 7

Classify the following into metals, non-metals and metalloids: (i) Germanium (ii) Boron (iii) Diamond (iv) Iodine (v) Copper (vi) Helium

Distinguish between physical change and chemical change.

Distinguish between metals and non-metals.

Distinguish between compounds and mixtures.

Class 9 Science Chapter 2 Important Extra Questions Set – 8