class 9 science chapter 2 assignment

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CBSE Worksheets for Class 9 Science

CBSE Worksheets for Class 9 Science: One of the best teaching strategies employed in most classrooms today is Worksheets. CBSE Class 9 Science Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and problem-solving capabilities. So in order to help you with that, we at WorksheetsBuddy have come up with Kendriya Vidyalaya Class 9 Science Worksheets for the students of Class 9. All our CBSE NCERT Class 9 Science practice worksheets are designed for helping students to understand various topics, practice skills and improve their subject knowledge which in turn helps students to improve their academic performance. These chapter wise test papers for Class 9 Science will be useful to test your conceptual understanding.

Board: Central Board of Secondary Education(www.cbse.nic.in) Subject: Class 9 Science Number of Worksheets: 30

CBSE Class 9 Science Worksheets PDF

All the CBSE Worksheets for Class 9 Science provided in this page are provided for free which can be downloaded by students, teachers as well as by parents. We have covered all the Class 9 Science important questions and answers in the worksheets which are included in CBSE NCERT Syllabus. Just click on the following link and download the CBSE Class 9 Science Worksheet. CBSE Worksheets for Class 9 Science can also use like assignments for Class 9 Science students.

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• CBSE Worksheets for Class 9 Science Assignment 1
• CBSE Worksheets for Class 9 Science Assignment 2
• CBSE Worksheets for Class 9 Science Assignment 3
• CBSE Worksheets for Class 9 Science Assignment 4
• CBSE Worksheets for Class 9 Science Assignment 5
• CBSE Worksheets for Class 9 Science Assignment 6
• CBSE Worksheets for Class 9 Science Assignment 7
• CBSE Worksheets for Class 9 Science Assignment 8
• CBSE Worksheets for Class 9 Science Assignment 9
• CBSE Worksheets for Class 9 Science Assignment 10

Advantages of CBSE Class 9 Science Worksheets

• By practising NCERT CBSE Class 9 Science Worksheet , students can improve their problem solving skills.
• Helps to develop the subject knowledge in a simple, fun and interactive way.
• No need for tuition or attend extra classes if students practise on worksheets daily.
• Working on CBSE worksheets are time-saving.
• Helps students to promote hands-on learning.
• One of the helpful resources used in classroom revision.
• CBSE Class 9 Science Workbook Helps to improve subject-knowledge.
• CBSE Class 9 Science Worksheets encourages classroom activities.

Worksheets of CBSE Class 9 Science are devised by experts of WorksheetsBuddy experts who have great experience and expertise in teaching Maths. So practising these worksheets will promote students problem-solving skills and subject knowledge in an interactive method. Students can also download CBSE Class 9 Science Chapter wise question bank pdf and access it anytime, anywhere for free. Browse further to download free CBSE Class 9 Science Worksheets PDF .

Now that you are provided all the necessary information regarding CBSE Class 9 Science Worksheet and we hope this detailed article is helpful. So Students who are preparing for the exams must need to have great solving skills. And in order to have these skills, one must practice enough of Class 9 Science revision worksheets . And more importantly, students should need to follow through the worksheets after completing their syllabus.  Working on CBSE Class 9 Science Worksheets will be a great help to secure good marks in the examination. So start working on Class 9 Science Worksheets to secure good score.

CBSE Worksheets For Class 9

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

NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure in PDF file format English and Hindi Medium updated for new academic session 2024-25 based on rationalised NCERT Books published for 2024-25 exams.

Class 9 Science Chapter 2 Question Answers

• Class 9 Science Chapter 2 Exercises
• Class 9 Science Chapter 2 Intext Questions
• Class 9 Science Chapter 2 MCQ
• Class 9 Science Chapter 2 Extra Questions
• Class 9 Science Chapter 2 Hindi Medium
• Class 9 Science Chapter 2 Notes in English
• Class 9 Science Chapter 2 Notes in Hindi
• Class 9 Science Chapter 2 NCERT Book
• Class 9 Science NCERT Solutions
• Class 9 all Subjects NCERT Solutions

In the NCERT (National Council of Educational Research and Training) Class 9 Science textbook, Chapter 2 Is Matter around Us Pure? explores various topics related to the purity of matter and the separation of mixtures. The main topics covered in Chapter 2 are definition of a pure substance and a mixture, difference between pure substances and mixtures, Types of Mixtures: Homogeneous mixtures (solutions) and Heterogeneous mixtures.

Solvent and solute expressing the concentration of solutions in terms of solute mass, volume of the solution, and mass by mass percentage. We have to learn the definitions of suspensions and colloids, Differences between suspensions, colloids, and true solutions. Class 9 Science chapter 2 explores various methods of separation of mixtures like Handpicking, Winnowing, Sieving, Magnetic separation, Sublimation, Separation by filtration, Separation by decantation, Separation by evaporation, Separation by centrifugation, Separation by chromatography, Separation by distillation.

UP Board students can download UP Board Solutions for class 9 Science chapter 2 in Hindi Medium. Class 9 Science Chapter 2 question-answers of Page 15, Page 18, Page 24 and Exercises in English Medium or Page 16 ke Uttar, Page 20 ke Uttar, Page 26 ke Uttar, Page 27 ke Uttar and Abhyaas ke Uttar in Hindi Medium. NCERT Solutions 2024-25 of other subjects are also available free to download in PDF file format. Download Class 9 Apps for offline use. Download NCERT Solutions Offline Apps based on latest NCERT Books 2024-25, which works offline without internet.

Physical and Chemical Changes Differentiating between physical and chemical changes Examples of physical and chemical changes. In class 9 science chapter we learn about types of Pure Substances like Elements and compounds Characteristics of elements and compounds. Differentiating between mixtures, compounds, and elements. Law of Conservation of Mass – Statement of the law Illustration through examples.

A recap of the main concepts and ideas presented in chapter 2, Is Matter Around Us Pure?, focuses on helping students understand the composition of matter, the classification of mixtures, methods for separating mixtures, and the distinction between pure substances and mixtures. These concepts are fundamental to chemistry and provide the basis for further exploration of chemical reactions and properties of matter in subsequent chapters of the textbook.

Preparing for NCERT Class 9 Science Chapter 2 requires a strategic approach to understanding the concepts related to the purity of matter and the separation of mixtures effectively. Here are some strategies to prepare for this chapter. Begin by reading the chapter carefully from the NCERT textbook. Pay attention to definitions, examples, and key points. While reading, make concise notes of important concepts, definitions, and the steps involved in various separation methods.

Understand the classification of matter into pure substances and mixtures. Learn the differences between homogeneous mixtures (solutions) and heterogeneous mixtures. Practice solving problems related to the concentration of solutions. Understand how to calculate mass by mass percentage, volume by volume percentage, and other concentration measures. Focus on the various separation methods mentioned in the chapter. Understand when and how to use methods like filtration, distillation, chromatography, and more.

NCERT Solutions for Class 9 Science Chapter 2 Is Matter Around Us Pure Intext Questions on different pages and Exercises question answers are given below in updated format. All the answers are checked by experts for academic session 2024-25. Download Class 9 all Subjects App for free use.

If possible, conduct simple experiments to understand the separation techniques practically. This can help in visualizing the processes and reinforcing your learning. Solve the exercise questions given at the end of NCERT books and examples provided in the textbook. Practice solving problems related to separation methods and concentration calculations. Consult additional resources such as reference books, online tutorials, and videos to clarify any doubts or delve deeper into specific topics if needed.

Discuss concepts and solve problems with classmates. Group study can be helpful in understanding different perspectives and learning from one another. Regularly review your notes and the key points of the chapter to reinforce your understanding. Utilize online educational platforms like websites or apps like Tiwari Academy offline and online apps that offer supplementary materials, such as videos, quizzes, and interactive simulations related to the chapter. Solve previous year’s question papers to become familiar with the types of questions that may appear in exams.

Extra Questions on 9th Science Chapter 2

Colloidal solution show tyndall effect but true solution do not. discuss..

In a colloidal solution, the particle size is such (1 nn to 100 nm), that these particles scatter the light rays as they fall on them. Because of scattering the path of the light as well as the particle become visible. But in a true solution, the particle size is so small (less than 1 nm) that these particles are not in a position to scatter the light. Therefore, true solution does not show any Tyndall effect.

Why do not the dispersed phase particles in a colloidal solution combine with one other?

They do not come closer because of the presence of either positive or negative charge on them. Due to mutual repulsion, these particles remain scattered in a colloidal solution.

What is the function of fractionating column in fractional distillation?

A fractionating column obstructs the upwards movement of the vapours of the liquids. As a result, the energy (latent heat of fusion) which is released by the high boiling liquid is taken by the low boiling liquid. It remains in the vapours state. The high boiling liquid by releasing energy condenses and falls back in the distillation flask. Thus, fractionating column helps in the separation of the components from a mixture.

Alloys are sometimes called substitutional solid solutions. Explain.

Alloys are the homogeneous mixtures of two or more metals. For example, brass is a mixture of copper and zinc. Actually, copper is a crystalline solid in which the atoms are closely packed to form a crystal lattice. Some of these atoms have been replaced or substituted by atoms of zinc. Therefore, brass is regarded as a substitutional solid solution.

All mixtures are homogeneous. Is this statement correct? Justify your answer.

No, this statement is wrong. Mixture because they have combined with each other to form ammonium chloride which is a new substance. A mixture is always formed by mixing non-reacting substances. Ammonia (gas) + Hydrogen chloride (gas) → Ammonium chloride (solid).

If you have any doubts or find certain topics challenging, don’t hesitate to seek help from your teacher or classmates. Clarify your doubts as soon as possible. Keep your notes, textbooks, and study materials organized for easy access during revision and exam preparation. Remember that Chapter 2 provides the foundational knowledge for understanding the composition of matter and various techniques used in chemistry. It is important to grasp these concepts thoroughly as they will be revisited and built upon in future science studies.

Tiwari Academy provides online support in all the subjects from class 1st to class 12. We can help students in preparation for NCERT Class 9 Science Chapter 2 in various ways. We offer detailed and step-by-step solutions to all the questions and exercises in the NCERT Class 9 Science textbook for Chapter 2. These solutions can help students understand how to approach and solve different types of problems related to the purity of matter and the separation of mixtures.

Concise and well-structured revision notes specific to Chapter 2 available on Tiwari Academy. These notes can help students review key concepts and important points quickly before exams. Some online platforms, including Tiwari Academy, provide online tests and mock exams that simulate the real exam environment. This is valuable for self-assessment and exam preparation. Tiwari Academy’s online resources are accessible 24/7, allowing students to study at their own pace and convenience.

This flexibility is particularly helpful for students who want to review the chapter multiple times. Online platforms offer options for students to submit their doubts or questions related to Chapter 2. Expert educators or tutors may respond to these doubts, providing clarification and assistance. Tiwari Academy provide access to previous years’ question papers and solutions. Solving these papers can help students become familiar with the exam pattern and practice time management. Students should explore the platform to determine which resources are most helpful for their individual learning needs and preferences.

Question 1: A diamond knife is quite often used for cutting glass. Why? Answer 1: Diamond is probably the hardest substance known. Therefore, a knife made from a special type of diamond is used for cutting the glass.

Question 2: Explain how does soap help in cleaning dirty clothes? Answer 2: In dirty clothes, the dust particles are present on oil are present on oil drops sticking to them. Simple water cannot remove these oil drops from the clothes because water and oil as such do not form a stable emulsion. Soap plays the role of emulsifier and helps in forming a stable emulsion between the two. This means that soap helps in removing these oil drops along with the dirt sticking to them. The dirty clothes get washed by soap solution.

From an examination point of view, NCERT Class 9 Science Chapter 2 holds significant importance for several reasons. Questions from Chapter 2 are often included in Class 9 Science examinations. This chapter is considered an integral part of the curriculum, and students can expect to see questions related to the purity of matter, types of mixtures, separation techniques, and concentration calculations in their exams.

Chapter 2 of 9th science delves into fundamental concepts related to the classification of matter into pure substances and mixtures. It also covers various methods of separating mixtures. A solid understanding of these concepts is crucial for scoring well in exams and for building a foundation in chemistry. Examinations include questions that require students to apply their knowledge of separation methods to real-world scenarios. This highlights the practical importance of the chapter’s content.

The chapter 2 of class 9 science includes numerical problems related to concentration calculations and other concepts. These problems test students’ ability to apply mathematical concepts to scientific situations, which is a valuable skill for exams. Understanding the separation techniques discussed in the chapter is not only useful for exams but also for practical applications in laboratory settings. Students may be asked to perform experiments related to these techniques during practical exams. Class 9 Science chapter 2 serves as a foundation for more advanced topics in chemistry that students will encounter in higher classes. A strong grasp of the concepts in this chapter is essential for success in future chemistry studies.

Question 3: How will you justify that rusting of iron is a chemical change? Answer 3: The rust is a brown chemical compound known as hydrated ferric oxide. It cannot be removed from the surface of the metal b any means. Formula of a rust shows that iron has undergone a chemical change.

Important Questions on 9th Science Chapter 2

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

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.

What do you understand by a Saturated solution?

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.

Describe Pure substance with example.

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.

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

Homogeneous mixture – soda water, air, vinegar, filtered tea. Heterogeneous mixture – wood, soil.

How would you confirm that a colourless liquid given to you is pure water?

Every liquid has a characteristic boiling point at 1 atmospheric pressure. If the given colourless liquid boils exactly at 373 K at 1 atmospheric pressure, then it is pure water. If the boiling point is different, then the water is contaminated.

The knowledge gained from chapter 2, of class 9 science, will be revisited and expanded upon in higher classes, particularly in Class 10 and beyond. Building a strong foundation in this chapter is, therefore, crucial for future science studies. In summary, NCERT Class 9 Science Chapter 2 is important both for scoring well in exams and for building a strong foundation in chemistry. Students should dedicate time and effort to comprehensively learn the concepts presented in this chapter, as they are fundamental to the understanding of matter and its properties.

« NCERT Solutions Class 9 Science Chapter 1

Ncert solutions class 9 science chapter 3 ».

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Is Matter Around Us Pure CBSE Class 9 Science Revision Notes Chapter 2

CBSE Class 9 Revision Notes for Chapter 2 Is Matter Around Us Pure are a study guide for students for understanding what mixtures and pure substances are. This is an important chapter from the exam perspective and includes many interesting topics that will help you in higher classes. askIITians Science teachers have created these revision notes based on the latest CBSE syllabus and exam pattern for Class 9 Science. So they are the perfect study material for exam preparation. The notes are pointwise so that you can revise or memorise the concepts quickly. With our online free revision notes for Class 9 Science Is Matter Around Us Pure, you can revise the whole chapter in just 20 minutes or less! 

CBSE Science Revision Notes for Class 9 Science Chapter 2 Is Matter Around Us Pure include all the topics introduced in the chapter such as mixtures, solutions, homogeneous solutions, the concentration of solutions, colloidal solutions, suspensions, separating the components of a mixture, physical and chemical changes, types of pure substances, and compounds. You can revise all these concepts easily with our revision notes. They are available for free for all students. 

askIITians Science experts have prepared different types of study resources for Class 9 Science. These include NCERT Solutions, NCERT Exemplar Solutions, mind maps, chapter-wise revision notes, flashcards, mnemonics, daily practice papers, tests and much more. We also provide class 6, 7, 8, 9, 10, 11, 12 notes for Science and Maths subjects. Check our study resources once you have done reading our revision notes for Class 9 Science Is Matter Around Us Pure. 

Online Revision Notes for Class 9 Science Chapter 2 Is Matter Around Us Pure

What is a substance.

• Anything that cannot be broken into further particles by applying any physical processes is called a Substance .
• Matter can be classified into two types of substances – Pure substances and Mixtures

What is a pure substance?

A substance that consists of only one type of particle is called a Pure Substance . For Example , Diamond, Salt, Sulphur, Tin. 

What is a mixture?

• When we combine different substances into each other a mixture is formed. For Example , Lemonade is a mixture of three substances, Lemon Juice, Sugar and Water.
• Which of these is a mixture or a pure substance?

Water, Copper, Chocolate cake, Hydrogen, Soil, Air

Mixture – Chocolate cake, Soil, Air

Pure substance – Water, Copper, Hydrogen

Types of Mixtures

There are two categories of mixtures: Homogeneous Mixtures and Heterogeneous Mixtures

Homogenous Mixtures

• When we add sugar, water and lemon juice together they all uniformly mix with each other. Now it is not possible to separate these substances from the mixture. Such mixtures in which the components mix with each other uniformly are called Homogeneous Mixtures .
• The ratio of compositions of homogeneous mixtures can be different. For Example , one may add two spoons of sugar in lemonade while someone else may add only one spoon of sugar in their lemonade. Still, lemonade is a homogeneous mixture.

Heterogeneous Mixtures

• The components in a heterogeneous mixture do not completely dissolve in each other and we can separate them by physical means. In other words, the composition of such mixtures is not uniform.
• For Example , If we mix sand in water the sand settles down in water after some time and we can separate it by filtration.

Here are a few differences between homogeneous and heterogeneous mixtures –

What is a solution?

A solution is nothing but a uniform mixture of two or more substances. Homogenous Mixtures are solutions.

Solution of -

• Liquid into a liquid:   Water and Ink
• Solid into solid: Alloys
• Gas into gas: Air
• Solid into a liquid: Sugar and Water
• Solid into gas: Hydrogen and Metals
• Liquid into gas: Carbon Dioxide and Water 

What is an alloy?

An alloy is a mixture of different metals or nonmetals and metals that cannot be separated from each other using physical methods. For Example

Brass – Copper with up to 50% zinc

Bronze – Copper with up to 12% tin

Solution consists of two types of substances, a solute and a solvent.

Solution = Solute + Solvent

Solvent – The substance in which another substance is mixed is called the Solvent . For Example , Water is a solvent in which we can mix different substances such as salt or sugar.

Solute – The substance that is added to the solvent to form a solution is called a Solute . For Example , Salt, when mixed in water, acts as a solute for the mixture. 

Properties of a Solution:

• A solution is a homogeneous mixture.
• We cannot see the particles of a solution through naked eyes as they are as small as 1 nanometer in diameter.
• The path of light is not visible through the solution. The particles of a solution do not scatter light through them as they are extremely small.
• We cannot separate the particles of a solution by methods of filtration. 

What is a stable solution?

A stable solution is a solution in which particles do not settle down if we leave the solution undisturbed for some time. This is because the particles of a stable solution are homogeneously spread.

Different Types of Solutions

• Dilute – A solution in which the concentration of the solute is much less than that of the solvent. For Example , If we mix 1gm of salt in 500 ml of water, the salt solution thus obtained will be diluted. If we keep on adding the solute in a solution there comes a point when no more solute dissolves in the solution. This is called the Saturation Point of a Solution .
• Unsaturated Solution – A solution, in which we can add more solute as it has not achieved its saturation level yet, is called an Unsaturated Solution. A dilute solution can be called an Unsaturated Solution .
• Concentrated Solution – A solution with a large amount of solute  is called a Concentrated Solution .
• Saturated Solution – A solution in which no more solute can be added since it has already dissolved the maximum amount of solute it can is called a Saturated Solution .

What is concentration?

C oncentration refers to the amount of a substance per defined space or can be defined as the ratio of solute in a solution to either solvent or total solution.

To calculate the concentration consider the formulae below:

• Percent by Mass = (Mass of solute / Mass of solution) X 100
• Percent by Volume = (Volume of solute / volume of solution) X 100
• Molarity (M) = number of moles of solute / Volume of Solution in litres 

Where, Moles of solute = Given mass/molar mass

• Molality (m) = Moles of solute / weight of solvent in kg
• Normality (N) = Number of mole equivalents/ volume of solution  in litres  

= Mass of solute / (equivalent mass * volume of solution in Litres)

• ppm (Parts Per Million) = ( Mass of Solute / Mass of Solvent ) * 10 6
• Mole Fraction SOLUTE = Moles of Solute / Total Moles of Solution
• Mole Fraction SOLVENT = Moles of Solvent / Total Moles of Solution
• Mole Fraction SOLUTE + Mole Fraction SOLVENT = 1

What is a suspension?

A suspension is formed when two or more substances are mixed in a non-uniform manner. Heterogeneous mixtures are suspensions. The solute does not mix with the solvent and can be viewed through naked eyes.

Properties of Suspensions:

• A suspension is a heterogeneous mixture.
• We can see the particles of suspensions through naked eyes.
• We can see the path of light through the particles of a suspension.
• The particles of suspension tend to settle down when left undisturbed. Then, they can be separated using filtration.

What are colloids or colloidal solutions?

A colloidal solution or a colloid is a uniform solution of two or more substances. The particles are relatively very small and the solution appears as a homogeneous mixture but it is not.

Properties of colloids:

• Colloids are heterogeneous in nature.
• The particles of a colloid cannot be seen through naked eyes.
• The particles scatter a beam of light passed through a colloid and produce the Tyndall effect.
• Colloids are stable in nature. The particles of colloids do not settle down if left uninterrupted.
• We cannot separate the particles of a colloid through filtration. We use a method called Centrifugation to separate the particles of a colloid.

What is the Tyndall Effect?

When a beam of light is passed through a colloid the particles of the colloid scatter the beam of light and we can see the path of light in the solution. For Example , when a ray of light enters a dark room it is scattered by the dust particles present in the air and we can see the path of light clearly.

Classification of Colloids

Dispersed Phase – The dispersed particles or the solute-like components in a colloid 

Dispersing Medium – The substance in which these solute-like particles are added

Based on the state of the dispersing medium colloids are classified as:

Types of Colloids

How to separate components of a mixture.

We can separate the heterogeneous mixtures into their constituents by means of physical methods like

• Hand-picking

The components of a mixture can be separated from each other using several other techniques like

• Evaporation
• Centrifugation
• Sublimation
• Chromatography
• Distillation
• Evaporation – For separating a mixture of a non-volatile and a volatile substance

• Separating coloured component from the ink
• Salt from water
• Sugar from Water
• Mix some ink into water and heat it. After some time the water will evaporate leaving behind the coloured substance.
• Centrifugation – Separating dense particles from lighter particles
• Separating milk from cream
• Separating butter from cream
• Squeezing out water from wet clothes
• Milk is put in a centrifuging machine or milk churner and the cream thus separates from milk.
• Using a Separating funnel – To separate two immiscible liquids

• Oil from water
• Iron and iron ore
• The immiscible liquids are allowed to settle in the funnel. They soon form separate layers due to varying densities. The first liquid is allowed to flow out of the funnel and as soon as it is completely poured out, the stopcock is closed thereby separating the two liquids from each other.  
• Sublimation – To separate a sublimable component from a non-sublimable component

• Ammonium chloride/camphor/naphthalene and salt
• Heat the mixture in an inverted funnel so that the sublimable component sublimes in the air and settles over the walls of the funnel and the non-sublimable component, on the other hand, is left behind.
• Chromatography – To separate solutes that can dissolve in the same solvent 

• Separating colour components of a dye
• Drugs from blood
• Take a filter paper or a blotting paper and place a drop of ink at the rear end. Dip the end in water. Since the ink is a mixture of two or more colours, the component of ink which is soluble in water mixes into it and then separates quickly from the other components that are less soluble in water.
• Distillation – To separate miscible liquids (the boiling points of the liquids must be sufficiently different)

• Acetone and water
• The mixture is heated in a distillation apparatus. The one substance with a lower boiling point evaporates first, condenses and gets separated from the one with a higher boiling point.
• Simple Distillation – when the miscible liquids have a satisfactory difference in their boiling points
• Fractional Distillation – when the difference between the boiling points of the liquids is less than 25 K

Separating different Gases from the Air

Method – Fractional Distillation

• Compress and cool the air by increasing the temperature and decreasing the pressure. The air turns to liquid air.
• Liquid air is warmed up slowly in a fractional distillation apparatus
• The several components of air get separated and are collected at various heights on the basis of their boiling points

Purifying Solids

Method used – Crystallisation

In the crystallisation method, we can obtain a pure solid in the form of crystals from its solution

• Salt from seawater
• Purification of copper sulphate
• The impurities of a substance are filtered out.
• Water is evaporated to obtain a saturated solution.
• The solution is covered with filter paper and left as it is.
• After some time, the crystals of pure solid are formed.
• Is evaporation better than crystallisation?

Simple evaporation is not better than crystallisation because

• Some solid substances decompose because of excess heat. For Example , Sugar gets charred on extra heating.
• If after filtration some impurities remain in the solution they can contaminate the solid and therefore we would not obtain a pure substance.

Physical Change and Chemical Change

Physical Property of a Substance:

Properties of a substance such as rigidity, colour, fluidity, boiling point, melting point, density and hardness which we can observe are called Physical Properties .

Physical Change:

When the physical properties of a substance change it is known as a Physical Change . When we convert a substance from one state to another, such as a solid into a liquid or vice-versa, it is also a physical change as only the physical nature of the substance changes without affecting its chemical nature.

For Example , Change of ice into water. The chemical properties of water remain the same.

Chemical Property of a Substance:

The chemical nature of a substance is known as its Chemical Property such as its odour or its chemical composition.

Chemical Change:

When the chemical properties or chemical composition of a substance gets altered it is called a chemical change. It is also called a Chemical Reaction .

For Example , Burning paper

Types of Pure Substances

Pure substances are classified as elements and compounds

An element is the simplest form of matter.  Elements cannot be broken down into further elements by chemical reactions. Elements are further characterised as Metals, Nonmetals and Metalloids

Metals – Silver, Mercury, Copper, Gold

• Metals are lustrous (shiny)
• Metals conduct heat and electricity
• Metals have a silver-grey or gold-yellow colour
• We can hammer metals and form thin sheets (Malleability)
• We can convert metals into wires (Ductility)
• Metals always produce a ringing sound if they are hit (Sonorous)

Non-Metals – Carbon, Iodine, Chlorine, Oxygen, Hydrogen

• Non-Metals do not conduct heat and electricity
• Non-Metals are not sonorous, lustrous or ductile
• Non-Metals have varied colours

Metalloids – Silicon, Germanium

They show some properties of metals and some of the non-metals.

Quick Facts –

• There are 100 elements known to us
• 92 elements out of them occur naturally
• Rest, 8 are man-made elements
• Most of the elements are solid in nature
• At room temperature, 11 elements exist in the gaseous state
• At room temperature, 2 elements exist in the liquid state – bromine and mercury
• At a temperature slightly higher than room temperature, 2 elements exist in the liquid state – calcium and gallium

It is a substance that consists of two or more substances. These substances are combined chemically with each other in fixed proportions. The properties of a compound are different from that of its constituents. For Example , Ammonium Sulphate, Sulphur Chloride, Water.

Mixtures vs. Compounds

Cbse class 9 science revision notes for is matter around us pure faqs.

• What is Class 9 Science Chapter 2 Is Matter Around Us Pure all about? 

This chapter helps you understand what pure and impure substances are found in the environment. This chapter teaches you important concepts related to mixtures, solvents, heterogeneous and homogeneous solutions, colloids and much more. All these topics help you in daily life as well as in higher studies. 

• How to study CBSE Class 9 Science Chapter 2 Is Matter Around Us Pure? 
• You must read the NCERT chapter carefully and try to understand every term mentioned in the chapter. 
• Make some notes side by side. You can also refer to our online revision notes for Is Matter Around Us Pure. 
• Once you have read the chapter thoroughly, practise the NCERT questions and NCERT Exemplar problems to test your understanding. 
• You can take the help of our NCERT Solutions and NCERT Exemplar problems solutions for Is Matter Around Us Pure chapter. 
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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 

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. 

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.

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

(c) Small pieces of metal in the engine oil of a car → Centrifugation or filtration or decantation

(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.

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.

(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) 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.

(c) Solubility of each salt at 393 K are as follows:

• 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

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

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• Important Questions for CBSE Class 9 Science Chapter 2 - Is Matter Around Us Pure

CBSE Class 9 Science Chapter-2 Important Questions - Free PDF Download

This chapter contains important questions for CBSE Class 9 Science Chapter 2 - Is Matter Around Us Pure in the form of free to download review notes. These questions will undoubtedly aid students in improving their understanding of Chapter 2 of Class 9 Science. Both homogeneous and heterogeneous states of mixtures are covered in Class 9 Science Chapter 2. In this chapter, students will also learn about matter and consider whether the surrounding stuff is pure.

Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. You can download Class 9 Science and Class 9 Maths NCERT Solutions to help you to revise the complete syllabus and score more marks in your examinations.

Download CBSE Class 9 Science Important Questions 2024-25 PDF

Also, check CBSE Class 9 Science Important Questions for other chapters:

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

Very Short Answer Questions (1 Mark)

1. Try segregating the things around you as pure substances or mixtures.

Ans: Try mixing soil and water then separate them. Where soil is a mixture as it is the mix of more than one substance. And water is a pure substance because it’s made up of one kind of substance.

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

Ans: Classification of the given into homogeneous or heterogeneous is enlisted below.

3. How would you confirm that a colorless liquid given to you is pure water?

Ans: Under the  atmospheric pressure one, the boiling point of water is \[{{100}^{{}^\text{o}}}C\]. and freezing point is \[{{0}^{{}^\text{o}}}C\]. When we boil the given colorless liquid, if it boils at \[{{100}^{{}^\text{o}}}C\]then it is pure water. If not boiling at \[{{100}^{{}^\text{o}}}C\] temperature, then there will be impurities mixed with it, hence not pure water.

4. Which of the following materials fall in the category of a “pure substance”?

d) Hydrochloric acid

e) Calcium oxide

Ans: Pure substances: ice, iron, hydrochloric acid, calcium oxide, mercury.

5. Identify the solutions among the following mixtures.

b) Sea water

e) Soda water.

Ans: Solutions: seawater, air, soda water.

6. Which of the following will show the “Tyndall effect”?

a) Salt solution

c) Copper sulfate solution

d) Starch solution.

Ans: (b)Milk and (d)starch solution 

7. Classify the following into elements, compounds, and mixtures.

c) Sugar solution

e) Calcium carbonate

l) Carbon dioxide

m) Blood 

Ans: Classification of the given into elements, compounds, and mixtures are enlisted below.

8. Which of the following are chemical changes?

(a) Growth of a plant

(b) Rusting of iron

(c) Mixing of iron filings and sand

(d) Cooking of food

(e) Digestion of food

(f) Freezing of water

(g) Burning of a candle.

Ans: Chemical changes: rusting of iron, cooking of food, digestion of food, burning of a candle.

9. Which of the following solutions scatter light?

(a) colloidal solution

(b) suspension

(d) none 

Ans: (c) both

10. Which of the following methods would you use to separate cream from milk?

(a) fractional distillation

(b) distillation

(c) centrifugation

(d) filtration

Ans: (c) Centrifugation

11. Cooking of food and digestion of food:

(a) are both physical processes, (b) are both chemical processes, (c) cooking is physical whereas digestion is chemical.

(d) Cooking is chemical whereas digestion physical 

Ans: (b) Are both chemical processes

12. Mercury and Bromine are both

(a) liquid at room temperature

(b) solid at room temperature

(c) gases at room temperature

(d) both (a) and (b)

Ans: (a) liquid at room temperature

13. Blood and seawater are:

(a) both mixtures

(b) both are compound

(c) blood is a mixture whereas seawater is a compound, (d) blood is a compound and seawater is a mixture , ans: (a) both mixtures.

14. Sol and Gel are examples of examples of

(a) Solid-solid colloids

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

(c) Sol is a solid-solid colloid and Gel is a solid-liquid colloid

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

Ans: (b) Sol is a solid-liquid colloid and Gel is a liquid-solid colloid

15. In a water-sugar solution:

(a) water is solute and sugar is solvent

(b) water is solvent and sugar is solute

(c) water is solute and water is also solute

(d) none of these

16. boron and carbon:.

(a) are metalloids

(b) boron is metalloid and carbon is non-metal

(c) boron is metallic and carbon is a metal

(d) boron is non-metal and carbon are a metalloid

Ans: (a) are metalloids

Short Answer Questions (2 Marks)

1. What is meant by a substance?

Ans: Substance will have similar chemical properties and can be defined as that kind of matter where constituent particles cannot be separated from each other by any physical process. 

2. How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than \[{{25}^{{}^\text{o}}}C\] ), which are miscible with each other?

Ans: We can use the distillation technique to separate a mixture containing kerosene and petrol since the difference in their boiling points is more than \[{{25}^{{}^\text{o}}}C\].

3. Name the technique to separate

(i) Butter from curd

Ans: Centrifugation method.

(ii) Salt from sea-water

Ans: Evaporation method.

(iii) Camphor from salt

Ans: Sublimation method.

4. What type of mixtures are separated by the technique of crystallization?

Ans: From liquid solutions of impure samples, pure solid crystals can be separated. This method is known as crystallization. 

Example: Pure sugar from impure sugar, salt from seawater.

5. What is a mixture? What are its various types?

Ans: A mixture is constituted by more than one element or compound or both mixed in any proportion. They are of two types:

(a) Homogenous mixture

(b) Heterogeneous mixture

6. Define solute, solvent, and solution?

Ans: Solute: It is the substance of the solution which is being added to the solvent.

Solvent: It dissolves the solute. The component of the solution to which the solute is added.

Solution: It is homogeneous, constituted by solute and solvent.

7. What is a solution? What are the properties of the solution?

Ans: A solution is a homogenous mixture of two or more substances. The various properties of the solution are: -

The particles of a solution cannot be seen by naked eyes because they are smaller than $1$ nm.

When the beam of light passes through a solution, it does not scatter.

Filtration cannot be used to separate the components of a solution from each other. 

8. Differentiate between elements and compounds.

Ans: The difference between elements and compounds is enlisted below.

9. What is the Tyndall effect? Which kinds of solutions show it?

Ans: Tyndall effect is a process in which the scattering of beams of light takes place in particles of a colloid when that is directed towards them. Heterogeneous mixtures like Suspension solution and colloidal solution show the Tyndall effect.

10. Differentiate between homogeneous and heterogeneous mixture?

Ans: The difference between homogeneous and heterogeneous mixtures are listed below.

11. What is centrifugation? Where it is used?

Ans: Centrifugation is a technique used for the separation of suspended particles of a substance from liquid and is based upon the principle that denser particles stay at the bottom and lighter particles stay at the top when rotated at a high speed in a centrifuge application . It is used in separate butter from milk, also in washing machines for squeezing out water from clothes.

12. What is a suspension? What are the properties of suspension?

Ans: A suspension is a heterogeneous mixture in which the solute particles do 

not dissolve in the solvent but remain suspended throughout the bulk of the medium. The suspension particle size is large enough to be visible from naked eyes.

Properties of suspension:

The particles are large so can be seen by naked eyes.

They scatter a beam of light passing through it.

When particles are left undisturbed, they settle down.

Short Answer Questions (3 Marks)

1. How are sol, solution, and suspension different from each other?

Ans: The difference between sol, solution, and suspension are enlisted below.

2. 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..

Ans: In the problem, it is given that to make a saturated solution, $36$ g of sodium chloride is dissolved in \[100\] g of water at \[293\] K. 

Mass of sodium chloride (solute) is $36$ g 

Mass of water (solvent) is \[100\] g

Mass of solution is the sum of solute and solvent 

\[\Rightarrow 36+100=136\]

Therefore, concentration percentage \[=\dfrac{mass\text{ }of\text{ }solute}{mass\text{ }of\text{ }solution}\times 100\]

$=\dfrac{36}{136}\times 100$

3. 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

the passing of electric current through water and the water breaking    down into hydrogen and oxygen gases

dissolving common salt in water

Making a fruit salad with raw fruits burning of paper and wood.

Ans: When the chemical properties of a substance change then it’s called a chemical change.

Chemical change: rusting of almirah, passing of electric current, through water and the water breaking down into hydrogen and oxygen gases, burning of paper and wood.

Physical properties of a substance such as a shape, size, color, state change then it’s called a physical change.

Physical change: cutting of trees, melting of butter in a pan, boiling of water to form steam, dissolving common salt in water, making a fruit salad with raw fruits.

4. Which separation techniques will you apply for the separation of the following?

a) Sodium chloride from its solution in water.

Ans: Evaporation method

b) Ammonium chloride from a mixture containing sodium chloride and 

Ammonium chloride., ans: sublimation method, c) small pieces of metal in the engine oil of a car..

Ans: Filtration method

d) Different pigments from an extract of flower petals.

Ans: chromatography.

e) Butter from curd.

Ans: Centrifugation method

f) Oil from water.

Ans: by using separating funnel.

g) Tea leaves from tea.

Ans: Filtration by using a strainer

h) Iron pins from sand.

Ans: magnetic separation.

i) Wheat grains from husk.

Ans: Winnowing method

j) Fine mud particles suspended in water.

5. write the steps you would use for making tea. use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate, and residue..

Ans: First, take the required amount of water as a solvent in a pan, and after boiling it add a little amount of sugar which is solute to the solvent. The solute will dissolve completely in the solvent forming the true solution, then add tea leaves that are insoluble along with another soluble liquid milk. After boiling the solution use the method of filtration with a sieve so that the filtrate obtained is tea while the residue has tea leaves that can be thrown away.

6. 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?

Ans: At temperature $313$ K the amount of potassium nitrate required was $62$ g in $100$ ml of water.

Therefore, in \[50\] g water we will need to dissolve $62\times \dfrac{50}{100}=31$ g potassium nitrate.

b) Pragya makes a saturated solution of potassium chloride in water at $353$ K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.

Ans: at $353$ k saturated solution preparation needs \[54\] g potassium nitrate and at room temperature (\[293\]k) saturation solution formation occurs with \[35\] g potassium nitrate hence \[5435=19\] g potassium nitrate will precipitate out as undissolved salt..

c) Find the solubility of each salt at \[293\] K. Which salt has the highest solubility at this temperature?

Ans: Solubilities are (in \[100\] mg of water) \[32,36,35,37\] respectively for the mentioned salts and the highest solubility is of ammonium chloride at this temperature.

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

Ans: Solubility of salts is directly proportional to the temperature i.e., if temperature increases then solubility will also increase, and if the temperature decreases then solubility will also decrease.

7. Explain the following examples.

(a) Saturated solution

Ans: It is a solution in which no more solute particles can be dissolved at a particular temperature.

(b) Pure substance

Ans: It is a substance that has a fixed composition and is made up of only one type of particle.

(c) Colloid

Ans: It is a substance that has a fixed composition and is made up of only one type of particle. It is a  kind of heterogeneous mixture/solution in which particle size is between $1$ nm and $1000$ nm that is intermediate between true solution and suspensions. Colloids have dispersion medium and dispersed phases.

(d) Suspension

Ans: It is a  kind of heterogeneous mixture, in which insoluble solid particles remain suspended in the medium and dispersion particles are visible to the bare eyes.

8. Write a method to separate different gases from the air.

Ans: Air is a homogeneous mixture of various gases.

Fractional distillation can be used to separate its various components.

(a) First, compress air by increasing the pressure and cool the air by decreasing the temperature.

(b) The obtained air is liquid air; now allow the liquid air to warm up slowly in a fractional distillation column.

(c) The various gases separate from each other according to their boiling points at various heights of the fractionating column.

9. What is a colloid? What are its various properties?

Ans: The heterogeneous mixture of substances are colloids, in which the particle size is too small and cannot be seen by naked eyes.

(1) It is a heterogeneous mixture but appears homogeneous.

(2) The size of particles is too small so cannot be seen by naked eyes.

(3) They make its path visible by scattering the beam of light passing through it.

(4) When the colloid is left undisturbed, the particles of it do not settle down.

10. A solution contains $60$ g of $NaCl$ in $400$ g of water. Calculate the concentration in terms of mass-by-mass percentage of the solution.

Ans: In the problem, it is given that, A solution contains $60$ g of $NaCl$ in 400g of water. 

Mass of solute ($NaCl$) is $60$ g

Mass of solvent (water) is $400$ g

Mass of solution $=$ Mass of solute $+$ Mass of solvent

$\Rightarrow 60+400=460$ g

Mass percentage of the solution is the percentage of the ratio of the mass of solute to the mass of solution.

\[\Rightarrow \dfrac{60}{460}\times 100=\dfrac{300}{23}~~\]

\[=13.4\%\]

11. Differentiate between metals and non-metal based upon the various properties that they show.

Ans : the difference between metals and non-metal based upon the various properties are enlisted below., 12. differentiate between mixtures and compounds by giving appropriate examples, ans : differences between mixtures and compounds are enlisted below., 13. write a method to separate a mixture of salt and ammonium chloride.

Ans: A mixture of salt and ammonium chloride can be separated by the process of sublimation. In this process, the solid substance is directly converted into a gaseous state.  Since ammonium chloride changes directly from a solid into a gaseous state on heating and salt does not have that property, this principle can be used to the mixture of two.

The mixture of $N{{H}_{4}}Cl$ (ammonium chloride) and salt is taken in a china dish inside an inverted funnel.

The mixture is then heated using a burner and because $N{{H}_{4}}Cl$ sublimates thus it changes into vapors directly.

Salt settles into the inverted funnel as it is a non-sublimely substance.

Separation of $N{{H}_{4}}Cl$ salt by sublimation

14. What is crystallization? Where is it used? Why is this better than the simple evaporation technique?

Ans: Crystallization is the process of the transformation of solution into pure solid in the form of crystals. It is used to purify solids. For example, salt from seawater is purified using crystallization. It is a better technique than simple evaporation because:

Some solids may decompose or get charred on heating to dryness during evaporation.

Some of the impurities will remain dissolved in the solution.

15. What is chromatography? What are its various applications and underline the basic principle involved?

Ans: A technique used for the separation of those components whose solubility is different in the same solvent is chromatography. The basic principle in chromatography is that different solutes have different solubility in the same solvent.

Its various applications are:

It is used to separate different colors in dye .

It is used to separate pigments from natural colors.

It is used to separate drugs from the blood.

16. A solution of \[{{H}_{2}}S{{O}_{4}}\] acid is labeled is \[95\%\]. What is the mass of this that must be diluted with water to get \[5\]L of a solution containing \[10\] g of \[{{H}_{2}}S{{O}_{4}}\] per litre?

Ans: In the problem, it is given that, A solution of \[{{H}_{2}}S{{O}_{4}}\] acid is labeled is \[95\%\].

\[1\] L of the diluted solution must contain \[10\] g of \[{{H}_{2}}S{{O}_{4}}\]. Therefore, \[5\] L of the diluted solution must contain \[50\] g of \[{{H}_{2}}S{{O}_{4}}\].

The concentration of the acid in the bottle is \[95\%\] as per the problem.

This means that,

\[95\] g of \[{{H}_{2}}S{{O}_{4}}\] is present in \[100\] g of the acid solution .

\[50\] g of \[{{H}_{2}}S{{O}_{4}}\] will be present in \[\dfrac{\left( 50\times 100 \right)}{95}=52.64\] g of the solution.

Chemistry plays an important role in everyone's life, we might not know much about it but it is present in our everyday life, that is what this chapter tries to show where it indicates the different types of matters that are present everywhere around us. In this chapter, students will learn how matter is composed of and how it differs from various substances. Regular practice of Chapter 2 Class 9 Important Questions can help students improve, become through the concepts and topics, and be efficient during preparation or revision. Students must learn to utilize the material given to them to get more marks. In this article, we will also look at Class 9 Science Chapter 2 extra questions .

Important Questions for CBSE Class 9 Science Chapter 2 - Is Matter Around Us Pure - Benefits of Class 9th Science Chapter 2 Important Questions

It is vital that students understand the importance of this subject and what it holds for students of Class 9. By utilising the important questions and with a rigorous practice regime, students will be able to score the most out of their exams. These exams can be a little difficult without the right guidance but by using Vedantu’s important questions on Chapter 2, students will be able to study in a more structured manner. Following is the list of benefits:

Students can use Vedantu to use their time wisely, it helps boost their confidence after consistent practice and students can plan their preparation accordingly. 

It provides students with a structure with which they can study for their upcoming examinations. 

This is a fundamental chapter for students and plays a crucial role in upcoming grades. 

Students don’t have to worry about the relevance of these questions as they are all cross-checked and updated according to the latest CBSE guidelines and rules. So, the information in this article is genuine and reliable.

Topics Covered under CBSE Class 9 Chapter 2 ‘Is Matter Around Us Pure’ 

Following are the topics that are covered in CBSE Class 9 Chapter 2 ‘Is Matter Around Us Pure’:

Matter and its Types

Metals, Non-metals, and Metalloids

Mixture and its Types

Solution and its Types

Factors Affecting the Solubility

Concentration of Solution

Key Takeaways of CBSE Class 9 Science Chapter 2 Is Matter Around Us Pure 

Students here will learn some of the basic elements of constituents of matters. This will help them in their future grades as it sets the groundwork. Constant practice of the essential questions should help students to tackle any difficult questions in their final examinations. Some of the topics that these chapter covers are as follows:

Chemistry 

This subject is known as the central science subject that connects all the science subjects. This subject is very detailed and helps students understand the chemical constituents in different materials. it is connected to a lot of the physical subjects together such as Chemistry with Applied Science and Life Sciences such as Engineering and Medicine. Chemistry is defined as the study of the interaction, composition, and properties of matter.

Homogeneous and Heterogeneous Mixture 

These are two very different mixtures as in a homogeneous mixture, it has a uniform composure of its constituents where heterogeneous is a nonuniform composure of its constituents.

Tyndall Effect 

The scattering of a beam of light by particles of a solution when light is passed through it is known as the Tyndall effect. The solution where the size of the particle is very small.

Matter is defined as anything that possesses mass, occupies space, and the presence that can be felt by the five senses. Matter exists in three forms, namely, a solid, liquid, and gas. Solids are substances that possess a definite structure and a definite shape like sugar, iron, etc. Liquids are substances that have a definite volume but lack a definite form and take the shape of the vessel in which they are put — for example, mercury, milk, water, etc. Gases are substances that can neither possess a definite shape or definite volume like hydrogen, oxygen, nitrogen, etc.

The Difference Between Mixture and Compound 

This chapter takes a detailed look into the difference between mixture and compounds. The mixture is basically the elements or compounds that are mixed together in a heterogeneous way. It has a variable composition and also shows us the properties of constituent elements and the various ways in which they can be mixed. The examples are air, blood, and water. In a compound when the elements react, they form new compounds. This new substance formed shows new properties and examples of this are sodium chloride.

Crystallisation 

This is a very important process where we can separate the pure solid in the form of its crystals from its solutions. This is an important process when forming crystals. Unlike many processes where the solids may decompose because of the heat during the process of decomposition. In the process of evaporations, some solids stay intact.

Substance 

In this situation, it is physically impossible to separate the constituent particles from one another on one's own. Chemical or electrochemical procedures can be used to separate them because their chemical properties are similar. A material possesses particular qualities or attributes. Physical properties and chemical properties are the two main categories into which properties of matter can be divided. Melting point, boiling point, colour, aroma, and other physical attributes can be observed or quantified without affecting the content or identification of the substance. Chemical characteristics, such as combustibility, basicity, or acidity, are the chemical transformations that result in a distinctive response.

Chromatography 

This is a process where substances used for the separation of different substances have different solubility in the same solvent. It is used to separate different colours in the dye. It is used to separate different pigments from natural colours and separate drugs from the blood. There are various ways in which they can be separated and we will learn that in this chapter.

Colloid 

These are heterogeneous mixtures of substances whose particles are too small for the naked eye and cannot be seen. It appears homogeneous but is actually a heterogeneous mixture. The particles are too small for the naked eye to see. They scatter a beam of light through it and make its path invisible. The particles of the colloid do not settle down when left undisturbed. 

Important Questions on CBSE Class 9 Science Chapter 2 - Is Matter Around Us Pure 

To get a better understanding of Class 9 Science Chapter 2 important questions , let's look at how the various essential questions are framed and how they can be beneficial to students. Using the following questions should help students in the long term. 

1. What is chromatography? What are its various applications and underline the basic principles involved?

2. What is crystallisation? Why is crystallisation used? 

3. Why is crystallisation a better technique than the evaporation process?

4. Write a method to separate salt from sodium chloride.

5. Differentiate between mixture and compound by giving appropriate examples.

6. Differentiate between metals and non-metals based on the various metal properties they show.

7. What is a colloid? 

8. What are the various properties of a colloid? 

9. Write a different method to separate gas from the air?

10. Explain and give the example of the following: 

a. Saturated solution 

b.Pure substance 

c. Colloid 

d.Suspension

Tips to Study Science Better

Following are some tips that will help the students to study science in a better way:

Follow the concepts and study them properly. 

Try out the experiments under the guidance of someone elderly, which will help you to understand the topic better.

Practise solving the questions and answers, this will increase your chance of getting better results.

Practise with reliable notes for this chapter, you can refer to Is Matter Around Us Pure Class 9 Notes CBSE Science Chapter 2 (Free PDF Download) of Vedantu, this is quite reliable.

Conclusion 

The Situation in the Matter Around Us Pure, students have found pure important questions to be incredibly helpful. As you can see from this article, this chapter is very important for students to begin their study of chemistry because it essentially lays the foundation for subsequent grades. This article can help students make the most of their time, build their confidence after constant practice, and manage their study sessions effectively. Students might aspire for higher grades by working more hard toward their goals. These significant questions ensure that students understand the chapter's numerous concepts, and with continued practice, they will develop the skills necessary to answer the challenging questions on exams.

Important Related Links for CBSE Class 9 

FAQs on Important Questions for CBSE Class 9 Science Chapter 2 - Is Matter Around Us Pure

1. What are the major differences between a mixture and a compound?

Mixture: Elements or compounds just mix together to form a mixture. It has a variable composition and shows the properties of constituent elements. These constituent particles can be separated by physical methods. For example, air, blood, etc.

Compound: When elements chemically react, they form new compounds. Compounds have a fixed composition. The new substance formed shows new properties. The constituent elements can only be separated by chemical methods. For example, Sodium Chloride, Calcium Sulphate, etc.

2. What is crystallisation? Why is it an important process?

Crystallisation is a process by which we can separate a pure solid in the form of crystals from its solutions. It is an important process which is used to purify solids. It is a better technique than many processes like simple evaporation as some solids may decompose on heating during evaporation. In evaporation, some impurities still remain dissolved in the solutions.

3. Where can I find Important Questions for CBSE Class 9 Science Chapter 2 - Is Matter Around Us Pure?

Students can find a set of important questions for Class 9 Chapter 2 Is Matter Around Us Pure on Vedantu, a reliable online learning site. Vedantu caters to solve important questions chosen from the exam perspective. These questions are answered by subject matter experts who have immense subject knowledge and expertise. The material is available in the free PDF format and can be downloaded at students’ convenience for learning and during exam preparation.

4. What is meant by a substance?

Substance can be defined as the kind of matter where constituent particles cannot be separated from one another with the help of any physical process. Since they are similar in chemical properties, they can also be separated by chemical or electrochemical methods. Examples are diamond, water, sulphur, etc.

5. What is the saturated solution according to Chapter 2 of Class 9 Science?

A saturated solution is a solution in which no more solute can be added without increasing the temperature of a solution. It is stable at room temperature.  Students study more about saturated solutions in their science books for Class 9. They will understand the concept of pure substances and matter too. All definitions related to different types of solutions and substances are given in detail on Vedantu.

6. What is a mixture and a compound according to Chapter 2 of Class 9 Science ?

A mixture and a compound are different from each other in many ways. A mixture is not a pure substance but a compound is a pure substance. A mixture has different constituents and the properties of the constituents do not change in a mixture but the properties of all constituents will change in a compound. In a mixture, two or more substances can be mixed together in any ratio whereas in a compound two or more substances are mixed together in a fixed ratio.

7. Write down the important properties of a colloidal solution?

Colloidal solutions are heterogeneous mixtures. The size of particles in a colloidal solution varies between 1nm and 100 nm. The particles are very small in size and cannot be seen easily. If the particles are left undisturbed, the solutions remain stable. Students of Class 9 can understand the concept of colloidal solution in Chapter 2. Students should read Chapter 2 of Class 9 carefully to understand the definitions of important terms given in the chapter.

8. Water acts as a compound and not a mixture. Give reasons.

Water acts as a compound because it is made of two components oxygen and hydrogen. The two constituents can be separated through the process of electrolysis. The two constituents are combined together in a fixed ratio of 1:2. The ratio of hydrogen and oxygen is fixed and does not change. It is not easy to separate the constituents of water easily. They can only be separated by a special process only called electrolysis.

9. What are the important features of important questions of Chapter 2 of Class 9 Science from Vedantu?

Important questions for Chapter 2 of Class 9 Science available on Vedantu include questions that can help students to score high marks in exams. These important questions are available at free of cost on Vedantu(vedantu.com) and mobile app. Important questions can also help students to understand the concepts and gain in-depth knowledge of all the concepts given in the chapter. Students can prepare for their exams by practising all-important questions given in Vedantu because they are prepared by expert and experienced teachers and professionals.

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From the perspective of CBSE, NCERT Solutions Class 9 Science Chapter 2 Is Matter Around Us Pure is crucial study material. Get solutions of all chapters of NCERT class 9 Science from Physics Wallah.

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

NCERT Solutions for class 9 Science  chapter 2 Matter around Us Pure is prepared and uploaded for reference by academic team of expert members of Physics Wallah. Get solutions of all chapters of NCERT class 9 Science from Physics Wallah. use as a reference of the following  NCERT solutions  of chapter 2 Matter around Us Pure is prepared by Physics Wallah.

NCERT Solutions for Class 9 Science Chapter 1

Question 1. What is meant by a substance? Solution :  Substance can be defined as that kind of matter where constituent particles cannot be separated from each other by any physical process since they are all similar in chemical properties.

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

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Question 5. 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. Solution : Mass of sodium chloride (solute) = 36 g Mass of water (solvent) = 100 g Mass of solution = 36 + 100 = 136 g Therefore, concentration percentage = mass of solute/mass of solution = 26.47 %

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Question 6. How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than), which are miscible with each other? Solution :  We can separate a mixture containing kerosene and petrol by distillation technique since difference in their boiling points is more than. So through distillation we can get them separated.

Question 7. Name the technique to separate (i) butter from curd, (ii) salt from sea-water, (iii) camphor from salt.

Solution : (i) centrifugation method. (ii) evaporation method. (iii) sublimation method.

Question 8. What type of mixtures are separated by the technique of crystallisation? Solution :  From impure samples of solids, pure solid crystals can be obtained by the method of crystallization for eg to obtain pure sugar from impure sample of the same.

Question 9. Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue. Solution :  Take more amount of solvent (water) in a pan and after heating it add little amount of solute (sugar) to the solvent. Solute will dissolve completely in the solvent forming true solution, then add tea leaves that are insoluble along with another soluble liquid milk. After boiling allow filtration with a sieve so the filtrate you obtain is tea while the residue has tea leaves that are thrown away.

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Question 10. Classify the following as chemical or physical changes : • cutting of trees, • melting of butter in a pan, • rusting of almirah, • boiling of water to form steam, • passing of electric current, through water and the water breaking down into hydrogen and oxygen gases, • dissolving common salt in water, • making a fruit salad with raw fruits, and • burning of paper and wood.

Solution : cutting of trees = chemical 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 the water breaking down into hydrogen and oxygen gases = 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

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Question 11. Try segregating the things around you as pure substances or mixtures. Solution : Pure substance: Water, salt, sugar Mixture: Saltwater, soil, wood, air, cold drink, rubber, sponge, fog, milk, butter, clothes, food

Question 12. 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. Solution : (a) Evaporation (b) Sublimation (c) Filtration (d) Chromatography (e) Centrifugation (f) Separating funnel (g) Filtration (h) Magnetic separation (i) Winnowing/ sedimentation (j) Decantation and filtration

Question 13. 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? Solution :  A mixture of kerosene and petrol which are miscible with each other can be separated by distillation. Method

• Take a mixture in a distillation flask. • Fit it with a thermometer. • Arrange the apparatus as shown in the figure. • Heat the mixture slowly. • Petrol vaporises first as it has lower boiling point. It condenses in the condenser and is collected from the condenser outlet. • Kerosene is left behind in the distillation flask.

Question 14. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below( results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).

(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K? (b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain. (c) Find the solubility of each salt at 293 K. What salt has the highest solubility at this temperature? (d) What is the effect of change of temperature on the solubility of a salt?

Solution : (a) At 313 K, 62 grams of Potassium nitrate dissolved in 100 grams of water. So to produce a saturated solution of potassium nitrate in 50 grams of water, we need 62 x 50 = 31 grams of potassium nitrate 100 (b) Some soluble potassium chloride will separate out in the form of crystals at room temperature because the solubility of potassium chloride will decrease with decrease in temperature. (c) (i)   Solubility of Potassium nitrate at 293 K is 32 grams. (ii)  Solubility of Sodium chloride at 293 K is 36 grams. (iii) Solubility of Potassium chloride at 293 K is 35 grams. (iv) Solubility of Ammonium chloride at 293 K is 37 grams. The solubility of Ammonium chloride is highest at this temperature. (d) The solubility of salt increases with increase in temperature.

Question 15. Explain the following giving examples. (a) saturated solution (b) pure substance (c) colloid (d) suspension

Solution : (a) saturated Solution :-  It is a solution in which no more solute particles can be dissolved at a particular temperature.

(b) pure substance :-  Such substance that has a uniform composition i.e. has particles with identical properties is called pure substance eg sugar, salt, water, nitrogen etc.

(c) 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.

(d) 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 16. Classify each of the following as a homogeneous or heterogeneous mixture. soda water, wood, air, soil, vinegar, filtered tea. Solution : Homogeneous mixtures: Soda water, air, vinegar Heterogeneous mixtures: Wood, soil, filtered tea

Question 17. How would you confirm that a colourless liquid given to you is pure water? Solution :  If we allow the given liquid to evaporate by heating it as in a clean china dish so: O any residue remaining in the china dish will indicate that water is not pure but contains impurities. O no residue in china dish will indicate that water is pure.

Question 18. Which of the following materials fall in the category of a “pure substance”? (a) Ice (b) Milk (c)Iron (d)Hydrochloric acid (e) Calcium oxide (f) Mercury (g) Brick (h) Wood (i) Air.

Solution :  Pure substances are: ice, iron, hydrochloric acid, calcium oxide, mercury.

Question 19. Identify the solutions among the following mixtures. (a) Soil (b) Sea water (c) Air (d) Coal (e) Soda water.

Solution :  Sea water, air and soda water are solutions.

Question 20. Which of the following will show “Tyndall effect”? (a) Salt solution (b)Milk (c) Copper sulphate solution (d) Starch solution.

Solution :  Milk and starch solution have larger particles since they are not true solutions so they will show tyndall effect.

Question 21. 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

Compounds :

Question 22. 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.

Solution : The following changes are chemical changes: (a) Growth of a plant (b) Rusting of iron (d) Cooking of food (e) Digestion of food (g) Burning of candle

NCERT Solutions Class 9 Science Chapter 2 FAQs

This chapter introduces the concept of pure substances and mixtures, which forms the foundation of understanding matter and its various forms in our surroundings.

Chapter 2 mainly focuses on pure substances, mixtures, types of solutions, elements, compounds, and methods of separation.

NCERT Solutions are comprehensive and cover the fundamental concepts. However, additional practice and reference from other sources can further enhance understanding.

Understanding the concepts thoroughly, practicing solved examples, solving exercises, and referring to NCERT Solutions for clarity can significantly aid in exam preparation.

Yes, NCERT Solutions generally offer detailed explanations, step-by-step solutions, and diagrams for better comprehension of concepts.

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NCERT Solutions for Class 9 Science Free PDF Download

Ncert solutions for class 9 science.

NCERT is on the priority list of recommendation of the Central Board of Secondary Education (CBSE). Here, you will find all the chapter-wise NCERT Solutions for Class 9 Science. It will help you to understand the subject and perform well in exams.

Class 9 Science is divided into 3 parts – Physics, Chemistry, Biology.

NCERT Solutions for Class 9 Science will help you to ace the unsolved problems in the Class 9 Science book prescribed by the NCERT for all the schools of CBSE. A thorough understanding of the NCERT Solutions for Class 9 Physics helps you in understanding Physics concepts to the point. The NCERT Solutions for Class 9 Science cover all the 5 chapters of the prescribed Physics syllabus and are the best alternative.

Chemistry is one of those subjects which are equally important from academics as well practical point of view. NCERT is specially made to clear most of the topics of the syllabus of chemistry. NCERT solutions help the students to clear all the doubts regarding any chapter.

Biology along with having a good weight in exams is equally important from experiment and practical point of view. In order to understand the chapters in biology, NCERT solutions play an important role.

Download NCERT Solutions for Class 9 here

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NCERT Solutions for Class 9 Science Chapterwise

Class 9 Science Chapter 1 – Matter in our Surroundings

Class 9 Science Chapter 2 – Is Matter Around Us Pure

Class 9 Science Chapter 3 – Atoms and Molecules

Class 9 Science  Chapter 4 – Structure of the Atom

Class 9 Science Chapter 5 – Fundamental Units of Life

Class 9 Science Chapter 6 – Tissues

Class 9 Science Chapter 7 – Diversity in Living Organisms

Class 9 Science Chapter 8 – Motion

Class 9 Science Chapter 9 – Force And Laws Of Motion

Class 9 Science Chapter 10 – Gravitation

Class 9 Science Chapter 11 – Work and Energy

Class 9 Science Chapter 12 – Sound

Class 9 Science Chapter 13 – Why do we fall ill?

Class 9 Science Chapter 14 – Natural Resources

Class 9 Science Chapter 15 – Improvement in Food Resources

NCERT Solutions for Class 9 Science Chapter 1: Matter in our Surroundings

Matter is an important part of our planet as our whole planet is made up of matter. This chapter provides a good amount of information about the matter. It includes the importance of matter based on the chemical and physical properties of matter. The physical nature of the matter will help the students to understand the concept of matter on the basis of its practical use in daily life. The chapter also explains some concepts like evaporation, boiling point, freezing point, etc. along with the factors affecting them.

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

In the present era, it believed that purity in anything is hardly seen. It is because of the presence of high level of pollutants everyone. In the context of the matter, it also holds true. In this chapter, the students will be able to learn about various concepts like solutions, pure substances, mixtures, suspensions, colloidal, etc. Moreover, this chapter talks about Homogenous substances and Heterogeneous Substances.

NCERT Solutions for Class 9 Science Chapter 3: Atoms and Molecules

In previous classes, students have already read about the presence of atoms as a part of the material. It showed them the divisibility of matter. Atom is the smallest thing or unit present on this planet. In this chapter, the students will be able to learn more about the molecules. This chapter highlights concepts like molecules of compounds, molecules of elements, etc. This chapter also talks about the presence of ions, Law of Conservation of Energy, Law of Chemical Combination, Dalton’s Atomic Theory, etc.

NCERT Solutions for Class 9 Science Chapter 4: Structure of the Atom

One of the most famous theories or models regarding atoms is given by J.J. Thomson. In this chapter, the students will learn about his model along with the drawbacks. It helps to understand the divisibility of atoms. It states that atoms are made of charged particles viz. electrons, protons, neutrons, etc. In the end, the students will learn about the atomic number, atomic mass, mass number, isotopes, isobars, etc

NCERT Solutions for Class 9 Science Chapter 5 Fundamental Units of Life

Robert Hooke was the first person to prepare a microscope and see a cell. A cell is the basic fundamental unit of life. In this chapter, students will learn the structures and functions of different cells. Moreover, the effect of hypotonic, isotonic, and hypertonic solutions on a cell.

NCERT Solutions for Class 9 Science Chapter 6 Tissues

This chapter helps the students to learn about the various types of Tissues in Plants like Merismatic Tissue and Permanent tissue. Moreover, in the end, the chapter highlights topics like Position of plant tissues, a section of a stem, guard cells, etc.

NCERT Solutions for Class 9 Science Chapter 7 Diversity in Living Organisms

There are various life forms on this planet and it becomes very essential to study all of them. In chapter helps students to know about the various groups of organisms. Aristotle has done the classification of animals on the basis of land, air, and water, etc. There are certain terms related to the classification which the students will confront are Kingdom, Phylum, Class, Order, Family, Genus, Species, etc.

NCERT Solutions for Class 9 Science Chapter 8 Motion

Here, you will learn about motion including motion along a straight line, types of motion, the difference between Vector & Scalar, Speed & Velocity, Distance & Displacement, Acceleration – Rate of change of velocity and average speed and velocity, Graphical representation of motion and derivation of three equation of motion.

NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion

In this chapter, you will learn the concept of balance and unbalance forces. Starting with the First law of motion, the Galileo’s concept, the law of inertia. You will also learn the Second law of motion and Third law of motion, momentum, rate of change of momentum. And applications on first, second and third laws of motion.

NCERT Solutions for Class 9 Science Chapter 10 Gravitation

This chapter gravitation takes you through the depths of motion of objects under the influence of gravitational force on the earth. Gravitational force and Newton’s universal law of gravitation.

NCERT Solutions for Class 9 Science Chapter 11 Work and Energy

In this chapter, you will learn about the relationship between work and energy, scientific conception of work and also different forms of energy such as Kinetic energy, Potential energy, application of kinetic and potential energy, and energy of an object at a certain height.

NCERT Solutions for Class 9 Physics Chapter 12 Sound

This chapter is a very interesting one as you will get to learn about the Reflection of Sound i.e ECHO, reverberation, and applications of multiple reflections of sound. All these concepts are taught by implementing various activities needed to be done in the Physics laboratory that makes the learning process more effective and interactive.

NCERT Solutions for Class 9 Science Chapter 13 Why do we fall ill?

Health and Diseases are the two most important topics of discussion in this present world. This chapter highlights Health and its significance, reasons why do we fall ill, etc. Along with this, this chapter discusses the distinction between a healthy person and disease-free person.

NCERT Solutions for Class 9 Science Chapter 14 Natural Resources

This chapter highlights the most important element of life: Air. Starting with the role of the atmosphere in climatic control, this chapter deals with topics like how air moves, winds, rain, and air pollution. While discussing the topic of Air pollution, the chapter states various pollutants of the same. Moreover, in the end, the chapter, states various solutions to the problem.

NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources

This chapter deals with concepts like Crop Yield. Along with that, the chapter discusses the improvement in Crop Yield and related topics. Improvement in livestock and food crops are also discussed. The chapter also discusses topics like the ways to improve crop yield, crop production improvement, etc.

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NCERT Solutions for Class 9

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NCERT Solutions for Class 9 Science

Ncert solutions for class 9 science updated for 2023-24 free chapterwise pdf download.

NCERT Solutions for Class 9 Science help students to clear any doubts instantly and efficiently. These NCERT Solutions guide students to learn the important concepts which are included in the CBSE Class 9 Science syllabus. Students are required to solve the exercise questions included in the textbook to create a proper understanding of the topics.

While solving the textbook questions, doubts arise among students. The NCERT Solutions for Class 9 come in handy at such times, as they include precise explanations and detailed answers to those questions. These CBSE Science NCERT Solutions for Class 9 cover solutions to all the important chapters included in the textbook, like Matter, Atoms, Tissues, Living Organisms, Motion, Force, Laws of Motion, Gravitation, Energy and Work, Sound, Natural Resources, etc.

Along with answers to the textbook questions, these solutions provide you with extra questions, exemplar problems, the important questions from previous year question papers, sample papers , worksheets, MCQs, short answering questions, descriptive type questions, their solutions, as well as tips and tricks.

NCERT Solutions for Class 9 Science Chapter-Wise PDFs

NCERT Solutions for Class 9 Science are provided in a format for better accessibility to the students. These solutions serve as an important studying tool for students who are preparing for their board examinations and assignments.

NCERT Solutions of Class 9 Science Book Chapter Brief

Chapter 1: matter in our surrounding.

The Class 9 science textbook starts with the chapter “Matter in Our Surroundings”. Everything that we see around us is made up of material which is called matter. These things occupy space and have mass. Earlier, Indian philosophers had classified matter into 5 basic elements called Panch Tatva – air, water, earth, sky and fire. Now, modern scientists have come up with 2 types of classification, i.e., based on physical property and chemical nature. In this chapter, students will learn about the physical properties of matter .

The matter is made up of particles and these particulars are very small. The particles of matter have space between them, they are continuously moving and attract each other. The matter around us exists in 3 different states – solids, liquids and gases. These states of matter arise due to variation in the characteristic of the particles of matter. All the 3 states of matters have been explained in-depth with the help of activities. Further, the textbook explains that the state of matter is inter-convertible. The state of matter can be changed by changing temperature or pressure. The phenomenon of change of a liquid into vapour at any temperature below its boiling point is called evaporation. The rate of evaporation depends upon the surface area exposed to the atmosphere, the temperature, the humidity and the wind speed. Evaporation also causes cooling.

Topics Covered in Class 9 Science Chapter 1 Matter in Our Surrounding :

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.

Also access the following resources for Class 9 Chapter 1 Matter in Our Surrounding at BYJU’S:

• Matter in Our Surroundings Class 9 Notes – Chapter 1
• CBSE Class 9 Science Chapter 1 Matter in Our Surroundings Important Questions
• NCERT Exemplar Class 9 Science Solutions for Chapter 1 – Matter in Our Surroundings

Chapter 2: Is Matter Around Us Pure?

This chapter of NCERT Class 9 Science students will teach about mixtures, solutions, properties of solutions, separation of mixtures, and physical and chemical changes. Along with this, they will also learn about compounds and their properties, the difference between mixtures and compounds, the classification of matter, etc.

Topics Covered in Class 9 Science Chapter 2 Is Matter Around Us Pure?  :

Nature of matter: Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Physical and chemical changes (excluding separating the components of a mixture).

Also, access the following resources for Class 9 Chapter 2 Is Matter Around Us Pure? at BYJU’S:

• CBSE Class 9 Science Chapter 2 Is Matter Around Us Pure? Notes
• Important Questions for CBSE Class 9 Science Chapter 2 – Is Matter Around Us Pure?
• NCERT Exemplar Class 9 Science Solutions for Chapter 2 – Is Matter Around Us Pure?

Chapter 3: Atoms and Molecules

In NCERT Class 9 Science Chapter 3, students will learn the laws of chemical combination , atoms and molecules. They will get to know how to write a chemical formula, molecular mass and mole concepts and some numerical problems related to these concepts.

Topics Covered in Class 9 Science Chapter 3 Atoms and Molecules  :

Particle nature and their basic units: Atoms and molecules, Law of Chemical Combination, Chemical formula of common compounds, Atomic and molecular masses.

Also, access the following resources for Class 9 Chapter 3 Atoms and Molecules, at BYJU’S:

• Atoms And Molecules Class 9 Notes – Chapter 3
• Important Questions for Class 9 Science Chapter 3 – Atoms and Molecules
• NCERT Exemplar Class 9 Science Solutions for Chapter 3 – Atoms And Molecules

Chapter 4: Structure of the Atom

This chapter of NCERT Class 9 Science deals with the various atomic models of atoms that were proposed by different scientists. In addition to it, this chapter also covers electrons’ distribution in different orbits, calculation of valency , atomic number and mass number.

Topics Covered in Class 9 Science Chapter 4 Structure of the Atom  :

Structure of atoms: Electrons, protons and neutrons, Valency, Atomic Number and Mass Number, Isotopes and Isobars.

Also, access the following resources for Class 9 Chapter 4 Structure of the Atom at BYJU’S:

• Structure of the Atom Class 9 CBSE Notes – Chapter 4
• Important Questions of CBSE Class 9 Science Chapter 4 – Structure of the Atom
• NCERT Exemplar Class 9 Science Solutions for Chapter 4 – Structure of the Atom

Chapter 5: The Fundamental Unit of Life

This chapter is related to Biology. In this chapter, students will get to know that the cell is the fundamental unit of life. The whole chapter revolves around the cell and its structural organisation, in which students will learn about the plasma membrane , cell wall, nucleus, cytoplasm and structure of an animal cell.

Topics Covered in Class 9 Science Chapter 5 The Fundamental Unit of Life  :

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.

Also, access the following resources for Class 9 Chapter 5 The Fundamental Unit of Life at BYJU’S:

• The Fundamental Unit of Life Class 9 CBSE Notes – Chapter 5
• Revision Notes For Class 9 Science Chapter 5 The Fundamental Unit of Life
• NCERT Exemplar Class 9 Science Solutions for Chapter 5 – The Fundamental Unit of Life

Chapter 6: Tissues

This chapter pertains to the basic definition of tissue and then elaborates on Plant and Animal tissue with proper diagrams. Students will get to know the different types of plant and animal tissues with a detailed explanation of each.

Topics Covered in Class 9 Science Chapter 6 Tissues  :

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).

Also, access the following resources for Class 9 Chapter 6 Tissues at BYJU’S:

• Tissues Class 9 CBSE Notes – Chapter 6
• Revision Notes For Class 9 Science Chapter 6 Tissues
• NCERT Exemplar Class 9 Science Solutions for Chapter 6 – Tissues

Chapter 7: Diversity in Living Organism

This chapter deals with the classification of plants and animals. Students get to know that all living organisms are divided into 5 kingdoms namely Monera, Protista, Fungi, Plantae and Animalia. It also describes the classification and evolution, the hierarchy of classification.

Also access the following resources for Class 9 Chapter 7 Diversity in Living Organism at BYJU’S:

• Diversity in Living Organisms Class 9 CBSE Notes – Chapter 7
• Important Questions for CBSE Class 9 Science Chapter 7 – Diversity in Living Organisms
• NCERT Exemplar Class 9 Science Solutions for Chapter 7 – Diversity In Living Organisms

Chapter 8: Motion

NCERT Class 9 Science Chapter 8 describes motion, the speed with direction, rate of change of velocity , and graphical representation of motion. Also, students will find the 3 equations of motion and numerical problems related to them. Overall, this chapter will be a mix of theory as well as the numerical part.

Topics Covered in Class 9 Science Chapter 8 Motion :

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, elementary idea of uniform circular motion.

Also, access the following resources for Class 9 Chapter 8 Motion at BYJU’S:

• CBSE Class 9 Physics Motion Notes
• NCERT Exemplar Class 9 Science Solutions for Chapter 8 – Motion

Chapter 9: Force and Law of Motion

This chapter explains the 3 laws of motion with the help of diagrams and examples. Below are the 3 laws of motion:

• First law of motion: An object continues to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force.
• Second law of motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force .
• Third law of motion: To every action, there is an equal and opposite reaction, and they act on two different bodies.

Topics Covered in Class 9 Science Chapter 9 Force and Law of 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.

Also, access the following resources for Class 9 Chapter 9 Force and Law of Motion at BYJU’S:

• Force and Laws Of Motion Class 9 CBSE Notes – Chapter 9
• Revision Notes For Class 9 Science Chapter 9 Force and Laws of Motion
• NCERT Exemplar Class 9 Science Solutions for Chapter 9 – Force and Laws Of Motion

Chapter 10: Gravitation

In chapter 10 of NCERT Class 9 Science, students will learn the universal law of gravitation and its importance, free fall, mass, weight, thrust and pressure, Archimedes’ principle and relative density. Students will also find numerical problems related to these topics.

Topics Covered in Class 9 Science Chapter 10 Gravitation :

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.

Also, access the following resources for Class 9 Chapter 10 Gravitation at BYJU’S:

• Gravitation Class 9 CBSE Notes – Chapter 10
• Revision Notes For Class 9 Science Chapter 10 Gravitation
• NCERT Exemplar Class 9 Science Solutions for Chapter 10 – Gravitation

Chapter 11: Work and Energy

In this chapter, the concept of work is defined with different activities, numerical and examples. The chapter also deals with energy and its different forms. Examples are given to explain all types of energy. The chapter ends with the topic “rate of doing work”.

Topics Covered in Class 9 Science Chapter 11 Work and Energy  :

Work, Energy and Power: Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy (excluding commercial unit of Energy).

Also, access the following resources for Class 9 Chapter 11 Work and Energy at BYJU’S:

• Work And Energy Class 9 CBSE Notes – Chapter 11
• Revision Notes For Class 9 Science Chapter 11 Work and Energy
• NCERT Exemplar Class 9 Science Solutions for Chapter 11 – Work And Energy

Chapter 12: Sound

Chapter 12 of CBSE Class 9 Science, deals with concepts such as the production of sound, propagation of sound, the reflection of sound, range of hearing, applications of ultrasound, and the structure of the human ear . A few numericals are also there in the chapter which can only be solved after understanding the concepts. So, students must grasp and study the chapter carefully.

Topics Covered in Class 9 Science Chapter 12 Sound :

Sound: Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Also access the following resources for Class 9 Chapter 12 Sound at BYJU’S:

• CBSE Class 9 Science Chapter 12 Sound Notes
• Important Questions for CBSE Class 9 Science Chapter 12 – Sound
• NCERT Exemplar Class 9 Science Solutions for Chapter 12 – Sound

Chapter 13: Why Do We Fall Ill?

NCERT Class 9 Science chapter 13 deals with health issues and different types of diseases. It covers topics like; Health and its failure , disease and its cause, infectious diseases. This chapter is added to make students aware of different types of diseases so that they take care of their health and be fit and healthy.

Also access the following resources for Class 9 Chapter 13 Why Do We Fall Ill? at BYJU’S:

• Why Do We Fall Ill? Class 9 CBSE Notes – Chapter 13
• Important Questions for CBSE Class 9 Science Chapter 13 – Why Do We Fall Ill?
• NCERT Exemplar Class 9 Science Solutions for Chapter 13 – Why Do We Fall Ill?

Chapter 14: Natural Resources

We are blessed with natural resources. These are essential to meet the basic requirements of all forms of life on Earth. So, this chapter has been included by NCERT to provide knowledge on the types of resources available on the Earth. This chapter deals with the topics like air, water, soil, the biogeochemical cycle and the ozone layer, its importance and how humans are polluting them.

Also access the following resources for Class 9 Chapter 14 Natural Resources at BYJU’S:

• Natural Resources Class 9 CBSE Notes – Chapter 14
• Revision Notes For Class 9 Science Chapter 14 Natural Resources
• NCERT Exemplar Class 9 Science Solutions for Chapter 14 – Natural Resources

Chapter 15: Improvement in Food Resources

The last chapter of NCERT Class 9 Science covers topics like Improvement in crop yields, manure, fertilizer, storage of grains, and animal husbandry . This chapter provides knowledge regarding agriculture, farming and dairy.

Also access the following resources for Class 9 Chapter 15 Improvement in Food Resources at BYJU’S:

• Improvement in Food Resources Class 9 CBSE Notes – Chapter 15
• Important Questions for CBSE Class 9 Science Chapter 15 – Improvement in Food Resources
• Revision Notes For Class 9 Science Chapter 15 Improvement in Food Resources
• NCERT Exemplar Class 9 Science Solutions for Chapter 15 – Improvement In Food Resources

To download the complete book in PDF format, visit NCERT Science Book Class 9 . For more practice tests and mock tests, sign up on BYJU’S the learning app. Apart from Science, get NCERT Solutions for all the subjects. Students can download worksheets, assignments, NCERT Books , notes and study materials for exam preparation and for a better understanding of the topics.

CBSE Class 9 Science Evaluation Scheme (Theory) –

Features of ncert 9 th class science book solutions.

The book has a wide variety of features which are listed below:

• CBSE Class 9 Science NCERT Solutions are available to everyone for free.
• Covers all the exercise problems from the Class 9 textbook.
• Consists of extra questions, exemplar problems, the important questions from previous year question papers and sample papers, worksheets, MCQs, short answering questions, descriptive type questions, their solutions and tips and tricks.
• CBSE Class 9 Science Solutions files are available for download in PDF format for easy access.
• Diagrams are included to help students visualize the topics.
• Most effective solutions are given which can help to score well in the exams.

Frequently Asked Questions on NCERT Solutions for Class 9 Science

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