Important Questions for CBSE 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”?

a) Ice

b) Milk

c) Iron

d) Hydrochloric acid

e) Calcium oxide

f) Mercury

g) Brick

h) Wood

i) Air

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

5. Identify the solutions among the following mixtures.

a) Soil

b) Sea water

c) Air

d) Coal

e) Soda water.

Ans: Solutions: seawater, air, soda water.

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

a) Salt solution

b) Milk

c) Copper sulfate solution

d) Starch solution.

Ans: (b)Milk and (d)starch solution 

7. 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 

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

(c) both

(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

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

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 \[=\dfrac{mass\text{ }of\text{ }solute}{mass\text{ }of\text{ }solution}\times 100\]

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

$=26.47\%$

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.

Ans: Centrifugation method

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.

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

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

• Elements

• Compounds

• Metals, Non-metals, and Metalloids

• Mixture and its Types

• Solution and its Types

• Solubility

• 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

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:

1. Follow the concepts and study them properly. 

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

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

4. 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.

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