# NCERT Solutions For Class 9 Science Chapter 2

## NCERT Solutions Class 9 Science Is Matter Around Us Pure

### Ncert Solutions For Class 9 Science Chapter 2 PDF Free Download

NCERT solutions for class 9 Science Is Matter Around Us Pure is an important study material for the students studying in class 9. In class 9 Science, there are many important topics that need to be learned in an effective way so as to score well in the examination. Along with this, these topics are also important for the competitive examination, thus it is important that one should solve NCERT Solution class 9 Science Chapter 2 Is Matter Around Us Pure Questions provided at the end of each chapter. Sometimes questions from NCERT are directly asked in the class 9 examination.

The detailed NCERT solutions for class 9 Science chapter 2 Is Matter Around Us Pure PDF are provided here so that students can refer to these solutions in offline mode for better understanding and clarification. NCERT class 9 science solutions for chapter 2 comprises of a wide range of topics like –

• Introduction
• What is mixture and its types
• Solution definition and its properties
• Concentration of solution
• Definition of suspension and its properties
• Meaning of colloidal solution and its properties
• Separation of components of mixtures
• Acquiring colored components form – black /blue ink
• Separation of cream from milk
• Separating 2 immiscible liquids
• Separating colors (dyes) in black ink
• Separating a mixture of 2 miscible liquids
• Acquiring different gases from the air
• Chemical and physical changes
• Types of pure substances
• Difference between compounds and mixtures

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

1) What does a pure substance mean?

Solution:

A pure substance is one, which contains only one type of atoms or molecules in a specific arrangement in any part of the sample taken. Example: water, diamond.

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

Solution:

 Homogeneous mixture Heterogeneous mixture Components are uniformly distributed throughout the mixture All the components are completely mixed and can be identified with the naked eye or under a microscope. No visible boundaries of separation Visible boundaries of separation. Has a uniform composition Has a non-uniform composition Example: rainwater, vinegar etc Example: seawater, pizza etc

3) How are sol, solution and suspension different from each other?

Solution:

 Property Sol Solution Suspension Nature Heterogeneous Homogeneous Heterogeneous Particle size 10-7 – 10-5cm Less than 1nm More than 100nm Stability Quite stable Very stable unstable Tyndall effect yes no yes/no Appearance Generally clear clear opaque Visibility Visible with an ultra microscope Not visible Visible with naked eye Diffusion Diffuses very slowly Diffuses rapidly Do not diffuse Settling Get settled in centrifugation Do not settle Settle on their own Example Milk, blood, smoke Salt and sugar in water Sand in water, dusty air

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

Solution:

Mass of solute (NaCl) = 36 g

Mass of solvent (H2O) = 100 g

Mass of solution (NaCl + H2O) = 136 g

Concentration = Mass of solute

Mass of solution

Concentration = 36 *100

136

= 26.47%

Hence, the concentration of the solution is 26.47%

5) How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25oC) which are miscible with each other?

Solution:

The mixture of miscible liquids whose boiling point difference is more than 25oC such as kerosene and petrol can be separated by a technique called simple distillation. The principle of separation is based on the volatility of the substances. The process of distillation is as follows:

(a) Take the mixture in a distillation flask.

(b) Fit it with a thermometer and heat the mixture.

(c) Petrol has a lower boiling point and evaporates first.

(d) As the vapors rise up and reach the condenser, the temperature is decreased and the vapor is condensed into liquid and is collected in a flask.

(e) The kerosene that has relatively higher boiling point remains in the flask in the liquid form.

(f) Hence, the liquids are separated.

6) Name the techniques used to separate the following:

(a) Butter from curd.

(b) Salt from seawater

(c) Camphor from salt

Solution:

(a) The butter is separated from the curd based on its density through the process of centrifugation.

(b) Simple evaporation technique is used in the separation of salt from seawater. Commercially, this process is producing salt.

(c) Camphor does not undergo liquid phase during the phase change. Therefore, the sublimation process is more suited for the separation of camphor from the other substance.

7) What type of mixtures are separated by the technique of crystallization?

Solution:

Crystallization is a technique of separation of solid from a liquid solution. It relates to precipitation but here, the precipitate obtained is in crystal form and has a very high purity. This technique is thus used in the purification of impure substances. Example: Salt from sea water after the evaporation.

8) Classify the following as physical or chemical change:

• 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
• Burning of paper and wood

Solution:

 Physical change Chemical change ●     Cutting the trees ●     Melting of butter in a pan ●     Boiling of water to form steam ●     Dissolving common salt in water ●     Making fruit salad with raw fruits ●     Rusting of almirah ●     Passing of electric current through water, and water breaking into hydrogen and oxygen gases ●     Burning of paper and wood

9) Try segregating the things around you as pure substances and mixtures.

Solution:

• Pure substance: water, salt, iron, diamond
• Mixture: sand, salad, concrete, air, steel

Exercises:

1) Which separation technique will you use 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)The process of Evaporation commercially does the sodium chloride separation from its solution in water.

(b) Ammonium chloride undergoes Sublimation. Hence, this technique is preferred.

(c) The metal scraps can be filtered manually.

(d) The delicate separation of different pigments from an extract of flower petal is done with the help of Chromatography.

(e)  The butter from milk is separated with the principle of difference in its density. Hence the Centrifugation technique is used.

(f) Oil and water are two immiscible liquids with a difference in their densities. Separation using Separating funnel is an effective way.

(g) The leaves can be manually separated by simple Filtration.

(h) The iron pins possess a magnetic quality which is the key character required for Magnetic separation.

(i) The wheat and husk vary in their mass thus with a little wind energy their moving distance varies. Winnowing/Sedimentation techniques are promoted in their separation.

(j) The mud particles being heavier settles at the bottom of the container when left undisturbed. The clear water is then separated by tilting it out. This process is known as Decantation/Sedimentation.

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

Solution:

(a) Take a cup of milk in a vessel that acts as a solvent and heat it.

(b) Drop in the tealeaves or the powdered tealeaves into the milk as solute and continue heating.

(c) The tealeaves or the powdered tea leaves used is insoluble in the milk and is visible even after the heating.

(d) Now, to the boiling solution, add sugar and stir it.

(e) The sugar acts yet another solute, but in this case, it is soluble in the solvent.

(f) Due to continued stirring, the sugar completely becomes soluble in the tea solution and a saturation level is reached.

(g) After enough heating, filter the solution using a medium. When done, the insoluble tea leaves stay behind as residue and the soluble essence and sugar passes through the filter medium and is collected as the filtrate.

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

 Substance dissolved Temperature in Kelvin and solubility 283 293 313 333 353 Potassium nitrate 21 32 62 106 167 Sodium chloride 36 36 36 37 37 Potassium chloride 35 35 40 46 54 Ammonium chloride 24 37 41 55 66

(a) What mass of potassium nitrate would be needed to produce a saturated solution of Potassium nitrate in 50 grams of water at 313K?

(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 down? 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?

Solution:

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 = ?

Required amount =$\frac{62 * 50}{100}=31$

Hence 31 g of potassium nitrate is required.

(b) When the solution cools down, Pragya observes salt crystals of potassium chloride precipitating.

(c) Solubility of each salt at 293 K is as follows:

• Potassium nitrate —-> 21 g
• Sodium chloride —-> 36 g
• Potassium chloride —-> 35 g
• Ammonium chloride —-> 24 g

It is observed that the potassium chloride salt has the highest amount of solubility when compared to any other salt at 293 K.

(d) Effect of change of temperature in the solubility of salts:

It is observed from the given table that the solubility of the salt increases with the increase in temperature. This means that when a salt reaches its saturation point at a particular temperature, more salt can be dissolved yet by increasing the temperature of the solution.

4) Explain the following by giving examples:

(a)        Saturated solution

(b)       Pure substance

(c)        Colloid

(d)       suspension

Solution:

(a) Saturated solution: At a given temperature, in a solution, if the solvent is no more soluble without increasing the temperature, then that state of solution is called the saturated solution.

(b) Pure substance: In a substance, if only one type of atoms or molecules or compounds is present with no contamination of other substance or any deviation of the arrangement, then the substance is known to be a pure substance.

(c) Colloid: A colloid, is a mixture in which one substance of microscopically dispersed insoluble particles is suspended throughout another substance. A colloid is a solution in which the size of solute particles are bigger than that of a true solution. These particles cannot be seen with naked eye, they are stable. Eg: ink, blood.

(d) Suspension:  A suspension is a heterogeneous mixture containing solid particles that are sufficiently large for sedimentation. Usually, they must be larger than one micrometer. A suspension is a heterogeneous mixture in which the solute particles do not dissolve but get suspended throughout the bulk of the medium. Particles of suspension are visible to the naked eye. It is called a suspenion when particles are left floating around freely in a solvent.

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

Solution:

Homogeneous: soda water, vinegar, filtered tea.

Heterogeneous: wood, air, soil.

6) How would you confirm that a colorless liquid given to you is pure water?

Solution:

Heat the given liquid to 100oC. If the liquid starts boiling, then it is confirmed that it is water. Else it is not water.

7) Which of the following materials fall into the category of pure substances?Ice, milk, iron, hydrochloric acid, calcium oxide, mercury, brick, wood, air.

Solution:

The pure substances from the above-mentioned materials are ice, iron, hydrochloric acid, calcium oxide and mercury.

8) Identify the solutions among the following mixtures.

(a) Soil

(b) Seawater

(c) Air

(d) Coal

(e) Soda water

Solution:

The solutions from the given mixtures are sea water, air and soda water.

9) Which of the following will show “Tyndall effect”?

(a) Salt solution

(b) Milk

(c) Copper sulphate solution

(d) Starch solution

Solution:

From the above list, only milk and the starch solution shows Tyndall effect.

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.

Solution:

 Elements Compounds Mixture Sodium Calcium carbonate Soil Silver Soap Sugar solution Tin Methane Coal Silicon Carbon dioxide Air, Blood

NCERT solutions for class 9 Science Is Matter Around Us Pure, briefly explains about all the concepts. Like –

• Introduction to the topic and other concepts of the matter. Like – matter is made up of 1 or more components. Collectively they are known as ‘substances’.
• A brief definition of the substance. Types and properties of substances and mixtures.
• The conclusion of this chapter comprises of 3 segments. They are – what have you learnt?, exercises and group activities.

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#### Practise This Question

A dehydration reaction is a reaction in which there is