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NCERT Exemplar Class 9 Science Chapter 2 Is Matter around Us Pure Solutions

NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter around Us Pure covers all the important questions and answers as well as advanced level questions. It helps in learning about the types of mixtures, homogeneous mixtures, heterogeneous mixtures, solution, suspension, concentration of solution, types of solution, Separating the Components of Mixtures, separation methods, miscible liquids, immiscible liquids and centrifugation.

The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter around Us Pure is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like physical and chemical changes, difference between the pure substances and a mixture, types of Pure substances, elements and compounds.


Chapter Name

Chapter 2 Is Matter around Us Pure

Book Title

NCERT Exemplar for Class 9 Science

Related Study

  • NCERT Solutions for Class 9 Science Chapter 2 Is Matter around Us Pure
  • Revision Notes for Class 9 Science Chapter 2 Is Matter around Us Pure
  • MCQ for Class 9 Science Chapter 2 Is Matter around Us Pure
  • Important Questions for Class 9 Science Chapter 2 Is Matter around Us Pure

Topics Covered

  • MCQ
  • Short Answers Questions
  • Long Answers Questions

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

Multiple Choice Questions

1. Which of the following statements are true for pure substances?
(i) Pure substances contains only one kind of particles.
(ii) Pure substances may be compounds or mixtures.
(iii) Pure substances have the same composition throughout.
(iv) Pure substances can be exemplified by all elements other than nickel.
(a) (i) and (ii)
(b) (i) and (iii)
(c) (iii) and (iv)
(d) (ii) and (iii)

Solution

(b) (i) and (iii)


2. Rusting of an article made up of iron is called :
(a) corrosion and it is a physical as well as chemical change.
(b) dissolution and it is a physical change.
(c) corrosion and it is a chemical change.
(d) dissolution and it is a chemical change.

Solution

(c) corrosion and it is a chemical change.
Rusting of iron is also known as corrosion and it is a chemical change because rust is a chemical compound called hydrated iron oxide Fe2O3. nH2O, iron(III) which is different from elemental iron.


3. A mixture of sulphur and carbon disulphide is :
(a) heterogeneous and shows Tyndall effect.
(b) homogeneous and shows Tyndall effect.
(c) heterogeneous and does not show Tyndall effect.
(d) homogeneous and does not show Tyndall effect.

Solution

(d) homogeneous and does not show Tyndall effect.
A mixture of sulphur and carbon disulphide is heterogeneous in nature and it shows Tyndall effect. This happens because the particles are big enough to scatter light.


4. Tincture of iodine has antiseptic properties. This solution is made by dissolving :
(a) Iodine in potassium iodide
(b) Iodine in vaseline
(c) Iodine in water
(d) Iodine in alcohol

Solution

(d) Iodine in alcohol.
Tincture is prepared by using2-7% elemental iodine, either potassium iodide or sodium dissolved in alcohol. Alcohol is a good solvent and iodine does not dissolve in water, so alcohol should be used.


5. Which of the following are homogeneous in nature?
(i) Ice
(ii) Wood
(iii) Soil
(iv) Air
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (i) and (iv)
(d) (iii) and (iv)

Solution

(c) (i) and (iv)
Air and ice are homogeneous mixture as their constituents are not visible and cannot be distinguished from one another.

6. Which of the following are physical changes?
(i) Melting of iron metal
(ii) Rusting of iron
(iii) Bending of an iron rod
(iv) Drawing a wire of iron metal
(a) (i), (ii) and (iii)
(b) (i), (ii) and (iv)
(c) (i), (iii) and (iv)
(d) (ii), (iii) and (iv)

Solution

(c) (i), (iii) and (iv)
During the process of rusting of iron, iron reacts with water and oxygen to produce iron oxide, hence it is a chemical change.


7. Which of the following are chemical changes?
(i) Decaying of wood
(ii) Burning of wood
(iii) Sawing of wood
(iv) Hammering of a nail into a piece of wood
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)

Solution

(a) (i) and (ii)
Decaying of wood and burning of wood are chemical changes as there will be a change of chemical composition and wood cannot be restored to its original form.


8. Two substances, A and B were made to react to form a third substance, A2B according to the following reaction
2A + B 
→ A2B
Which of the following statements concerning this reaction are incorrect?
(i) The product A2B shows the properties of substances A and B.
(ii) The product will always have a fixed composition.
(iii) The product so formed cannot be classified as a compound.
(iv) The product so formed is an element.
(a) (i), (ii) and (iii),
(b) (ii), (iii) and (iv)
(c) (i), (iii) and (iv)
(d) (ii), (iii) and (iv)

Solution

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

A2B is a compound made up of two elements A and B in a fixed ratio. The properties of a compound (e.g., A2B) are entirely different from those of its constituent elements (i.e. A and B). The composition of a compound is fixed.


9. Two chemical species X and Y combine together to form a product P which contains both X and Y
X + Y 
→ P
X and Y cannot be broken down into simpler substances by simple chemical reactions. Which of the following concerning the species X, Y and P are correct?
(i) P is a compound.
(ii) X and Y are compounds.
(iii) X and Y are elements.
(iv) P has a fixed composition.
(a) (i), (ii) and (iii)
(b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)
(d) (i), (iii) and (iv)

Solution

(d) (i), (iii) and (iv)
As X and Y cannot be further broken down into simpler substances, they are elements. P can be broken down to simpler substances hence it is a compound with a fixed composition.


Short Answer Questions

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

(i) Mercury and water
(ii) Potassium chloride and ammonium chloride
(iii) Common salt, water and sand
(iv) Kerosene oil, water and salt

Solution

(i) Separation by using separating funnel (used for separating two immiscible liquids). The principle is that immiscible liquids separate out in layers depending on their densities.
(ii) Sublimation : It is a process by which solid changes directly into gas and vice versa without passing
through the liquid state. Ammonium chloride is sublime.
(iii) Filtration followed by evaporation or centrifugation followed by evaporation/distillation.
(iv) Separation by using separating funnel to separate kerosene oil followed by evaporation or distillation.


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

Solution

Salt can be recovered from its solution by crystallization. Crystallization removes soluble impurities which are not removed in case of evaporation. Hence it is a more efficient process.


12. Which of the tubes in figure given here (a) and (b) will be more effective as a condenser in the distillation apparatus?

Solution

Beads in tube A increase the surface area for the cooling of vapours that come in contact with them. Hence, tube A is a good condenser.


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

Solution

The sea water on the surface is a homogenous mixture as it comprises of water and salts. Whereas the sea water from deep sea consists of salts, water, mud, decayed plants etc. hence it will be a heterogeneous mixture.


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

Solution

Fractional distillation can be used to separate acetone from the mixture of salt and water. There is a considerable difference in the boiling points of acetone (56°C) and water (100°C). Acetone evaporates first when the solution is heated. The water is collected in the distillation flask. The vapours of acetone are then condensed to obtain acetone.


15. What would you observe when :
(i) A saturated solution of potassium chloride prepared at 60°C is allowed to cool at room temperature?
(ii) An aqueous sugar solution is heated to dryness?
(iii) A mixture of iron filings and sulphur powder is heated strongly?

Solution 

(a) Solid potassium chloride will separate out.

(b) The water will evaporate initially and then sugar will get charred.

(c) Iron sulphide will be formed.


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

Solution

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


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

Solution

Gas forms the dispersion medium for both fog and smoke. The only difference is that the dispersed phase in fog is liquid while in case of smoke it is a solid.


18. Classify the following as physical or chemical properties :
(i) The composition of a sample of steel is : 98% iron, 1.5% carbon and 0.5% other elements.
(ii) Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.
(iii) Metallic sodium is soft enough to be cut with a knife.
(iv) Most metal oxides form alkalis on interacting with water.

Solution

(i) Composition of a sample of steel is 98% iron, 1.5% carbon and 0.5% other elements. It is a chemical property as no new compound is formed. Steel is an alloy.

(ii) Zinc reacts with HCl to give out zinc chloride and hydrogen gas hence it is chemical property.

(iii) Cutting metallic sodium with knife will not involve any chemical reaction. Also, no new compounds are formed hence it is a physical property.

(iv) A new compound is form by the interaction of metal oxides with alkalis hence it is chemical property.


19. The teacher instructed three students ‘A’, ‘B’ and ‘C’ respectively to prepare a 50%. (mass by volume)
solution of sodium hydroxide (NaOH). ‘A’ dissolved 50 g of NaOH in 100 ml of water, ‘B’ dissolved 50 g
of NaOH in 100 g of water while ‘C’ dissolved 50 g of NaOH in water to make 100 ml of solution. Which one of them has made the desired solution and why?

Solution

Student ‘C’ has made the desired solution.


20. Name the process associated with the following:
(i) Dry ice is kept at room temperature and at one atmospheric pressure.
(ii) A drop of ink placed on the surface of water contained in a glass spreads throughout the water.
(iii) A potassium permanganate crystal is in a beaker and water is poured into the beaker with stirring.
(iv) A acetone bottle is left open and the bottle becomes empty.
(v) Milk is churned to separate cream from it.
(vi) Settling of sand when a mixture of sand and water is left undisturbed for some time.
(vii) Fine beam of light entering through a small hole in a dark room, illuminates the particles in its paths.

Solution

(i) Sublimation
(ii) Diffusion
(iii) Dissolution/diffusion
(iv) Evaporation, diffusion
(v) Centrifugation
(vi) Sedimentation
(vii) Tyndall effect (Scattering of light)


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

Solution

Sample ‘B’ is not pure water and hence it will not freeze at 0°C. At a pressure of 1 atm, the boiling point of pure water is 100°C and the freezing point of pure water is 0°C.


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

Solution

Pure gold (24 karats) is very soft and it does not have enough strength. Silver and copper are alloyed with gold, to give it strength. 24 karat gold is an alloy that has 20 parts of gold and 4 parts of silver.


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

Solution

The element is a metal. Other characteristics possessed by the element are:

  • Lustre
  • Malleability
  • Heat
  • Electrical conductivity


24. Give an example each for the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures.
(i) A volatile and a non-volatile component.
(ii) Two volatile components with appreciable difference in boiling points.
(iii) Two immiscible liquids.
(iv) One of the components changes directly from solid to gaseous state.
(v) Two or more coloured constituents soluble in some solvent.

Solution

(i) Mixture of acetone and water can be separated by distillation.

(ii) Mixture of petrol and kerosene can be separated by distillation.

(iii) Mixture of oil and water can be separated by fractional distillation.

(i)v Separating naphthalene by filtration and then separation of ammonium chloride from water by evaporation.

(v) Mixture of pigments from a flower petal extract can be separated by chromatography.


25. Fill in the blanks
(i) A colloid is a _____ mixture and its components can be separated by the technique known as_____
(ii) Ice, water and water vapour look different and display different _____ properties but they are_____ the same.
(iii) A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for some time. The upper layer in the separating funnel will be of _____ and the lower layer will be that of _____
(iv) A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25K can be separated by the process called _____
(v) When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the _____ of light by milk and the phenomenon is called _____ This indicates that milk is a _____ solution.

Solution

(i) A colloid is a heterogeneous mixture and its components can be separated by the technique known as centrifugation.

(ii) Ice, water and water vapour look different and display different physical properties but they are chemically the same. 

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

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

(v) When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the scattering of light by milk and the phenomenonis called Tyndall effect. This indicates that milk is a colloidal solution.


26. Sucrose (sugar) crystals obtained from sugarcane and beetroot are mixed together. Will it be a pure substance or a mixture? Give reasons for the same.

Solution

As the chemical composition of sugar crystals remains the same irrespective of its source, the mixture is a pure substance.


27. Give some examples of Tyndall effect observed in your surroundings.

Solution

When light passes through a heterogeneous mixture, Tyndall effect can be seen. Example, when sunlight passes through the canopy of a dense forest.


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

Solution

Alcohol dissolved in water cannot be separated by separating funnel because both are miscible solvents.


29. On heating calcium carbonate gets converted into calcium oxide and carbon dioxide.
(i) Is this a physical or a chemical change?
(ii) Can you prepare one acidic and one basic solution by using the products formed in the above process?
If so, write the chemical equation involved.

Solution

(i) It is a chemical change Calcium carbonate → Calcium oxide + Carbon dioxide
CaCO3 → CaO + CO2
(ii) Acidic and basic solutions can be prepared by dissolving the products of the above process in water
CaO + H2O → Ca(OH)2 (basic solution)
CO2 + H2O → H2CO3(acidic solution)


30. Non-metals are usually poor conductors of heat and electricity. They are non-lustrous, non-sonorous, non-malleable and are coloured.
(a) Name a lustrous non-metal.
(b) Name a non-metal which exists as a liquid at room temperature.
(c) The allotropic form of a non-metal is a good conductor of electricity. Name the allotrope.
(d) Name a non-metal which is known to form the largest number of compounds.
(e) Name non-metals other than carbon which show allotropy.
(f) Name a non-metal which is required for combustion.

Solution

(a) Iodine
(b) Bromine
(c) Graphite
(d) Carbon
(e) Sulphur, Phosphorus
(f) Oxygen


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

Cu

Sand

H2O

CaCO3

O3

Zn

NaCl(aq)

F2

Hg

Diamond(C)

Wood

Solution 

Elements

Compounds

Cu

CaCO3

Zn

H2O

F2

O2

Diamond (c)

Hg


32. Which of the following are not compound?
(a) Chlorine gas
(b) Potassium chloride
(c) Iron
(d) Iron sulphate
(e) Aluminium
(f) Iodine
(g) Carbon
(h) Carbon monoxide
(i) Sulphur powder

Solution

Chlorine gas, iron, aluminium, iodine, carbon and sulphur powder are not compounds.


Long Answer Questions

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

    Solution

    The most important part of fractional distillation apparatus is the fractionating column. It has glass beads in it. This prevents the upward movement of the vapours of the two liquids. The vapours of the liquid having high boiling point get condensed earlier at the lower levels of the column. The latent heat released helps to take the vapours of low boiling liquid to a greater height in the fractionating column.

    The advantages of fractional distillation are as follows:
    1. The liquids with a boiling point difference of about or less than 25 K can be separated by this method.
    2. Both evaporation and condensation take place simultaneously during the process.
    3. A mixture (like petroleum) which contains several components can also be separated by this process.

    34. (i) Under which category of mixtures will you classify alloys and why?
    (ii) A solution is always a liquid. Comment.
    (iii) Can a solution be heterogeneous?

    Solution

    (i) An alloy is a homogenous mixture of two or more elements. The constituent elements of an alloy can be two metals or a metal with a non-metal. An alloy is classified as a homogenous mixture as it exhibits the properties of its constituent elements.

    Examples: Brass is an alloy which shows characteristics of copper and Zinc and their composition vary from 20 to 35 %.

    (ii) The solutions are usually liquids, but not always. Solutions can also include a mixture of two solids like in alloys, and a mixture of gases, such as air.

    (iii) Yes, solutions can be heterogeneous also.


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

    Solution

    Part A
    When iron fillings and sulphur is heated it will give following reaction

    Part B

    Fe (s) + S (s)→ Mixture of iron filings and sulphur
    When dilute HCl is added to it
    Fe (s) + S (s) + 2 HCl (aq) → FeCl2+ H2 gas
    Sulphur remains unreacted.H2S gas formed has a foul smell and on passing through lead acetate solution, it turns the solution black. Hydrogen gas burns with a pop sound.


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

    (i) What would you expect to see, if the ink contains three different coloured components?
    (ii) Name the technique used by the child.
    (iii) Suggest one more application of this technique.

    Solution

    (i) Three different bands are observed on the paper if the ink contains three different coloured components.

    (ii) Child uses the technique of paper chromatography.

    (iii) The different pigments present in the chlorophyll can be separated by paper chromatography.


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

    (a) Explain why the milk sample was illuminated. Name the phenomenon involved.
    (b) Same results were not observed with a salt solution. Explain.
    (c) Can you suggest two more solutions which would show the same effect as shown by the milk solution?

    Solution

    (a) Milk is a colloidal substance. The particulate matter present in the milk scatter the light scatter thereby resulting in Tyndall effect. Because of this, the milk gets illuminated.

    (b) Salt solution is a homogenous solution. Small particles present in salt solution do not scatter light. This is the reason why there is no Tyndall effect in salt solution. The salt solution does not get illuminated as it does not exhibit the Tyndall effect.

    (c) Detergent solution and sulphur solution will exhibit the Tyndall effect.


    38. Classify each of the following, as a physical or a chemical change. Give reasons.
    (i) Drying of a shirt in the Sun.
    (ii) Rising of hot air over a radiator.
    (iii) Burning of kerosene in a lantern.
    (iv) Change in the colour of black tea on adding lemon juice to it.
    (v) Churning of milk cream to get butter.

    Solution

    (i) There is no chemical reaction or any chemical changed involved in drying a shirt in the sun, hence it is a physical phenomenon.

    (ii) In a radiator water is converted to vapours. Hot air being lighter rises upwards. Thus, rising of hot air over radiator is a physical change.

    (iii) Kerosene burns by using the atmospheric oxygen and produces carbon dioxide, this is the reason why burning of kerosene in a lantern is a chemical change.

    (iv) Lemon juice is a source of citric acid, ascorbic acid and malic acid. These acids react with the flavin antioxidants present in black tea and change the colour of the tea. Thus, the change in the colour of black tea on adding lemon juice is a chemical change.

    (v) There is no chemical reaction taking place when milk is churned to make butter. This is the reason why churning of milk cream to get butter is a physical change.


    39. During an experiment, the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution.
    (i) Are the two solutions of the same concentration?
    (ii) Compare the mass % of the two solutions.

      Solution 

      (i) The two solutions are not of the same concentration.

      (b) Solution prepared by Ramesh:


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

      Solution

      The procedures to be used to separate the constituents from the mixture are:

      • Step-1:With the help of a magnet separate iron filings.
      • Step-2:Sublimation of the remaining mixture separates ammonium chloride.
      • Step-3:Add water to the remaining mixture, stir and filter.
      • Step-4:The filtrate can be evaporated to get back sodium chloride.


      41. Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
      (i) 1.00 g of NaCl + 100 g of water
      (ii) 0.11 g of NaC1 + 100 g of water
      (iii) 0.01 g of NaC1 + 99.99 g of water
      (iv) 0.10 g of NaC1 + 99.90 g of water

      Solution

      Option (iii) correctly represents the composition of the solutions.


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

      Solution

      In a 20% solution containing 100 g water; the mass percentage of water = 100 — 20 = 80%
      80% of solution is 100 gram
      100% of solution is 100/80 gram
      20% of solution is 100/80 × 20 = 25 gram
      Hence, to prepare 20% (w/w) solution in 100 gram of water 25 gram of sodium sulphate is needed.

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