NCERT Exemplar Class 9 Science Chapter 2 Is Matter Around Us Pure are part of NCERT Exemplar Class 9 Science. Here we have given NCERT Exemplar Class 9 Science Solutions Chapter 2 Is Matter Around Us Pure.

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

Multiple Choice Questions

Question 1.
Which of the following statements are true for pure substances?
(i) Pure substances contain 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) Pure substances are made up of only one kind of particles and they have same composition throughout. Mixtures are not pure substances. Only elements and compounds are pure substances.

Question 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) Rusting of an article made up of iron is called corrosion. It is a chemical change because a new substance hydrated iron oxide called rust is formed.

Question 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) Sulphur is soluble in carbon disulphide hence, a solution is formed when sulphur is mixed with carbon disulphide. Solution is homogeneous and does not show Tyndall effect.

Question 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 dissolved in alcohol is known as tincture of iodine and has antiseptic properties.

Question 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) Ice and air are homogeneous in nature since they have same composition throughout and there are no visible boundaries between the components.

Question 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) Melting of iron metal, bending of an iron rod and drawing a wire of iron metal are physical changes since no new substances are formed during these changes. Only rusting of iron is a chemical change since a new substance rust is formed.

Question 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) Decaying of wood and burning of wood are chemical changes since there is a change in chemical composition of wood. Sawing of wood and hammering of a nail into a piece of wood are physical changes since there is no change in the composition of the wood during these changes.

Question 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) (iii) and (iv)
Solution:
(c)  2A + B → A₂B
The product A2B is a new compound formed hence, it does not show properties of A and B. The product formed is a compound and not an element.

Question 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) Pis a compound.
(ii) X and Y are compound.
(iii) X and Y are elements.
(iv) P has a fixed composition.
(a) (1), (ii) and (iii),
(b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)
(d) (i), (iii) and (iv)
Solution:
(d) X + Y→P
X and Y are elements hence, cannot be broken down into simpler substances. P is a compound hence it has fixed composition.

Short Answer Type Questions

Question 10.
Suggest separation technique(s) one would need to employ to separate the following mixtures.
(a) Mercury and water
(b) Potassium chloride and ammonium chloride
(c) Common salt, water and sand
(d) Kerosene oil, water and salt
Solution:
(a) Mercury and water – since there is a difference in densities of mercury and wate’ and both liquids are insoluble in each other hence they can be separated by separating funnel.
(b) Potassium chloride and ammonium chloride can be separated by sublimation. Ammonium chloride being volatile will be converted into vapours. KC1 does not sublime.
(c) Filtration will separate salt solution (salt and water) and sand. Evaporation of salt solution will separate salt from the solution.
(d) Separating funnel will separate kerosene oil and salt solution. Evaporation of salt solution will separate salt and water.

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

Solution:
The tube (a) containing beads will be more effective as a condenser because the surface area of the tube increases.

Question 12.
Salt can be recovered from its solution by evaporation. Suggest some other technique for the same?
Solution:
Salt solution can be concentrated by heating to make a supersaturated solution. Crystallisation will occur when the solution is left for cooling and salt will separate out from the solution.

Question 13.
The ‘sea-water’ can be classified as a homogeneous as well as heterogeneous mixture. Comment.
Solution:
Sea-water can be classified as a homogeneous mixture because it contains salts dissolved in water. It can be classified as a heterogeneous mixture also since it contains mud, sand and decayed parts of plants.

Question 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:
The mixture of acetone and salt solution in water can be separated by distillation since a difference in their boiling points is more than 25°C. Acetone will evaporate and get condensed first leaving behind the salt solution.

Question 15.
What would you observe when
(a) a saturated solution of potassium chloride prepared at 60°C is allowed to cool to room temperature
(b) an aqueous sugar solution is heated to dryness
(c) a mixture of iron filings and sulphur powder is heated strongly?
Solution:
(a) Crystals of potassium chloride will separate out.
(b) On heating sugar solution, water will evaporate first. Once the solution dries up, it would turn black and sugar will get charred.
(c) Iron sulphide is formed when a mixture of iron filings and sulphur is heated strongly.

Question 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:
The size of colloidal particles in a colloidal solution is smaller than suspension. These particles are in a random motion hence do not settle down when left undisturbed. The particles of suspension are bigger and they tend to settle down under the effect of gravity.

Question 17.
Smoke and fog both are aerosols. In what way are they different?
Solution:
In smoke, the dispersed phase is solid and the dispersion medium is gas. In fog, the dispersed phase is liquid and the dispersion medium is gas.

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

Question 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:
By definition, 50% mass by volume percent solution means 50 grams of a solute dissolved in 100 mL of solution. Therefore, student C made the desired solution. Student A dissolved 50 g of NaOH in 100 ml of water, So the solution is diluted and it is not a desired solution. By definition, 50% mass by mass percent solution means 50 grams of a solute dissolved in 100 grams of solution.
Student B dissolved 50 g of NaOH in 150 g of solution so, it is not the desired solution.
‘ C’ has made the desired solution by dissolving 50 g NaOH in water to make the volume of the solution 100 mL

Question 20.
Name the process associated with the following:
(a) Dry ice is kept at room temperature and at one atmospheric pressure.
(b) A drop of ink placed on the surface of water contained in a glass spreads throughout the water.
(c) A potassium permanganate crystal is in a beaker and water is poured into the beaker with stirring.
(d) An acetone bottle is left open and the bottle becomes empty.
(e) Milk is churned to separate cream from it.
(f) Settling of sand when a mixture of sand and water is left undisturbed for some time.
(g) Fine beam of light entering through a small hole in a dark room, illuminates the particles in its paths.
Solution:
(a) Sublimation of dry ice (solid) to C02 (gas)
(b) Diffusion of ink into water
(c) Diffusion or dissolution of solid into liquid
(d) Evaporation, diffusion of acetone in air
(e) Centrifugation
(f) Sedimentation
(g) Tyndall effect – Scattering of light

Question 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’ which boils at 102°C contains impurities. It will not freeze at 0°C. There will be a depression in freezing point.

Question 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 is highly malleable and soft. When it is alloyed with copper or silver it becomes hard and strong and can be moulded into various shapes.

Question 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:
It is a metal. The element is expected to be lustrous, malleable and good conductor of heat and electricity.

Question 24.
Give an example each for the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures.
(a) A volatile and a non-volatile component.
(b) Two volatile components with appreciable difference in boiling points.
(c) Two immiscible liquids.
(d) One of the components changes directly from solid to gaseous state.
(e) Two or more coloured constituents soluble in some solvent.
Solution:
(a) Evaporation or distillation – Common salt solution
(b) Distillation – Acetone – water mixture
(c) Separation using separating funnel – Oil – water mixture
(d) Sublimation – Mixture of common salt and ammonium chloride
(e) Chromatography – Ink

Question 25.
Fill in the blanks.
(a) A colloid is a _______ mixture and its components can be separated by the technique known as _______.
(b) Ice, water and water vapour look different and display different _______ properties but they are _______ the same.
(c) A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for some time. The upper layer in the separating funnel will be of _______ and the lower layer will be that of _______.
(d) A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25 K can be separated by the process called _______.
(e) When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the _______ of light by milk and the phenomenon is called _______. This indicates that milk is a _______ solution.
solution:
(a) heterogeneous; centrifugation
(b) physical, chemically
(c) water, chloroform
(d) fractional distillation
(e) scattering, Tyndall effect, colloidal

Question 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:
The composition of the sugar (sucrose) will remain constant irrespective of the source of its preparation. Hence sugar or sucrose is a. pure substance with fixed composition.

Question 27.
Give some examples of Tyndall effect observed in your surroundings?
solution:
Examples of Tyndall effect:
(i) When light rays enter into a dark room through a hole or a small window.
(ii) Sunlight passing through a group of trees in the forest.
(iii) Path of light rays seen in front of the projector in a cinema hall.

Question 28.
Can we separate alcohol dissolved in water by using a separating funnel?
If yes, then describe the procedure. If not, explain.
solution:
No, mixture of water and alcohol cannot be separated since both are miscible and they form a solution. Only immiscible liquids can

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

(a) It is a chemical change in which new substances are formed.
(b) Calcium oxide when dissolved in water, forms a basic solution.
CaO + H₂O → Ca(OH)₂
Carbon dioxide when dissolved in water, forms an acidic solution
CO₂ + H₂O →  H₂CO₃

Question 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 a non-metal other than carbon which shows 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

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

solution:
Elements – Cu, Zn, F₂,O₂,diamond (C), Hg
Compounds – CaCO₃, H₂O,

Question 32.
Which of the following are not compounds?
(a) Chlorine gas
(b) Potassium chloride
(c) Iron
(d) Iron sulphide
(e) Aluminium
(f) Iodine
(g) Carbon
(h) Carbon monoxide
(i) Sulphur powder
solution:
Chlorine gas, iron, aluminium, iodine, carbon and sulphur powder are not compounds.

Long Answer Type Questions

Question 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:
In fractional distillation, a fractionating column is used which is packed with glass beads or small plates. It increases the surface area for the vapours and they quickly loose energy when they come in contact with beads or plates and can be quickly condensed. The length of the column would increase the efficiency of the process.

Question 34.
(a) Under which category of mixtures will you classify alloys and why?
(b) A solution is always a liquid. Comment.
(c) Can a solution be heterogeneous?
solution:
(a) Alloys are homogeneous mixtures because they have uniform composition throughout.
(b) No, a solution can be solid (alloys) or gaseous (air) also.
(c) No, a solution is a homogeneous mixture.

Question 35.
Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘6’. Part ‘A’ was heated strongly while Part ‘S’ 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 – Iron sulphide is formed which gives out hydrogen sulphide gas with HCI.

H₂S gas can be identified by its smell. It has a foul smell and it turns lead acetate solution black.
Part B – Fe and S will not react, when HC1 is added to this mixture, only Fe will react with HCI to give out H₂ gas.
Fe + 2HC1 > FeCl₂+ H₂
Hydrogen gas burns with a pop sound hence, can be identified by bringing a burning matchstick near it.

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

(i) What would you expect to see, if the ink contains three different coloured components?
(ii) Name the technique used by the child.
(iii) Suggest one more application of this technique.
solution:
(i) The components of the ink will travel with water and we would see three bands on the filter paper at various lengths.

(ii) The technique is called chromatography.
(iii) Separation of pigments present in chlorophyll.

Question 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. 2.4. They were amazed to see that milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it?

(a) Explain why the milk sample was illuminated. Name the phenomenon involved.
(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 solution hence shows Tyndall effect.
(b) True solutions do not show Tyndall effect because they do not scatter light.
(c) Detergent solution, sulphur solution.

Question 38.
Classify each of the following, as a physical or a chemical change. Give reasons.
(a) Drying of a shirt in the sun.
(b) Rising of hot air over a radiator.
(c) Burning of kerosene in a lantern.
(d) Change in the colour of black tea on adding lemon juice to it.
(e) Churning of milk cream to get butter.
solution:
(a, b, e) : Physical changes because there is no change in chemical composition, (c), (d) : Chemical changes because new substances are formed.

Question 39.
During an experiment the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution.
(a) Are the two solutions of the same concentration?
(b) Compare the mass % of the two solutions.
solution:
(a) No, Sarika has higher mass percentage.

Question 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:
Maim Mixture – Sand + Fe + NH₄Cl + NaCl
Step I – Separate iron filings using magnet.
Step II – Separate NH₄Cl by sublimation.
Step III – Add,water, stir and filter to separate sand.
Step IV – Evaporate filtrate to get salt (NaCl).

Question 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?
(a) 1.00 g of NaCl + 100 g of water
(b) 0.11 g of NaCl + 100 g of water
(c) 0.01 g of NaCl + 99.99 g of water
(d) 0.10 g of NaCl + 99.90 g of water
solution:
(c)

Question 42.
Calculate the mass of sodium sulphate required to prepare its 20% (mass percent) solution in 100 g of water?
solution:
Let the mass of sodium sulphate required be = x g
The mass of solution would be = (x + 100) g x g of solute in (x + 100) g of solution