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Is Matter Around Us Pure

Class 9th Science NCERT Exemplar Solution
Multiple Choice Questions
  1. Which of the following statements are true for pure substances? (i) Pure substances…
  2. Rusting of an article made up of iron is calledA. corrosion and it is a physical as well…
  3. A mixture of sulphur and carbon disulphide isA. heterogeneous and shows Tyndall effect B.…
  4. Tincture of iodine has antiseptic properties. This solution is made by dissolvingA. iodine…
  5. Which of the following are homogeneous in nature? (i) ice (ii) wood (iii) soil (iv) airA.…
  6. Which of the following are physical changes? (i) Melting of iron metal (ii) Rusting of…
  7. Which of the following are chemical changes? (i) Decaying of wood (ii) Burning of wood…
  8. Two substances, A and B were made to react to form a third substance, A2B according to the…
  9. Two chemical species X and Y combine together to form a product P which contains both X…
Short Answer Type
  1. Suggest separation technique(s) one would need to employ to separate the following…
  2. Which of the tubes in Fig. 2.1 (a) and (b) will be more effective as a condenser in the…
  3. Salt can be recovered from its solution by evaporation. Suggest some other technique for…
  4. The sea-water can be classified as a homogeneous as well as a heterogeneous mixture.…
  5. While diluting a solution of salt in water, a student by mistake added acetone (boiling…
  6. What would you observe when (a) a saturated solution of potassium chloride prepared at 60C…
  7. Explain why particles of a colloidal solution do not settle down when left undisturbed,…
  8. Smoke and fog both are aerosols. In what way are they different?
  9. Classify the following as physical or chemical properties. (a) The composition of a sample…
  10. The teacher instructed three students A, B and C respectively to prepare a 50% (mass by…
  11. Name the process associated with the following. (a) Dry ice is kept at room temperature…
  12. You are given two samples of water labelled as An and B. Sample A boils at 100C and sample…
  13. What are the favourable qualities given to gold when it is alloyed with copper or silver…
  14. An element is sonorous and highly ductile. Under which category would you classify this…
  15. Give an example each of the mixture having the following characteristics. Suggest a…
  16. Fill in the blanks. (a) A colloid is a mixture and its components can be separated by the…
  17. Sucrose (sugar) crystals obtained from sugarcane and beetroot are mixed together. Will it…
  18. Give some examples of Tyndall effect observed in your surroundings?…
  19. Can we separate alcohol dissolved in water by using a separating funnel? If yes, then…
  20. On heating calcium carbonate gets converted into calcium oxide and carbon dioxide. (a) Is…
  21. Non-metals are usually poor conductors of heat and electricity. They are non-lustrous,…
  22. Classify the substances given in Fig. 2.2 into elements and compounds.…
  23. Which of the following are not compounds? (a) Chlorine gas (b) Potassium chloride (c) Iron…
Long Answer Type
  1. Fractional distillation is suitable for separation of miscible liquids with a boiling…
  2. Under which category of mixtures will you classify alloys and why?…
  3. A solution is always a liquid. Comment.
  4. Can a solution be heterogeneous?
  5. Iron filings and sulphur were mixed together and divided into two parts, A and B. Part A…
  6. A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a…
  7. A group of students took an old shoe box and covered it with a black paper from all sides.…
  8. Classify each of the following, as a physical or a chemical change. Give reasons. (a)…
  9. During an experiment, the students were asked to prepare a 10% (Mass/Mass) solution of…
  10. You are provided with a mixture containing sand, iron filings, ammonium chloride and…
  11. Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the…
  12. Calculate the mass of sodium sulphate required to prepare its 20% (mass percent) solution…

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)


Answer:

Pure substances contain only one kind of particles. Pure substances have the same composition throughout.
Substances that are made of only one type of atom or only one type of molecule are called pure substances. These substances have the same composition throughout. All elements and compounds are examples of pure substances. Tin, diamond, water, sugar and table salt 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.


Answer:

Rusting of an article made up of iron is called corrosion and it is a chemical change.


Corrosion is a chemical change because in such a change chemical properties or composition of the original substance changes resulting in the formation of at least one new substance. Formation of rust is a chemical change as a new compound called hydrated iron oxide (Fe203. H2O) is formed which is totally different from element iron (Fe). Chemical changes are irreversible.


The chemical reaction showing rusting of iron is shown below.



Other examples of chemical change are burning of paper and burning of magnesium wire.


In physical changes, no new substances are formed. In such a change, chemical properties and composition of the original substance do not change. Physical changes are reversible. Examples of physical changes are melting of ice, boiling of water and breaking of a glass tumbler.


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


Answer:

A mixture of sulphur and carbon disulphide is heterogeneous and shows Tyndall effect.


A mixture of sulphur and carbon disulphide is a heterogeneous colloid* and shows Tyndall effect. The Tyndall effect is scattering of light by particles in a colloid. Colloids appear to be homogeneous but they are actually heterogeneous in nature. Milk and fog are some other examples of colloids.

*Colloids are the substances that have some other substances dispersed in the solution and these particles do not settle down at the bottom.


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


Answer:

Tincture of iodine solution is made by dissolving iodine in alcohol. It contains around 2–7% iodine dissolved in a mixture of ethanol and water.


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)


Answer:

Ice and air are homogeneous in nature.


Ice and air are homogeneous in nature as they have a uniform composition throughout its mass. It does not have a variable composition of its constituents. Examples are air, sugar solution and brass.
A heterogeneous mixture, on the other hand, does not have a uniform composition throughout its mass. It has a variable composition of its constituents. Examples are soil, wood and blood.


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)


Answer:

(i) Melting of iron metal, (iii) Bending of an iron rod and (iv) Drawing a wire of iron metal are physical changes because in all these processes, no new substance is formed and chemical properties of the original substance remain same. Shape and size of a substance can change in a physical change.


In case of rusting of iron the chemical composition of iron changes.


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)


Answer:

(i) Decaying of wood and (ii) Burning of wood are chemical changes because in these processes, the chemical composition of wood is changed and new substances are formed. These changes are irreversible which cannot be converted back into their original form because chemical reactions are irreversible in nature.
In (iii) Sawing of wood and (iv) Hammering of a nail into a piece of wood, chemical composition and properties of wood are not changed, therefore, these are physical changes.


Question 8.

Two substances, A and B were made to react to form a third substance, A2B according to the following reaction

2 A + 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)


Answer:

A2B is a compound made up of two elements A and B which are present 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 product will always have a fixed composition.


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) 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)


Answer:

In this reaction, X and Y are elements as they cannot be broken down into simpler substances by chemical reactions.


P is a compound as it is made up of two or more elements that combine in a fixed proportion. Thus P has a fixed composition.



Short Answer Type
Question 1.

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


Answer:

(a) Mercury and water are separated by using a separating funnel.

Mercury and water are two immiscible liquids having different densities. Thus they are separated by the process of decantation* using separating funnel. Mercury is heavier than water, thus forms the lower layer and is separated from water.

*Decantation: It is a process in which the mixtures are. This separation is done by removing a layer of liquid in which a precipitate has settled.


(b) Potassium chloride and ammonium chloride are separated by sublimation because ammonium chloride sublimes (i.e. changes into vapours) leaving behind the potassium chloride. Sublimation is the process in which solid changes directly into vapours on heating and vapours change into solid on cooling without going in liquid state.


(c) Common salt, water and sand are separated by (i) decantation (or filtration) (ii) evaporation Decantation (or filtration) is used to separate sand from common salt solution in
water. Common salt is soluble in water whereas sand is insoluble in water. Therefore, sand is separated as residue and filtrate is a common salt solution in water.



(ii) Evaporation is used to separate common salt from water. Water evaporates and common salt remains as residue.


(d) Separation of kerosene oil from salt solution is done by using separating funnel to separate kerosene oil followed by evaporation or distillation.
(i) Decantation separates kerosene oil from salt solution in water as these are immiscible and form separate layers.
(ii) Evaporation is used to separate salt from water (as shown above).



Question 2.

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



Answer:

The tube in figure (a) will be more effective as a condenser in the distillation apparatus.
This is because due to the presence of beads in the tube (a) will provide more surface area as compared to the tube in which there are no beads. And, the tube with large surface area leads to effective cooling.
Thus, tube (a) will be more effective condenser in the distillation apparatus.
* A condenser condenses steam to water in a distillation process. It turns gases to liquids by cooling them.



Question 3.

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


Answer:

Other than evaporation, crystallization can be done to recover salt from its solution.

In crystallization, continuous adding salt is done until no more salt will dissolve in the solution as it becomes a saturated solution. Crystallization is a better technique than evaporation because it does not require high temperature and boiling of solvent.



Question 4.

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


Answer:

The ‘sea-water’ can be a homogeneous mixture as it contains salts and water only.

It can also be a heterogeneous mixture as it contains other insoluble components such as mud, decayed plant, calcium carbonate shells along with salts and water.



Question 5.

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.


Answer:

Distillation can be employed to get back the acetone as it is more volatile and it will separate out first.
Acetone forms a homogeneous mixture as it is soluble in water. Acetone can get back by simple distillation because the difference in the boiling points of acetone and water are different.
Boiling point of acetone is 56°C
Boiling point of water is 100°C
In distillation flask, acetone will boil at 56°C and convert into vapours. Then, it can then be collected in the flask after condensation.



Question 6.

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.


Answer:

(a) When a saturated solution of potassium chloride prepared at 60°C is allowed to cool to room temperature, crystals of potassium chloride will be formed and get separated out.
(b) Initially, when sugar solution is heated, water gets evaporated. But when heated to dryness, sugar gets charred and turns black.

The chemical equation to show that sugar turns black on continuous dry heating.



(c) A black coloured compound called iron sulphide is formed when a mixture of iron filings and sulphur powder is heated strongly.


The chemical equation showing the reaction between iron filings and sulphur powder is as follows:



Question 7.

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


Answer:

Particles of a colloidal solution are smaller and lighter than that of a suspension. The range of diameter of particle of colloids are: 1 to 1000nm. They are always in a state of random motion. This random movement is called the Brownian movement, which counters the force of gravity acting on colloidal particles and hence, it does not allow the particles to settle down. Colloidal particles are charged and repel each other. That is another reason contributing to the fact that the particles of colloidal solution do not settle down.

On the contrary, particles of suspension are larger, heavy and have less movement. Thus, they settle down due to gravity.



Question 8.

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


Answer:

An aerosol is a colloid of fine solid particles or liquid droplets, in the air or another gas. Smoke is artificial aerosol and fog is a natural aerosol. Smoke is a mixture of gases. Fog is a mixture of liquid and gases. In both fog and smoke, the dispersion medium is same, i.e. gas or air. The only difference is that the dispersed phase in fog is liquid (water particles) and in smoke, it is a solid (carbon particles).



Question 9.

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.


Answer:

Options (a) and (c) show physical properties. Options (b) and (d) show chemical properties.

(a) The composition of a sample of steel shows a physical property because steel is an alloy and alloy is a homogeneous mixture of two or more metals or of metallic elements with non-metallic. No new compound is formed.
(b) Such a reaction shows a chemical property because chemical reaction takes place between zinc and hydrochloric acid with the evolution of hydrogen gas and a compound zinc chloride is formed.
The chemical equation shows the following reaction:


Zn(s)+ 2HCI (dil)→ ZnCI2(aq) + H2(g)
(c) Cutting of metallic sodium shows a physical property because no new substance is formed in the process. It is just cutting a big portion of metal into a small portion.
(d) Such a reaction shows a chemical property because chemical reaction takes place between metal oxide and water to form metal hydroxide, which is a new substance.


M2O (s) (Metal oxide) + H2O (Z) (Water) → 2MOH (aq) (Metal hydroxide)



Question 10.

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


Answer:

Student ‘C’ has made the desired solution as 50% (mass by volume)
means 50 g of solute for every 100 mL of solution and not in 100 mL of solvent.

Mass by volume % = Mass of solute (in g)/Volume of solution (in mL) x 100


50/100 x 100 = 50% mass by volume


Student ‘A’ dissolved 50 g of NaOH in 100 mL of water (solvent).
Student ‘B’ dissolved 50 g of NaOH in 100 g of water (solvent).
Both A and B cannot make the desired solution.



Question 11.

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) A fine beam of light entering through a small hole in a dark room illuminates the particles in its paths.


Answer:

(a)Sublimation.

When dry ice (solid CO2) is kept at room temperature at one atmospheric pressure, it sublimes (changes into vapours) leaving no residue behind.


(b) Diffusion.


When a drop of ink is placed on the surface of water contained in a glass, it spreads throughout the water. This process of mixing of one substance (ink) into another substance (water) takes place by the process of diffusion. This mixing goes on until a uniform mixture is formed.


(c)Dissolution/diffusion.


When a potassium permanganate crystal is kept in a beaker and water is poured into the beaker with stirring potassium permanganate crystal is dissolved in water the process of dissolution/diffusion.


(d)Evaporation, diffusion.


When an acetone bottle is left open, the bottle becomes empty. This happens because acetone evaporates when kept open in the bottle.


(e)Centrifugation.


Milk is churned to separate cream from it by the process of centrifugation. Milk is put in a centrifuge machine. Milk is rotated at a very high speed in the centrifuge machine. As a result, milk separates into ‘cream’ and ‘skimmed milk’. The cream being lighter floats over the skimmed milk and can be removed.


(f)Sedimentation.
Sand settles down when a mixture of sand and water is left undisturbed for some time by the process called sedimentation. This happens because sand is insoluble in water and forms a suspension. Thus, it settles down at the bottom when left undisturbed for some time.


(g)Scattering of light (Tyndall effect).
A fine beam of light entering through a small hole in a dark room illuminates the particles in its paths due to scattering of light (Tyndall effect) by particles in a fine suspension. Dust particles are suspended in air that scatters the light coming from the small hole.



Question 12.

You are given two samples of water labelled as ‘An’ 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.


Answer:

Sample ‘B’ will not freeze at 0°C because it is not pure water. It must have contained impurities. At 1 atm, the boiling point of pure water is 100°C and the freezing point of pure water is 0°C. The only pure substance has such sharp melting point.



Question 13.

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


Answer:

Pure gold is a very soft metal and is very ductile and malleable. Thus, it can be given any shape and it can also break by applying a little force. Therefore, it is not suitable for making ornaments. Pure gold, when alloyed with copper or silver, becomes harder and stronger it will not break on applying even a little force. Thus, it becomes suitable for making ornaments.



Question 14.

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?


Answer:

This element is a metal. Metals are sonorous and highly ductile. Some other characteristics of metals are as follows:
(i) Metallic lustrous and can be polished.
(ii) Metals are good conductors of heat and electricity.
(iii) Ductile
(iv) Malleable
(v) High tensile strength
(vi) High densities and melting point/boiling point
(vii) Hard (except sodium and potassium, which are soft metals)
(viii) Solid at room temperature (except mercury, which is liquid at room temperature)
Non-metals are neither malleable nor ductile and do not conduct electricity. Metalloids show some properties of metals and some other properties of non-metals.



Question 15.

Give an example each of 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.


Answer:

(a) Example: Mixture of acetone (volatile) and water (non-volatile)
Method of separation: Simple distillation can be used to separate a mixture of volatile and non-volatile components. Components will separate out based on different boiling points.
(b) Example: Mixture of kerosene (volatile) and petrol (volatile)
Method of separation: Simple distillation can be used to separate two volatile components with different boiling points.
(c) Example: Mixture of mustard oil and water (immiscible in each other)
Method of separation: A separating funnel is used to separate a mixture of immiscible liquids.
(d) Example: Mixture of ammonium chloride and common salt
Method of separation: Sublimation can be used to separate the mixture in which one component changes directly from solid to gas (sublime substance).
(e) Example: A mixture of different pigments from an extract of flower petals
Method: Chromatography can be used to separate two different substances present in the same solution. Paper chromatography is generally used to separate coloured substances.



Question 16.

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.


Answer:

(a)A colloid is a heterogeneous mixture and its components can be separated by the technique known as centrifugation.
Heterogenous mixture: It is the mixture which in not uniform in composition.
Centrifugation: It is a separation technique which uses the concept of centrifugal force to settle the particles at the bottom of the container.
(b) Ice, water and water vapour look different and display different physical properties but they are chemically the same.
Ice, water and vapour all the forms of water but are physically different. Because, ice is a solid, water is liquid and vapour is gas.

(c) A mixture of chloroform and water is 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.
It is because water is lighter than chloroform. Chloroform consists of chlorines molecules that are denser than water molecules and it make the chloroform nearly 50% heavier than water.
(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 fractional distillation.
(e) 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 phenomenon is called Tyndall effect. This indicates that milk is a colloidal solution.
Colloidal Solution: A colloidal solution is also known as suspension. It is a solution in which a material is evenly suspended in a liquid



Question 17.

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.


Answer:

The law of constant composition or definite proportions states that, “when two elements are combined together in the same fixed proportion by mass, it has the same composition irrespective of the source a chemical compound”. Thus, sucrose (sugar) crystals obtained from the mixing of sugarcane and beetroot will be a pure substance and will have the same composition.



Question 18.

Give some examples of Tyndall effect observed in your surroundings?


Answer:

Tyndall effect is usually observed in colloids. It is called the light scattering by the particles in a colloid. It is named after the physicist John Tyndall. The other name is Tyndall scattering or Tyndall effect.


is observed in our surroundings when
(i) Sunlight passes through a heterogeneous mixture such as the canopy of a dense forest.
(ii) a fine beam of light enters a dark room through a small hole.



Question 19.

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


Answer:

No, we cannot directly separate alcohol dissolved in water using a separating funnel because alcohol is completely miscible in water.
A separating funnel is used to separate only those liquids which are immiscible.
But, we can use the process of distillation to separate alcohol dissolved in water.

Distillation is used to separate different compounds. The concept of distillation is the different boiling points of the liquids in a mixture. That means a more volatile compound will evaporate at a lower temperature than a less volatile compound at the same temperature.



Question 20.

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.


Answer:

(a) Heating of calcium carbonate to get calcium oxide and carbon dioxide is a chemical change as new substance with different composition and chemical properties are formed.

CaCO3 (Calcium carbonate) → CaO (Calcium oxide) + CO2 (Carbon dioxide)
(b) Acidic and basic solutions can be prepared by dissolving the products of the above process in water.


When CaO dissolves in water it forms calcium hydroxide which is basic solution.


CaO+H2O → Ca(OH)2 (basic solution)


When CO2(g) dissolves in water it forms carbonic acid which is an acidic solution.


CO2 + H2O→H2CO3 (acidic solution)



Question 21.

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.


Answer:

(a) Iodine is a lustrous non-metal. Graphite and diamond are also lustrous non-metals. These are allotropes of carbon.
(b) Bromine is a non-metal which exists as a liquid at room temperature.
(c) Graphite is the allotropic form of carbon and it is a good conductor of electricity.
(d) Carbon is a non-metal which is known to form the largest number of compounds.
(e) Sulphur and phosphorus are non-metals other than carbon which shows allotropy. Phosphorus exists as white phosphorus, red phosphorus and black phosphorus
(f) Oxygen is a non-metal which is required for combustion.



Question 22.

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


Answer:



Wood is neither an element nor a compound. It is a mixture.




Question 23.

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


Answer:

These are not compounds:
(a) Chlorine gas
Chlorine gas consists of chlorine molecules. Chlorine is a non-metal.

(c) Iron


Iron is a metal.
(e) Aluminium


Aluminum is a metal


(f) Iodine
Iodine is a non-metal.
(g) Carbon


Carbon is a non-metal.


(i) Sulphur powder


Suphur is a non-metal.


Others are compounds because they are formed by the combination of different elements.




Long Answer Type
Question 1.

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.


Answer:

Fractional distillation is a process to separate the volatile liquids. In the apparatus, a fractionating column is the most important part of the fractional distillation apparatus. This column has glass beads in it which provides a large surface area for the vapours to collide and lose energy so that they can be quickly condensed and distilled. Apart from the presence of bead, if the length of the column is increased, then it would also increase the efficiency. The vapours of high boiling liquid get condensed earlier (at a lower level).

The advantages of a fractionating column are as follows:
(i) The purity of the mixture is the main advantage of fractional distillation.
(ii) This method can separate the liquids with a boiling point difference about or less than 25 K.
(iii) During the process, both evaporation and condensation take place simultaneously.
(iii) A mixture (such as petroleum containing several components) can also be separated by the fractional distillation process.



Question 2.

Under which category of mixtures will you classify alloys and why?


Answer:

Alloys are homogenous mixture because:
(i) It shows the properties of its constituents materials
(ii) It has variable composition, for example, brass is considered a mixture because it shows the properties of its constituents – copper and zinc. Also, it has a variable composition (amount of zinc in brass can vary from 20 to 35 per cent).
(iii) The elements that make up the mixture are only physically together and not chemically bonded.



Question 3.

A solution is always a liquid. Comment.


Answer:

No, solutions are generally liquid but not always necessary. Solutions can be solids and gases. Examples are brass and air, respectively. Alloys are a solution of solid in solid. Air is a solution of gases in gases.



Question 4.

Can a solution be heterogeneous?


Answer:

No, a solution cannot be heterogeneous. A solution is a homogeneous mixture that is made up of two or more substances. It has a uniform composition. The air we breathe is an example of a homogeneous solution.

Though colloidal solutions are heterogeneous in nature, they appear to be homogeneous.



Question 5.

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?


Answer:

Part A was heated strongly.

The equation given below shows the reaction:


Fe (s) + S (s) FeS (s)


When dilute hydrochloric acid was added to Part A, following reaction took place:


FeS + 2 HCl (aq) → FeCl2+ H2S


H2S gas is evolved.


H2S gas formed has a foul smell (as that of rotten eggs) and on passing through lead acetate solution, it turns the solution black.


(CHCOO)2Pb(ag) + H2S(g) → PbS(s) + 2CH3COOH
Lead acetate Black ppt Acetic acid


Part B is not heated.


Fe (s) + S (s) (mixture of iron filings and sulphur): Not heated


When dilute HCl is added to Part B, following reaction took place:


Fe (s) + S (s) + 2 HCl (aq) → FeCl2 + H2


Hydrogen gas is evolved and sulphur remains unreacted.


Hydrogen gas burns with a pop sound if burning match stick is brought near it.
Fe (s)+ 2HCI (dil.) → FeCI2(ag) + H2(g)



Question 6.

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.

(a) What would you expect to see, if the ink contains three different coloured components?

(b) Name the technique used by the child.

(c) Suggest one more application of this technique.



Answer:

(a) Three different coloured bands are obtained on the strip at different heights if the ink contains three different coloured components.
Chromatography is a method to separate components of the mixture.


(b) Chromatography method (paper chromatography) is used by the child.


(c) Chromatography is used to separate the pigments present in chlorophyll. It is also used to separate drugs from the blood.



Question 7.

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?


Answer:

(a) The milk sample was illuminated because milk is a colloidal solution and its particles are big and hence, scatter the light passing through it. The phenomenon observed is called the “Tyndall effect” or “Tyndall scattering”.

(b) As the salt solution is a true solution i.e., the solute particle size is too small and hence they cannot scatter the light. That is why the salt solution does not show “Tyndall effect”.

(c) Examples of colloid are gold sol (colloidal gold), arsenic sulphide (AS2S3) sol, blood, detergent solution and sulphur solution. Sols have good stability and show the Tyndall effect.



Question 8.

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.


Answer:

(a) Drying of a shirt in the sun is a physical change because evaporation of water takes place but no change occurs in the composition of the substance.
(b) Rising of hot air over a radiator is a physical change because it is involving the only movement of air, no change in the composition of air.
(c) Burning of kerosene in a lantern is a chemical change because kerosene burns to form carbon dioxide and water vapour.
(d) Change in the colour of black tea on adding lemon juice to it is a chemical change as a new substance with chemical properties are formed.
(e) Churning of milk cream to get butter is a physical change as there is no change in composition. Only the separation of components takes place by the physical phenomenon called centrifugation.



Question 9.

During an experiment, the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10g of sugar in 100g of water while Sarika prepared it by dissolving 10g of sugar in water to make 100g of the solution.

(a) Are the two solutions of the same concentration?

(b) Compare the mass % of the two solutions.


Answer:

(a) No, the two solutions do not have the same concentration.

(b) Mass percentage of solution prepared by Ramesh =




Mass percentage of solution prepared by Sarika =
The solution prepared by Ramesh has less percentage (9.09%) by mass than that of Sarika (10%). Thus, the solution prepared by Sarika has a higher mass % than that prepared by Ramesh.



Question 10.

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?


Answer:

Step 1: Remove iron filings with the help of magnet
Keep the mixture on a paper. Move a bar magnet many times over the mixture. Iron filings get attracted to the magnet and hence get separated.
Step 2: Remove ammonium chloride from sand and sodium chloride by sublimation
The remaining mixture is transferred to China dish and subjected to sublimation. Ammonium chloride will get vaporized and change into vapours and on condensation will form NH4CI(s). Sand and sodium chloride will be left in China dish.
Step 3: Remove sand from sodium chloride by filtration after dissolution
Dissolve the sand and sodium chloride in water. Sodium chloride will dissolve. Filter the solution. Sand will be left as residue and is separated.
Step 4: Get sodium chloride by evaporation or crystallization
Sodium chloride is present in the filtrate. So, evaporate the filtrate to dryness to get sodium chloride back or use the crystallization method.



Question 11.

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 + 100g of water

(b) 0.11g of NaCl + 100g of water

(c) 0.0l g of NaCl + 99.99g of water

(d) 0.10 g of NaCl + 99.90g of water


Answer:

(c) Here,




which is equal to the percentage of sodium chloride in water prepared by Arun. So, option (c) is correct.


In option (a), mass% =


In option (b), mass% =


In option (d), mass% =


Thus, other three representations are incorrect.



Question 12.

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


Answer:

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




Mass of sodium sulphate = 25 g