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

NCERT Exemplar Class 9 Science Solutions Chapter 2 (MCQ)

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: B

Solution: A pure substance consists of a single type of particles. Element or compound can be considered as pure form of matter. A cylinder of oxygen gas can be considered as pure substance. Milk cannot be considered as pure substance as it is a mixture of water fat protein etc. The composition or density of pure substance will be same throughout its distribution.

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: C

Solution: In a chemical change, either one of the substance changes and comes with new properties. Corrosion is a natural phenomenon, in which any metal tries to attain chemically stable form such as oxide for hydroxide. When any article made up of iron get rusted, it gets converted in iron oxide it looks reddish brown in colour. As it is a change of the substance, it is a chemical change.

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: D

Solution: Solution will be homogeneous at molecular level it means if you take any part of the solution percentage of substances will be same. The Tyndall effect is shown by mixture, where mixture has tendency to scatter the light of shorter wavelength and transmit the light of larger wavelength. The Tyndall effect is shown by heterogeneous mixture such as colloid. Sulphur is soluble in carbon disulphide. A solution is formed when sulphur is mixed with carbon disulphide. Therefore, it 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.
Answer: D

Solution: Tincture of iodine is also known as iodine tincture. It is a very weak solution of iodine in ethanol and water. It is used as an anti-septic. It was used since start of 20th century by surgeons.

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: C

Solution: A mixture is called homogeneous, if the percentage of constituent substances will be same in any part of that mixture. The density of any sample will also be same in homogeneous mixture. If a sample is not homogeneous, it is called heterogeneous. All the pure substances will be homogeneous. Ice is a pure substance; hence it is homogeneous. Air is a homogeneous mixture of multiple gases. Soil and wood or not homogeneous.

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: C

Solution: If a matter goes under a physical change its appearance might change but the substance of matter does not change. In a chemical change in a chemical change either one of the substance changes and comes with new properties. In melting, the state of metal will change but the substance will not change. In bending or drawing a wire appearance of the metal change but basic properties of substance will not change. When any article made up of iron get rusted, it gets converted in iron oxide it looks reddish brown in colour. As it is a change of the substance, it is a chemical change

$4Fe+3O_2\to 2F{e}_2{O}_3.x{H}_{2}O$

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: A

Solution: In the decaying or burning of wood, there will be a chemical change in the composition of wood hence it is a chemical process. Sowing of wood and hammering of nail into the piece of wood does not change the properties of wood chemically, hence it is a physical change.

Question:8

Two substances, A and B were made to react to form a third substance, $A_{2}B$ according to the following reaction:$2A+B\to A_{2}B$. 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)

Answer: C
Solution: $2A+B\to A_{2}B$, A and B will combine to make a product $A_{2}B$ . The product $A_{2}B$ is called compound and it will not have any property of elements A and B. The composition of elements in product will be fixed: If N moles of element A is present that means N/2 moles of element B will be present. This product will be pure. Hence statement (i), (iii) and (iv) are wrong statements.

Question:9

Two chemical species X and Y combine together to form a product P which contains both X and Y.
$X+Y\to P$
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 compound.
(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: D

Solution: $X+Y\to P$
X and Y will combine to make a product P. X and Y are elements because they cannot be broken down into simpler substances. The product P is called compound and it will not have any property of elements X and Y.
The composition of elements in product will be fixed: If N moles of element X is present that means moles of element Y will be present. This product will be pure.

NCERT Exemplar Class 9 Science Solutions Chapter 2 (Short Answer)

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 – Mercury and water both are liquids but they are insoluble with each other.
The density of mercury is 13.6 times higher than the density of water.
Due to this difference in density and insolubility we can separate them by separating funnel.
(b) Potassium chloride and ammonium chloride – Ammonium chloride can be sublimated that means it can convert in
vapour without having the intermediate phase of liquid. Potassium chloride does not sublime.
Hence, we can separate this mixture by the process of sublimation.
(c) Common salt, water and sand – Salt is soluble in water but sand is not, hence we can separate sand from salty water by filtration. Now by the process of evaporation we can find dry salt as the water gets evaporated but salt remain there.
(d) Kerosene oil, water and salt – Due to difference in densities of kerosene oil and salty water, we can use the method of separating funnel to separate kerosene oil from salty water. Now by the process of evaporation we can find dry salt as the water gets evaporated but salt remain there.

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 use of condenser tube in distillation Apparatus is to liquefy separated vapour.
This liquefaction is done by continuous flow of cold water.
If the area of condenser is increased, cooling will be faster.
Due to uneven shape of tube (A), its surface area is more than tube (B).
Therefore, it is advisable to use tube (A) for condenser.

Question:12

Salt can be recovered from its solution by evaporation. Suggest some other technique for the same?
Solution:
By the process of evaporation, we can find dry salt as the water gets evaporated but salt remain there. This process is a slow process.
We can think of another method:
The name of this process is distillation. It is similar to evaporation but we will not lose water in this case.
The boiling point of salt is very large in comparison of water.
On heating, the water will start boiling and we can collect the steam and condense it back in water.
Finally in the pot, we will left with only salt.
Note: The solubility of salt in water is independent of temperature, hence we can use crystallization method.

Question:13

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

Solution:

If we consider Sea water as salty water only then it will be homogeneous.
Salt is soluble in water and makes a solution. All solutions are homogeneous.
But due to presence of mud, sand and other insoluble compounds: Sea water is also treated as heterogeneous mixture.

Question:14

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

Solution:

If we mix two liquids of significant different boiling points, we can separate them by process of distillation.
Generally, this technique is used if difference in boiling point is more than ${25}^{\circ }C\sim {30}^{\circ }C$.
Here the difference in boiling point of salty water and acetone is more than ${44}^{\circ }C$.
On heating, the acetone will start boiling and we can collect the vapour and condense it back in acetone liquid.
Finally in the pot, we will left with only salty water.

Question:15

What would you observe, when
(a) a saturated solution of potassium chloride prepared at ${60}^{\circ }C$ 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) The solubility of potassium chloride decrease on decreasing temperature. Therefore on cooling, Crystals of potassium chloride will separate out.
(b) On heating, the water will start boiling and completely evaporate to dryness.
Finally in the pot, we will left with only black sugar.
(c) At High temperature, they will react and 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 are small, apparently they look like homogeneous solution but they are heterogeneous at molecular level.
These particle move randomly (Brownian motion) within the solution and do not settle down.
The particles in suspension are heavy and can be seen by naked eye.
When it is left undisturbed, due to gravity, these particles settle down.

Question:17

Smoke and fog both are aerosols. In what way are they different?
Solution:
A colloidal solution of particle dispersed in air or gas is called aerosol.
Therefore, Smoke and Fog; both are aerosol as both of them are dispersed in air.
The difference in both of them is on the basis of solute particle.
In Smoke: solid carbon is the solute.
In Fog: liquid water is the solute.

Question: 18

Classify the following as physical or chemical properties.
(a) The composition of a sample of steel is: $98\mathrm{\%}$ iron, $1.5\mathrm{\%}$ carbon and $0.5\mathrm{\%}$ 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:

Chemical property: When two substance combine to make a new compound, which does not show the properties of any reacting substances.
Physical property: Appearance, composition, state, melting point, elasticity, malleability etc.
(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 $50\mathrm{\%}$ (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:
Mass by volume: X% mass by volume solution means X grams of a solute dissolved in 100 mL of solution.
Mass by volume $(\mathrm{\%})=\frac{massofsolute\times 100}{Volumeofsolution}$
Student A: dissolved 50 g of NaOH in 100 mL of water.
Therefore, volume of solution will be more than 100 ml as only solvent is 100 mL.
This solution will have less than $50\mathrm{\%}$ (mass by volume) solution
Student B: dissolved 50 g of NaOH in 100 g of water. (100 gram water =100mL of water)
Therefore, volume of solution will be more than 100 ml as only solvent is 100 mL.
This solution will have less than $50\mathrm{\%}$ (mass by volume) solution.
Student C: dissolved 50 g of NaOH in water to make 100 mL.
It means, Volume of solution is 100mL.
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 CO₂ (gas): Dry Ice will convert in gas without going through intermediate liquid state.
(b) Diffusion of ink into water: The ink flow from higher concentration to lower concentration.
(c) Dissolution of solid into liquid: Slowly potassium permanganate crystal will dissolve in water to make a solution.
(d) Evaporation of acetone in air : As the boiling point of acetone is very small , it get evaporated and diffused in the atmosphere.
(e) Centrifugation of cream: In the process of churning, a centrifugal force acts away from the central line. Thus, cream get collected on the walls of vessel.
(f) Sedimentation
(g) Tyndall effect – Heterogeneity of air causes bending and scattering of light.

Question:21

You are given two samples of water labelled as ‘A’ and ‘B’. Sample ‘A’ boils at ${100}^{\circ }C$ and sample ‘B’ boils at ${102}^{\circ }C$. Which sample of water will not freeze at $0^{\circ }C$? Comment.

Solution:
When any non-volatile solute (impurities) is added in water, It causes elevation in boiling point as well as depression in freezing point.
Therefore impure water will have boiling point more than ${100}^{\circ }C$ and freezing point will be less than $0^{\circ }C$,
Sample ‘B’ which boils at ${102}^{\circ }C$ contains impurities. It will not freeze at $0^{\circ }C$.

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:

If we use pure gold (24 carat), it is difficult to make ornaments.
Pure Gold is highly malleable and soft in nature, which will not give any strength to the ornament and it will be de-shaped easily.
When Pure is alloyed with copper or silver (less than 24 carat), it becomes hard and strong. Now it can be used to make ornaments.

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:
Sonorous: It means, substance makes ringing sound when we strike it.
Ductile: We can create wire, as it can be deformed without losing strength.
These two properties indicate that the element must be metal.
The metal 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) A volatile (can be easily boiled) and a non-volatile (cannot be boiled) component.–
We can use process of distillation or evaporation
Example: Common salt solution
(b) Two volatile components with appreciable difference in boiling points.
We can use process of distillation.
Example: Acetone (boiling point = ${56}^{\circ }C$) and water (boiling point = ${100}^{\circ }C$)
(c) Two immiscible liquids (of different densities)
We can use process of Separation using separating funnel
Example: Oil Water mixture
(d) One of the components changes directly from solid to gaseous state.
We can use process of Sublimation
Example: Ammonium chloride (sublimate) and potassium chloride (will not sublimate)
(e) Two or more coloured constituents soluble in some solvent.
We can use process of chromatography in which mixture is dissolved in solvent to form mobile phase and on passing through a fixed phase they separate
Example: ink or plant dyes.

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
Colloid: It is heterogeneous mixture of tiny particles which perform random motion and cannot be separated by gravity.
Centrifugation: on rotating container particle are repelled towards wall.
(b) physical, chemically
Physical Property: appearance, state etc.
Chemical Property: capability to react with some other substance.
(c) water, chloroform
Density of water (1 gram per cc) is less than density of chloroform ( 1.49 gram per cc)
(d) fractional distillation
Example: Acetone (boiling point = ${56}^{\circ }C$) and water (boiling point = ${100}^{\circ }C$)
(e) scattering, Tyndall effect, colloidal
Milk is a heterogeneous colloidal of water, fat and protein etc.

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 – which is a compound) will remain constant irrespective of the source of its preparation.
It means sugar is pure substance and its composition will be same.

Question:27

Give some examples of Tyndall effect observed in your surroundings?
Answer:

Scattering of light through heterogeneous mixture is called Tyndall effect.
Examples of Tyndall effect:
(i) In Fog, Beam of head light of car is visible.
(ii) Sunlight passing through dust particle will be visible.
(iii) Path of light rays seen in front of the projector in a cinema hall.
(iv) Shining of flash light when pass through milk.

Question:28

Can we separate alcohol dissolved in water by using a separating funnel? If yes, then describe the procedure. If not, explain.
Solution:
We can separate immiscible liquids of different densities by using a separating funnel. Mixture of water and alcohol cannot be separated, since both are miscible and they form a solution.

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:
If a matter goes under a physical change its appearance might change but the substance of matter does not change.
In a chemical change in a chemical change either one of the substance changes and comes with new properties.
$CaC{O}_3\to CaO+C{O}_2$
(a) It is a chemical change as new substances are formed.
(b) Calcium oxide when dissolved in water, forms a basic solution.

$CaO+H_{2}O\to Ca(OH{)}_2$
Carbon dioxide when dissolved in water, forms an acidic solution
$C{O}_2+H_{2}O\to H_{2}C{O}_3$

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 and diamond: They shine as they are lustrous
(b) Bromine: fuming red brown liquid at room temperature
(c) Graphite: Allotrope of carbon (non-metal) good electric conductor
(d) Carbon: each carbon can combine with four different atoms
(e) Sulphur: Yellow rhombic sulphur and monoclinic sulphur are allotropic form
(f) Oxygen: it is required for combustion.

Question:31

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

Solution:
Elements – pure substance which cannot be broken down in simpler substance
$Cu(metal),Zn(metal),F_2(gas),O_2(gas)$, diamond (Carbon-non-metal), Hg (metal)
Compounds – They are also pure, but can be broken down in constituents.
$CaC{O}_3,H_{2}O$, NaCl_(aq), Wood and Sand

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:

Elements – pure substance which cannot be broken down in simpler substance.
Chlorine, Iron, Aluminium, Iodine, Carbon and Sulphur Powder

Compounds – They are also pure, but can be broken down in constituents.
Potassium Chloride (KCl)
Iron Sulphide (FeS)
Carbon monoxide (CO)

NCERT Exemplar Class 9 Science Solutions Chapter 2 (Long Answer)

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:
If we mix two liquids of significant different boiling points, we can separate them by process of distillation.
Generally, this technique is used if difference in boiling point is more than ${25}^{\circ }C\sim {30}^{\circ }C$.
But Fractional Distillation is effective even the boiling point difference is less than ${25}^{\circ }C$
In fractional distillation Method, a fractionating column is used and it makes it efficient. This column is packed with glass beads or small plates. It increases the surface area. The increase in surface area will increase the rate of condensation for the vapours. Vapour quickly loose energy, when they come in contact with beads or plates and can be condensed easily.
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:
Any Homogeneous mixture of two substances is called Solution.
The substance in large quantity is called solvent and substance in small quatity is called Solute.
Solution can be gaseous (For example; Air), Liquid (For example; Salt solution with water) and Solid (For example; Alloy
(a) Alloys they have uniform composition throughout hence they are homogeneous solid solution.
(b) No, Any Solution can be gaseous (For example; Air), Liquid (For example; Salt solution with water) and Solid (For example; Alloy)
(c) No, a solution is a homogeneous mixture and does not show Tyndall Effect.

Question: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 and sulphur are heated they react with each other and form iron-sulphide.
$Fe+S\to FeS$
This iron sulphide can react with acid Hydrochloride and hydrogen Sulphide gas will come out from the mixture.
$FeS+2HCl\to FeC{l}_2+H_{2}S(gas)$
Hydrogen Sulphide has smell like rotten egg and can be identified easily.
This gas can be chemically tested by reaction with lead Salts.
Part B –
When iron and sulphur are heated they react with each other and form iron-sulphide but without heat they will remain unaffected.
Sulphur does not react with hydrochloride but iron will react and hydrogen gas will come out.
$Fe+2HCl\to FeC{l}_2{H}_2(gas)$
Hydrogen gas is highly inflammable and produces sound on burning.
If we bring any spark near the mixture, we can identify the hydrogen gas.

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:
Chromatography is a method, which is used to separate components of any mixture.
In this process different component travels with different speed when dissolved in solution.
Due to different speed, all the components get separated from each other
(i) The components of the ink will travel with water with different speeds and we would see three bands on the filter paper at different levels.
(ii) Chromatography.
(iii) This technique is also used for separating 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:
Scattering of light through heterogeneous mixture is called Tyndall effect.
Examples of Tyndall effect:
(i) In Fog, Beam of head light of car is visible.
(ii) Sunlight passing through dust particle will be visible.
(iii) Path of light rays seen in front of the projector in a cinema hall.
The Tyndall effect is shown by mixture, where mixture has tendency to scatter the light of shorter wavelength and transmit the light of larger wavelength.
The Tyndall effect is shown by heterogeneous mixture such as colloid.
(a) Milk is a colloidal solution hence shows Tyndall effect.
(b) Salt solution is homogeneous therefore do not show Tyndall effect and they do not scatter light.
(c) Detergent mixture and air with dust will show Tyndall effect.

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:
If a matter goes under a physical change its appearance might change but the substance of matter does not change.
In a chemical change in a chemical change either one of the substance changes and comes with new properties.
a) Drying of a shirt in the sun $\Rightarrow$Substance of shirt does not change$\Rightarrow$- Physical change
(b) Rising of hot air over a radiator$\Rightarrow$Substance in air does not change$\Rightarrow$- Physical change
(c) Burning of kerosene in a lantern$\Rightarrow$ Kerosene goes under chemical reaction with oxygen$\Rightarrow$- Chemical change
(d) Change in the colour of black tea on adding lemon juice to it$\Rightarrow$Black tea goes under chemical reaction$\Rightarrow$-Chemical change
(e) Churning of milk cream to get butter$\Rightarrow$Substance in milk does not change$\Rightarrow$-Physical change
(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 $10\mathrm{\%}$ (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 $\mathrm{\%}$ of the two solutions.
Solution:
Mass Percent: amount of solute in 100 gram of solution.
Mass percent(%) $=\frac{massofSolute\times 100}{massofSolution}$
Ramesh’s Solution:
Mass percent(%) $=\frac{10gram\times 100}{10gram+100gram}=9.0909\mathrm{\%}$
Sarika’s Solution:
Mass percent(%) $=\frac{10gram\times 100}{100gram}=10\mathrm{\%}$
Part a: Sarika has higher mass percentage than Ramesh.
Part b: The mass percentage ratio of Sarika and Ramesh is 10: 9.0909

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:
Given Mixture – Sand + Iron filings + Ammonium chloride +Sodium chloride
Iron filing can be separated by using magnet which will attract all the iron filings.
Now by the process of sublimation we can separate ammonium chloride from the mixture as it is the only constituent which get sublimate.
Now we mix the remaining salt and sand in water. The sodium chloride will dissolve in water and we can collect the sand which will settle in the base.
Now we are left with salt solution. By the process of evaporation water will vaporise and we are left with salt .

Question:41

Arun has prepared $0.01\mathrm{\%}$ (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
(a) $1.00gofNaCl+100gofwater$

(b) $0.11gofNaCl+100gofwater$

(c) $0.01gofNaCl+99.99gofwater$

(d) $0.10gofNaCl+99.90gofwater$
Answer: C
Mass Percent: amount of solute in 100 gram of solution.
Mass percent (%) $=\frac{massofsolute\times 100}{massofSolution}$
Option a:
Mass percent (%) $=\frac{1gram\times 100}{1gram+100gram}=0.99$
Option b:
Mass percent (%) $=\frac{0.11gram\times 100}{0.11gram+100gram}=0.1099$
Option c:
Mass percent (%) $=\frac{0.1gram\times 100}{0.1gram+99.99gram}=0.01$
Option d:
Mass percent (%) $=\frac{0.1gram\times 100}{0.1gram+99.90gram}=0.1$
Hence the correct answer is C.

Question:42

Calculate the mass of sodium sulphate required to prepare its 20% (mass percent) solution in 100 g of water?
Solution:
Mass Percent: amount of solute in 100 gram of solution.
Mass Percent (%) $=\frac{massofsolute\times 100}{massofSolution}$
Let the mass of sodium sulphate required be = m gram
Mass of the solvent (water) = 100 gram
The mass of solution = (m + 100) gram (sum of mass of solute and solvent)
$20=\frac{m\times 100}{m+100}$
$\Rightarrow 20m+2000=100m$
$\Rightarrow 2000=80m$
$\Rightarrow m=25g$