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Page No 12:

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 particle and they have the same composition throughout. Pure substances may be compounds but not mixtures. All elements and compounds are pure substances.

Hence, the correct answer is option (b).

Page No 12:

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. It is a chemical change. Iron is converted to hydrated ferric oxide.

Hence, the correct answer is option (c).

Page No 12:

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:

Sulphur is soluble in carbon disulphide and hence forming a true solution. A true solution is homogeneous and does not show Tyndall effect.

Hence, the correct answer is option (d).

Page No 12:

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:

Iodine dissolved in alcohol is known as iodine tincture.

Hence, the correct answer is option (d).

Page No 12:

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.

Hence, the correct answer is option (c).



Page No 13:

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:

A physical change involves a change in the physical properties (like shape, size, state, etc) of the substance. Melting of iron metal, bending of iron rod, drawing a wire of iron metal are physical changes.

Hence, the correct answer is option (c).
 

Page No 13:

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 chemical change is the one in which a new substance is formed. Decaying of wood and burning of wood are chemical changes.

Hence, the correct answer is option (a).

Page No 13:

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) (ii), (iii) and (iv)

Answer:

A2B is a compound made up of elements A and B in a fixed ratio with entirely different properties from those of its constituent elements A and B. The composition of a compound is fixed.

Hence, the correct answer is option (c).

Page No 13:

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:

Since X and Y cannot be broken down into simpler substances and hence they are elements.
A compound is a substance made up of two or more elements chemically combined in a fixed proportion by mass. Therefore, P is a compound, having a fixed composition.

Hence, the correct answer is option (d).



Page No 14:

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

Answer:

(a) Mercury and water are immiscible liquids and can be separated using separating funnel.
(b) Ammonium chloride is sublimable compound and Potassium chloride is not. They can be separated using sublimation.
(c) Common salt is soluble in water and sand is not. Sand can be separated from an aqueous solution of common salt by using a funnel. A mixture of common salt and water can be separated using evaporation.
(d) Salt is soluble in water. Kerosene and an aqueous solution of salt can be separated using separating funnel. Salt and water can be separated using evaporation. 

Page No 14:

Question 11:

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 containing beads will be more effective as it has more surface area available. 
Hence, the correct answer is (b).

Page No 14:

Question 12:

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

Answer:

Salt can be recovered from its solution by crystallisation. It is a better technique than 'evaporation' because it removes soluble impurities also, which do not get removed in the process of evaporation.

Page No 14:

Question 13:

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

Answer:

Sea-water is considered homogeneous because it has salts dissolved in it. It is considered a heterogeneous solution as it contains sand, mud and parts of plants. 

Page No 14:

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.

Answer:

A mixture of acetone and salt solution can be separated using simple distillation since their boiling points differ by more than 25°C. Acetone will evaporate and salt solution will remain.

Page No 14:

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.

Answer:

(a) When a saturated solution of potassium chloride prepared at 60°C is allowed to cool at room temperature, crystals of potassium chloride will be formed. 
(b) Initially, sugar will be obtained as water will get evaporated. But on dry heating sugar gets charred and it turns black.
(c) The black coloured compound, iron sulphide (FeS) is formed when a mixture of iron filings and sulphur powder is heated strongly. It is a new substance that has properties entirely different from Fe and S.

Page No 14:

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.

Answer:

The size of particles of a colloidal solution is smaller than that of suspension and hence they don't settle down. The particles of suspension are bigger in size and they settle down under the effect of gravity.

Page No 14:

Question 17:

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

Answer:

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.

Page No 14:

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.

Answer:

(a) Physical Property
(b) Chemical Property
(c) Physical Property
(d) Chemical Property



Page No 15:

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 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:

50% mass by volume means 50 grams of solute is dissolved in 100 mL of solution. 
Student 'A' dissolved 50 g of NaOH in 100 mL of water (solvent) which is incorrect. Student 'B' dissolved 50 g of NaOH in 100 g of water (solvent), which is incorrect. Student 'C' dissolved 50g of NaOH in 100 mL of solution.
Therefore, student 'C' made the desired solution. 

Page No 15:

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) A 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.

Answer:

(a) Sublimation of dry ice (solid to gas)
(b) Diffusion of ink into water
(c) Diffusion or dissolution of solid into liquid
(d) Evaporation
(e) Centrifugation
(f) Sedimentation
(g) Tyndall effect 

Page No 15:

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.

Answer:

Sample 'B' which boils at 102°C contains impurities and will not freeze at 0°C. The freezing point will be less than 0°C.

Page No 15:

Question 22:

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 soft and malleable. When it is alloyed with copper or silver it hardens and can be moulded into different shapes.

Page No 15:

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?

Answer:

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

Page No 15:

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.

Answer:

(a) Mixture of acetone and water: Simple distillation
(b) Mixture of kerosene and petrol: Distillation
(c) Mixture of oil-water: Separating funnel
(d) Mixture of ammonium chloride and sodium chloride: Sublimation
(e) Mixture of pigments from a flower petal extract: Chromatography

Page No 15:

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.

Answer:

(a) heterogenous; centrifugation
(b) physical; chemically
(c) water; chloroform
(d) fractional distillation
(e) scattering; Tyndall effect; colloidal



Page No 16:

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.

Answer:

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 a fixed composition.

Page No 16:

Question 27:

Give some examples of Tyndall effect observed in your surroundings?

Answer:

Examples of Tyndall effect:
(i) When light rays enter into a dark room through 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 the cinema hall.

Page No 16:

Question 28:

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 separate the alcohol-water solution using separating funnel since both are miscible and form a solution. Immiscible liquids can be separated using separating funnel.

Page No 16:

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.

Answer:

Calcium carbonate on heating gives calcium oxide and carbon dioxide.
CaCO3 → CaO + CO2

(a) It is a chemical change as there is the formation of new compounds.

(b) Calcium oxide, when dissolved in water, gives a basic solution. Carbon dioxide, when dissolved in water, gives an acidic solution.

CaO + H2O → Ca(OH)2
                       (Basic)    

CO2 +  H2O → H2CO3
                      (Acidic)
                                
 

Page No 16:

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.

Answer:

(a) Iodine
(b) Bromine
(c) Graphite
(d) Carbon
(e) Phosphorous
(f) Oxygen

Page No 16:

Question 31:

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

Answer:

Elements- Cu, Zn, F2, O2, Diamond, Hg
Compounds- CaCO3, H2O



Page No 17:

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

Answer:

Chlorine gas, Iron, Aluminium, Iodine, Carbon and Sulphur powder are not compounds.

Page No 17:

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.

Answer:

In fractional distillation, a fractionating column that is packed with glass beads is used. The glass beads increase the surface area available for the vapours. The vapours on coming in contact with the glass beads lose energy and are condensed quickly. The length of the column increases the efficiency of the process.


Page No 17:

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?

Answer:

(a) Alloys are homogeneous mixtures because they have a uniform composition throughout.
(b) No, a solution is not always liquid. It can be solid or gaseous also.
(c) No, a solution is a homogeneous mixture.

Page No 17:

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?

Answer:


Part A: When iron is mixed with sulphur and heated then it forms iron sulphide.
Fe + S → FeS

On treatment with dilute HCl, it produces ferrous chloride and hydrogen sulphide gas.
FeS + HCl → FeCl2 + H2S↑

Hydrogen sulphide can be identified by its smell. It has a smell of rotten egg and it turns lead acetate solution black.

Part B: Iron and Sulphur do not react when HCl is added to the mixture. Iron will react with HCl and it gives out H2 gas.
Fe + 2HCl → FeCl2 + H2

Hydrogen gas burns with a pop sound and hence can be identified by bringing a burning candle near it.

Page No 17:

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.
 

Answer:

(i) The components of the ink will travel with water and we would see three bands on the filter paper at various lengths.
(ii) This technique is known as chromatography.
(iii) Separation of pigments present in chlorophyll is done using chromatography.

Page No 17:

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?
 

Answer:

(a) Milk is a colloidal solution and hence shows the Tyndall effect.
(b) True solutions don't show the tyndall effect because their particles are very small in size and they don't scatter light.
(c) Detergent solution and sulphur solution also show the Tyndall effect.



Page No 18:

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.

Answer:

(a) Drying of a shirt in the sun: Physical Change
Water converts from liquid state to gaseous state. There is no new substance formed.

(b) Rising of hot air over a radiator: Physical Change
There is no new substance formed.

(c) Burning of kerosene in a lantern: Both Physical and Chemical change
First, kerosene vapourises, and then it burns to produce new substances.

(d) Change in the colour of black tea on adding lemon juice to it: Physical change
No new substance is formed.

(e) Churning of milk cream to get butter: Physical change
No change in the composition. Only the separation of components takes place by the physical phenomenon, centrifugation.

Page No 18:

Question 39:

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, Sarika has a higher mass by mass percentage.

(b) Solution made by Ramesh:
Mass by mass % = 10(10+100)×100=10110×100=9.09%

Solution made by Sarika:
Mass by mass %=10100×100=10%

Page No 18:

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?

Answer:

Mixture: Sand + Iron + Ammonium Chloride + Sodium Chloride
Step 1: Separate the iron fillings from the given mixture by using a magnet.
Step 2: Separate ammonium Chloride by sublimation.
Step 3: Dissolve the remaining mixture into the water. Sodium Chloride will be dissolved completely. Sand can be separated using filter paper.
Step4: Evaporate the filtrate to get sodium chloride.

Page No 18:

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 + 100g of water
(b) 0.11g of NaCl + 100g of water
(c) 0.01 g of NaCl + 99.99g of water
(d) 0.10 g of NaCl + 99.90g of water

Answer:

Mass by mass %=mass of solutemass of solution×100(a) Mass by mass % =1.00(1+100)×100=0.99(b) Mass by mass %=0.110.11+100×100=0.1098(c) Mass by mass %=0.0199.99+0.01×100=0.01(d) Mass by mass %=0.100.10+99.90×100=0.1

Hence, the correct answer is option (c).

Page No 18:

Question 42:

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 be g.
The mass of the solution would be (100 + x)g 

Mass by mass% = Mass of soluteMass of solution×10020 =xx+100×10020x + 2000 = 100x80x=2000x=200080=25 g
Hence, 25g of sodium sulphate is required to prepare its 20%  (mass percent) solution in 100g of water.



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