Chapter 2: Is Matter Around Us Pure?

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Chapter 2: Is Matter Around Us Pure?

Class: IX

Exemplar sheet 2

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 1

(b)

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 2

(c)

 

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 3

(d)       

 

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 4

(d)

 

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 5

(c)

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 6

(c)       

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 7

(a)       

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 → A2 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) (ii), (iii) and (iv)

Answer 8

(c)       

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 9:

(d)

Short Answer Questions

Question 10:

Suggest separation technique(s) one would need to employ to

separate the following mixtures.

(a) Mercury and water

(b) Potassium chloride and ammonium chloride

(c) Common salt, water and sand

(d) Kerosene oil, water and salt

Answer 10:

(a) Separation by using separating funnel

(b) Sublimation

(c) Filtration followed by evaporation  or

     Centrifugation followed by evaporation/distillation

(d)  Separating funnel is used to separate kerosene oil followed by evaporation or

     distillation.

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 11:

Hint— Look for the larger surface area. The presence of beads in tube (a) would provide a larger surface area for cooling.

 

Question 12:

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

Answer 12:

Crystallization

 

Question 13:

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

Answer 13:

Homogeneous— mixture of salts and water only

Heterogeneous— contains water,salts, decayed plant, mud etc.

 

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 14:

Hint— Distillation methd is used, since acetone is more volatile it will separate out first.

 

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 15:

(a) Solid potassium chloride will separate out.

(b) Initially the water will evaporate and then sugar will get charred.

(c) Iron sulphide will be formed.

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 16:

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

Question 17:

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

Answer 17:

Both fog and smoke have gas as the dispersion medium. The only difference is that the dispersed phase in fog is liquid and in smoke it is a solid.

 

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 18:

Physical properties – (a) and (c)

Chemical properties – (b) and (d)

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 19:

‘C’ has made the desired solution

Mass by volume % = (Mass of solute ×100) / Volume of solution

= (50 ×100) / 100

= 50 % mass by volume

 

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 20:

(a) Sublimation                        (e) Centrifugation

(b) Diffusion    (f) Sedimentation

(c) Dissolution/diffusion          (g) Scattering of light (Tyndall effect)

(d) Evaporation, diffusion

 

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 21:

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

Answer 22:

As compared to gold alloyed with silver or copper, pure gold is very soft. So in order to   provide strength to gold, it is alloyed.

 

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 23:

This element is a metal. Other characteristics expected to be possessed by the element are–lustre, malleability, heat and electrical conductivity.

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 24:

(a) Evaporation or distillation

(b) Distillation

(c) Separation by using separating funnel

(d) Sublimation

(e) Chromatography

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 25:

(a) heterogenous, centrifugation

(b) physical, chemically

(c) water, chloroform (hint– density of water is less than that of chloroform)

(d) fractional distillation

(e) scattering, Tyndall effect, colloidal

 

Question 26:

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

Answer 26:

It is a pure substance because chemical composition of sugar crystals is

same irrespective of its source.

Question 27:

Give some examples of Tyndall effect observed in your surroundings?

Answer 27:

Hint— We can see tyndall effect, when we passes light through a heterogeneous mixture. 

      Example, when sunlight passes through dense forest.

Question 28:

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

Answer 28:

Hint— Water and alcohol are miscible.

 

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 29:

(a) Chemical change

 (b) Acidic and basic solutions can be prepared by dissolving in water

      CaO+H₂O →Ca(OH)₂ (basic solution)

      CO₂ + H₂O→H₂CO₃     (acidic solution)

 

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 30:

(a) Iodine

(b) Bromine

(c) Graphite

(d) Carbon

(e) Sulphur, phosphorus

(f) Oxygen

 

Question 31:

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

                          

Answer 31:

Elements      Compounds

      Cu                      CaCO3

      Zn           H2O

      F2

      O2

      Diamond (carbon)

      Hg

 

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 32:

Chlorine gas, Iron, Aluminium, Iodine, Carbon, Sulphur powder.

Long Answer Questions

Question 33:

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

Answer 33:

The fractionating column packed with glass beads provides a surface for the vapours to collide and lose energy so that they can be quickly condensed and distilled. Also length of the column would increase the efficiency.

 

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 34:

Hint– (a) Homogenous mixture, because they have a uniform composition throughout

(b) No, solid solutions and gaseous solutions are also possible. Examples brass and air

(c) No, solution is a homogenous mixture of two or more substances

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 35:

Part A

      Fe (s) + S (s) Heat FeS (s)

      FeS + 2 HCl (aq) → FeCl2 + H2 S

Part B

      Fe (s) + S (s)→ Mixture of iron filings and sulphur

      When dilute HCl is added to it

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

      Sulphur remains unreacted

      H2S gas formed has a foul smell and on passing through lead acetate solution, it turns the    

      solution black. Hydrogen gas burns with a pop sound.

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 36:

Hint– (i) Three different bands will be observed.

             

         (ii) Chromatography

         (iii) To separate the 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?

        

Answer 37:

(a) Milk is a colloid and would show Tyndall effect.

(b) Salt solution is a true solution and would not scatter light.

(c) Detergent solution, sulphur solution.

 

Question 38:

Classify each of the following, as a physical or a chemical change. Give reasons.

(a) Drying of a shirt in the sun.

(b) Rising of hot air over a radiator.

(c) Burning of kerosene in a lantern.

(d) Change in the colour of black tea on adding lemon juice to it.

(e) Churning of milk cream to get butter.

Answer 38:

 Hint — Physical changes  — (a), (b), (e)

             Chemical changes — (c), (d)

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 39:

(a) No.

      Mass % = ( Mass of solute ×100) / (Mass of solute + Mass of solvent )

(b) Solution made by Ramesh

      Mass % = (10 *100) / (10 +100)

      = (10 / 110) ×100 = 9.09%

      Solution made by Sarika0

      Mass % = (10 / 100) * 100  = 10%

      The solution prepared by Sarika has a higher mass % than that prepared by Ramesh.

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 40:

Hint–

Step-1  Separate iron filings with the help of a magnet

Step-2  Sublimation of the remaining mixture separates ammonium chloride

Step-3  Add water to the remaining mixture, stir and filter

Step-4  The filtrate can be evaporated to get back sodium chloride.

 

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) 0.0l g of NaCl + 99.99g of water

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

Answer 41:

(c)

Mass % = Mass of solute / (Mass of solute+ Mass of solvent) * 100

= 0.01 / (0.01 + 99.99) * 100

= (0.01 / 100) * 100

= 0.01 g

Question 42:

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

Answer 42:

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

 20% = x / (x+100) * 100

  20 x + 2000= 100 x

  80 x = 2000

  x = 2000 / 80

     = 25 g