NCERT Exemplar Solutions Class 9 Science Is Matter Around us Pure

Multiple Choice Questions (MCQs)

Question 1:  Which of the following statements are true for pure substances?
(i)   Pure substances contain only one kind of particles.
(ii)  Pure substances may be compounds or mixtures.
(iii) Pure substances have the same composition throughout.
(iv) Pure substances can be exemplified by all elements other than nickel,
(a) (i) and (ii)           (b) (i) and (iii)            (c) (iii) and (iv)             (d) (ii) and (iii)

Solution 1:   (b) (i) and (iii). 

A pure substance is one which consists of atoms or molecules of only one kind. Everywhere, they have the same composition.

Question 2:  Rusting of an article made up of iron is called
(a) corrosion and it is a physical as well as chemical change
(b) dissolution and it is a physical change
(c) corrosion and it is a chemical change
(d) dissolution and it is a chemical change

Solution 2:  (c) corrosion and it is a chemical change.

Corrosion is a chemical shift because rust is a chemical compound that is entirely different from the iron component (hydrated iron oxide, Fe₂O₃-*H₂0) (Fe). Corrosion is

Physical changes are those changes in which no new substances are produced. The substances do not alter their identity and certain physical processes, such as ice melting, glowing of an electric bulb, breaking of glass tumbler, etc., can easily be restored to their original form.

Chemical changes, on the other hand, are those modifications in which new substances are created. The substances involved modify their identity and are transformed into completely new substances that cannot be restored to their original shape, such as paper burning, iron rusting, magnesium wire burning, etc.

Question 3:  A mixture of sulfur and carbon disulfide is
(a) heterogeneous and shows Tyndall effect
(b) homogeneous and shows Tyndall effect
(c) heterogeneous and does not show Tyndall effect
(d) homogeneous and does not show Tyndall effect

Solution 3:   (a) heterogeneous and shows Tyndall effect.

A heterogeneous colloid is a mixture of sulfur and carbon disulfide that displays the Tyndall effect. The particles are large enough in a colloidal solution to scatter light.

The dispersion of light into colloidal particles is referred to as the Tyndall effect. Colloids, although they tend to be homogeneous, are actually heterogeneous in nature.

Question 4:  Tincture of iodine has antiseptic properties. This solution is made by dissolving

(a) iodine in potassium iodide       

(b) iodine in Vaseline

(c) iodine in water                      

(d) iodine in alcohol

Solution 4:   (d) iodine in alcohol.

Iodine tincture is formed by the dissolution of iodine into alcohol.

Question 5:  Which of the following are homogeneous in nature?

(i) Ice                         

(ii) Wood                     

(iii) Soil                 

(iv) Air

(a) (i) and (iii)            (b) (ii) and (iv)              (c) (i) and (iv)           (d) (iii) and (iv)

Solution 5:    (c) (i) and (iv).

Options (i) Ice and (iv) Air are homogeneous in nature since they are not clearly visible to their particles. There is a uniform composition of a homogeneous mixture across its mass. It has no visible separation limits between its different components, such as air, sugar solution, brass, etc.

Throughout its mass, a heterogeneous mixture does not have a uniform composition, it has visible boundaries of separation between its different constituents, e.g. soil, wood, blood etc.

Question 6:  Which of the following are physical changes?
(i)   Melting of iron metal
(ii)  Rusting of iron
(iii) Bending of an iron rod
(iv) Drawing a wire of iron metal
(a) (i), (ii) and (iii)                                            (b) (i), (ii) and (iv)
(c) (i), (iii) and (iv)                                            (d) (ii), (iii) and (iv)

Solution 6:   (c) (i), (iii) and (iv).

Options I Iron metal melting, (iii) Iron rod bending, and (iv) Iron metal wire drawing are physical changes because iron changes its shape in three processes, not its chemical composition. Its chemical composition is altered during the rusting of iron.

Question 7:  Which of the following are chemical changes?
(i)   Decaying of wood
(ii)  Burning of wood
(iii) Sawing of wood
(iv) Hammering of a nail into a piece of wood
(a) (i) and (ii)                                          (b) (ii) and (iii)
(c) (iii) and (iv)                                       (d) (i) and (iv)

Solution 7:   (a) (i) and (ii).

Options I Wood decay and (ii) Wood combustion are chemical changes because the chemical composition of wood is altered in these processes and new substances are created which cannot be transformed back into their original form.
In (iii) the sawing of wood and (iv) the hammering of a nail into a piece of wood, there is no change in the chemical composition of the wood, so (iii) and (iv) 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 2A + 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)

Solution 8:   (c) (i), (iii) and (iv). 

A₂B is a compound composed of a fixed ratio of two elements, A and B. A compound's (e.g., A2B) properties are completely different from those of its constituent elements (e.g. A and B). A compound's composition is set.

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)

Solution 9:   (d) (i), (iii) and (iv).

In this reaction, X and Y cannot be broken down by chemical reactions into simpler compounds, so X and Y are elements. A compound is a substance consisting of two or more elements combined chemically by mass in a fixed proportion, so P is a compound with a fixed composition.

Short Solution Type Questions......................

Question 10:  Suggest separation technique(s) one would need to employ to separate the following mixtures.
(a) Mercury and water
(b) Potassium chloride and ammonium chloride
(c) Common salt, water and sand
(d) Kerosene oil, water and salt

Solution 10:  (a) Mercury and water are separated by decantation by means of a separating funnel, because the separation by a separating funnel of two immiscible liquids depends on the difference in their mass.

Consequently, mercury is heavier than water, forms the lower layer and is isolated from water.

(b) Potassium chloride and ammonium chloride are separated by a process of sublimation because ammonium chloride is a sublimate that leaves potassium chloride behind it.

Sublimation is the phase in which a solid transforms directly into heating vapours and vapours transform into cooling solids without going into a liquid state.

(c) Common salt, water and sand are separated by-

The following are divided by traditional salt, water and sand:

(i) The method of decantation (or filtration) is used to isolate the sand from the common salt solution in the water, since the common salt is water soluble, while the sand is water insoluble. Sand can then be isolated as an insoluble material by filtration, as residue and filtrate will be a natural salt solution in water.

(ii) The method of evaporation is used for the separation of common salt from water. Water evaporates, and the trace of common salt remains.

Evaporation separates common salt dissolved in water. 

(d) Kerosene oil, water and salt are separated by

(i) Using a separating funnel, decantation is used to isolate petroleum oil from salt solution in water as it forms separate layers (salt is soluble in water).

(ii) To isolate salt from water, evaporation is used. 

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

Solution 11: The condenser in the distillation apparatus would be more powerful in Figure (a) since the beads present would have more surface area for cooling the vapor that move through it.

Question 12:  Salt can be recovered from its solution by evaporation. Suggest some other technique for the same?

Solution 12:   Salt can be extracted by 'crystallization' from its solution as well.

Crystallization is a better approach than 'evaporation' because it also eliminates soluble impurities that are not extracted in the evaporation process.

Question 13:  The ‘sea-water’ can be classified as a homogeneous as well as heterogeneous mixture. Comment.

Solution 13:   As it contains dissolved salts in it, 'sea-water' is called homogeneous. As it also includes various insoluble components such as sand, bacteria, shells made of calcium carbonate and so many other things, it can be considered heterogeneous.

Question 14:  While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56°C). What technique can be employed to get back the acetone? Justify your choice.

Solution 14:   Acetone is water soluble, a homogeneous mixture is obtained and isolation can therefore not be used by separating the funnel. By simple distillation, acetone can be recovered since the difference in the boiling points of acetone and water is more than 25°C.

Acetone Boiling Point - 56°C

Water boiling point -100°C

Acetone boils at 56°C in the distillation flask and converts into vapors and can be collected after condensation in the flask. 

Question 15:  What would you observe when
(a) a saturated solution of potassium chloride prepared at 60°C is allowed to cool at room temperature?
(b) an aqueous sugar solution is heated to dryness?
(c) a mixture of iron filings and sulfur powder is heated strongly?

Solution 15:  (a) If a saturated potassium chloride solution prepared at 60 °C is allowed to cool at room temperature, potassium chloride crystals can form.

Initially, sugar is obtained by the evaporation of water. But the sugar gets burned on dry heat and it turns black.                        

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 16:    The colloidal particles are not heavy and  are smaller. They always stay in a zig-zag movement state, called the Brownian movement, which counters the force of gravity acting on colloidal particles and thus helps to provide colloidal soles with stability by not allowing them to settle down. Colloidal particles, apart from this, are charged and repel each other.

This reality also does not permit the colloidal solution particles to settle down. Whereas suspension particles are bigger, heavier and have less motion, they settle down due to gravity.

Question 17:  Smoke and fog both are aerosols. In what way are they different?

Solution 17:    The dispersion medium i.e. air, is the same for smoke and fog, but they vary in the dispersed phase. Solid carbon particles are dispersed in the air in smoke, while liquid water particles are dispersed in the air in fog.

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

Solution 18:

a)      Since no new compound is made, it is a physical property because steel is an alloy and alloy is a homogeneous mixture of two or more metals or with non-metallic elements of metallic elements.

b)      It is a chemical property since, with the evolution of hydrogen gas, there is a chemical reaction between zinc and hydrochloric acid and the compound zinc chloride is formed.
Zn(s)+ 2HCI (dil)→ ZnCI₂(aq) + H₂(g)

c)       It is a physical property since no new material is created by cutting with a knife.

d)      It is a chemical property, because the reaction of metal oxide and water produces a new compound.

                         M₂0 (s) + H₂0 (Z) → 2MOH (aq)
                    Metal oxide Water                           Metal hydroxide

Question 19:  The teacher instructed three students ‘A’, ‘B’ and V 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 lOOg of water while V dissolved 50g of NaOH in water to make 100 mL of solution. Which one of them has made the desired solution and why?

Solution 19:  Student 'C' did it correctly in the given question because 50 percent (mass by volume) means 50 g of solution for every 100 mL of solution and not in 100 mL of solvent.

Mass/volume percent =      mass of solute (in g) /Volume of solution (in mL)  x 100                                           

                                         =    50/100   x 100 = 50%

  Student A in 100 mL of water (solvent) dissolved 50 g of NaOH, which is wrong.

 In 100 g of water (solvent), Student 'B' dissolved 50 g of NaOH, which is incorrect.

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. Illuminate the particles in its paths.

Solution 20:

a)      The process is sublimation because it sublimates, leaving no trace while dry ice (solid C02) is held at room temperature at a single atmospheric pressure.

b)      This method is diffusion since the mixing of one substance (ink) into another substance (water) continues until a uniform mixture is produced during diffusion.

c)       Dissolution/diffusion is this method as the crystal of potassium permanganate is dissolved in water.

d)      As acetone evaporates when left open in the bottle, this process is evaporation.

e)      Centrifugation is this operation. Milk is placed into a large centrifuge machine in a closed jar. The milk rotates at a very high speed when the pump is turned on. Milk splits into 'cream' and 'skimmed milk' because of this. The cream is lighter and floats over the skimmed milk and can be removed afterwards.

f)       This phase is sedimentation, since sand does not dissolve entirely in water and forms a suspension and settles on the bottom for a period of time while it is left undisturbed.

g)      This implies the Tyndall effect, i.e. the dispersion of light by colloidal solution or fine suspension particles. In the air, dust particles are suspended, dispersing the light coming from a tiny opening.

Question 21:  You are given two samples of water labeled as ‘A’ and ‘B’. Sample A’ boils at 100°C and sample ‘B’ boils at 102°C.Which sample of water will not freeze at 0°C? Comment.

Solution 21:   At 0°C, sample '6' will not freeze as it is impure water. It is because sample 'B' boils at 102°C while the pure water's boiling point is 100°C. That means there are impurities in this sample. There is only pure matter with a sharp melting point.

Question 22:  What are the favorable qualities given to gold when it is alloyed with copper or silver for the purpose of making ornaments?

Solution 22:  Gold is a soft metal that can change its shape easily with a little force. It is not, therefore, appropriate for making ornaments. But gold becomes harder and stronger, when it is alloyed with copper or silver, and its brittleness decreases. Thus, it becomes ideal for making 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 23:   Therefore, since the given element is sonorous and highly ductile, it is classified as a metal. Any other anticipated features are

i)        It should have metallic lustre and it can be polished.

ii)       The conductors for heat and electricity should be fine.

iii)     It needs to be ductile.

iv)     It needs to be malleable.

v)      High tensile strength should be present.

vi)     It should also have high densities and a boiling point/melting point.

vii)   It ought to be hard (except sodium and potassium which are soft metals).

viii)  At room temperature, it should be solid (except mercury, which is liquid at room temperature).

Although the characteristics of non-metals are:

i)        Non-metals do not conduct electricity and are neither malleable nor ductile.

ii)       Certain properties of metals and some other properties of non-metals are shown by metalloids.

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 colored constituents soluble in some solvent.

Solution 24:

a)      Example: acetone and water mixture.

         Method: For separating a mixture of volatile and non-volatile materials, simple distillation can be used.

b)      Example: kerosene and petrol mixture.

           Method: Two volatile components with an appreciable difference in boiling points can be separated using simple distillation.

c)       Example: Mixture of water and mustard oil.

           Method: A mixture of immiscible liquids is separated by the Separating Funnel.

d)      Example: ammonium chloride mixture and common salt.

         Method: The mixture in which one part transitions directly from solid to gas may be distinguished by sublimation.

e)      Example: a mixture of various pigments from a flower petal extract.

         Method: It is possible to use the chromatography method to isolate two different substances found in the same solution.

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 vapor 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 25:

a)      A colloid is a heterogeneous mixture and its components can be separated by the technique known as centrifugation.

b)      Ice, water and water vapor look different and display different physical properties but they are chemically 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 water and the lower layer will be that of chloroform (It is because water is lighter than chloroform).

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.

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 26:   According to the law of constant composition or definite proportions; it is often considered, irrespective of the source of a chemical substance, to be composed of the same elements mixed in the same fixed proportion by mass.

Thus, it would be a pure material in the light of the law above. That is because sugar obtained from various sources, such as sugar cane and beetroot, would have the same structure.

Question 27:   Give some examples of Tyndall effect observed in your surroundings?

Solution 27:    Examples of the Tyndall effect:

i)        As when the sun passes through a dense forest canopy.

ii)       When, through a tiny hole, a fine beam of light enters a dark space.

Question 28:  Can we separate alcohol dissolved in water by using a separating funnel? If yes, then describe the procedure. If not, explain.

Solution 28:   No, by using a separating funnel, alcohol cannot be isolated from water because alcohol in water is totally 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.

Solution 29:

(a) it is a chemical alteration in the given phenomenon in question since the composition of the product produced differs from the substance taken. The reaction that is involved is
                                            CaC0₃   →    CaO         +         C0₂
                                            Calcium carbonate      Calcium oxide    Carbon dioxide
(b) Yes,
(i) When CaO dissolves in water, forms the basic solution that is calcium hydroxide.
                                          CaO(s) + H₂0(Z) → Ca(OH)₂(ag)
                                                                         Calcium hydroxide
(ii) When CO2(g) dissolves in water, forms  the acid solution that is carbonic acid.
                                         C0₂(g)+ H₂0(/) → H₂C0₃(ag)
                                                                       Carbonic acid

Question 30:  Non-metals are usually poor conductors of heat and electricity. They are non-lustrous, non-sonorous, non-malleable and are colored.
(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 30:

a)      The lustrous non-metal is iodine.

b)      Bromine is a non-metal that, at room temperature, remains as a liquid.

c)       The allotropic form of carbon is graphite and it is a strong conductor of electricity.

d)      Carbon is a non-metal of which the largest numbers of compounds are known to form.

e)      Phosphorus exhibits allotropy as a non-metal other than carbon.

f)       A non-metal that is needed for combustion is oxygen.

Note: wood is neither an element nor a compound. It is a mixture of both.

Question 32:  Which of the following are not compounds?
(a) Chlorine gas                                            (b) Potassium chloride
(c) Iron                                                         (d) Iron sulfide
(e) Aluminum                                               (f) Iodine
(g) Carbon                                                   (h) Carbon monoxide
(i) Sulfur powder

Solution 32: The followings are not compounds:

(a) Chlorine gas                            

(c) Iron

(e) Aluminum                                      

(f) Iodine

(g) Carbon                                             

(i) Sulfur powder

Long Solution Type Questions.........................

Question 33:  Fractional distillation is suitable for separation of miscible liquids with a boiling point difference of about 25 K or less. What part of fractional distillation apparatus makes it efficient and possess an advantage over a simple distillation process. Explain using a diagram.

Solution 33:  The most significant component of the fractional distillation apparatus is the fractionating column. With some glass beads in it, this column is given.

It helps to obstruct the upward motion of the two liquids' vapors. The high boiling liquid vapors are condensed earlier (at lower level). The energy released (latent heat) helps to carry the vapours in the fractionating column of low boiling liquid to a height.

The benefits are as presented below:

i)        With a boiling point difference of around or less than 25 K, this method can separate the liquids.

ii)       The Both evaporation and condensation take place simultaneously during the process.

iii)     It is also possible to isolate a mixture (like petroleum) through a fractional distillation process that involves many components.

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 34:

a)      Alloys are homogeneous metal mixtures because: (i) they display the properties of their components and (ii) they are of variable composition. Brass, for example, is considered a blend because it demonstrates the properties of its components, copper and zinc, and has a variable composition (amount of Zn in brass can vary from 20 to 35 per cent).

b)      A solution is, in general, not always a liquid. Solids and gases may also be involved, e.g. alloys are solid in solid solution. Air is a gas solution contained in gases.

c)       Colloidal solutions, though they appear to be homogeneous, are heterogeneous in nature.

Question 35:  Iron filings and sulfur were mixed together and divided into two parts, ‘A’ and ‘S’. Part ‘A’ was heated strongly while part ‘S’ was not heated. Dilute hydrochloric acid was added to both the parts and evolution of gas was seen in both the cases. How will you identify the gases evolved?

Solution 35: A compound FeS is formed by the reaction between iron filings and sulphur as part 'A' is heated. FeS will react with dil HCl to form FI2S gas that smells of rotten eggs when dilute HCI is added to part A and will render lead acetate paper black.
Fe (s) + S(s) → FeS(s)
FeS(s) +2HCI(dil.) → FeCI₂(ag) + H₂S(g)
(CHjCOO)₂Pb(ag) + H₂S(g) → PbS(s) + 2CH₃COOH
       Lead acetate                  Black ppt       Acetic acid
as It is not heated  part 'S', so S is a mixture of iron filings and sulphur powder. When dil is added to it with HCL, iron filings respond with dill. HLI to form H₂(g), which burns with a 'pop' sound if the burning match stick is brought near it .
Fe (s) + 2HCI (dil.)             FeCI₂ (ag) + H₂(g)

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 figure. 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 colored components?
(b) Name the technique used by the child.
(c) Suggest one more application of this technique.

Solution 36:

(a) Three different spots of color are obtained at different heights on a strip.

(b) The technique of chromatography (paper chromatography) is used.

(c) For the isolation of medicines from the blood, the chromatography process is also used.

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 Figure. 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 37:

a)      Since milk is a colloidal solution, the milk sample has been illuminated and its particles are large enough to distribute the light, so they disperse the light that passes through it. "The phenomenon observed is called the "effect of Tyndall".

b)      Because the salt solution is a true solution, i.e. the particle size of the liquid is too small to spread the light; it does not exhibit the "Tyndall effect.

c)       Gold sol, arsenius sulphide (As2S3) sol., blood etc. are examples of colloids.

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 color of black tea on adding lemon juke to it.
(e)  Churning of milk cream to get butter, f Thinking Process

Solution 38:

a)      Physical change, since water evaporation takes place, but the composition of the substance does not change.

b)      Physical change, as it often includes just air movement, not a change in air composition.

c)       First physical change, when vaporizing kerosene. After that, when new products are produced, the burning of kerosene is a chemical change.

d)      Physical changes because only dissolution occurs.

e)      Physical change, because the structure does not change. The physical phenomenon, centrifugation takes place only by the separation of materials.

Question 39:  During an experiment the students were asked to prepare a 10% (mass/mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution.
(a) Are the two solutions of the same concentration?
(b) Compare the mass % of the two solutions.

Solution 39:

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

The Ramesh solution has a lower percentage (9.09 percent) by mass than that of Sarika (10 percent ).

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 40:

i)        Using a magnet, remove the iron filings.

ii)       Place the mixture on a petri dish paper and move a bar magnet over the mixture several times. Iron filings are bound to and are isolated from the magnet.

iii)     By sublimation, remove ammonium chloride from sand and sodium chloride. The remaining mixture is sent to China dish  and sublimated. Ammonium chloride can be vaporized and converted into vapours, producing NH4CII on condensation (s). There will be sand and sodium chloride left in China dish.

iv)     Extract sand from sodium chloride after dissolution by filtration.

v)      In water, dissolve the sand and sodium chloride. This will dissolve sodium chloride. Solution is filtering. Sand would be left and separated as residue.

vi)     By evaporation or crystallization, get sodium chloride. There is sodium chloride present in the filtrate. So, to get back sodium chloride evaporates the filtrate to dryness or use crystallization.

Question 41:  Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
(a) 1.00 g of NaCl+ 100 g of water
(b) 0.11 g of NaCl+ 100 g of water
(c) 0.01 g of NaCl+ 99.99 g of water
(d) 0.10 g of NaCl+ 99.90 g of water

Solution 41:

While these remaining percentages do not equate to the percentage of the sodium chloride solution prepared by Arun in water. So, that's wrong.

Question 42:  Calculate the mass of sodium sulphate required to prepare its 20% (mass per cent) solution in 100 g of water?

Solution 42:   In the given question:

The mass percentage of sodium sulphate solution = 20 percent

Solvent Mass = 100 g

Let the solute (sodium sulphate) mass = x g