CBSE - Is Matter Around Us Pure

A material that is composed of only one type of particles is called pure substance. All the constituent particles of a pure substance have same chemical nature.

Homogeneous mixtures	Heterogeneous mixtures
Homogeneous mixtures have uniform composition.	Heterogeneous mixtures have non uniform composition.
It has no visible boundaries of separation between its constituents.	It has visible boundaries of separation between its constituents.

A homogeneous mixture is a mixture having a uniform composition throughout the mixture. For example, mixtures of salt in water, sugar in water, copper sulphate in water, iodine in alcohol, alloy, and air have uniform compositions throughout the mixtures.

On the other hand, a heterogeneous mixture is a mixture having a non-uniform composition throughout the mixture. For example, composition of mixtures of sodium chloride and iron fillings, salt and sulphur, oil and water, chalk powder in water, wheat flour in water, milk and water are not uniform throughout the mixtures.

Sol	Solution	Suspension
They are heterogeneous in nature.	They are homogeneous in nature.	They are heterogeneous in nature.
They scatter a beam of light and hence show Tyndall effect.	They do not scatter a beam of light and hence do not show Tyndall effect	They scatter a beam of light and hence show Tyndall effect.
They are quite stable.	Examples of solution are: salt in water, sugar in water.	Examples of suspension are: sand in water, dusty air

Mass of solute (sodium chloride) = 36 g (Given)

Mass of solvent (water) = 100 g (Given)

Then, mass of solution = Mass of solute + Mass of solvent

= (36 + 100) g

= 136 g

Therefore, concentration (mass by mass percentage) of the solution

A mixture of two miscible liquids having a difference in their boiling points more than 25�C can be separated by the method of distillation. Thus, kerosene and petrol can be separated by distillation.



In this method, the mixture of kerosene and petrol is taken in a distillation flask with a thermometer fitted in it. We also need a beaker, a water condenser, and a Bunsen burner. The apparatus is arranged as shown in the above figure. Then, the mixture is heated slowly. The thermometer should be watched simultaneously. Kerosene will vaporize and condense in the water condenser. The condensed kerosene is collected from the condenser outlet, whereas petrol is left behind in the distillation flask.

(i) butter from curd

-> By Centrifugation

(ii) salt from sea-water

-> By Evaporation

(iii) camphor from salt

-> By Sublimation

The crystallisation method is used to purify solids.

• Cutting of trees
-> Physical change

• Melting of butter in a pan
-> Physical change

• Rusting of almirah
-> Chemical change

• Boiling of water to form steam
► Physical change

• Passing of electric current through water, and water breaking down into hydrogen and oxygen gas
-> Chemical change

• Dissolving common salt in water
-> Physical change

• Making a fruit salad with raw fruits
-> Physical change

• Burning of paper and wood
-> Chemical change

(a) Sodium chloride from its solution in water.
► Evaporation

(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
► Sublimation

(c) Small pieces of metal in the engine oil of a car.
► Filtration or Centrifugation or decantation

(d) Different pigments from an extract of flower petals.
► Chromatography

(e) Butter from curd.
► Centrifugation

(f) Oil from water.
► Using separating funnel

(g) Tea leaves from tea.
► Filtration

(h) Iron pins from sand.
► Magnetic separation

(i) Wheat grains from husk.
► Winnowing

(j) Fine mud particles suspended in water.
► Centrifugation

First, water is taken as a solvent in a saucer pan. This water (solvent) is allowed to boil. During heating, milk and tea leaves are added to the solvent as solutes. They form a solution. Then, the solution is poured through a strainer. The insoluble part of the solution remains on the strainer as residue. Sugar added to the filtrate, which dissolves in the filtrate. The resulting solution is the required tea.

Pragya tested the solubility of three different substances at different temperatures and collected the data as given below( results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).

(a)  What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?

(b)  Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.

(c)  Find the solubility of each salt at 293 K. What salt has the highest solubility at this temperature?

(d)  What is the effect of change of temperature on the solubility of a salt? 

Answer

(a) Since 62 g of potassium nitrate is dissolved in 100g of water to prepare a saturated solution at 313 K, 31 g of potassium nitrate should be dissolved in 50 g of water to prepare a saturated solution at 313 K.

(b) The amount of potassium chloride that should be dissolved in water to make a saturated solution increases with temperature. Thus, as the solution cools some of the potassium chloride will precipitate out of the solution.

(c) The solubility of the salts at 293 K are:
Potassium nitrate – 32 g
Sodium chloride – 36 g
Potassium chloride – 35 g
Ammonium chloride – 37 g

Ammonium chloride has the highest solubility at 293 K.

(d) The solubility of a salt increases with temperature.

(a) Saturated solution
(b) Pure substance
(c) Colloid
(d) Suspension

Answer

(a) Solution in which no more solute can be dissolved at a particular temperature is known as saturated solution. For example in aqueous solution of sugar no more sugar can be dissolved at room temperature.

(b) A pure substance is a substance consisting of a single type of particles i.e., all constituent particles of the substance have the same chemical properties. For example water, sugar, salt etc.

(c) A colloid is a heterogeneous mixture whose particles are not as small as solution but they are so small that cannot be seen by naked eye. When a beam of light is passed through a colloid then the path of the light becomes visible. For example milk, smoke etc.

(d) A suspension is a heterogeneous mixture in which solids are dispersed in liquids. The solute particles in suspension do not dissolve but remain suspended throughout the medium. For example Paints, Muddy water chalk water mixtures etc.

Homogeneous mixtures: Soda water, air, vinegar, filtered tea
Heterogeneous mixtures: Wood, soil
Note: Pure air is homogeneous mixture but Polluted air is heterogeneous mixture.

Take a sample of colourless liquid and put on stove if it starts boiling exactly at 100 ºC then it is pure water. Any other colourless liquid such as vinegar always have different boiling point. Also observe carefully that after some time whole liquid will convert into vapour without leaving any residue.

Which of the following materials fall in the category of a "pure substance"?

(a) Ice
(b) Milk
(c) Iron
(d) Hydrochloric Acid
(e) Calcium oxide
(f) Mercury
(g) Brick
(h) Wood
(i) Air

Answer

The following materials fall in the category of a "pure substance":
(a) Ice
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury

(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water

Answer

The following mixtures are solutions:
(b) Sea water
(c) Air
(e) Soda water

(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution

Answer

Tyndall effect is shown by colloidal solution. Here milk and starch solution are colloids therefore milk and starch solution will show Tyndall effect.

(a) Sodium
(b) Soil
(c) Sugar solution
(d) Silver
(e) Calcium carbonate
(f) Tin
(g) Silicon
(h) Coal
(i) Air
(j) Soap
(k) Methane
(l) Carbon dioxide
(m) Blood

Answer

Elements: Sodium, Silver, Tin and Silicon.

Compounds: Calcium carbonate, Methane and carbon dioxide.

Mixtures: Soil, Sugar, Coal, Air, Soap and Blood.

(a) Growth of a plant
(b) Rusting of iron
(c) Mixing of iron fillings and sand
(d) Cooking of food
(e) Digestion of food
(f) Freezing of water
(g) Burning of candle

Answer

The following changes are chemical changes:
(a) Growth of a plant
(b) Rusting of iron
(d) Cooking of food
(e) Digestion of food

(g) Burning of candle

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)
Ans. (b) (i) and (iii)
Explanation: Mixture is not a pure substance, hence option (iii) is ruled out. Nickel is an element and hence pure substance; so, option (iv) is incorrect.

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
Ans. (c) corrosion and it is a chemical change
Explanation: Rusting of iron is a chemical change and it leads to corrosion.

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
Ans. (d) homogeneous and does not show Tyndall effect
Explanation: Solution is a homogenous mixture and does not show Tyndall effect. Colloid and suspension are heterogeneous mixtures and show Tyndall effect. Mixture of Sulphur and carbon disulphide is a solution.

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
Ans. (d) iodine in alcohol

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)
Ans. (c) (i) and (iv)
Explanation: Soil is a heterogeneous mixture and wood too has heterogenous composition.

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)
Ans. (c) (i), (iii) and (iv)
Explanation: A new substance is formed only in case of rusting of iron. In all other cases, no new substance is formed and hence they are physical changes.

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)
Ans. (a) (i) and (ii)
Explanation: Burning and decay result in formation of new substances and they cannot be reversed. Hence, these are chemical changes.

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) 
Ans. (c) (i), (iii) and (iv)

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)

Ans. (d) (i), (iii) and (iv) 
 

(a) Mercury and water
Ans. Separation by using separating funnel
(b) Potassium chloride and ammonium chloride
Ans. Sublimation
(c) Common salt, water and sand
Ans. Filtration followed by evaporation
or
Centrifugation followed by evaporation/distillation
(d) Kerosene oil, water and salt
Ans. Separation by using separating funnel to separate kerosene oil followed by
evaporation or distillation. 

Ans. The tube (a) will be more affective as a condenser. Presence of marbles increases surface area. This allows more time for condensation and hence it would be more effective than the column without marbles

Crystallization

Homogeneous— mixture of salts and water only

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

Acetone can be separated from this mixture by distillation. The boiling points of acetone is much lower than that of water. So, it will evaporate and can be collected after condensation.

(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.

Ans. (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

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.

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.

(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.
Ans. Physical properties – (a) and (c)
Chemical properties – (b) and (d) 

‘C’ has made the desired solution

(a) Dry ice is kept at room temperature and at one atmospheric pressure.
Ans. Sublimation
(b) A drop of ink placed on the surface of water contained in a glass spreads throughout
the water.
Ans. Diffusion
(c) A potassium permanganate crystal is in a beaker and water is poured into the beaker
with stirring.
Ans. Dissolution/diffusion
(d) A acetone bottle is left open and the bottle becomes empty.
Ans. Evaporation, diffusion
(e) Milk is churned to separate cream from it.
Ans. Centrifugation
(f) Settling of sand when a mixture of sand and water is left undisturbed for some time.
Ans. Sedimentation
(g) Fine beam of light entering through a small hole in a dark room, illuminates the
particles in its paths.
Ans. Scattering of light (Tyndall effect)

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

Pure gold is very soft as compared to gold alloyed with silver or copper. Thus for providing strength to gold, it is alloyed.

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

(a) A volatile and a non-volatile component.
Ans. Evaporation or distillation
(b) Two volatile components with appreciable difference in boiling points.
Ans. Distillation
(c) Two immiscible liquids.
Ans. Separation by using separating funnel
(d) One of the components changes directly from solid to gaseous state.
Ans. Sublimation
(e) Two or more coloured constituents soluble in some solvent.
Ans. Chromatography

(a) A colloid is a __________ mixture and its components can be separated by the technique known as __________.
Ans. heterogeneous, centrifugation
(b) Ice, water and water vapour look different and display different __________ properties but they are __________ the same.
Ans. physical, chemically
(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 __________.
Ans. water, chloroform (hint– density of water is less than that of 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 __________.
Ans. 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 __________ of light by milk and the phenomenon is called __________. This indicates that milk is a __________ solution.
Ans. scattering, Tyndall effect, colloidal

It is a pure substance because chemical composition of sugar crystals is same irrespective of its source.

Following are some examples of Tyndall effects.
(a) A beam of light coming through ventilation near ceiling.
(b) Beam of light coming through canopy of trees.

Water and alcohol are miscible.

(a) Is this a physical or a chemical change?
Ans. 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. 

Ans. Acidic and basic solutions can be prepared by dissolving the products of the above
process in water
CaO + H2O → Ca(OH)₂ (basic solution)
CO₂ + H2O → H₂CO₃ (acidic solution)

(a) Name a lustrous non-metal.
Ans. Iodine
(b) Name a non-metal which exists as a liquid at room temperature.
Ans. Bromine
(c) The allotropic form of a non-metal is a good conductor of electricity. Name the
allotrope.
Ans. Graphite
(d) Name a non-metal which is known to form the largest number of compounds.
Ans. Carbon
(e) Name a non-metal other than carbon which shows allotropy.
Ans. Sulphur, phosphorus
(f) Name a non-metal which is required for combustion.
Ans. Oxygen

Ans. Elements: Cu, Zn, O2, F2, Hg, Diamond
Compounds: H2O, CaCO3

(a) Chlorine gas
(b) Potassium chloride
(c) Iron
(d) Iron sulphide
(e) Aluminium
(f) Iodine
(g) Carbon
(h) Carbon monoxide
(i) Sulphur powder

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

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.

(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?

Ans. (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 
 

Following reaction takes place when part A is heated:
Fe+S →  FeS
When dil. Hydrochloric acid is added to iron sulphide, following reaction takes place and hydrogen sulphide gas is evolved.

FeS + 2HCI →  FeCI₂  + H₂S
Hydrogen sulphide is a foul smelling gas and smells like rotten egg. When dil. Hydrochloric acid is added to the mixture of iron and Sulphur, following reaction takes place and hydrogen gas is evolved.

Fe + S → 2HCI → FeCI + H_{2 +} S

In this case, Sulphur does not participate in the reaction.
When a burning matchstick is brought near the evolved gas the matchstick burns with a pop sound. This confirms the evolution of hydrogen gas.

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. 

Ans. (i) Three different bands will be observed.
(ii) Chromatography
(iii) To separate the pigments present in Chlorophyll.

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?

Ans. (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. 
 

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.

Ans. Physical changes —(a), (b), (e)
Chemical changes— (c), (d) 

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.

ANS (a) No

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

Components of given mixture can be separated by following methods:
(i) By using a magnet: Hovering a magnet over the mixture will result in iron fillings getting stuck to the magnet. Thus, iron will be separated.
(ii) Sublimation: The remaining mixture is heated in a china dish. Ammonium chloride is a sublime and hence it will evaporate without undergoing the liquid phase. Crust of ammonium chloride can be collected by placing and inverted funnel on top of china dish.
(iii) Sedimentation, decantation and filtration: The remaining mixture is dissolved in water. The mixture is allowed to settle for some time. Sand, being insoluble in water, settles at the bottom. Liquid is decanted in another beaker. Then, the liquid is filtered to remove any trace of sand in it.
(iv) Evaporation: The liquid is now a solution of salt in water. This is heated in a beaker so that water evaporated. Once all the water evaporates, we get salt in the beaker. 

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

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

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