Goa Board Class 9 Solutions for Chemistry – Is Matter Around Us Pure?
( English Medium)
Page No. 15:
Question 1:
What is meant by a substance?
Solution:
A substance is a pure single form of matter. It cannot be separated into other kinds of matter by any physical process. For example: sugar, sodium chloride.
Concept insight: In this answer, you should write the definition of substance and give at least two examples.
Question 2:
List the points of differences between homogeneous and heterogeneous mixtures.
Solution:
Homogeneous mixtures | Heterogeneous mixtures |
1. They have uniform compositions. 2. The components of homogeneous mixtures are not physically distinct. 3. They have no visible boundaries of separation between the constituents. |
1. They have non-uniform compositions. 2. They contain physically distinct parts. 3. They have visible boundaries of separation between the constituents. |
Concept insight: This is an important question from exam point of view.
Page No. 18:
Question 1:
Differentiate between homogeneous and heterogeneous mixtures with examples.
Solution:
Homogeneous mixtures | Heterogeneous mixtures |
The substances are completely mixed together in these mixtures. | The substances remain separate in these mixtures. |
They have uniform compositions throughout their mass. | They have non-uniform compositions throughout their mass. |
They have no visible boundaries of separation between the constituents. | They have visible boundaries of separation between the constituents. |
Salt in water, sugar in water are examples of homogeneous mixtures. | Mixtures of sodium chloride and iron filings and oil and water are examples of heterogeneous mixtures. |
Concept insight: This is an important question from exam point of view. It can come for 2 or 3 marks.
Question 2:
How are sol, solution and suspension different from each other?
Solution:
Solution | Sol | Suspension |
A solution is a homogeneous mixture of two or more substances. |
A sol is a heterogeneous mixture.
|
Suspension is a heterogeneous mixture. |
The particles of a solution are smaller than 1 nm (10-9 metre) in diameter. So, they cannot be seen by naked eyes. |
The size of particles of a sol is too small to be individually seen by naked eyes. It is between 1 nm and 100 nm in diameter.
|
The particles of a suspension can be seen by the naked eye. Their size is larger than 100 nm in diameter. |
Due to small particle size, they do not scatter a beam of light passing through them. Thus, the path of light is not visible in a solution. |
Sols are big enough to scatter a beam of light passing through them and make its path visible.
|
The particles of a suspension scatter a beam of light passing through it and make its path visible. |
The solute particles do not settle down when left undisturbed, i.e., a solution is stable. | They do not settle down when left undisturbed, i.e., a sol is quite stable. | The solute particles settle down when a suspension is left undisturbed, i.e., a suspension is unstable. |
The solute particles cannot be separated from the mixture by the process of filtration. | The solute particles cannot be separated from the mixture by the process of filtration. | They can be separated from the mixture by the process of filtration. |
Concept insight: This is an improtant question from exam point of view. It can come for 2 or 3 marks.
Question 3:
To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.
Solution:
Page No. 24-I:
Question 1:
How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25°C), which are miscible with each other?
Solution:
The process of distillation is used for the separation of a mixture containing kerosene and petrol since the difference in their boiling points is more than 25°C.
(i) Take the mixture in the distillation flask and fit it with a thermometer.
(ii) Arrange the apparatus as shown in the given figure.
(iii) Heat the mixture slowly keeping a close watch on thermometer.
(iv) Petrol has a lower boiling point than kerosene. So, petrol vaporizes first and condenses in the condenser which is finally collected in the beaker.
(v) Kerosene is left behind in the distillation flask.
Concept insight: Write the name of the process first and then followed by detailed steps
Question 2:
Name the technique to separate:
(i) butter from curd
(ii) salt from sea-water
(iii) camphor from salt
Solution:
(i) Centrifugation
(ii) Evaporation
(iii) Sublimation
Concept insight: For answering this question you should first analyse what type of mixture is given and then accordingly write the name of the process.
Question 3:
What type of mixtures are separated by the technique of crystallisation?
Solution:
The crystallisation method is used to purify solids.
Concept insight: This question can be asked for 1 mark in the exam.
Page No. 24-II:
Question 1:
- cutting of trees,
- melting of butter in a pan,
- rusting of almirah,
- boiling of water to form steam,
- passing of electric current, through water and the water breaking down into hydrogen and oxygen gases,
- dissolving common salt in water,
- making a fruit salad with raw fruits, and
- burning of paper and wood.
Solution:
(a) Cutting of trees – Physical change
(b) Melting of butter in a pan – Physical change
(c) Rusting of almirah – Chemical change
(d) Boiling of water to form steam – Physical change
(e) Passing of electric current through water and the water breaking down into hydrogen and oxygen gases – Chemical change
(f) Dissolving common salt in water – Physical change
(g) Making a fruit salad with raw fruits – Physical change
(h) Burning of paper and wood – Chemical change
Concept insight: Recall the definition of physical and chemical change. If the chemical composition is changing then it is a chemical change and if the chemical composition is remaining the same, then it is not a chemical change.
Question 2:
Try segregating the things around you as pure substances or mixtures.
Solution:
Pure substances: water, iron nail, pencil lead, sugar, etc.
Mixtures: Air, blood, butter, milk, steel, paper, etc.
Concept insight: Recall the definition of pure substances and mixtures and then apply it in all the things around you.
Page No. 28:
Question 1:
Which separation techniques will you apply for the separation of the following?
(a) Sodium chloride from its solution in water.
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride.
(c) Small pieces of metal in the engine oil of a car.
(d) Different pigments from an extract of flower petals.
(e) Butter from curd.
(f) Oil from water.
(g) Tea leaves from tea.
(h) Iron pins from sand.
(i) Wheat grains from husk.
(j) Fine mud particles suspended in water.
Solution:
(a) Evaporation
(b) Sublimation
(c) Filtration
(d) Chromatography
(e) Centrifugation
(f) Separating funnel
(g) Sieving
(h) Magnetic separation
(i) Sieving and winnowing
(j) Sedimentation, decantation and filtration
Concept insight: For answering this question you should first analyse what type of mixture is given and then accordingly write the name of the process.
Question 2:
Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Solution:
1. Take some water as solvent in a pan.
2. Add some insoluble tea leaves into it.
3. Add some sugar and milk into it which act as solutes.
4. Boil the mixture.
5. Filter the prepared tea through a sieve.
6. The tea leaves are left on the seive.
7. The filtrate tea is collected in a cup.
Concept insight: Recall the steps to prepare tea and then apply the terms given in the question to it.
Question 3:
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. Which salt has the highest solubility at this temperature?
(d) What is the effect of change of temperature on the solubility of a salt?
Solution:
(a) Since 62 g of potassium nitrate is dissolved in 100 g of water to prepare a saturated solution at 313 K so, 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 crystals of 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 rate of solubility of a salt increases with increase in temperature.
Concept insight: For answering this question, you should know what is solubility, saturated solution and variation of solubility with temperature.
Page No. 29:
Question 4:
Explain the following giving examples:
(a) Saturated solution
(b) Pure substance
(c) Colloid
(d) Suspension
Solution:
(a) At any particular temperature, a solution that has dissolved as much solute as it is capable of dissolving is said to be a saturated solution. Examples of saturated solutions are soft drinks and nitrogen in Earth’s soil.
(b) A pure substance is one which is made up of only one kind of particles. It cannot be separated into different constituents by physical or chemical processes. Some examples of pure substances are bromine, nitric acid, calcium oxide, etc.
(c) Colloids are heterogeneous mixtures in which the size of solute particles is intermediate between those in true solutions and those in suspensions and are big enough to scatter light. Some examples are: Milk, blood, paint, etc.
(d) A suspension is a heterogeneous mixture in which the small particles of a solid are spread throughout a liquid without dissolving in it. The particles of a suspension can be seen by the naked eye. Some examples of suspension are Chalk – water mixture, flour in water, milk of Magnesia etc.
Concept insight: In this answer, you should write the definitions and give at least two examples.
Question 5:
Classify each of the following as a homogeneous or heterogeneous mixture.
soda water, wood, air, soil, vinegar, filtered tea.
Solution:
Homogeneous – soda water, air, vinegar, filtered tea
Heterogeneous – wood, soil
Concept insight: For answering this question, you should recall what homogeneous and heterogeneous mixtures are. If the mixture has one phase, it is homogeneous otherwise it is heterogeneous.
Question 6:
How would you confirm that a colourless liquid given to you is pure water?
Solution:
Evaporate the colourless liquid on a low flame. If no residue is left, then it is pure water otherwise not.
Concept insight: You should recall here that a pure substance is one which does not have any other substance or impurity present in it.
Question 7:
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
Solution:
Ice, iron, hydrochloric acid, calcium oxide, and mercury are pure substances.
Concept insight: You should recall here that a pure substance is one which does not have any other substance or impurity present in it and cannot be separated into other kinds of matter by physical processes. It can be an element or a compound.
Question 8:
Identify the solutions among the following mixtures.
(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water.
Solution:
Sea water, air, and soda water are solutions.
Concept insight: Recall that solutions are homogeneous mixtures. Homogeneous mixtures have only one phase.
Question 9:
Which of the following will show “Tyndall effect”?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution.
Solution:
Milk and starch solution are colloids. Thus, they will show tyndall effect since colloids show tyndall effect.
Concept insight: Recall that only colloids and suspensions will show Tyndall effect. Solutions do not show Tyndall effect.
Question 10:
Classify the following into elements, compounds and mixtures.
(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
Solution:
Elements: sodium, silver, tin, and silicon.
Compounds: calcium carbonate, methane, and carbon dioxide.
Mixtures: soil, sugar solution, coal, soap, air, and blood.
Concept insight: For answering this questions, recall the definitions of elements, compounds and mixtures and then apply the concepts to the substances given in the question.
Page No. 30:
Question 11:
Which of the following are chemical changes?
(a) Growth of a plant
(b) Rusting of iron
(c) Mixing of iron filings and sand
(d) Cooking of food
(e) Digestion of food
(f) Freezing of water
(g) Burning of a candle.
Solution:
Growth of a plant, rusting of iron, cooking of food, digestion of food, and burning of candle are chemical changes.
Concept insight: Recall the definition of physical and chemical change. If the chemical composition is changing then it is a chemical change and if the chemical composition is remaining the same, then it is a physical change.