NCERT Solutions Class 9 Science Chapter 2 Is Matter around Us Pure

NCERT Solutions Class 9 Science 

The NCERT Solutions in English Language for Class 9 Science Chapter – 2 (Is Matter around Us Pure) has been provided here to help the students in solving the questions from this exercise.

Chapter – 2 (Is Matter around Us Pure)

Question

1. What is meant by a substance?
Answer – It is a pure single form of matter. A substance has definite properties and compositions. Example – Iron

2. List the points of difference between homogeneous and heterogeneous mixtures.
Answer –

Question

1. Differentiate between homogenous and heterogeneous mixtures with examples.
Answer – The following are the differences between heterogeneous and homogenous mixtures.

2. How are sol, solution and suspension different from each other?
Answer –

3. To make a saturated solution, 36g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.
Answer –
Mass of solute (NaCl) = 36 g
Mass of solvent (H₂O) = 100 g
Mass of solution (NaCl + H₂O) = 136 g
Concentration = Mass of solute/Mass of solution x 100
Concentration = = 26.47%
Hence, the concentration of the solution is 26.47%

Question

1. How will you separate a mixture containing kerosene and petrol (the difference in their boiling points is more than 25°C), which are miscible with each other?
Answer – The mixture of two miscible liquids such as kerosene and petrol whose boiling points differ by more than 25°C can be easily separated by the technique of simple distillation.

The separation is based upon the principle that the boiling point of more volatile (low building liquid of the mixture. The vapour almost exclusively consists of the more volatile liquid Likewise at the boiling of the less volatile (high boiling) liquid, vapours almost entirely consists  of the less volatile liquid since the more volatile liquid has already distilled over.      

2. Name the techniques used to separate the following:
(i) Butter from curd
(ii) Salt from seawater
(iii) Camphor from salt
Answer –

(i) Butter from curd can be separated by the technique of centrifugation.

(ii) Salt from sea water can be separated by the technique of crystallization or by the evaporation.

(iii) Camphor is sublimable but salt is not. So, camphor can be separated from salt sublimation technique.

3. What types of mixtures are separated by the technique of crystallisation?
Answer – Homogeneous mixtures such as common salt solution and copper sulphate solution separated by technique of crystallization. 

Question

1. Classify the following as physical or chemical changes:

• 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 water breaking into hydrogen and oxygen gases.
• Dissolving common salt in water
• Making a fruit salad with raw fruits, and
• Burning of paper and wood

Answer – The following is the classification into physical and chemical change:

2. Try segregating the things around you as pure substances and mixtures.
Answer – Listed below are the classifications based on pure substances and mixtures:

Exercise 

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.
Answer –
(a) Evaporation : Water will evaporate leaving behind sodium chloride. 

(b) Sublimation : Ammonium chloride will be collected as sublime. 

(c) Filtration : Pieces of metal can be separated by filtration. 

(d) Chromatography: Pigments (coloured components) from the extract of flower plants can be separated by chromatography. 

(e) Centrifugation : Butter will get separated upon centrifugation. 

(f) Separating funnel: Oil and water can be separated by the use of separating funnel. 

(g) Filtration : Upon filtration through a sieve, tea leaves will be collected on the sieve. 

(h) Magnetic separation : A magnet will attract iron pins and not sand particles. 

(i) Sieving : Wheat grains from husk can be separated with the help of sieves. 

(j) Sedimentation : As a result of sedimentation, mud particles will settle as precipitate. It can be separated later on by filtration. 

2. Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate, and residue.
Answer – Steps for making tea –
(i) Take 100ml of water as solvent and boil it for few minutes.
(ii) When solvent boil, add solutes, i.e., milk, tea leaves, sugar.
(iii) Now again boil the solution for few minutes. Sugar is soluble in water thus it will be dissolve.
(iv) Colour of tea leaves goes into solution as filtrate. The remaining tea leaves being insoluble remains as residue.
(v) Now filter the solution. Collect the filtrate in cup. The insoluble tea leaves will be left behind as residue.

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 a 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 313K?
(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?
Answer –
(a) Solubility of potassium nitrate at  313K = 62 g
100 g of water contains potassium nitrate = 62 g
50 g of water contains potassium nitrate =   × 50 = 31 g
Hence, 31 g of potassium nitrate is required.

(b) Pragya would observe crystals of potassium chloride as the solubility of potassium chloride in water decreases when a saturated solution of potassium chloride loses its heat during the cooling process below 353 K.

(c) Listed below is the solubility of each salt at 293 K
Solubility of Potassium nitrate = 32 g
Solubility of Sodium chloride = 36 g
Solubility of Potassium chloride = 35 g
Solubility of Ammonium chloride = 37 g
We can observe from this data that ammonium chloride has the highest solubility at 293K.

(d) Solubility of a salt is directly proportional to the change of temperature. Solubility of salt will increase when the temperature increases and solubility will decrease when the temperature decreases .

4. Explain the following, giving examples.
(a) Saturated solution
(b) Pure substance
(c) Colloid
(d) suspension
Answer –
(a) Saturated solution: It is the state of a solution at a specific temperature when a solute is no more soluble in the solvent at that present temperature. Solute will be again soluble in solvent only with an increase in the temperature. Example: Excess carbon leaves off as bubbles from the carbonated water solution which is already saturated with carbon.

(b) Pure substance: A substance that comprises of only one kind of molecules, atoms or particles is called as pure substance. Example: Sulphur, diamond etc. 

(c) Colloid: A colloid is a homogenous non-crystalline substance in which the size of solute particles are bigger than that of the true solution. Example: Blood, Ink.

(d) Suspension: It is a heterogeneous mixture that comprises of solute particles that are large enough to settle down. E.g. paints, etc.

5. Classify each of the following as a homogeneous or heterogeneous mixture.
soda water, wood, air, soil, vinegar, filtered tea.
Answer – The following is the classification of the given substances into homogenous and heterogenous mixtures.

6. How would you confirm that a colourless liquid given to you is pure water?
Answer – We can confirm if a colourless liquid is pure by setting it to boil. If it boils at 100°C, it is said to be pure. But if there is a decrease or increase in the boiling point, we infer that water has added impurities, hence not pure.

7. Which of the following materials fall into the category of “pure substance”?
(a) Ice
(b) Milk
(c) Iron
(d) Hydrochloric acid
(e) Calcium oxide
(f) Mercury
(g) Brick
(e) Wood
(f) Air.
Answer – Following substances from the above-mentioned list are pure substances:

• Iron
• Ice
• Hydrochloric acid
• Calcium oxide
• Mercury

8. Identify the solutions among the following mixtures.
(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water
Answer – The following are the solutions from the above-mentioned list of mixtures:

• Sea water
• Air
• Soda water

9. Which of the following will show the “Tyndall effect”?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution
Answer – Tyndall effect is exhibited by only milk and starch solution from the above-mentioned list of solutions.

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
Answer –

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 candle
Answer – Out of the given list, the following are chemical changes:
Growth of a plant, rusting of iron, cooking of food, digestion of food and burning of candle.

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