Haryana State Board HBSE 9th Class Science Solutions Chapter 2 Is Matter Around Us Pure Textbook Exercise Questions and Answers.
Haryana Board 9th Class Science Solutions Chapter 2 Is Matter Around Us Pure
HBSE 9th Class Science Is Matter Around Us Pure Intext Questions and Answers
Questions from Sub-section 2.1
Question 1.
What is meant by a substance?
Answer:
A substance is a pure single form of matter. It consists of a single type of particles i.e. all the constituent particles in the substance are identical in their chemical nature.
Question 2.
List the points of differences between homogeneous and heterogeneous mixtures.
Answer:
Homogeneous mixtures: Mixtures that have uniform composition are called homogeneous mixtures. Heterogeneous mixtures: Mixtures that have non-uniform composition are called heterogeneous mixtures.
Questions from Sub-section 2.2
Question 1.
Differentiate between homogeneous and heterogeneous mixtures with examples.
Answer:
Homogeneous mixtures: Those mixtures that have uniform composition throughout their masses are called homogeneous mixtures, for example, a mixture of sugar in water, a mixture of salt in water, a mixture of alcohol in water.
Heterogeneous mixtures: Those mixtures that do not have uniform composition throughout their masses are called as heterogeneous mixtures, for example, a mixture of sand and salt, a mixture of salt and sugar.
Question 2.
How are the solution, suspension and colloid (sol) different from each other?
Answer:
Following are the differences among solution, suspension, colloid (sol.):
Solution:
1. This solution is homogeneous and transparent, e.g. solution of salt in water.
2. Here, the size of the solute particle is 10-9m.
3. Here, the particles of solute cannot be seen under the microscope.
4. Here the solute particles cannot be separated by filtration.
5. Due to the small size, the particles of solution cannot scatter the rays of light passing through it. So that in the solution the path of light is not visible.
Suspension:
1. This solution is heterogeneous and opaque, e.g., muddy water, paint.
2. Here, the size of solute parti-cles used to be 107 m or more than of that.
3. Here, the solute particles can be seen with naked eyes too.
4. Here, the solute particles can be separated by the filtration method.
5. Suspended particles scattered the rays of light, by which its path become visible
Colloid (Sol.):
1. This, the solution is homogeneous but little transparent e.g., milk, blood, ink, tooth-paste.
2. Here, the size of solute particles is between 109 to 107 m, i.e., their size is bigger than the size of solution particles.
3. Here, the solute particles can only be seen through the powerful microscope.
4. Here, the solute particles also cannot be separated by filtration.
5. Colloidal particles are as big as these scattered the rays of light and make its path, visible.’
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:
At temperature 293K –
Mass of solute substance (Sodium Chloride) = 36g
Mass of solvent (Water) = 100g
Mass of solution = Mass of solute substance + Mass of solvent = 36g + 100g = 136g
= \(\frac {36}{136}\) × 100 = 26.47
Questions from Sub-section 2.3
Question 1.
How will you separate a mixture containing kerosene and petrol, which are miscible with each other ? The difference in their boiling points is more than 25°C.
Answer:
Petrol and kerosene oil which are miscible to each other, their mixture is separated by the fractional distillation method. This method is based upon the fact that different components have different boiling points. Because the difference of boiling points of petrol and kerosene is more than 25°C, liquid with low boiling point will separate first and after some intervals liquid with high boiling point will be separated after becoming distilled.
Question 2.
Name the technique to separate:
(i) butter from curd
(ii) salt from sea-water
(iii) camphor from salt
Answer:
(i) Butter is separated from curd by centrifugation method.
(ii) Salt from sea water is separated by an evaporation method.
(iii) Camphor is separated from salt by the sublimation method.
Question 3.
What type of mixtures are separated by the technique of crystallization?
Answer:
Crystallization is a method by which pure solid is separated from a solution in the form of a pure crystal for example obtaining of salt from seawater.
Questions from Sub-section 2.4
Question 1.
Classify the following as chemical and physical 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 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.
Answer:
Chemical change: Rusting of almirah; passing of electric current, through the water and the water breaking down into hydrogen and oxygen gases; burning of paper and wood.
Physical change: Cutting of trees, melting of butter in a pan; boiling of water to form steam, dissolving of common salt in water; making of fruit salad with raw fruits.
Question 2.
Try segregating the things around you as pure substances or mixtures.
Answer:
Pure Substances: Iron, gold, silver, copper, aluminium, sugar, salt etc.
Mixture: Sea-water, minerals, soil, air, beverages etc.
HBSE 9th Class Science Is Matter Around Us Pure Textbook Questions and Answers
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) Butter from curd.
(e) Oil from water.
(f) Tea leaves from tea.
(g) Iron pins from sand.
(h) Wheat grains from husk.
(i) Fine mud particles suspended in water.
(j) hues from the essence of the crushed flower petals.
Answer:
(a) To separate sodium chloride from the solution of water, the evaporation method is adopted.
(b) To separate ammonium chloride from a mixture of sodium chloride and ammonium chloride, a sublimation method is adopted since ammonium chloride is a volatile substance.
(c) To separate a piece of metal from the engine oil of car, the filtration method is adopted.
(d) In order to separate butter from curd centrifugation method is adopted.
(e) To separate oil from water separating funnel is used, since water and oil both are immiscible liquids.
(f) To separate tea leaves from tea, the filtration method is brought in use. For filtration, tea strainer is used.
(g) To separate iron from sand, the magnetic separation method is adopted, because iron is attracted towards the magnet.
(h) To separate wheat grains from chaff, the threshing method is adopted, because with the threshing method chaff being lighter in weight, flies away with the wind and the wheat grains being heavier in weight, directly falls bn the ground.
(i) Tiny particles of soil floating in water can be separated by means of loading method for the particulates of soil get heavier by alum and thus, they settle down at the bottom.
(j)To separate various hues (dyes) from the essence of the crushed flower petals chromatography method is followed.
Question 2.
Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Answer:
We will prepare tea by using the given words in the following manner:
1. Solvent: Take water in the pan in the form of solvent and keep it on the burner.
2. Solute: Add sugar to water in the form of solute.
3. Solution: The mixture of water and sugar will become the solution.
4. Dissolve: Sugar will dissolve in water and make a solution.
5. Soluble: Sugar dissolves in water being miscible and after boiling, it is also a soluble substance in milk.
6. Insoluble: In the mixture of water and sugar, add tea leaves in the form of an insoluble substance and boil it.
7. Filtrate and residue: After boiling of tea leaves, filter the tea with a filtrate strainer. Use filtrate tea to drink arid and throw away the residue remaining in the strainer.
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:
| Substance Dissolved | Temperature in Kelvin (K) | ||||
| 283 | 293 | 313 | 333 | 353 | |
| Solubility | |||||
| Potassium nitrate | 21 | 32 | 62 | 106 | 167 |
| Sodium chloride | 36 | 36 | 36 | 37 | 37 |
| Potassium chloride | 35 | 35 | 40 | 46 | 54 |
| Ammonium chloride | 24 | 37 | 41 | 55 | 66 |
(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in SO 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 a change of temperature on the solubility of a salt?
Solution:
(a) According to the Question:
The essential quantity of potassium nitrate for a saturated solution of potassium nitrate in 100 g of water at 313 K= 62 g
The required quantity of potassium nitrate for a saturated solution of potassium nitrate in 1 g of water at 313 K = \(\frac {62}{100}\)g
The required quantity of potassium nitrate for saturated solution in 50 g of water at 313 K = \(\frac {62}{100}\) × 50g = 31 g
(b) Pragya obtains a saturated solution of potassium chloride at 353 K and leaves the solution at room temperature (293 K) to cool down when the solution will cool down, then it will be a most saturated solution because at room temperature, it will have (54-35) 19 g more potassium chloride than saturation.
(c) At 293 K temperature in 100 g of water the solubility of potassium nitrate, sodium chloride, potassium chloride and ammonium chloride are 32 g, 36 g, 35 g, and 37 g, respectively. So, at this temperature, the solubility of ammonium chloride salt will be the most.
(d) On changing the temperature, the solubility of the salt change positively, i.e. the solubility of salt increases with the increase in temperature.
Question 4.
Explain the following with examples:
(a) saturated solution
(A) pure substance
(c) colloid
(d) suspension.
Answer:
(a) Saturated Solution:
If at the given fixed temperature, the solute does not dissolve in solution, it is called a saturated solution, i.e., at the given temperature when in a solution the solute dissolves more than the capacity of the solution, it is called saturated solution. For instance-take 50 ml of water in a beaker, now add little quantity of salt into it gradually and stir it, when the salt stops dissolving any more, then it will be called a saturated solution.
(b) Pure Substance:
Matter formed of molecules Of a similar type is called as a pure substance. Either element or compound is pure like iron, gold, silver, sugar, water etc.
(c) Colloid:
Colloid is a heterogeneous mixture whose molecules are of the size in between lnm to 100 nm. These molecules cannot be seen with naked eyes and they diverge the rays of light; like – milk, shaving cream, toothpaste, jelly, face cream etc.
(d) Suspension:
Suspension is a heterogeneous mixture in which a solute substance does not dissolve, rather than they remain suspended in the medium. Suspended molecules are bigger in size than 100 nm (107m) and can be seen with naked eyes like contaminated water of a river, the mixture of thick lime mortar stones and water, etc.
Question 5.
Classify each of the following as a homogeneous and heterogeneous mixture: soda water, wood, ice, air, soil, vinegar, filtered tea.
Answer:
Homogeneous Mixture: Soda water, ice, vinegar, filtered tea.
Heterogeneous Mixture: Wood, air, soil.
Question 6.
How would you confirm that a colorless liquid given to you is pure water?
Answer:
We will find out the boiling point of the given colorless liquid. If that boiling point comes out to 373 K, we will approve that the given colorless liquid is pure water, but if it does not fulfill the condition then it is not pure water.
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.
Answer:
In the given substances following are the pure substances :
(a) ice
(c) iron
(d) hydrochloric acid
(e) calcium oxide
(f) Mercury.
Question 8.
Identify the solutions among the following mixtures :
(a) soil
(b) Seawater
(c) Air
(d) Coal
(e) Soda water.
Answer:
Soda water is a solution.
Question 9.
Which of the following will show the “Tyndall effect”?
(a) Salt solution
(A) Milk
(c) Copper sulfate solution
(d) Starch solution.
Answer:
Milk exhibits the 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.
Answer:
Element:
(a) Sodium
(d) Silver
(f) Tin
(g) Silicon.
Compound:
(e) Calcium carbonate
(j) Soap
(A) Methane
(f) Carbon dioxide.
Mixture:
(A) Soil
(c) Sugar solution
(h) Coal
(i) Air
(m) Blood.
Question 11.
Which of the following are chemical changes?
(a) Growth of a plant
(h) 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.
Answer:
Chemical changes are as follows:
(a) growth of a plant
(h) rusting of iron
(d) cooking of food
(e) digestion of food
(g) burning of a candle.
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