2. Is Matter Around us Pure

Assam Class 9 Science Class 9 Science Chapter 2. Is Matter Around us Pure

Chapter 2. Is Matter Around us Pure ?

Class 9 Science Chapter 2 Is Matter Around us Pure ? Internal / Example Questions and Answers

Internal Questions :

1. What is meant by a substance?

Answer: A substance is a pure single form of matter.
2. List the points of differences between homogeneous and heterogeneous mixtures.

Answer: The differences between homogeneous and heterogeneous mixtures are :

(i). Homogeneous mixtures have a uniform composition , while heterogeneous mixtures has non- uniform composition.

(ii). Homogeneous mixtures have a single phase, while heterogeneous mixtures have multiple phases.

(iii). Homogeneous mixtures have constant properties throughout, while properties can vary in different regions of a heterogeneous mixture.

Example 2.1 : A solution contains 40 g of common salt in 320 g of water. Calculate the concentration in terms of mass by mass percentage of the solution.

Solution: Mass of solute (salt) = 40 g

Mass of solvent (water) = 320 g

We know that ,

Mass of solution = Mass of solute + Mass of solvent

= 40 g + 320 g

= 360 g

Mass percentage of solution

Internal Questions :

1. Differentiate between homogeneous and heterogeneous mixtures with examples.

Answer: Difference between homogeneous and heterogeneous mixtures are :

Homogeneous mixtures

Heterogeneous mixtures

Uniform composition

Non-uniform composition

For example :

(i) sugar in in water ,

(ii) Sulphar in carbon disulphide ,

(iii) water in alcohol .

For example :

(i) sand and salt ,

(ii) sugar and salt ,

(iii) water in oil .

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

Answer:

Sol (colloid)	Solution	Suspension
(i) The size of particles of a colloid is too small to be individually seen with naked eyes.	The particles of a solution are smaller than 1 nm (10-9 metre) in diameter. So, they cannot be seen by naked eyes.	The particles of a suspension can be seen by the naked eye.
(ii) Colloids are big enough to scatter a beam of light passing through it and make its path visible.	The path of light is not visible in a solution.	The particles of a suspension scatter a beam of light passing through it and make its path visible.
(iii) They 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.

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

Mass of water (solvent) = 100 g

Mass of solution = Mass of solute + Mass of solvent

Mass of solution = 36 g + 100 g

Mass of solution = 136 g

The concentration of the solution is the mass of the solute divided by the mass of the solution, multiplied by 100 to express it as a percentage:

Concentration

Therefore, the concentration of the saturated solution of sodium chloride at 293 K is 26.47%.

Internal Questions :

1. How will you separate a mixture containing kerosene and petrol (difference in their boiling points is more than 25ºC), which are miscible with each other?

Answer: To separate a mixture of kerosene and petrol, which are miscible (able to mix) with each other, based on their difference in boiling points, a process called fractional distillation can be employed. By heating the mixture, the components with lower boiling points will vaporize first, allowing their separation from the mixture.
2. Name the technique to separate
(i) butter from curd,
(ii) salt from sea-water,
(iii) camphor from salt.

Answer: (i) The technique to separate butter from curd is known as churning or centrifugation.

(ii) The technique to separate salt from sea-water is known as evaporation.

(iii) The technique to separate camphor from salt is known as sublimation.

3. What type of mixtures are separated by the technique of crystallization ?

Answer: The technique of crystallization is primarily used to separate homogeneous mixtures, specifically solutions, where the solute can form crystals. It is commonly employed to purify substances and obtain them in a solid, purified form.

Internal Questions :

1. Classify the following as chemical or 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: The classification of the given processes as chemical or physical changes:

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 the water breaking down into hydrogen and oxygen gases: 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

2. Try segregating the things around you as pure substances or mixtures.

Answer: Segregating the things around me as pure substances or mixtures:

Pure Substances:

(i). Distilled water

(ii). Salt (NaCl)

(iii). Sugar (Sucrose)

(iv). Iron rod

Mixtures:

(i). Air (a mixture of gases)

(ii). Coffee (a mixture of water and coffee grounds)

(iii). Plastic pen (made of polymer blend)

(iv). Wooden table (a mixture of cellulose fibers and other compounds)

Class 9 Science Chapter 2 Is Matter Around us Pure ? Exercise Questions and Answers

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: The appropriate separation techniques are :

(a) Sodium chloride from its solution in water: The technique used for this separation is called evaporation.

(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride: The technique used for this separation is sublimation.

(d) Different pigments from an extract of flower petals: The technique used for this separation is chromatography.

(e) Butter from curd: The technique used for this separation is centrifugation.

(f) Oil from water: The technique used for this separation is called decantation or separation by density difference.

(g) Tea leaves from tea: The technique used for this separation is filtration.

(h) Iron pins from sand: The technique used for this separation is magnetic separation.

(i) Wheat grains from husk: The technique used for this separation is winnowing.

(j) Fine mud particles suspended in water: The technique used for this separation is sedimentation and decantation.

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). Boil water (solvent) to create a hot liquid.

(ii). Add tea leaves (solute) to the hot water.

(iii). Stir the mixture to allow the tea leaves to dissolve in the water. Tea leaves are soluble in water.

(iv). Let the tea steep for a few minutes to extract flavor.

(v). Filter the mixture to separate the tea leaves (residue) from the tea solution (filtrate).

(vi). Serve the tea solution in a cup, discarding the tea leaves.

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)

Substances Dissolved	Temperature in 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 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 ?

Answer: (a) Let grams of potassium nitrate are needed to produce a saturated solution in 50 grams of water at 313 K.

We have,

grams

So, 31 grams of potassium nitrate would be needed to produce a saturated solution in 50 grams of water at 313 K.

(b) When Pragya makes a saturated solution of potassium chloride in water at 353 K and then cools the solution to room temperature, she would observe the formation of crystals of potassium chloride. As the solution cools, its temperature decreases, leading to a decrease in the solubility of potassium chloride. Since the solubility of potassium chloride at 353 K is given as 54 grams of substance dissolved in 100 grams of water, it means that the solution can hold a maximum of 54 grams of potassium chloride at this temperature.

Solubility of Potassium nitrate = 32 grams of substance dissolved in 100 grams of water.

Solubility of Sodium chloride = 36 grams of substance dissolved in 100 grams of water.

Solubility of Potassium chloride = 35 grams of substance dissolved in 100 grams of water.

Solubility of Ammonium chloride = 37 grams of substance dissolved in 100 grams of water.

Among these salts, ammonium chloride has the highest solubility at 293 K.

(d) The effect of temperature on the solubility of a salt depends on the specific salt. In general, for many salts, an increase in temperature leads to an increase in solubility. This is because higher temperatures provide more energy for the solvent molecules, allowing them to break the intermolecular forces holding the solute particles together. As a result, more solute particles can dissolve in the solvent.

4. Explain the following giving examples.
(a) saturated solution
(b) pure substance
(c) colloid
(d) suspension

Answer: (a) When no more solute can be dissolved in a solution at a given temperature, it is called a saturated solution.

For example : If no more sugar can dissolve in a cup of hot tea, it forms a saturated solution.

(b) A pure substance consists of a single type of particle .

For example : Silver (Ag) , gold (Au), and oxygen ().

(c) The particles of a colloid are uniformly spread throughout the solution. Due to the relatively smaller size of particles, as compared to that of a suspension, the mixture appears to be homogeneous.

For example : milk, fog, and blood .

(d) A suspension is a heterogeneous mixture in which the solute particles do not dissolve but remain suspended throughout the bulk of the medium.

For example : muddy water, sand in water, and dust in air.

5. Classify each of the following as a homogeneous or heterogeneous mixture : soda water, wood, air, soil, vinegar, filtered tea.

Answer: The classification of the given substances as homogeneous or heterogeneous mixtures are :

Soda water: Homogeneous mixture. Soda water is a uniform mixture of carbonated water and dissolved substances (such as sugar and flavorings).

Wood: Heterogeneous mixture. Wood is a complex mixture of various substances, including cellulose fibers, lignin, and other organic compounds. It contains distinct regions with different compositions and structures.

Air: Homogeneous mixture. Air is a homogeneous mixture of gases, primarily nitrogen, oxygen, carbon dioxide, and trace amounts of other gases. It appears uniform throughout.

Soil: Heterogeneous mixture. Soil is a complex mixture of minerals, organic matter, water, and air. It contains various particles of different sizes, textures, and compositions, resulting in a non-uniform appearance.

Vinegar: Homogeneous mixture. Vinegar is a homogeneous mixture of acetic acid and water, along with other minor components. It appears uniform throughout.

Filtered tea: Homogeneous mixture. Once tea has been filtered, the resulting liquid is a homogeneous mixture of water, dissolved tea compounds, and any added sweeteners or flavorings. It appears uniform throughout

6. How would you confirm that a colourless liquid given to you is pure water?

Answer: To confirm that a colorless liquid is pure water, we can perform several tests. First, we can check its boiling point, which should be close to 100°C at sea level. Second, we can test its freezing point, which should be around 0°C. Additionally, we can use a conductivity meter to verify that it has low electrical conductivity.
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: (a) Ice , (c) Iron , (e) Calcium oxide and (f) Mercury are pure substances .

(b) Milk, (d) Hydrochloric acid , (g) Brick, (h) Wood, and (i) Air are not pure substances.

8. Identify the solutions among the following mixtures.
(a) Soil
(b) Sea water
(c) Air
(d) Coal
(e) Soda water.

Answer: (a) Soil is not a solution. Soil is a heterogeneous mixture composed of various components such as minerals, organic matter, water, and air.

(b) Sea water: Sea water is a solution. It is a homogeneous mixture of water and various dissolved salts, such as sodium chloride, magnesium chloride, and potassium chloride.

(c) Air: Air is not a solution. It is a mixture of gases, primarily nitrogen, oxygen, carbon dioxide, and trace amounts of other gases. While it is a homogeneous mixture .

(e) Soda water: Soda water is a solution. It is a homogeneous mixture.
9. Which of the following will show “Tyndall effect”?
(a) Salt solution
(b) Milk
(c) Copper sulphate solution
(d) Starch solution.

Answer: The substances that will show the Tyndall effect among the given options are:

(b) Milk (d) Starch solution

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: The classification of the given substances is as follows:

Elements: (a) Sodium (d) Silver (f) Tin (g) Silicon

Compounds: (e) Calcium carbonate (j) Soap (k) Methane (l) Carbon dioxide

Mixtures: (b) Soil (c) Sugar solution (h) Coal (i) Air (m) Blood

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.

Answer: (a) Growth of a plant --- Chemical Changes .

(b) Rusting of iron --- Chemical Changes .

(d) Cooking of food --- Chemical Changes .

(e) Digestion of food --- Chemical Changes .

(f) Freezing of water --- No Chemical Changes (Physical Changes) .

(g) Burning of a candle --- Chemical Changes .