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NCERT Solution Class 9 Science Exploring Mixtures and their Separation (In-text Exercise)

Refer to the detailed NCERT Solutions of Class 9 Chapter 5 Exploring Mixtures and their Separations. These solutions help students understand the concepts easily.

Class 9 Science Chapter 5 Exploring Mixtures and their Separations NCERT Solutions (Think it over)

Refer to the solutions of tink it over questions given in the Exploring Mixtures and their Separations Class 9 Science Chapter 5 NCERT. These questions helps students develop curiosity.

Page 72

Question: Why do suspended particles settle in muddy water over time but not in milk?

Answer:

Muddy water is a suspension; hence, suspended particles settle in it due to the effect of gravity. Milk is a colloid with very small particles, and these particles remain uniformly spread throughout; suspended particles do not settle down in milk.

Question: How is evaporation different from boiling?

Answer:

Question: Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree?

Answer:

Bright rays of sunlight are visible when it passes through small gaps between the leaves of a dense tree because of the scattering of light. In air small dust particles and water droplets are present that cause light to scatter. This is also known as the Tyndall effect.

Class 9 Science Chapter 5 Exploring Mixtures and their Separations NCERT Solutions (Pause and Ponder)

Refer to the detailed solutions of pause and ponder questions given in the NCERT Class 9 Science Chapter 5 Exploring Mixtures and their Seperations. These questions are given to check the knowledge and memorising capacity of students.

Page-79

Question 1. A common talcum powder contains 4 % m/m zinc oxide, which acts as an antiseptic. How much zinc oxide is present in 300 g of the talcum powder?

Answer:

Mass percentage (m/m) of Zinc Oxide: 4%

Total mass of the talcum powder: 300 g

Mass = (Mass percentage/100) x Total mass of powder

Mass of Zinc Oxide =(4/100) X 300

Mass of Zinc Oxide = 0.04 x 300

Mass of Zinc Oxide = 12 g

Question 2. Your mother gives you a bottle of orange juice concentrate to mix with water and serve it to your visiting friends. She asks you to mix two tablespoons of the concentrate with water in a glass tumbler. If each tablespoon measures 15 mL and you make 150 mL of juice per person, what is the % v/v of orange juice concentrate in the mixture you prepared?

Answer:

Volume of concentrate per tablespoon: 15 mL

Number of tablespoons used: 2

Total volume of the final mixture: 150 mL

Volume of Solute = 2 x 15

Volume of Solute = 30 mL

%v/v = (Volume of Solute/Total Volume of Solution) x 100

= (30/150) x 100

= 0.2 x 100

= 20% v/v

Question 3. Vinegar, used as a food preservative and additive, contains 5 % v/v acetic acid. Glacial acetic acid is a liquid, i.e., 100% acetic acid. If you want to make vinegar from glacial acetic acid, how would you proceed?

Answer:

Vinegar contains 5 % v/v acetic acid.

Means 5ml acetic acid in 100 ml of solution

Glacial acetic acid is 100% pure acetic acid. So, to prepare vinegar we need 5 ml glacial acetic acid and 95 ml distilled water.

Page 79

Question 4: Refer to the solubility curves given in Activity 5.2. If equal masses of hot, saturated solutions of compounds ‘A’ and ‘B’ are cooled from 80 °C to 60 °C, which solution is likely to deposit more solid?

Answer:

On cooling the solution whose solubility decreases more sharply from 80°C to 60°C will deposit more solid, the solubility of compound B decreases more than compound A. Hence, saturated solutions of compounds B will deposit more solid

Question 5. Will there be any change in the size of common salt crystals if the rate of evaporation is increased or decreased? Explain

Answer:

Yes, because when evaporation occurs slowly the particles get more time to arrange and thus form larger crystals while if evaporation occurs fastly the particles get less time to arrange leads to small crystal size.

Page- 82

Question 6. State whether the following statements are True or False. Also, correct the False statements.

(i) Salt can be separated from a salt solution by evaporation or distillation.

(ii) Distillation can be used for separation of two liquids even when these have the same boiling point.

(iii) In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment.

(iv) Evaporation and crystallization are the same processes.

Answer:

(i) True

(ii) False

Distillation can be used for separation of two liquids when these have the difference in boiling point.

(iii) False

In paper chromatography, the solvent level should be below the sample spot at the beginning of the experiment.

(iv) False

Evaporation and crystallization are the different processes.

Page-84

Question 7. Why do immiscible liquids form two separate layers in a separating funnel?

Answer:

Immiscible liquids form two separate layers because they cannot mix due to difference in densities.

Question 8. Is sublimation different from evaporation? Justify

Answer:

Yes, sublimation is different from evaporation because in sublimation solid directly converts into gas while in evaporation liquid converts into gas.

Page -88

Question 9. Clouds are made up of tiny water droplets or ice crystals floating in the air. Based on what you know about solutions, suspensions and colloids, what type of mixture do you think clouds are and why?

Answer:

Clouds are colloids because tiny water droplets and ice crystals are there in clouds. Both of these particles are small and remain suspended in air.

Question 10. Why do cities with a lot of smoke and dust in the air often look hazy?

Answer:

Cities look hazy because small particles present in air scatters sunlight in all directions.

Class 9 Science Exploring Mixtures and their Separation Question Answer (Revise, Reflect, Refine)

The detailed end of exercise questions of chapter 5 exploring mixtures and their separations are given below. Prepare this chapter effectively with the help of detailed NCERT Solutions for Class 9 Science given below:

Question 1. Which of the following mixtures are correctly classified as homogeneous (Hm) and heterogeneous (Ht)? Choose the correct option.

(i) Air-Hm , Milk-Ht , Sugar solution-Hm , Smoke-Hm

(ii) Brass-Ht , Fog-Ht , Vinegar-Ht , Muddy water-Hm

(iii) Copper sulfate solution-Hm , Salt solution-Hm , Milk-Hm , Bronze-Hm

(iv) Muddy water-Ht , Milk-Ht , Blood-Ht , Brass-Hm

Answer:

Homogeneous mixture is a mixture which has uniform composition whereas heterogeneous mixture is a mixture which has non uniform composition.

(i) Smoke is a heterogeneous mixture. Hence, this is incorrect

(ii) Brass is homogeneous and muddy water is heterogeneous. Hence, this is incorrect

(iii) Milk is heterogeneous. Hence, this is incorrect

(iv) In this all are correctly classified

Hence, the correct answer is option (iv)

Question 2. Choose the correct options, and explain the reason for the correct and incorrect options.

Which among the following mixtures show the Tyndall Effect? A mixture of:

(a) air and dust particles

(b) copper sulfate and water

(c) starch and water

(d) acetone and water

(i) a and b

(ii) b and d

(iii) a and c

(iv) c and d

Answer:

Scattering of light by colloidal particles is called the Tyndall effect.

(a) Air and dust are colloidals because dust particles are large enough to scatter light. Hence, it will show the Tyndall effect.

(b) Copper Sulfate and Water is a true solution, because ions are very small to scatter the light. Hence, it will not show the Tyndall effect.

(c) Starch and water is a colloidal mixture, because particles are large enough to scatter light. Hence, it will show the Tyndall effect.

(d) Acetone and Water is a true solution, because both are miscible liquids which form uniform solutions. Particles are very small. Hence, it will not show the Tyndall Effect.

Option (a) and (c) show the Tyndall effect.

Hence, the correct answer is option (iii)

Question 3. A mixture can be categorised as a solution, a suspension, or a colloid, each possessing distinct properties. Utilise the words or phrases provided in the box to fill in the Table 5.2. Words and phrases may be used more than once.

Answer:

Question 4. Solve the following problems:

(i) A cake recipe uses dry ingredients, namely 75 g of sugar for 420 g of all-purpose flour and 5 g of sodium hydrogencarbonate. Express the concentration of each component in the mixture using an appropriate method.

(ii) A brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass

Answer:

(i)

Weight of Sugar= 75g

Weight of flour=420g

Weight of sodium hydrogencarbonate=5g

Total Weight=75g+420g+5g = 500g

Concentration= (Weight of quantity/Total Weight)100

Concentration of Sugar = (Weight of sugar/Total weight)100

= (75/500)100

=15%

Concentration of Flour=(Weight of flour/Total weight)100

= (420/500)100

=84%

Conc. of sodium hydrogencarbonate=(Weight of sodium hydrogencarbonate/Total weight)100

= (5/500)100

=1%

(ii)

Weight of Brass = 120g

Copper Percentage = 70% by weight

Weight of Copper = 70% x 120

= (70/100) x 120

= 84 g

Weight of zinc = weight of Brass - Weight of copper

=120 - 84

= 36g

Question 5: The label on a cooking oil pack says one litre (910 g). If this oil is mixed with water, will it form a separate layer? If so, which substance will be on top? How will you separate the two layers? Also, draw the diagram of the apparatus used.

Answer:

Yes, when oil is mixed with water, it forms a separate layer because water and oil are immiscible liquids.

The substance with less density floats over other

Weight of Oil = 910g

Volume of Oil = 1 litre = 1000ml

Density = Weight/ Volume

=910/1000

=0.91g/ml

The density of water is approximately 1.

The cooking oil will be on top, because the density of cooking oil is less than that of water hence, it will float over it.

Two layers can be separated by separating funnel

Question 6: Assertion (A): Solutions do not exhibit the Tyndall effect.

Reason (R): The particles in solutions are larger than 100 nm, so they cannot scatter light. Choose the correct option:

(i) Both A and R are true, and R is the correct explanation of A.

(ii) Both A and R are true, but R is not the correct explanation of A.

(iii) A is true, but R is false.

(iv) A is false, but R is true

Answer:

The solution does not show the Tyndall effect because particles are dissolved and very small to scatter light. Hence, assertion is correct

Particle size in solution is less than 1 nm not larger than 100 nm. Hence, the reason is incorrect.

Assertion is true, but Reason is false

Hence, the correct answer is option (iii)

Question 7: How would you separate the mixtures given in Table 5.3? Mention the reason for choosing your method. If a mixture cannot be separated, explain why

Answer:

Question 8. Two miscible liquids, A and B, are present in a mixture. The boiling point of A is 60 °C and the boiling point of B is 90 °C. Suggest a method to separate them. Also, draw a labelled diagram of the method suggested.

Answer:

Both A and B are miscible and the difference between their boiling points is 30°C hence, they can be separated by simple distillation.

Liquid A with 60°C boiling point will vaporise first and then its vapours are cooled in a condenser and collected separately while Liquid B due to its high temperature will remain in the distillation flask.

Question 9: Compare evaporation, crystallization and distillation. In which situation, would you prefer each of these over the others?

Answer:

Evaporation

It is preferred when dissolved solid is stable on heating and solvent is not required.

Example: Preparing Salt from Saltwater

Crystallization

When pure solid is required.

Example: Purification of Copper sulfate crystals

Distillation

When two miscible liquids are required to separate.

Example: Separation of Alcohol and Water

Question 10: Blood is an example of a colloidal mixture. (i) What would happen if blood behaved like a true suspension inside the body? (ii) In a blood sample, identify the dispersed phase and the dispersion medium.

Answer:

(i) If blood behaves like a true suspension, while standing the particles of blood settle down, due to this blood flow in vessels would be interrupted, it will also affect transport of oxygen, nutrients and waste materials.

(ii) Dispersed phase: Blood Cells

Dispersion medium: Plasma

Question 11: You are given a mixture of sand, common salt and naphthalene (Fig. 5.25a). The Fig. 5.25b depicts various steps used to separate the components of this mixture. Identify and write down the correct sequence of separation techniques

Answer:

To separate mixture of sand, common salt and naphthalene the correct sequence of separation techniques is:

1. Sublimation for separation of naphthalene

2. Dissolution to separate common salt and sand because common salt dissolved in water while sand does not.

3. Filtration for the separation of sand from salt water

4. Evaporation to obtain salt from solution.

Question 12: Why is distillation an effective method for separating a mixture of water and acetone?

Answer:

Water and acetone both are miscible liquids with differences in boiling points. Hence, distillation is the most effective method to separate them. The boiling point of acetone is 56 °C and boiling point of water is 100 °C when heated acetone boils first and changes into vapour then cooled in a condenser and collected separately while water due to its high temperature will remain in the distillation flask.

Question 13: Answer the following questions with the help of the data given in Table 5.4.

(i) What mass of potassium nitrate would be needed to prepare its saturated solution in 50 g of water at 40 °C?

(ii) A student makes a saturated solution of potassium chloride in water at 80 °C and leaves the solution to cool at room temperature (25 °C). What would she observe as the solution cools? Explain.

(iii) What is the effect of a change in temperature on the solubility of salts? Also, compare the changes in the solubility of the four given salts with increasing temperature from 10 °C to 80 °C.

Answer:

(i) Solubility of potassium nitrate at 40°C = 62 g for 100 g of water

For 50 g

Solubility=62/2

=31g

(ii) Potassium chloride has high solubility at 80°C but at room temperature 25°C, its solubility decreases. Potassium chloride crystals will separate on cooling.

(iii) With the increase in temperature solubility increases because at higher temperatures more thermal energy is required to break the solute-solute bonds.

Question 14: Three students, A, B and C, are preparing sugar solutions for an experiment:

• Student A dissolves 20 g of sugar in 80 g of water.
• Student B dissolves 20 g of sugar in 100 g of water.
• Student C dissolves 30 g of sugar in 80 g of water.

(i) Calculate the mass percentage (% m/m) concentration of sugar in each student’s solution.

(ii) Whose solution is the most concentrated? Explain why.

Answer:

(i)

For student A

Mass of Sugar = 20 g

Mass of Water = 80 g

Mass of Total solution = 100 g

Mass $\mathrm{\%}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100$

=(20/100)x100

=20%

For Student B

Mass of Sugar = 20 g

Mass of Water = 100 g

Mass of Total solution = 120 g

Mass%=(20/120)x100

=16.67%

For Student C

Mass of Sugar = 30 g

Mass of Water = 80 g

Mass of Total solution = 110 g

Mass%=(30/110)x100

=27.27%

(ii) Solution of Student C is most concentrated, because the mass percentage of sugar is 27.27% which is highest.

Question 15: Examine Fig. 5.26.

(i) Identify the separation technique marked as ‘S’.

(ii) Label the apparatus A, B and C.

(iii) Which of the following mixtures can be separated by the technique identified above? Use the data given in Table 5.5. Mixtures:

(a) water-acetone

(b) water-salt

(c) acetone- alcohol

(d) sand-salt

(e) alcohol-chloroform

(f) alcohol-benzene

Answer:

(i) The separation technique marked as S is Simple distillation

(ii) A is Distillation flask

B is Condenser

C is Receiving flask

(iii) Distillation is used to separate miscible liquid with sufficient difference in boiling points.

Boiling point of Water = 100°C

Boiling point of Acetone = 56°C

Boiling point of Alcohol = 78°C

Boiling point of Chloroform = 61°C

Boiling point of Benzene = 80°C

(a) water-acetone

Boiling point of Water = 100°C

Boiling point of Acetone = 56°C

Difference = 44°C

Yes, Can be separated by distillation

(b) water-salt

Water can be distilled and salt remains in flask

Yes, Can be separated by distillation

(c) acetone- alcohol

Boiling point of Acetone = 56°C

Boiling point of Alcohol = 78°C

Difference = 22°C

No, Cannot be separated by distillation because less difference in boiling points

(d) sand-salt

Both are solids

No, Cannot be separated by distillation

(e) alcohol-chloroform

Boiling point of Alcohol = 78°C

Boiling point of Chloroform = 61°C

Difference = 17°C

No, Cannot be separated by distillation because less difference in boiling points

(f) alcohol-benzene

Boiling point of Alcohol = 78°C

Boiling point of Benzene = 80°C

Difference = 2°C

No, Cannot be separated by distillation because less difference in boiling points

Topics Covered in Class 9 Science Chapter 5

Given below are all the topics and subtopics that are covered in this chapter.

5.1 How Can We Classify Mixtures?

5.2 Solutions

• 5.2.1 Concentration of a solution
• 5.2.2 How do we express concentration?
• 5.2.3 Solubility of substances

5.3 Methods of Separation of Homogeneous Mixtures

• 5.3.1 Crystallization
• 5.3.2 Distillation
• 5.3.3 Paper Chromatography

5.4 How Can We Separate the Components of Heterogeneous Mixtures?

• 5.4.1 Separation of two immiscible liquids
• 5.4.2 Sublimation
• 5.4.3 Suspensions
• 5.4.4 Colloids

5.5 Tyndall Effect

Also Check

NCERT Notes for Class 9 Science

NCERT Syllabus for Class 9 Science

NCERT Solutions for Class 9 Science

Refer to the links given below for detailed solutions of NCERT Class 9 Science. These solutions are prepared by subject experts to help you understand the chapters well and prepare effectively for exams.