Exploring Mixtures and Their Separation Class 9 Question Answer Science Exploration Chapter 5

Chapter-wise NCERT Solutions for Class 9 Science Exploration Chapter 5 Exploring Mixtures and Their Separation Question Answer NCERT Solutions are useful for focused study.

Class 9 Science Exploration Chapter 5 Question Answer

Class 9 Science Ch 5 Exploring Mixtures and Their Separation Question Answer

Exploring Mixtures and Their Separation Class 9 Questions and Answers (Exercise)

Revise, Reflect, Refine (NCERT Textbook Page No. 90)

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:
(iv) Muddy water has large visible particles that settle quickly, making it heterogeneous. Milk shows Tyndall effect (light scattering) with suspended particles, appearing non-uniform. Blood contains cells that separate by centrifugation, behaving as heterogeneous. Brass is a uniform metal alloy throughout, making it homogeneous.

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
(e) starch and water
(d) acetone and water
(i) (a) and (b) (ii) (b) and (d)
(iii) (a) and (e) (iv) (e) and (d)
Answer:
(iii) Air with dust particles shows the Tyndall effect because the suspended dust particles scatter light, similar to how smoke or fog scatters light rays. Starch and water forms a colloid called starch sol, where the colloidal particles are large enough to scatter light but do not settle down over time.

Copper sulfate and water is a true solution where solute particles are smaller than 1 nm and do not scatter light at all. Acetone and water is also a true solution that appears completely uniform throughout with no light scattering.

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.

Words and Phrases: Large-sized particles; Particles remain evenly distributed; Small- sized particles (less than 1 nm diameter); Moderate-sized particles (1-1000 nm); Settles down when left undisturbed (more than 1000 nm in diameter); Does not settle down; Scatters light; Separates by filtration; Transparent; Salt solution; Milk; Sand in water; Smoke; Heterogeneous mixture; Cannot be separated by filtration; Mud; Butter; Brass. Complete the Table 5.2.

Answer:

Solution	Suspension	Colloid
Properties	Properties	Properties
Small-sized particles (less than 1 nm in diameter); Particles remain evenly distributed;	Large-sized particles (more than 1000 nm in diameter);	Moderately-sized particles (1-1000 nm); Does not settle down; Heterogeneous mixture;
Does not Settles down, transparent, cannot be separated by filtration; Does not scatter light	settle down when left undisturbed; Heterogeneous mixture; Scatters light; Separates by filtration	Cannot be separated by filtration; Scatters light
Examples Salt solution; Brass	Examples Mud; Sand in water	Examples Milk; Butter; Smoke

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) Total dry mixture = 420g flour + 75 g sugar + 5 g sodium hydrogencarbonate = 500 g
Sugar = 75/500 × 100 = 15%
All-purpose flour = 420/500 × 100 = 84%
Sodium hydrogencarbonate = 5/500 × 100 = 1%

(ii) Copper = 70% of 120 g = 84 g
Zinc = 120 g- 84 g = 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, cooking oil and water will form separate layers. Cooking oil and water are immiscible liquids that do not mix. Cooking oil has lower density than water, so oil floats on water, forming two separate layers. The cooking oil will be on the top, while the water will be at the bottom.

A separating funnel can be used to separate the mixture of oil and water. The following image is the diagram of the apparatus to be used:

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:
(iii) Solutions do not exhibit the Tyndall effect because their particles are smaller than 1 nm, making them too small to scatter visible light. Thus, the assertion that solutions do not exhibit Tyndall effect is true but the reason that the particles in solutions are larger than 1oo nm is wrong.

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.

Table 5.3

Mixture	Method of separation	Reason for selection
Mud from muddy water
Plasma from other components in the blood sample
Naphthalene and sand
Chalk powder and common salt
Common salt and water
Oil from water
Pigments of the flower

Answer:

Mixture	Method of separation	Reason for selection
Mud from muddy water	Filtration	Mud particles (>1000 nm) settle and can be trapped by filter paper
Plasma from other components in the blood sample	Centrifugation	Blood cells heavier than plasma; centrifugation separates by density
Naphthalene and sand	Sublimation	Naphthalene sublimes on heating sand does not
Chalk powder and common salt	Dissolution + Filtration	Salt dissolves in water, chalk powder does not (insoluble)
Common salt and water	Evaporation	Salt left behind when water evaporates
Oil from water	Separating funnel	Oil (less dense) floats on water; separates into layers
Pigments of the flower	Chromatography	Different pigments have different solubilities and travel different distances

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:
The method to separate miscible liquids A (b.p. 60°C) and B (b.p. 90°C) is simple distillation. Since they have a significant boiling point difference of 30 °C, heat the mixture to around 60°c in a round-bottom flask over a Bunsen burner.

Liquid A vaporises first while liquid B remains in the flask, then the vapour travels through a condenser cooled by water, where it condenses back to pure liquid A and collects in a receiving flask. The setup includes a still head with thermometer to monitor temperature, ensuring only liquid A distils over before liquid B starts vaporising at 90°c.

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

Method	Process	What you get	Best situation
Evaporation	Heat the solution → solvent evaporates	Solid solute	When the solvent is not required (e.g., obtaining salt from seawater)
Crystallisation	Prepare hot saturated solution → cool slowly to form crystals	Pure solid crystals	When a pure solid is required (e.g., copper sulfate crystals)
Distillation	Heat to boil → Vapour condenses separately	Pure liquid (and residue left behind)	When the solvent or liquid needs to be recovered (e.g., separating water from a solution or separating miscible liquids)

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) Blood is a colloidal mixture in which particles remain uniformly dispersed and do not settle. If blood behaved like a suspension, the heavier components would settle down over time. This could block blood vessels, disrupt circulation, and interfere with the transport of oxygen and nutrients, which would be harmful to the body.

(ii) In a blood sample, the dispersed phase consists of cells (red blood cells, white blood cells, and platelets, while the dispersion medium is plasma (the liquid component).

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:
The correct sequence of separation techniques is:
Step 1: Sublimation (Fig. 5.25b, image 1). The mixture is heated in a china dish. Naphthalene sublimes (changes directly from solid to vapour) and is collected on a cool surface, while sand and common salt remain behind.

Step 2: Dissolution and Filtration (Fig. 5.25b, image 3). Water is added to the remaining mixture. Common salt dissolves in water, whereas sand does not. The mixture is filtered to separate sand (residue) from the salt solution (filtrate).

Step 3: Evaporation (Fig. 5.25b, image 2). The salt solution is heated. Water evaporates, leaving behind solid common salt.

Complete sequence:
Sublimation (image 1) → Filtration (after dissolution) (image 3) →Evaporation (image 2)

Question 12.
Why is distillation an effective method for separating a mixture of water and acetone?
Answer:
Distillation can be used to separate the two miscible liquids because they have a sufficient difference in boiling points. It is effective for separating water (b.p. 100 °C) and acetone (b.p. 56 uq due to their 44 °C boiling point difference.

Acetone has a lower boiling point than water. When the mixture is heated, acetone vaporises first. Its vapour is then cooled in a condenser and collected as a pure liquid, while water remains in the distillation flask. Thus, the two liquids are separated based on their different boiling points.

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) Mass of potassium nitrate for 50 g in water at 40 °C:
Solubiity = 62 g/100g water
For 50g water = (62 × 50) ÷ 100 = 31 g
Potassium nitrate needed for saturated solution.

(ii) As the solution cools, its solubility decreases. The excess potassium chloride that can no longer remain dissolved separates out in the form of crystals.

(iii) As the temperature increases from 10°C to 80°C, the solubility of all the four salts – potassium nitrate, sodium chloride, potassium chloride and ammonium chloride increases, but at significantly different rates;

Potassium nitrate: Shows the most dramatic increases in solubility.
Sodium chloride: Shows the least change in solubility.
Potassium chloride: Shows moderately and steady increase with temperature.
Ammonium chloride: Shows moderate and steady increase in solubility. Potassium nitrate shows maximum solubility increase, sodium chloride shows minimum increase with temperature.

Question 14.
Three students, A, B and C, are preparing sugar solutions for an experiment:
Student A dissolves 20g of sugar in 80g of water.
Student B dissolves 20g 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:
Mass percentage (% m/m) calculations for sugar solutions:
(i) Mass % (m/m) for each student:
Student A: 20 g sugar + 80 g water = 100 g total
Mass % (20/100) × 100 = 20%
Student B: 20 g sugar + 100 g water = 120 g total
Mass % (20/ 120) × 100 = 16.67%
Student C: 30 g sugar + 80 g water = 110 g total
Mass % (30/110) × 100 = 27.27%

(ii) Most concentrated solution: Student C (27. 27%) Student C has the highest mass percentage of sugar; hence, it is the most concentrated solution.

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

Table 5.5: Boiling points of some compounds

Solvent	Water	Acetone	Alcohol	Chloroform	Benzene
Temperature (°C)	100 °C	56 °C	78 °C	61 °C	80 °C

Answer:
(i) The setup in the picture marked as ‘S’ shows simple distillation.
(ii) The labelled parts are:
A = distillation flask,
B = condenser, and
C = receiving flask.
(iii) Simple distillation separates a liquid from a dissolved solid or two miscible liquids with a large difference in boiling points (greater than 25 °C).
(a) water – acetone’: yes, because their boiling points are different enough.
(b) water – salt: yes, by distilling off the water and leaving salt behind.
(c) acetone – alcohol: no, their boiling points are too close for simple distillation.
(d) sand – salt: no, this is not a distillation mixture.
(e) alcohol – chloroform: no, because their boiling points are too close.
(f) alcohol – benzene: no, because their boiling points are too close.
So, the correct choices are (a) and (b).

Class 9 Science Chapter 5 Exploring Mixtures and Their Separation Question Answer (InText)

Think it Over (NCERT Textbook Page No. 72)

Question 1.
Why do suspended particles settle in muddy water over time but not in milk?
Answer:
In muddy water, the suspended particles are large and heavy, so they settle down after some time due to gravity. However, milk is a colloid in which the particles are very small and remain uniformly dispersed without settling down.

Question 2.
How is evaporation different from boiling?
Answer:
Evaporation is a surface phenomenon that occurs at all temperatures, whereas boiling is a bulk phenomenon that occurs throughout the liquid at a fixed temperature called the boiling point.

Question 3.
Why do you see bright rays of sunlight when it passes through small gaps between the
leaves of a dense tree?
Answer:
We can see the path of sunlight because tiny dust or smoke particles present in air scatter light. This scattering of light is called the Tyndall effect.

Pause and Ponder (NCERT Textbook Page No. 76)

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:
Zinc Oxide in Talcum Powder:
4% m/ m means 4 g zinc oxide present in 100 g of talcum powder.

In 300 g of talcum powder:
\(\frac{4}{100} \times 300\) = 12 g
So, 12 g of zinc oxide is present.

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 % u/v of orange juice concentrate in the mixture you prepared?
Answer:
2 tablespoons 2 × 15 ml =30 ml concentrate
Total volume = 150 ml
% u/v (30/150) × 100 = 20% u/u concentrate

Question 3.
Vinegar, used as a food preservative and additive, contains 5% ν/ν 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:
A 5% (ν/ ν) solution means 5 g of solute in 1oo ml of solution. Therefore, to prepare vinegar containing 5% acetic acid, take 5 ml of glacial acetic acid and add water to make the total volume 100 ml.

Pause and Ponder (NCERT Textbook Page No. 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:
From the solubility curves in Fig. 5.6 (Activity 5.2), Compound ‘B’ (blue, steeper curve) shows a larger decrease in solubility between 80°C and 60°C compared to Compound ‘A’ (red, gentler slope).

When equal masses of hot saturated solutions are cooled, the compound with a larger decrease in solubility deposits more solid. Therefore, compound ‘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, crystal size changes with evaporation rate. Faster evaporation leads to the formation of small crystals because particles do not get enough time to arrange themselves properly. Slower evaporation leads to the formation of large, well-shaped crystals as particles get sufficient time to arrange in a regular pattern.

Pause and Ponder (NCERT Textbook Page No. 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 crystallisation are the same processes.
Answer:
(i) True: Salt can be separated from a salt solution by evaporation or distillation.
(ii) False: Correction: Distillation cannot separate two liquids with the same boiling point. Fractional distillation is needed for liquids with close boiling points (difference >25°C for simple distillation).
(iii) False: Correction: In paper chromatography, the solvent level should be below the sample spot at the beginning of the experiment.
(iv) False: Correction: Evaporation and crystallisation are different processes. Evaporation is used to remove the solvent, leaving the solute behind (often impure). Crystallisation is used to obtain pure crystals of a substance from its saturated solution by cooling it slowly.

Pause and Ponder (NCERT Textbook Page No. 84)

Question 7.
Why do immiscible liquids form two separate layers in a separating funnel?
Answer:
Immiscible liquids do not mix with each other and have different densities. The denser liquid settles at the bottom, while the lighter liquid forms the upper layer.

Question 8.
Is sublimation different from evaporation? Justify.
Answer:
Yes, sublimation and evaporation are different processes. Sublimation is the change of a solid directly into vapour without passing through the liquid state, as seen in substances like naphthalene or camphor. However, evaporation is the process in which a liquid changes into vapour from its surface at temperatures below its boiling point.

Pause and Ponder (NCERT Textbook Page No. 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. They consist of tiny water droplets or ice crystals dispersed in air. These particles remain suspended and do not settle down. They also scatter light (Tyndall effect), which makes them visible.

Question 10.
Why do cities with a lot of smoke and dust in the air often look hazy?
Answer:
Smoke and dust particles present in air scatter light. This scattering of light is called the Tyndall effect. Due to this scattering, the air appears hazy, and visibility is reduced.

Think as a Scientist (NCERT Textbook Page No. 79)

Question 1.
If a hot, saturated solution of copper sulfate is cooled rapidly in ice-cold water, smaller and less well-formed crystals will form than if it is cooled slowly at room temperature. How would you design and perform an experiment to test this hypothesis?
Hint: Prepare a hot saturated solution of copper sulfate and divide it into two equal parts.
Answer:
To test the hypothesis, the following experiment can be designed:
Aim: To study the effect of the rate of cooling on the size and shape of copper sulfate crystals.
Materials Required: Copper sulfate, water, beaker, heat source, glass rod, two clean beakers, filter paper, ice bath.

Procedure:

1. Prepare a hot saturated solution of copper sulfate by dissolving copper sulfate in hot water with continuous stirring.
2. Filter the solution to remove any impurities.
3. Divide the hot saturated solution into two equal parts in separate beakers.
4. For rapid cooling: Place one beaker in an ice-cold water bath to cool it quickly.
5. For slow cooling: Leave the second beaker undisturbed at room temperature to cool slowly.
6. After some time, observe the crystals formed in both beakers.

Observations:

• The solution cooled rapidly, forms small and irregular crystals.
• The solution cooled slowly, forms large, well-shaped crystals.

Conclusion: The experiment shows that slow cooling produces larger and better-formed crystals, whereas rapid cooling results in smaller and poorly formed crystals. Thus, the hypothesis is verified.

Reason: During slow cooling, particles get enough time to arrange themselves in a
regular pattern, forming well-defined crystals. In rapid cooling, particles do not get sufficient time, leading to smaller crystals.

What if… (NCERT Textbook Page No. 83)

Question 1.
Two immiscible liquids of the same density are mixed in a separating fùnnel, how will The layers form?
Answer:
When two immiscible liquids having the same density are mixed in a separating funnel, then no distinct layers are formed.

Normally, the denser liquid sinks to the bottom while the lighter one floats on the top. But when densities are equal, gravity cannot separate them, so they stay mixed as an emulsion or suspension (cloudy or milky mixture) with no clear boundary of separation.

Class 9 Science Chapter 5 Question Answer (Activities)

Activity 5.1:
Let Us Experiment-Group Activity (NCERT Textbook Page No. 73)

Aim: To identify whether the given mixtures are a true solution, a suspension, or a colloid.

Observation:

• Group A (Salt + Water):
• The mixture appears clear and transparent.
• No particles are visible to the naked eye.
• The path of the laser beam is not visible.
• On standing, no particles settle.
• No residue is left on the filter paper after filtration.

Group B (Chalk Powder + Water):

• The mixture appears cloudy.
• Particles are clearly visible.
• The path of the laser beam is visible due to scattering.
• On standing, particles settle at the bottom.
• Residue is left on the filter paper after filtration.

Group C (Milk + Water):

• The mixture appears uniform but slightly cloudy.
• Particles are not visible to the naked eye.
• The path of the laser beam is visible (Tyndall effect).
• On standing, particles do not’ settle.
• No residue is left on the filter paper.

Conclusion:
The three mixtures are different types of mixtures:

• Salt + water is a true solution (homogeneous, no light scattering, no settling).
• Chalk powder + water is a suspension (heterogeneous, visible particles, settles, canbe filtered).
• Milk + water is a colloid (appears homogeneous, shows Tyndall effect, does not settle).
• The activity shows that mixtures differ based on particle size, visibility, stability, and light scattering properties.

Activity 5.2:
Let Us Represent Solubility Graphically (NCERT Textbook Page No. 77)

Aim: To study the solubility curves of compounds A and B.

Question 1.
Based on the information from the above graph, predict which of the two compounds, ‘A’ or ‘B will dissolve more in a given amount of water at a given temperature?
Answer:
Compound ‘B’ will dissolve more in a given amount of water at any given temperature. This is because the solubility curve of compound B lies above that of compound A, indicating higher solubility at the same temperature.

Question 2.
Observe Fig. 5.6 and fill in the blanks of the following statements:
(i) The solubility of compound ‘A’ in water at 20°c is …………………… (less than/more than/similar to) its solubility at 60°c.
(ii) The solubility of compound ‘B’ at 20°c is ………………. (less than/more than/ similar to) its solubility at 60°c.
(iii) The solubility of ………………….. increases more than that of with an increase in the temperature.
Answer:
(i) The solubility of compound ‘A’ in water at 20°c is less than its solubility at 60°c.
(ii) The solubility of compound ‘B’ at 20 °c is less than its solubility at 60 °c.
(iii) The solubility of compound ‘B’ increases more than that of compound ‘A’ with an increase in temperature.

Question 3.
What do you think will happen if you make a saturated solution at a higher temperature and cool it slowly? Let us find out!
Answer:
If you make a saturated solution at a higher temperature and cool it slowly, crystals of the solute will form and separate out. This occurs because solubility decreases with temperature for most solids, making the solution supersaturated upon cooling. The slow cooling allows solute particles to organize into well-defined crystals at the bottom.

Activity 5.3:
Let Us Prepare (NCERT Textbook Page No. 78)

Aim: To obtain crystals of copper sulfate from its solution.

Observation:
On cooling the hot saturated solution, large, shiny, well-shaped blue crystals of copper sulfate are formed.

Conclusion:
Pure crystals can be obtained from a saturated solution by the process of crystallisation. Crystallisation is a useful method for the purification of solids and is based on the change in solubility with temperature.

Activity 5.4:
Let Us Describe A Process (NCERT Textbook Page No. 79)

Aim: To study how salt crystals are obtained from seawater by evaporation.

Seawater is collected in large, shallow evaporation ponds. The heat of the sun and the action of wind cause the water to evaporate gradually, making the solution more concentrated. As evaporation continues, the solution becomes saturated, and salt begins to crystallise out. These salt crystals settle at the bottom of the ponds. The crystals are then collected, washed, dried, and used.

Activity 5.5:
Let Us Investigate (NCERT Textbook Page No. 82)

Aim: To separate the different coloured components present in black ink using paper chromatography.

Observation:

• As the water rises up the paper, it carries the ink along with it.
• The black ink spreads and separates into different coloured spots on the paper.
• Different colours move at different speeds and reach different heights on the paper.

Conclusion:

• Black ink is a mixture of different coloured substances (dyes).
• Paper chromatography separates the components of a mixture based on their different rates of movement in a solvent.

Activity 5.6:
Let Us Separate (NCERT Textbook Page No. 83)

Aim: To separate two immiscible liquids (mustard oil and water) using a separating funnel.

Observation:

• When the mixture is left undisturbed in the separating funnel, two distinct layers are formed.
• The mustard oil forms the upper layer, while water forms the lower layer.
• On opening the stopcock, the lower layer of water flows out first, leaving the oil behind.
• Both liquids can be collected separately.

Conclusion:

• Immiscible liquids like oil and water do not mix and form separate layers due to difference in density.
• A separating funnel can be used to separate such liquids effectively based on their density difference.

Activity 5.7:
Let Us Explore (NCERT Textbook Page No. 84)

Aim: To separate camphor from sand by sublimation.

Observation:

• On heating the mixture, camphor changes directly from solid to vapour.
• The vapours of camphor rise and deposit as white solid on the inner walls of the inverted funnel.
• Sand remains in the China dish and does not undergo any change.

Conclusion:

• Sublimation is used to separate a sublimable
• substance (camphor) from a non-sublimable substance (sand).

Activity 5.8:
Let Us Make A Model (NCERT Textbook Page No. 86)

Aim: To make a simple centrifuge model using a cardboard disc and thick thread.

Observation:
On rapid spinning, the mixture moves outwards. Heavier particles move away from the centre and settle at the outer side, while the lighter liquid remains closer to the centre.

Conclusion:
Centrifugation helps in separating fine suspended particles from liquids based on differences in density.

Activity 5.9:
Complete Table 5.1 And Review What You Have Learnt About Solutions, Suspensions And Colloids (NCERT Textbook Page No. 88)

Table 5.1: Properties of different types of mixtures

Property	Solution	Suspension	Colloid
1. Nature (homogeneous/heterogeneous)
2. Particle size
3. Visibility
4. Separation by filtration
5. Settling
6. Tyndall effect