Chap 5 : Exploring Mixtures and their Separation

Think It Over

Q – Why do suspended particles settle in muddy water over time but not in milk? A – In muddy water, the mud particles are quite large and heavy (it is a suspension). Gravity pulls these heavy particles down to the bottom over time. Milk, however, is a colloid. The fat and protein particles in milk are much smaller and lighter than mud particles, so they remain floating and spread evenly throughout the liquid without settling down.

 Q – How is evaporation different from boiling? A – Evaporation happens only at the surface of a liquid and can take place at any temperature (like wet clothes drying on a sunny or windy day). Boiling happens throughout the entire liquid (forming bubbles everywhere) and only takes place at a specific fixed temperature called the boiling point (like water boiling at 100°C).

 Q – Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree? A – This happens because of the Tyndall effect. The air in a forest or under a tree contains tiny dust particles, mist, and smoke. When sunlight passes through the small gaps, these tiny particles scatter (spread out) the light in all directions, making the bright beam of light clearly visible to our eyes.

 Section 5.1 & Activity 5.1: Classifying Mixtures

Q – Is the mixture of oil and water homogeneous or heterogeneous? Can you think of some other heterogeneous mixtures? A – The mixture of oil and water is a heterogeneous mixture because the oil and water do not mix evenly and form two separate visible layers. Other examples of heterogeneous mixtures include sand in water, iron filings mixed with sulfur, chalk powder in water, and dust particles in air.

 Q – Activity 5.1 (Step 4): Are the particles visible in each mixture? (Group A: salt + water; Group B: chalk powder + water; Group C: milk + water)

A –

• Group A (Salt + Water): No, the salt particles completely dissolve and are not visible.
• Group B (Chalk powder + Water): Yes, the chalk particles are clearly visible floating in the water.
• Group C (Milk + Water): No, individual milk particles are too small to be seen with the naked eye, though the water looks cloudy or milky.

 Q – Activity 5.1 (Step 5): Direct the light from a laser pointer through the beakers containing the mixtures and observe it from the side. What are your observations?

A –

• Group A (Salt + Water): The path of the laser light is not visible inside the beaker.
• Group B (Chalk powder + Water): The path of the laser light is clearly visible because the larger chalk particles scatter the light.
• Group C (Milk + Water): The path of the laser light is clearly visible because the colloid particles in milk scatter the light.

 Q – Activity 5.1 (Step 6): Predict what you would observe in each of the beakers if you leave them undisturbed for a few minutes.

A –

• Group A (Salt + Water): No change occurs; the salt remains completely dissolved.
• Group B (Chalk powder + Water): The heavy chalk particles settle down at the bottom of the beaker over time.
• Group C (Milk + Water): No change occurs; the milk particles do not settle down at the bottom.

 Q – Activity 5.1 (Step 7): Set up a filtration apparatus and filter each mixture separately. Is there any residue left on the filter paper?

A –

• Group A (Salt + Water): No residue is left on the filter paper (the salt water passes straight through).
• Group B (Chalk powder + Water): Yes, solid chalk powder is left behind as residue on the filter paper.
• Group C (Milk + Water): No residue is left on the filter paper (the milky water passes completely through).

 Q – Activity 5.1 (Step 8): Based on your observations, do you think these are the same types of mixtures or are they different? A – They are three different types of mixtures:

• Group A is a solution (a true homogeneous mixture).
• Group B is a suspension (a heterogeneous mixture with large particles that settle).
• Group C is a colloid (a mixture that looks homogeneous to the eye but scatters light like a heterogeneous mixture).

 Section 5.2: Solutions & Concentrations

Q – In what proportion are a solute and a solvent present in a solution? Can these be expressed quantitatively? A – A solvent is usually present in a larger amount, while the solute is present in a smaller amount. Yes, the exact proportion can be expressed quantitatively, and this measurement is called the concentration of the solution.

 Q – When farmers spray pesticides on their crops, they must mix the right amount of pesticide with a fixed amount of water to prepare a solution. If they do not do so, what is likely to happen? Can you think of other such examples? A – If farmers add too little pesticide, it will not be strong enough to protect the crops from pests. If they add too much, the strong chemical can damage or burn the crops, harm the soil, and pollute the environment. Other everyday examples where exact concentration matters:

• Making Oral Rehydration Solution (ORS) by mixing precise amounts of salt and sugar in water to treat dehydration safely.
• Preparing baby milk formula with the correct amount of milk powder and water.
• Adding the right amount of salt while cooking dish so it is neither tasteless nor too salty.

 Q – Example 5.1: If 10 g of salt is dissolved in 90 g of water, calculate the mass by mass percentage of the solution formed.

A –

• Mass of solute (salt) = 10 g
• Mass of solvent (water) = 90 g
• Total mass of solution = Mass of solute + Mass of solvent = 10 g + 90 g = 100 g
• Mass by mass percentage () =
• Calculation: .
• The concentration is 10% m/m.

 Q – Example 5.2: If 5 g of glucose is dissolved in water to make 100 mL of solution, calculate its concentration in mass by volume percentage.

A –

• Mass of solute (glucose) = 5 g
• Volume of solution = 100 mL
• Mass by volume percentage () =
• Calculation: .
• The concentration is 5% m/v.

 Q – Example 5.3: If 1 mL of a liquid pesticide is mixed with a sufficient amount of water to form 100 mL of a pesticide spray for rice crop, calculate its volume by volume percentage.

A –

• Volume of solute (pesticide) = 1 mL
• Total volume of solution = 100 mL
• Volume by volume percentage () =
• Calculation: .
• The concentration is 1% v/v.

 Q – Why do we mention the temperature when defining the solubility of a substance? A – We mention temperature because the amount of solute that can dissolve in water changes when the liquid gets hotter or colder. For solid solutes (like sugar or salt), solubility generally increases as temperature rises. For gases (like oxygen or carbon dioxide in water), solubility decreases as temperature rises.

 Pause and Ponder (Page 76)

Q – 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?

A –

• Concentration of zinc oxide =
• Total mass of talcum powder = 300 g
• Formula: Mass of zinc oxide =
• Calculation: .
• There are 12 g of zinc oxide present in 300 g of the talcum powder.

Q – 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?

A –

• Volume of orange juice concentrate (solute) = 2 tablespoons = .
• Total volume of juice prepared (solution) = 150 mL.
• Formula: Volume by volume percentage () =
• Calculation: .
• The concentration of the juice is 20% v/v.

 

Q – 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? A – To make 100 mL of 5% v/v vinegar:

1. Measure exactly 5 mL of glacial (100%) acetic acid using a measuring cylinder.
2. Pour this 5 mL of acetic acid into a clean container.
3. Add 95 mL of clean water to it and mix well so that the total volume of the mixture becomes 100 mL.

 Section 5.3 & Activities: Separation of Homogeneous Mixtures

Q – Activity 5.2: Based on the information from the graph (Fig. 5.6), predict which of the two compounds, ‘A’ or ‘B’, will dissolve more in a given amount of water at a given temperature? A – Compound ‘B’ will dissolve much more than Compound ‘A’ at any given temperature because its solubility curve is located much higher on the graph than the curve for Compound ‘A’.

Q – Activity 5.2: Fill in the blanks of the following statements based on Fig. 5.6: (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. A – (i) less than (At 20°C it is about 37g, while at 60°C it is about 57g). (ii) less than (At 20°C it is about 200g, while at 60°C it is 287g). (iii) Compound ‘B’, Compound ‘A’ (The curve for B rises steeply, while A rises very slowly).

Q – What do you think will happen if you make a saturated solution at a higher temperature and cool it slowly? A – When a hot saturated solution cools down slowly, the liquid cannot hold as much dissolved solid as it did when it was hot. As a result, the extra dissolved substance comes out of the solution and forms pure, shiny solid crystals.

Q – If you take a saturated solution of compound ‘B’ in water at 60 °C prepared by dissolving 287 g of compound ‘B’ in 100 g of water, what will happen if you gradually cool it to 40 °C? A – At 40°C, 100 g of water can only dissolve 241 g of compound ‘B’. Therefore, when cooled from 60°C to 40°C, the extra amount () of compound ‘B’ will separate out from the water and settle down as solid crystals.

Q – Activity 5.3: Place 1-2 mL of the saturated copper sulfate solution on a small glass plate or a lamination sheet. Leave it for some time. What do you observe? Did you get crystals? If yes, is this a good way to experiment? Explain.

A –

• Observation: As the water evaporates quickly from the flat plate, tiny blue solid crusts or small crystals of copper sulfate appear. Yes, we get crystals.
• Is it a good way? No, this is not the best method for obtaining clean, well-formed crystals. Because evaporation happens very quickly on an open plate, the crystals formed are very small, powdery, and not well-shaped. Slow cooling in a beaker covered with a watch glass is a much better method to get large, shiny, and pure crystals.

Q – Think as a Scientist: 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?

A –

1. Prepare: Make a hot, saturated solution of copper sulfate in a beaker by dissolving copper sulfate in hot water with a drop of dilute sulfuric acid.
2. Divide: Pour equal amounts of this hot solution into two separate clean beakers (Beaker A and Beaker B).
3. Test: Place Beaker A undisturbed on a table to cool down slowly at normal room temperature. Place Beaker B immediately into a large bowl filled with ice cubes and cold water to cool it down rapidly.
4. Observe & Compare: After some time, filter the crystals from both beakers. You will see that Beaker A (slow cooling) has large, well-defined, shiny crystals, while Beaker B (rapid ice cooling) has tiny, powdery, and poorly formed crystals. This proves the hypothesis is correct!

Pause and Ponder (Page 79)

Q – 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? A – Compound ‘B’ will deposit much more solid crystals. Looking at the graph in Fig. 5.6, the curve for Compound B drops sharply from 80°C to 60°C (meaning its solubility decreases by a large amount), whereas the curve for Compound A drops very slightly.

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

A – Yes!

• If the rate of evaporation is increased (for example, by boiling quickly over high heat), the salt crystals formed will be very small and powdery because the particles do not get enough time to arrange themselves neatly.
• If the rate of evaporation is decreased (allowing water to evaporate slowly over days), the particles get plenty of time to join together in regular patterns, forming larger, well-shaped crystals.

More In-Text Questions & Chromatography

Q – Activity 5.4: Describe the process of obtaining salt crystals from seawater (shown in Fig. 5.9) in your own words.

A –

1. Seawater is collected and trapped in large, shallow open ponds or beds along the seashore.
2. The heat from the sun gradually evaporates the water, making the trapped seawater more and more concentrated until it becomes a saturated solution.
3. As more water evaporates by solar heat and wind, solid salt crystals begin to separate out and settle at the bottom.
4. These salt crystals are then gathered, cleaned, and refined for daily use.

Q – When we allow a solvent to evaporate to get a solute, such as salt from a salt solution, the solvent disappears into the air. However, we may also want to recover the solvent (water). What should we do then? A – To save and collect both the dissolved solid (solute) and the liquid (solvent), we should use the process of distillation instead of simple evaporation. In distillation, the liquid evaporates into vapour, but instead of escaping into the air, the vapour is passed through a cooling tube (condenser) where it turns back into liquid water and is collected in a separate flask.

Q – How can we separate two miscible liquids like acetone and water? Is it possible to separate the mixture of two miscible liquids by evaporation and obtain both the liquids?

A –

• We can separate a mixture of acetone and water using distillation. Acetone boils at a much lower temperature (about 56°C) than water (100°C). When heated gently, the acetone turns into vapour first, travels through the condenser, cools down, and is collected as pure liquid acetone, leaving the water behind in the distillation flask.
• No, simple open evaporation cannot separate them if you want to keep both liquids. In simple evaporation, the vapour escapes into the atmosphere and is lost. To obtain and save both liquids, distillation must be used.

Q – Activity 5.5 (Steps 5 & 6): Observe the paper as the water rises through the paper. What do you notice? As the water rises, the ink starts to separate into different colour spots. What can you infer from this?

A –

• Observation: As water climbs up the strip of paper, the black sketch pen spot dissolves and spreads upwards, splitting into streaks of different colours (like blue, pink, or yellow).
• Inference: This tells us that black ink is not just a single pure colour. It is actually a mixture of two or more different coloured dyes. The colours that dissolve better in water move up the paper faster and reach higher spots, separating from each other. This process is called paper chromatography.

Q – Will water work as a solvent in every case for paper chromatography? A – No, water will not work in every case. If the ink or substance does not dissolve in water (like permanent markers or certain plant pigments), water cannot carry it up the paper. In such cases, we must use a different liquid solvent, such as alcohol or a mixture of different solvents.

Pause and Ponder (Page 82)

Q – 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.

A –

• (i) True. (Both methods work; evaporation leaves salt behind, while distillation leaves salt behind and also saves the water).
• (ii) False. Correction: Distillation can only separate two liquids if there is a sufficient difference (at least about 25°C) between their boiling points. If they boil at the exact same temperature, they will vaporise together and cannot be separated by simple distillation.
• (iii) False. Correction: In paper chromatography, the solvent level must always be below the sample ink spot. If the spot is submerged in water, the ink will dissolve directly into the beaker of water instead of climbing up the paper.
• (iv) False. Correction: Evaporation simply removes liquid by turning it into vapour, often leaving behind powdery or impure solids. Crystallization is a controlled process where a hot saturated solution cools down slowly to form pure, well-defined geometric solid crystals.

Section 5.4: Separation of Heterogeneous Mixtures

Q – How can you separate the components of a mixture of two immiscible liquids? A – Two immiscible liquids (liquids that do not mix together, like oil and water) can be separated easily using a separating funnel. Because they have different densities, they form two distinct layers. The heavier liquid settles at the bottom and can be drained out through a stopcock valve, leaving the lighter liquid behind in the funnel.

Q – Activity 5.6: When you mix mustard oil and water in a separating funnel and let it stand undisturbed, what do you observe? Why does mustard oil form the upper layer and water form the lower layer?

A –

• Observation: The mixture separates into two distinct, clear layers.
• Reason: This happens because oil and water are immiscible. Mustard oil is less dense (lighter) than water, so it floats on top forming the upper layer. Water is more dense (heavier), so it settles at the bottom forming the lower layer.

Q – Can you think of any heterogeneous mixtures with a gas as one of the components? A – Yes! Here are common examples of heterogeneous mixtures containing a gas:

• Smoke: Solid ash and carbon particles suspended in air (gas).
• Fog or Mist: Tiny liquid water droplets suspended in air (gas).
• Dusty air: Solid dust particles floating in the atmosphere.

Q – What if… two immiscible liquids of the same density are mixed in a separating funnel, how will the layers form? A – If two immiscible liquids have the exact same density, neither one is heavier than the other, so they will not separate into a clear neat top layer and bottom layer by gravity alone. Instead, they will remain suspended as clumps or droplets within each other (like an emulsion), making it impossible to separate them simply by opening the stopcock of a separating funnel.

Q – Activity 5.7: When heating a mixture of crushed camphor and sand in a china dish covered with an inverted plugged funnel, do you notice any solid deposits on the inner wall of the funnel? A – Yes, white, shiny solid crusts of pure camphor get deposited on the cool inner walls of the inverted glass funnel, while the sand remains left behind at the bottom of the china dish. This separation method is called sublimation.

Q – Can you think of any other mixtures where sublimation can be used to separate the components? A – Sublimation can be used whenever one substance in the mixture sublimes (turns directly from solid to gas) and the other does not. Examples include:

• A mixture of naphthalene balls (mothballs) and sand.
• A mixture of ammonium chloride (nausadar) and common salt.
• A mixture of iodine crystals and chalk powder.

Q – Is it possible to dissolve one metal in another? A – Yes! While metals cannot dissolve into each other at normal room temperature, they can be easily dissolved into one another when they are melted at very high temperatures. When this liquid mixture of melted metals cools down and solidifies, it forms a homogeneous solid mixture called an alloy (such as brass, bronze, or stainless steel).

Pause and Ponder (Page 84)

Q – 7. Why do immiscible liquids form two separate layers in a separating funnel? A – Immiscible liquids form separate layers because they do not dissolve in each other and they have different densities. Gravity pulls the heavier (more dense) liquid down to the bottom, while the lighter (less dense) liquid naturally floats to the top.

Q – 8. Is sublimation different from evaporation? Justify. A – Yes, sublimation is very different from evaporation:

• Sublimation is the change of a solid directly into a gas/vapour without ever melting into a liquid state (like camphor or dry ice heating up).
• Evaporation is the change of a liquid into a gas/vapour at its surface (like liquid water turning into water vapour).

Section 5.4.3 & 5.4.4: Suspensions, Centrifugation, Coagulation & Colloids

Q – How can we separate mud from water? If the muddy water is still not clear even after keeping for some time, how can it be cleaned?

A –

• We can first let the water stand so heavy mud settles at the bottom (sedimentation) and then pour off the top water (decantation) or pass it through filter paper (filtration).
• However, very fine mud particles can pass through filters and keep the water cloudy. To clean it completely, we use coagulation: we swirl a piece of alum (fitkari) in the water. Alum acts as a coagulant, causing the tiny suspended clay particles to stick together into heavy clumps. These heavy clumps settle rapidly to the bottom, leaving clear water on top that can be filtered easily. Another advanced method is centrifugation, where spinning at high speed forces the mud down.

Q – Activity 5.8: Which mixture would you like to separate using a mini centrifuge (paperfuge)?

A – You can easily test your mini cardboard centrifuge on everyday mixtures like:

• Chalk powder suspended in water (spinning forces the chalk to pack tightly at the bottom of a small tube).
• Muddy water (to force fine clay particles to settle rapidly).
• Milk (to see if a thin cream layer begins to separate from the watery whey).

 Q – Can you think of any other coagulation processes used in everyday life? A – A very common everyday example of coagulation is making paneer (cottage cheese) from milk! When we add lemon juice or vinegar (an acid) to boiling milk, it acts as a coagulant. This causes the milk proteins and fats to clump together into solid white masses (curd/paneer), separating out from the watery liquid (whey). Another example is the natural clotting of blood over a scrape or wound to stop bleeding.

Q – If the components of blood can be separated by centrifugation and the blood coagulates, is it a suspension? However, we cannot see blood cells with the naked eye. Is it a solution? What are colloids, and how are they different from solutions and suspensions? Can you think of some other substances that could be colloids?

A –

• Blood is neither a solution nor a suspension; it is a colloid.
• What are colloids? Colloids are mixtures where the particle size (between 1 and 1000 nm) is larger than the tiny particles in a true solution (less than 1 nm) but smaller than the heavy visible particles in a suspension (more than 1000 nm).
• How are they different? Unlike suspensions, colloidal particles never settle down at the bottom by themselves when left undisturbed, and they cannot be separated by simple filter paper. Unlike true solutions, colloidal particles are large enough to scatter a beam of light (Tyndall effect).
• Other examples of colloids: Milk, tomato sauce, ice cream, butter, face creams, fog, and smoke.

Section 5.5: Tyndall Effect

Q – Have you ever observed light scattering by particles in your surroundings? Think of some more examples of the Tyndall effect! A – Yes! Here are common everyday examples of the Tyndall effect:

• A beam of sunlight entering a dark or dusty room through a small window or keyhole.
• The bright cones of light coming from the headlights of a car or motorbike driving through fog, mist, or smoke at night.
• The beams of floodlights shining through the night air in a sports stadium or cinema hall.
• Sunlight rays shining down through the gaps of dense tree branches in a foggy forest.

Pause and Ponder (Page 88)

Q – 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? A – Clouds are a colloid (specifically, an aerosol colloid where liquid water droplets or ice crystals are dispersed in a gas medium, air). Why? Because the water droplets are very small and light enough to stay floating in the sky without settling down immediately by gravity (unlike a suspension), yet they are large enough to scatter sunlight and block clear vision (unlike a transparent solution).

Q – 10. Why do cities with a lot of smoke and dust in the air often look hazy? A – This happens because of the Tyndall effect. Smoke and dust particles suspended in the city air act as colloidal particles. As sunlight travels through the atmosphere, these tiny floating particles scatter the light in all directions, making the air look cloudy, blurry, and hazy.

Activity 5.9: Table 5.1

Q – Complete Table 5.1 to review the properties of solutions, suspensions, and colloids. A – Here is the completed comparison table:

S. No.	Property	Solution	Suspension	Colloid
1.	Nature	Homogeneous	Heterogeneous	Heterogeneous (looks homogeneous)
2.	Particle size	Very small (less than 1 nm)	Very large (more than 1000 nm)	Medium size (1 to 1000 nm)
3.	Visibility	Particles cannot be seen	Particles easily visible to naked eye	Particles cannot be seen with naked eye
4.	Separation by filtration	Cannot be separated by filtration	Can be separated by filtration	Cannot be separated by simple filtration
5.	Settling	Does not settle down	Settles down on standing	Does not settle down
6.	Tyndall effect	Does not show Tyndall effect	Shows Tyndall effect	Shows Tyndall effect

End of Section In-Text Question

Q – Imagine trying to separate all the ingredients in a lemonade once they have been mixed. Can you do it?

A – You can separate some ingredients, but not all easily in their original form!

• By using distillation, you can boil and condense the mixture to separate and collect pure water.
• After distillation, solid sugar and salt will be left behind in the flask as a dry mixture.
• However, separating the sugar and salt from each other, or getting back the fresh lemon juice flavor compounds exactly as they were, is extremely difficult using simple school laboratory methods because they are intimately mixed and chemically altered by heat!

Revise, Reflect, Refine (Chapter 5 Exercises)

Q – 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 A – Option (iv) is the correct choice.

• Explanation: Muddy water, Milk, and Blood are all heterogeneous mixtures (Ht) (milk and blood are colloids, which are scientifically heterogeneous). Brass is an alloy, which is a uniform, solid homogeneous mixture (Hm).

Q – 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 A – Option (iii): a and c is the correct answer.

• Reason for correct option: The Tyndall effect (scattering of light) is shown only by colloids and suspensions. A mixture of air and dust is a suspension/colloid, and starch in water forms a colloid. Their particles are large enough to scatter light.
• Reason for incorrect option: Copper sulfate in water and acetone in water form true, transparent solutions. The particles in true solutions are smaller than 1 nm, which is too tiny to scatter light.

Q – 3. Utilise the words or phrases provided in the box to fill in Table 5.2 for Solution, Suspension, and Colloid. A – Here is the completed Table 5.2 using the provided terms:

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

Q – 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.

A –

(i) Calculation for cake mixture:

• Mass of sugar = 75 g
• Mass of flour = 420 g
• Mass of sodium hydrogencarbonate = 5 g
• Total mass of mixture = .
• Using Mass by mass percentage ():
  • Concentration of sugar: .
  • Concentration of flour: .
  • Concentration of sodium hydrogencarbonate: .

(ii) Calculation for brass alloy:

• Total mass of brass = 120 g.
• Percentage of copper = 70% by mass.
• Mass of copper: .
• Mass of zinc: Total mass minus mass of copper =  (or 30% of 120 g).

Q – 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.

A –

• Will it form a separate layer? Yes, because cooking oil and water are immiscible (they do not mix).
• Which substance is on top? Cooking oil will float on top because it is lighter (less dense) than water (1 litre of oil weighs 910 g, while 1 litre of water weighs 1000 g).
• How to separate them: We use a separating funnel. Pour the oil-water mixture into the funnel and let it stand undisturbed until two clear layers form. Open the stopcock slowly to drain the lower water layer into a beaker, and then close the valve. The oil remains trapped inside the separating funnel.
• Diagram: (Please refer to Fig. 5.16 on page 83 of your textbook for the exact drawing of the separating funnel setup). It consists of a pear-shaped glass separating funnel clamped to a stand, with a stopcock at the bottom draining liquid into a conical flask.

Q – 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. A – Option (iii): A is true, but R is false.

• Explanation: The assertion is true: true solutions are completely transparent and do not show the Tyndall effect. But the reason is false: particles in a solution are actually extremely tiny (smaller than 1 nm, not larger than 100 nm), which is why they are too small to scatter light.

Q – 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. A – Here is the completed separation table with methods and scientific reasons:

Mixture	Method of separation	Reason for selection
Mud from muddy water	Coagulation (using alum) followed by Filtration	Alum makes fine mud particles clump together into heavy grains that settle down. Filtration easily removes these large settled particles.
Plasma from other components in the blood sample	Centrifugation	When spun at high speed, centrifugal force pushes the heavier red and white blood cells down to the bottom, leaving the lighter liquid plasma on top.
Naphthalene and sand	Sublimation	Naphthalene sublimes (changes directly from solid to gas when heated and condenses back to solid on cool surfaces), while sand does not.
Chalk powder and common salt	Dissolving in water, followed by Filtration and Evaporation	Salt dissolves in water but chalk powder is insoluble. Filter out the solid chalk residue first. Then evaporate the filtrate water to recover the dry salt.
Salt and water	Distillation	Salt is non-volatile while water evaporates easily. Distillation boils off the water vapour and condenses it back into a flask (saving pure water), leaving dry salt behind.
Oil from water	Separating funnel	Oil and water are immiscible liquids with different densities. They form two separate layers that can be drained one by one.
Pigments of the common flower	Paper Chromatography	Different flower pigments dissolve at different rates in a solvent (like water or alcohol). As the solvent moves up the filter paper, it separates the pigments based on their solubilities.

 

Q – 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.

A –

• Method of separation: Simple Distillation. Because both liquids are miscible and their boiling points differ by 30°C (which is greater than the required 25°C difference), distillation will separate them easily. When heated to 60°C, liquid A will boil into vapour first, travel through the water condenser, cool down into liquid, and get collected in the receiving flask. Liquid B (boiling point 90°C) will stay behind in the distillation flask.
• Diagram: (Please refer to Fig. 5.12 on page 80 of your textbook). It includes a distillation flask sitting over a burner on a tripod stand, a thermometer at the top, a sloping water condenser with water inlet/outlet tubes, and a conical receiving flask collecting the distillate.

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

A –

• Evaporation: Heating a liquid to turn it into vapour until only solid solute remains.
  • When preferred: Use this when you only want to collect a heat-stable solid (like common salt) quickly and do not care about saving the solvent (water).
• Crystallization: Dissolving a solid in hot solvent to make a saturated solution, then cooling it slowly so pure geometrical solid crystals form.
  • When preferred: Use this when you need to purify a solid containing impurities (like copper sulfate or sugar candy), or when the solid might burn or decompose if heated to dryness by simple evaporation.
• Distillation: Boiling a liquid mixture into vapour and then cooling (condensing) the vapour back into liquid in a separate container.
  • When preferred: Use this when you want to recover and save both the solvent and the solute (like getting drinking water from seawater), or when separating two miscible liquids with different boiling points (like acetone and water).

 Q – 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. A – (i) If blood behaved like a true suspension, the heavy red blood cells and platelets would settle down at the bottom of our blood vessels and heart whenever we rested or slept! This would block blood circulation, stop oxygen from reaching organs, and make survival impossible. (ii) In a blood colloid:

• Dispersed phase (the floating particles): Red blood cells, white blood cells, and platelets.
• Dispersion medium (the liquid medium): Plasma (the pale yellow liquid).

 Q – 11. You are given a mixture of sand, common salt and naphthalene. Figure 5.25b depicts various steps used to separate the components of this mixture. Identify and write down the correct sequence of separation techniques. A – The correct sequence of steps to separate all three components is:

1. Step 1: Sublimation (Diagram 1) — Heat the dry mixture in a china dish with an inverted funnel. The naphthalene sublimes into vapour and collects as solid crusts inside the funnel, leaving behind a mixture of sand and salt.
2. Step 2: Dissolving & Filtration (Diagram 3) — Add water to the remaining sand-salt mixture and stir so the salt dissolves. Pour it through a filter funnel. The insoluble sand stays behind on the filter paper as residue, while the clear salt water (filtrate) passes through into the beaker.
3. Step 3: Evaporation (Diagram 2) — Heat the clear salt water in a china dish over a burner until all the water evaporates into the air, leaving behind pure, dry common salt.

• (Correct chronological order of diagrams from Fig 5.25b: 1 → 3 → 2).

 Q – 12. Why is distillation an effective method for separating a mixture of water and acetone? A – Distillation is effective because acetone and water are miscible liquids with a large difference in their boiling points. Acetone boils at a low temperature of about 56°C, whereas water boils at 100°C. When the mixture is gently heated to around 56°C, acetone rapidly vaporises while water remains liquid. The acetone vapours pass through the cooling condenser and are collected safely as pure liquid acetone.

 Q – 13. Answer the following questions with the help of the solubility 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. A – (i) From Table 5.4, the solubility of potassium nitrate at 40°C is 62 g per 100 g of water. Therefore, to make a saturated solution in only 50 g of water (which is half of 100 g), we need:  of potassium nitrate. (ii) She will observe white, shiny crystals of potassium chloride appearing and settling down at the bottom of the beaker.

• Explanation: At 80°C, water can dissolve 54 g of potassium chloride per 100 g. When cooled to room temperature (25°C), its solubility drops to around 36 g. Because cool water cannot hold as much dissolved salt, the extra solid salt () is forced out of the solution as crystals. (iii)
• Effect of temperature: In general, the solubility of solid salts in water increases as the temperature increases.
• Comparison of the four salts (from 10°C to 80°C):
  • Potassium nitrate: Shows the largest and fastest increase in solubility (leaping from 21 g at 10°C up to 167 g at 80°C).
  • Ammonium chloride: Shows a moderate increase in solubility (rising steadily from 24 g to 66 g).
  • Potassium chloride: Shows a slow and steady increase (rising from 35 g to 54 g).
  • Sodium chloride (common salt): Shows almost no change at all (staying nearly flat from 36 g at 10°C to only 37 g at 80°C).

 Q – 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. A – (i) Mass percentage calculations ():

• Student A: Total mass = . Concentration = .
• Student B: Total mass = . Concentration = .
• Student C: Total mass = . Concentration = .

(ii) Who has the most concentrated solution?

• Student C’s solution is the most concentrated (27.27% m/m).
• Why? Because Student C dissolved the largest amount of sugar (30 g) in a smaller amount of water (80 g), resulting in the highest ratio of solute to total solution.

 Q – 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 A – (i) The separation technique marked as ‘S’ is Distillation. (ii) Labels for apparatus:

• A: Distillation flask (containing the liquid mixture).
• B: Water condenser (the sloping tube that cools and condenses the vapour).
• C: Conical flask / Receiving flask (collecting the condensed pure liquid or distillate).

(iii) Which mixtures can be separated using distillation? To separate two liquids by simple distillation, they must be miscible and have a boiling point difference of at least 25°C. Let’s check the differences using Table 5.5:

• (a) water (100°C) – acetone (56°C): Difference is 44°C  Can be separated.
• (b) water – salt: Salt is a non-volatile dissolved solid  Can be separated (water boils off and condenses in flask C, salt stays in flask A).
• (c) acetone (56°C) – alcohol (78°C): Difference is 22°C (less than 25°C)  Cannot be easily separated by simple distillation.
• (d) sand – salt: Both are dry solids  Cannot be separated by distillation (needs filtration & evaporation).
• (e) alcohol (78°C) – chloroform (61°C): Difference is 17°C (less than 25°C)  Cannot be easily separated by simple distillation.
• (f) alcohol (78°C) – benzene (80°C): Difference is only 2°C  Cannot be separated by simple distillation.

Final Answer for (iii): Only mixtures (a) water – acetone and (b) water – salt can be successfully separated using the simple distillation setup shown in Figure 5.26.