Exploring Mixtures and their Separation takes a deeper look at mixtures. It sorts them into homogeneous and heterogeneous, and then into solutions, suspensions and colloids by particle size. It shows how to express concentration as % m/m, % m/v and % v/v, explains solubility and saturated solutions, and covers the main separation methods: crystallization, distillation, paper chromatography, separating funnel, sublimation, centrifugation and coagulation. It ends with the Tyndall effect. School exams draw percentage calculations, method-and-reason questions and solution-suspension-colloid comparisons from it.
Key Concepts
1. Homogeneous and Heterogeneous Mixtures
A well-stirred mixture of sugar and water is equally sweet in the first sip and the last sip. Its composition is uniform throughout, so it is a homogeneous mixture, also called a solution. A solution always remains homogeneous. Vinegar (acetic acid in water) and soda (carbon dioxide in water) are other examples.
A stirred mixture of sand and water is not uniform: the sand particles are easily visible and settle with time. This is a heterogeneous mixture. Oil and water, which form two separate layers, are heterogeneous too.
Mixtures with a gas: gas particles move freely in all directions, so they mix uniformly with other gases. Most gas mixtures are therefore homogeneous, such as hydrogen and oxygen used as a rocket fuel. Smoke (solid particles suspended in air), fog (tiny water droplets in air) and dust in the air are heterogeneous mixtures with a gas as one component.
Solid-solid mixtures: most are heterogeneous, such as iron filings and sulfur. You separate them by a property of one solid: a magnet pulls iron nails out of sawdust, and sublimation removes camphor from sand.
Alloys: metals do not dissolve into each other at room temperature, but molten metals can be mixed to form a solution that cools into a material that looks like a single metal. A homogeneous mixture of two or more metals, or of a metal and a non-metal, is an alloy. Physical methods cannot separate its components. Alloys are made to be stronger, more rigid or more corrosion-resistant.
| Alloy | Approximate composition (NCERT) |
|---|---|
| Brass | About 80% copper and 20% zinc |
| Bronze | About 80% copper and 20% tin |
| Stainless steel | Iron with carbon (about 0.03 to 0.8%), chromium (16 to 18%), nickel (10.0 to 14.0%) and molybdenum (2.0 to 3.0%) |
Activity 5.1: three beakers, three kinds of mixture
Salt (A), chalk powder (B) and a few drops of milk (C) are each stirred into 50 mL of water. A laser beam is shone through each beaker and viewed from the side. (Never look directly into a laser beam.)
| Observation | A: salt in water | B: chalk in water | C: milk in water |
|---|---|---|---|
| Particles visible? | No | Yes | No, looks uniform |
| Path of laser beam | Not visible | Visible (particles scatter light) | Visible (particles scatter light) |
| On standing | Nothing settles | Chalk settles | Nothing settles |
| Residue on filter paper | None | Chalk left behind | None |
| Type of mixture | Solution | Suspension | Colloid |
2. Concentration of a Solution as a Percentage
A solution has a solute (the substance that gets dissolved) and a solvent (the substance that dissolves the solute). In sugar water, sugar is the solute and water is the solvent.
The proportion matters. ORS needs specified amounts of salt and sugar in a fixed amount of water; change them and you get a salt and sugar solution that is no longer ORS. A pesticide spray with too little pesticide may not protect crops, and too much can damage crops, soil and the environment. (Dilip Mahalanabis, an Indian paediatrician, formulated the ORS that the World Health Organization popularised worldwide.)
The concentration of a solution is the amount of solute dissolved in a given amount of solvent or solution. This chapter expresses it as a percentage.
A. Mass by mass percentage (% m/m or % w/w)
Grams of solute in 100 g of the total solution.
Mass by mass percentage = (Mass of solute / Mass of solution) × 100
It is used for homogeneous mixtures and for heterogeneous ones such as milk powder and spice mixtures, and on packaged food labels. Industries write % w/w; numerically, % m/m and % w/w are equal.
NCERT Example 5.1: 10 g of salt in 90 g of water. Mass of solution = 10 + 90 = 100 g, so the concentration = (10 / 100) × 100 = 10% m/m. The denominator is the mass of the solution, never the 90 g of water alone.
B. Mass by volume percentage (% m/v or % w/v)
Grams of solute in 100 mL of solution. It is used where measuring a liquid’s volume is easier than weighing it, as in medicines and laboratories.
Mass by volume percentage = (Mass of solute / Volume of solution) × 100
NCERT Example 5.2: 5 g of glucose is dissolved in water to make 100 mL of solution. Mass by volume percentage = (5 g / 100 mL) × 100 = 5% m/v. This is the common 5% glucose solution.
Saline drip: usually 0.9% m/v sodium chloride in water, that is, 0.9 g of salt in 100 mL of solution. It is safe for blood and replaces lost body fluids.
C. Volume by volume percentage (% v/v)
Millilitres of solute in 100 mL of solution. It is used when two miscible liquids are mixed, as in perfumes, cosmetics and vinegar.
Volume by volume percentage = (Volume of solute / Volume of solution) × 100
NCERT Example 5.3: 1 mL of a liquid pesticide is mixed with enough water to form 100 mL of spray for a rice crop. Volume by volume percentage = (1 mL / 100 mL) × 100 = 1% v/v.
Choosing the right method
| Solute | Solution measured by | Method | Example |
|---|---|---|---|
| Solid (weighed) | Mass | % m/m | Milk powder label, spice mix, alloy |
| Solid (weighed) | Volume | % m/v | 5% glucose, 0.9% saline |
| Liquid (measured) | Volume | % v/v | Vinegar, perfume, pesticide spray |
Making vinegar (NCERT Pause and Ponder 3): vinegar is 5% v/v acetic acid; glacial acetic acid is 100% acetic acid. Add 5 mL of glacial acetic acid to some water, then add water until the total volume is 100 mL.
3. Solubility, Saturated Solutions and Crystallization
The solubility of a substance is the maximum amount of it that dissolves in a fixed quantity of the solvent (100 mL or 100 g) at a given temperature. A solution that cannot dissolve any more solute at that temperature is a saturated solution.
The temperature must always be stated. The solubility of a solid in a liquid generally increases with temperature; the solubility of a gas in a liquid generally decreases as temperature rises.
Solubility curve: a graph of solubility (g per 100 g of water, y-axis) against temperature (°C, x-axis). In NCERT Activity 5.2, compound ‘B’ has a solubility of 287 g per 100 g of water at 60 °C and 241 g at 40 °C. On such a graph the higher curve at a given temperature is the more soluble compound, and the steeper curve is the one whose solubility changes more with temperature.
Cooling a hot saturated solution: dissolve 287 g of ‘B’ in 100 g of water at 60 °C and cool it gradually to 40 °C. Only 241 g can stay dissolved, so the rest separates out as a pure solid, often as crystals.
A crystal is a solid made up of particles arranged in a regular geometric pattern: rock salt, candy sugar (mishri), snowflakes and frost are examples.
Crystallization is the process of forming crystals from a saturated solution. It separates two solids when one is present in small quantity and both dissolve in the same solvent, and it purifies solids, such as a newly prepared compound mixed with impurities. Its principle is the difference in the solubility of a substance at different temperatures.
Activity 5.3: growing copper sulfate crystals, step by step
- Place 1 g of copper sulfate (blue vitriol) in a 100 mL beaker with 25 mL of water. The teacher adds a drop of dilute sulfuric acid, which helps make pure crystals by preventing unwanted reactions (only some salts need it).
- Heat the mixture gently in a water bath while stirring constantly.
- Add more copper sulfate gradually until the solution becomes saturated.
- Filter the hot solution to remove insoluble impurities; cover the filtrate with a watch glass.
- Let it cool slowly without disturbance. The particles have time to come together into large, shiny, well-shaped blue crystals.
- Filter out the crystals, rinse them with cold water, and dry them on a watch glass.
Copper sulfate is toxic: do not touch it with bare hands, and work under adult supervision.
Testing the cooling rate: NCERT’s hypothesis is that rapid cooling in ice-cold water gives smaller, less well-formed crystals than slow cooling at room temperature. To test it, divide one hot saturated solution into two equal parts, cool one fast and one slowly, and compare the crystals.
Salt from seawater (Activity 5.4): seawater, then a saturated solution, then salt crystals, as water evaporates. Coastal Indian communities made panga salt by boiling concentrated sea brines and karkatch salt by evaporating sea water, giving crystals of different sizes. Evaporation loses the solvent; to recover it, you need distillation.
4. Distillation and Paper Chromatography
Distillation
Distillation separates a homogeneous mixture of two miscible liquids: heat it until the liquid with the lower boiling point vaporises, then cool the vapour back into liquid (the distillate). It needs boiling points that differ by at least about 25 °C. It also recovers a liquid from a solution of a solid, such as water from salt water; the solid stays in the flask.
The set-up in words: the mixture sits in a distillation flask on a wire gauze and tripod stand over a burner, held by a laboratory stand. A thermometer at the top reads the vapour temperature. The vapour passes into a water condenser, cooled by water flowing in through a water inlet and out through a water outlet (usually in at the lower end, out at the upper end), and condenses into a conical flask.
Acetone and water: acetone boils at about 56 °C and water at 100 °C, a difference of 44 °C. Acetone vaporises before water vapour forms in significant amount, so acetone collects in the conical flask and water stays in the distillation flask.
In Kannauj, Uttar Pradesh, the perfume capital of India, the earthy smell of the first rain is captured as Mitti ka Ittar by the traditional Deg-Bhapka distillation method.
Fractional distillation separates components whose boiling points differ by relatively small amounts (less than 25 °C). A petroleum refinery uses it to separate crude oil into petroleum gas, petrol, aviation fuel (kerosene), diesel, lubricating oil and bitumen. The gaseous fraction, filled into steel cylinders under high pressure, liquefies and is used as domestic fuel (LPG).
Paper chromatography
From the Greek chroma (colour) and graphein (to write), because it was first used to separate coloured substances such as dyes and inks.
Activity 5.5 set-up:
- Take a 3 cm wide strip of chromatographic paper (or filter paper). With a pencil, draw a straight horizontal line 2 cm from the bottom.
- Put a spot of black sketch-pen ink at the centre of the line.
- Pour enough water to make a thin layer at the bottom of a gas jar, measuring cylinder or beaker.
- Hang the strip vertically so its lower end dips into the water. The water level must be below the ink spot. A watch glass covers the jar.
- As the water rises through the paper, the ink separates into different colour spots.
Principle: the components interact differently with the solvent and the paper, so the rising liquid carries them up at different rates. A black ink that splits into several colours is a mixture of dyes. Pencil is used because it does not run in the solvent; the spot sits above the solvent so the ink is carried up the paper instead of dissolving into the jar.
The method also separates pigments in spinach leaf extract or flower petals, sometimes with alcohol or a mixture of solvents instead of water. Green food colour can be run with a 2% m/v salt solution.
5. Separating Heterogeneous Mixtures: Separating Funnel and Sublimation
Two immiscible liquids: the separating funnel
Immiscible liquids, such as oil and water, do not mix and form separate layers. A separating funnel separates them by their different densities.
Activity 5.6 (mustard oil and water):
- Pour 5 mL mustard oil and 20 mL water into a 50 mL separating funnel on a laboratory stand (glass stopper on top, stopcock at the bottom).
- Let it stand. Yellow mustard oil forms the upper layer and water the lower layer, because the less dense liquid floats.
- Open the stopcock slowly to drain the water; close it when the water is almost fully drained.
- Discard the next small portion, which may contain both liquids, then collect the oil separately.
Sublimation
Activity 5.7 (camphor and sand): put a spatula full of crushed camphor and sand in a clean, dry china dish on a wire gauze over a tripod stand. Cover it with an inverted glass funnel whose nozzle is plugged with cotton. Heat gently for a few minutes. White solid camphor deposits on the inner wall of the funnel, and sand remains in the dish.
- Sublimation: camphor, on heating below its melting point, changes directly from solid to vapour without passing through the liquid state.
- Deposition: on cooling, the vapour changes back into a solid without becoming a liquid. This happens on the cooler funnel wall.
Camphor separates because sand does not sublime. Naphthalene works in place of camphor, and dry ice (solid carbon dioxide) also sublimes. Evaporation, by contrast, turns a liquid into vapour.
6. Suspensions, Centrifugation and Coagulation
A suspension is a heterogeneous mixture in which solid particles do not dissolve but remain suspended throughout the bulk of the medium. Its particles are larger than 1000 nm, visible to the naked eye, settle when left undisturbed, can be filtered, and scatter light. Examples: sand, chalk powder, sawdust or tea leaves in water, and muddy water.
Cleaning muddy water: left undisturbed, the heavier mud settles but the water may stay cloudy, and filtering through cloth or filter paper often leaves it cloudy too. Tiny particles need centrifugation and/or coagulation.
A. Centrifugation
Spin with a partner holding crossed hands (the folk game phugadi in Marathi, kikli in Punjabi) and you feel pulled outwards. In centrifugation a mixture in a tube is spun at high speed. The tubes become horizontal, and the centrifugal force (the outward force on a body moving in circular motion) moves the heavier particles outwards, so they settle at the bottom of the tube while the lighter liquid stays on top.
Laboratories use it to separate blood into plasma (top), platelets and white blood cells (middle band) and red blood cells (bottom); many chemical industries use it too. The paperfuge, a hand-powered cardboard disc spun on strings, does the same job without electricity and helps detect malaria and anaemia in remote areas.
B. Coagulation
Powdered alum (fitkari), a white crystalline substance used in water purification, makes the fine suspended particles in muddy water clump together. This is coagulation, and alum is the coagulant. The clumps settle by gravity (sedimentation), and the water is then separated by decantation or filtration. Paneer making is also coagulation: lemon juice or vinegar coagulates the milk proteins. Sewage treatment uses sedimentation, coagulation and filtration.
7. Colloids and the Tyndall Effect
Blood can be centrifuged and it coagulates, yet its cells are invisible to the naked eye. Blood is neither a solution nor a true suspension: it is a colloid, like milk, tomato sauce and ice cream.
The three types differ mainly in particle size: solution particles are less than 1 nm in diameter, colloid particles 1 to 1000 nm, and suspension particles more than 1000 nm. Colloid particles do not settle over time; they stay uniformly dispersed, as in a solution, which is why milk looks uniform. NCERT’s exercise classes milk and blood as heterogeneous.
Tyndall effect
In Activity 5.1 the beam was invisible in the salt solution, and visible in the chalk suspension and in the milk mixture, although the milk mixture looked homogeneous. The scattering of light by particles is the Tyndall effect, named after John Tyndall, who first explained it. It occurs in colloids and suspensions, and not in a transparent solution. Examples: a beam entering a dark room through a small hole (dust and smoke scatter it), stadium floodlight beams, and sunbeams between the leaves of a dense tree. Smoky, dusty cities look hazy for the same reason.
Dispersed phase, dispersion medium and emulsions
The solute-like dispersed particles of a colloid form the dispersed phase; the component they are suspended in is the dispersion medium. In blood, the cells are the dispersed phase and plasma is the medium. A colloid with both components liquid is an emulsion:
| Type | Dispersed phase in medium | Examples (NCERT) |
|---|---|---|
| Oil-in-water | Oil droplets spread in water | Milk, vanishing cream |
| Water-in-oil | Water droplets spread in oil | Butter, body lotions, cold cream |
Emulsifying agents, such as the proteins in milk and butter, stabilise emulsions. Shaking a few drops of cooking oil with water containing a little soap solution makes an emulsion.
Solution, suspension and colloid compared (NCERT Table 5.1 completed)
| Property | Solution | Suspension | Colloid |
|---|---|---|---|
| Nature | Homogeneous | Heterogeneous | Heterogeneous (looks homogeneous) |
| Particle size | Less than 1 nm | More than 1000 nm | 1 to 1000 nm |
| Visibility of particles | Not visible; transparent | Visible to the naked eye | Not visible to the naked eye |
| Separation by filtration | Cannot be separated | Separates by filtration | Cannot be separated by filtration |
| Settling | Does not settle | Settles when left undisturbed | Does not settle; particles remain evenly distributed |
| Tyndall effect | Not shown | Shown | Shown |
| Examples | Salt solution, sugar solution, vinegar, brass | Sand in water, mud, chalk in water | Milk, blood, butter, tomato sauce |
Separation relies on properties such as size, density and solubility. It happens in nature too (kidneys remove waste from blood), and sorting dry waste for recycling and wet waste for composting is a separation task at home.
Formula and Reaction Sheet
| Quantity or rule | Formula or statement | Unit or note |
|---|---|---|
| Mass by mass percentage | (Mass of solute / Mass of solution) × 100 | % m/m (= % w/w) |
| Mass by volume percentage | (Mass of solute / Volume of solution) × 100 | % m/v; solute in g, solution in mL |
| Volume by volume percentage | (Volume of solute / Volume of solution) × 100 | % v/v; both in mL |
| Mass of solute from a percentage | Mass of solute = (% m/m × Mass of solution) / 100 | 4% of 300 g = 12 g |
| Solute in a saturated solution | Solute = Solubility × (Mass of water / 100 g) | Solubility in g per 100 g water |
| Solid deposited on cooling | Solubility at higher temperature - Solubility at lower temperature | Per 100 g of water |
| Simple distillation | Boiling point difference of at least about 25 °C | Less than 25 °C: fractional distillation |
| Particle size | Solution < 1 nm; colloid 1 to 1000 nm; suspension > 1000 nm | Diameter |
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Important Definitions
- Homogeneous mixture (solution): a mixture with a uniform composition throughout. A solution always remains homogeneous.
- Heterogeneous mixture: a mixture whose composition is not uniform, such as sand and water.
- Concentration of a solution: the amount of solute dissolved in a given amount of solvent or solution.
- Mass by mass percentage: grams of solute present in 100 g of solution.
- Mass by volume percentage: grams of solute present in 100 mL of solution.
- Volume by volume percentage: millilitres of solute present in 100 mL of solution.
- Solubility: the maximum amount of solute that dissolves in a fixed quantity of solvent (100 mL or 100 g) at a given temperature.
- Saturated solution: a solution that cannot dissolve any more solute at that temperature.
- Crystallization: the process of forming crystals from a saturated solution.
- Distillation: heating a mixture until the liquid with the lower boiling point vaporises, then cooling the vapour back into liquid.
- Paper chromatography: separating components by their different rates of movement up the paper, caused by different interactions with the solvent and the paper.
- Sublimation: the change of a solid directly into vapour, below its melting point, without passing through the liquid state.
- Deposition: the change of a vapour directly back into a solid on cooling, without becoming a liquid.
- Alloy: a homogeneous mixture of two or more metals, or of a metal and a non-metal.
- Suspension: a heterogeneous mixture in which solid particles do not dissolve but remain suspended throughout the bulk of the medium.
- Centrifugation: spinning a mixture in a tube at high speed so that heavier particles move outwards and settle at the bottom of the tube.
- Coagulation: adding a substance (a coagulant) that makes small particles clump together and settle down.
- Colloid: a mixture with particles 1 to 1000 nm in size that stay uniformly dispersed and do not settle.
- Tyndall effect: the scattering of light by particles.
- Dispersed phase and dispersion medium: the dispersed particles of a colloid, and the component they are suspended in.
- Emulsion: a colloid in which both the dispersed phase and the dispersion medium are liquids.
Solved Examples (NCERT-Based)
Example 1: Zinc oxide in talcum powder (NCERT Pause and Ponder 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?
Solution: 4% m/m means 4 g of zinc oxide in every 100 g of powder.
Mass of zinc oxide = (4 × 300) / 100 = 1200 / 100 = 12 g.
Example 2: Orange juice concentrate (NCERT Pause and Ponder 2)
You mix two tablespoons of orange juice concentrate with water in a glass tumbler. 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?
Solution: Volume of concentrate (solute) = 2 × 15 mL = 30 mL. Volume of juice (solution) = 150 mL.
% v/v = (30 / 150) × 100 = 20% v/v.
Example 3: Three students, three sugar solutions (NCERT Exercise 14)
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.
Solution:
- A: mass of solution = 20 + 80 = 100 g. % m/m = (20 / 100) × 100 = 20%.
- B: mass of solution = 20 + 100 = 120 g. % m/m = (20 / 120) × 100 = 16.67% (to two decimal places).
- C: mass of solution = 30 + 80 = 110 g. % m/m = (30 / 110) × 100 = 27.27% (to two decimal places).
(ii) Student C’s solution is the most concentrated, because it has the largest mass of sugar in every 100 g of solution.
Example 4: Dry cake mix (NCERT Exercise 4 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.
Solution: All three are solids that are weighed, so use mass by mass percentage. Total mass of mixture = 75 + 420 + 5 = 500 g.
- Sugar = (75 / 500) × 100 = 15% m/m
- Flour = (420 / 500) × 100 = 84% m/m
- Sodium hydrogencarbonate = (5 / 500) × 100 = 1% m/m
Check: 15 + 84 + 1 = 100%.
Example 5: Copper and zinc in brass (NCERT Exercise 4 ii)
A brass alloy contains 70% copper by mass. Calculate the quantities of copper and zinc present in 120 g of brass.
Solution: Copper = (70 × 120) / 100 = 84 g.
Brass here is copper and zinc, so zinc = 120 - 84 = 36 g (that is, 30% of 120 g).
Example 6: Choosing a method for each mixture (NCERT Exercise 7, Table 5.3)
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.
| Mixture | Method of separation | Reason for selection |
|---|---|---|
| Mud from muddy water | Coagulation with alum, then sedimentation and decantation or filtration (centrifugation also works) | Fine mud particles are too small to remove by filtration alone; alum clumps them so they settle |
| Plasma from other components in the blood sample | Centrifugation | Spinning moves the heavier cells to the bottom; the lighter plasma stays on top |
| Naphthalene and sand | Sublimation | Naphthalene sublimes on heating; sand does not |
| Chalk powder and common salt | Add water, filter, then evaporate or crystallize the filtrate | Salt dissolves in water and chalk does not, so chalk stays on the filter paper |
| Common salt and water | Evaporation (to get the salt) or distillation (to get both) | Water has a much lower boiling point than salt; distillation recovers the water as well |
| Oil from water | Separating funnel | The liquids are immiscible and have different densities, so they form two layers |
| Pigments of the flower | Paper chromatography | The coloured components move up the paper at different rates |
Example 7: Sand, common salt and naphthalene (NCERT Exercise 11)
You are given a mixture of sand, common salt and naphthalene. Write the correct sequence of separation techniques.
Solution:
- Sublimation: heat the dry mixture in a china dish under an inverted funnel with a cotton-plugged nozzle. Naphthalene sublimes and deposits on the cool funnel wall. Sand and salt remain.
- Dissolving and filtration: add water to the remaining solid and stir. Salt dissolves, sand does not. Filter: sand stays on the filter paper as the residue.
- Evaporation or crystallization: heat the filtrate (salt solution) to remove the water and obtain the salt.
Sublimation comes first, while the mixture is still dry.
Example 8: Which pairs can simple distillation separate? (NCERT Table 5.5 data)
Boiling points: water 100 °C, acetone 56 °C, alcohol 78 °C, chloroform 61 °C, benzene 80 °C. Which of these mixtures can be separated by distillation: (a) water and acetone, (b) water and salt, (c) acetone and alcohol, (d) sand and salt, (e) alcohol and chloroform, (f) alcohol and benzene?
Solution: Miscible liquids need a gap of at least about 25 °C.
| Mixture | Difference | Distillation? |
|---|---|---|
| (a) Water and acetone | 100 - 56 = 44 °C | Yes |
| (b) Water and salt | Solid dissolved in a liquid | Yes, water is collected and salt stays in the flask |
| (c) Acetone and alcohol | 78 - 56 = 22 °C | No, less than 25 °C (needs fractional distillation) |
| (d) Sand and salt | Two solids | No, dissolve in water, filter and evaporate |
| (e) Alcohol and chloroform | 78 - 61 = 17 °C | No, less than 25 °C |
| (f) Alcohol and benzene | 80 - 78 = 2 °C | No, less than 25 °C |
Answer: (a) and (b).
Competency-Based Questions (with answers)
1. Case-based: The hospital saline drip
A nurse prepares saline drips. The standard drip is 0.9% m/v sodium chloride in water. It is safe for blood and replaces lost fluids. A patient also needs 5% glucose solution.
(a) How much salt is present in a 250 mL drip? 0.9 g per 100 mL, so 0.9 × 250 / 100 = 2.25 g.
(b) Why is concentration written as % m/v here and not % m/m? The solute (salt or glucose) is weighed, and the solution is a liquid whose volume is easier to measure than its mass. NCERT notes that % m/v is used in medicines and laboratories for this reason.
2. Case-based: Salt pans on the coast
Workers let seawater into shallow pans. As the sun heats the pans, the water level falls. After some days the liquid becomes saturated and white crystals appear.
(a) Name the stages in the order they occur. Seawater, then a saturated solution, then salt crystals.
(b) Why do crystals appear only after some days? Salt crystallizes only after the solution becomes saturated. Until enough water has evaporated, all the salt stays dissolved.
3. Source-based: Solubility data (NCERT Table 5.4)
Solubility in g per 100 g of water:
| Salt | 10 °C | 20 °C | 30 °C | 40 °C | 60 °C | 80 °C |
|---|---|---|---|---|---|---|
| Potassium nitrate | 21 | 32 | 45 | 62 | 106 | 167 |
| Sodium chloride | 36 | 36 | 36.3 | 36.5 | 37 | 37 |
| Potassium chloride | 35 | 35 | 37.4 | 40 | 46 | 54 |
| Ammonium chloride | 24 | 37 | 41 | 41 | 55 | 66 |
(a) 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. Crystals of potassium chloride appear. Its solubility falls from 54 g per 100 g of water at 80 °C to between 35 and 37.4 g at 25 °C (the table gives 20 °C and 30 °C). The extra salt, about 16.6 to 19 g for every 100 g of water (54 - 37.4 to 54 - 35), can no longer stay dissolved and crystallizes out.
(b) What is the effect of a change in temperature on the solubility of salts? Compare the four salts from 10 °C to 80 °C. The solubility of all four increases with temperature, by very different amounts. Potassium nitrate rises from 21 to 167 g (an increase of 146 g), ammonium chloride from 24 to 66 g (42 g), potassium chloride from 35 to 54 g (19 g), and sodium chloride only from 36 to 37 g (1 g).
4. Assertion-Reason (NCERT Exercise 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.
(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). A is true: a transparent solution does not scatter light. R is false: solute particles in a solution are the smallest, less than 1 nm in diameter. They are too small to scatter light, which is the real reason.
5. Assertion-Reason
Assertion (A): A mixture of acetone and water can be separated by simple distillation. Reason (R): Acetone boils at about 56 °C and water at 100 °C.
Answer: Both A and R are true, and R is the correct explanation of A. The boiling points differ by 44 °C, more than the minimum of about 25 °C, so acetone vaporises before the vapours of water form in significant amount.
6. Assertion-Reason
Assertion (A): The components of brass cannot be separated by a separating funnel or by filtration. Reason (R): Brass is a heterogeneous mixture of copper and zinc.
Answer: A is true, but R is false. Brass is an alloy, a homogeneous mixture. NCERT states that physical methods cannot separate the components of an alloy.
7. Case-based: The paperfuge in a village clinic
A health worker in a remote village has no electricity. She places a blood sample in a thin tube, tapes it to a cardboard disc threaded with string, and pulls the handles so that the disc spins very fast.
(a) Name the separation technique. Centrifugation.
(b) Where do the red blood cells collect and why? At the bottom of the tube. They are the heavier components, and the outward (centrifugal) force during spinning moves them outwards, which is the bottom of the tube. The lighter plasma stays at the top.
8. Case-based: Water for a small town
A town treats river water in three tanks. In the first, muddy water stands still for a day. In the second, alum is added. In the third, the water passes through layers of sand.
(a) Name the process in each tank. Tank 1: sedimentation. Tank 2: coagulation (alum is the coagulant). Tank 3: filtration.
(b) Why is the second tank needed if the first tank already lets mud settle? The heavier mud settles in tank 1, but the finest particles stay suspended and the water stays cloudy. Alum makes these fine particles clump together into larger clumps that settle by gravity.
Important Questions for Exams
1-Mark Questions
Q1. Which of the following mixtures are correctly classified as homogeneous (Hm) and heterogeneous (Ht)? (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). Smoke and milk are heterogeneous; brass and vinegar are homogeneous; muddy water is heterogeneous.
Q2. 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: (iii) a and c. Dust in air and starch in water have particles large enough to scatter light; copper sulfate and acetone form true solutions in water.
Q3. State the particle size range of a colloid.
Answer: 1 to 1000 nm in diameter.
2-Mark Questions
Q4. Why is distillation an effective method for separating a mixture of water and acetone?
Answer: The two are miscible liquids whose boiling points differ by 44 °C (acetone about 56 °C, water 100 °C), which is more than the minimum of about 25 °C. On heating, acetone vaporises first, before water vapour forms in significant amount. Its vapour is condensed and collected separately, while water stays in the flask.
Q5. Why do immiscible liquids form two separate layers in a separating funnel?
Answer: Immiscible liquids do not mix with each other. They have different densities, so the less dense liquid floats as the upper layer and the denser liquid forms the lower layer. For mustard oil and water, oil forms the upper layer.
Q6. Is sublimation different from evaporation? Justify.
Answer: Yes. In sublimation a solid changes directly into vapour, below its melting point, without passing through the liquid state (camphor, naphthalene, dry ice). In evaporation a liquid changes into vapour.
3-Mark Questions
Q7. 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?
Answer: Yes, oil and water are immiscible, so they form separate layers. Density of the oil = 910 g / 1000 mL = 0.91 g/mL, less than water (about 1 g/mL), so oil is on top. Separate them with a separating funnel: let the mixture stand, open the stopcock to drain the lower water layer, close it when water is almost drained, discard the small mixed portion, then collect the oil. Diagram: separating funnel on a stand, glass stopper on top, oil above water, stopcock below, conical flask underneath.
Q8. 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.
Answer: Distillation. The difference in boiling points is 90 - 60 = 30 °C, which is at least about 25 °C. Heat the mixture in a distillation flask; A vaporises first. Its vapour passes through a water condenser, condenses, and is collected in a conical flask. B stays in the distillation flask. Label: burner, wire gauze, distillation flask, thermometer, water condenser (inlet and outlet), conical flask.
Q9. Describe how you would obtain pure copper sulfate crystals from an impure sample.
Answer: Dissolve the sample in water with a drop of dilute sulfuric acid, heating in a water bath and stirring, and keep adding the solid until the solution is saturated. Filter the hot solution to remove insoluble impurities, cover the filtrate with a watch glass, and let it cool slowly without disturbance. Large blue crystals form. Filter them, rinse with cold water and dry on a watch glass. The soluble impurities, present in small amount, stay in the solution.
Q10. 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) Suspension particles settle when left undisturbed. The blood cells would settle out of the plasma instead of staying evenly spread, so blood could not carry nutrients, gases and hormones evenly to all parts of the body. (ii) Dispersed phase: the blood cells (and platelets). Dispersion medium: the plasma.
Q11. Compare evaporation, crystallization and distillation. In which situation would you prefer each of these over the others?
Answer:
| Method | What it does | Prefer it when |
|---|---|---|
| Evaporation | Removes the solvent completely and leaves the solid; the solvent is lost | You want only the dissolved solid and its purity or crystal shape does not matter, such as salt from seawater |
| Crystallization | Forms pure crystals from a saturated solution, leaving soluble impurities in solution | You want a pure solid, such as purifying copper sulfate |
| Distillation | Vaporises and condenses a liquid, so the liquid is recovered | You want the liquid back, or need to separate miscible liquids with boiling points at least about 25 °C apart |
5-Mark Questions
Q12. Differentiate between a solution, a suspension and a colloid on the basis of nature, particle size, visibility, filtration, settling and the Tyndall effect. Give two examples of each.
Answer: Solution: homogeneous; particles less than 1 nm; not visible, transparent; cannot be separated by filtration; does not settle; no Tyndall effect. Examples: salt solution, vinegar. Suspension: heterogeneous; particles more than 1000 nm; visible to the naked eye; separates by filtration; settles when left undisturbed; shows the Tyndall effect. Examples: sand in water, mud in water. Colloid: heterogeneous though it looks uniform; particles 1 to 1000 nm; not visible to the naked eye; cannot be separated by filtration; does not settle, particles remain evenly distributed; shows the Tyndall effect. Examples: milk, blood.
Common Mistakes and Examiner Tips
- Dividing by the mass of solvent. 20 g sugar in 100 g water is 20/120 × 100 = 16.67%, not 20%.
- Mixing up the three percentages. Solid in a measured liquid volume is % m/v; two liquids is % v/v; everything weighed is % m/m. Always write the unit, such as 10% m/m.
- Saying all solubilities rise with temperature. Solids in liquids generally become more soluble when heated; gases in liquids generally become less soluble.
- Treating evaporation and crystallization as the same. Evaporation removes the solvent completely; crystallization grows pure crystals from a saturated solution and leaves soluble impurities behind.
- Using distillation for liquids with close boiling points. Simple distillation needs a difference of at least about 25 °C. For less than 25 °C, write fractional distillation.
- Chromatography set-up errors. The start line is drawn in pencil, and the solvent level must be below the spot at the start.
- Calling milk or blood a solution. They look uniform but are colloids. NCERT’s exercise classes milk and blood as heterogeneous, and they show the Tyndall effect.
- Calling an alloy heterogeneous or “separable by physical methods”. Brass, bronze and stainless steel are homogeneous mixtures; physical methods cannot separate their components.
- Writing that the heavier liquid rises in a separating funnel. The less dense liquid is on top. Drain the lower layer first.
- Defining sublimation without “directly” or without “no liquid state”. Write all three parts: directly, below its melting point, no liquid state.
Quick Revision Points
- Homogeneous mixture = uniform composition = solution; it always remains homogeneous.
- Heterogeneous mixture = composition not uniform: sand in water, oil and water, smoke, fog.
- Alloy = homogeneous mixture of metals, or a metal and a non-metal; brass about 80% Cu + 20% Zn, bronze about 80% Cu + 20% Sn.
- % m/m = (mass of solute / mass of solution) × 100; % m/m and % w/w are numerically equal.
- % m/v = (mass of solute / volume of solution) × 100; used in medicines, for example 0.9% saline and 5% glucose.
- % v/v = (volume of solute / volume of solution) × 100; used for two miscible liquids, for example vinegar at 5% v/v acetic acid.
- Solubility = maximum solute in 100 g (or 100 mL) of solvent at a given temperature; saturated solution holds no more.
- Crystallization works on the difference in solubility at different temperatures.
- Distillation: boiling point difference at least about 25 °C; acetone (56 °C) from water (100 °C).
- Fractional distillation: difference less than 25 °C; crude oil into petroleum gas, petrol, kerosene, diesel, lubricating oil, bitumen.
- Paper chromatography: pencil line, spot above the solvent, components move at different rates.
- Separating funnel: immiscible liquids separate by density; less dense liquid on top.
- Sublimation: solid directly to vapour below its melting point; deposition is the reverse. Camphor, naphthalene, dry ice.
- Suspension: particles more than 1000 nm, visible, settle, filterable, scatter light.
- Centrifugation: fast spinning; heavier particles go to the bottom of the tube; blood into plasma and cells; paperfuge works without electricity.
- Coagulation: alum makes fine particles clump and settle; lemon juice coagulates milk proteins into paneer.
- Colloid: particles 1 to 1000 nm, do not settle, cannot be filtered, scatter light; milk, blood, tomato sauce, ice cream.
- Tyndall effect = scattering of light by particles; shown by colloids and suspensions, never by true solutions.
Weightage and Exam Pattern
Class 9 has no board exam. This chapter is assessed in school exams (periodic tests, half-yearly and annual papers), and the NCERT source gives no fixed mark allocation for it. It is the foundation for Class 10 chemistry.
| Question type | What this chapter typically supplies |
|---|---|
| MCQ and one-word | Homogeneous or heterogeneous, which mixtures show the Tyndall effect, particle size ranges, naming a process (sublimation, deposition, coagulation) |
| Assertion-Reason | Tyndall effect in solutions, boiling point gap for distillation, alloys as homogeneous mixtures |
| Short answer | % m/m, % m/v and % v/v calculations, sublimation versus evaporation, why oil floats, choosing a method with a reason |
| Data and case-based | Reading solubility tables and curves, saline and ORS, water treatment, blood centrifugation |
| Long answer | Solution, suspension and colloid comparison; crystallization steps; distillation with a labelled diagram; separating a three-component mixture |
Class 9 Science · Chapter 5 – swipe through all 10 cards to understand the whole chapter.
Homogeneous vs Heterogeneous
A homogeneous mixture has the same composition everywhere; a heterogeneous one does not.
A homogeneous mixture is also called a solution and always stays homogeneous.
- Homogeneous: sugar water, vinegar, soda, air, brass.
- Heterogeneous: sand in water, oil and water, smoke, fog.
- Most gas mixtures are homogeneous because gas particles move freely.
Mass by Mass Percentage
Grams of solute present in 100 g of the solution.
Mass of solution = solute + solvent. % m/m equals % w/w.
- 10 g salt in 90 g water = 10% m/m.
- 20 g sugar in 100 g water = 16.67% m/m.
- Used for food labels, milk powder, spice mixes and alloys.
% m/v and % v/v
Use m/v for a weighed solid in a liquid, v/v for two miscible liquids.
% v/v = (mL solute ÷ mL solution) × 100
- 5 g glucose in 100 mL solution = 5% m/v.
- 0.9% m/v saline = 0.9 g salt in 100 mL.
- 1 mL pesticide in 100 mL spray = 1% v/v.
Solubility and Saturation
The most solute a fixed amount of solvent can dissolve at a given temperature.
Solids usually dissolve more when hot; gases dissolve less.
- Potassium nitrate at 40 °C: 62 g per 100 g water, so 31 g for 50 g.
- Sodium chloride barely changes: 36 g to 37 g from 10 °C to 80 °C.
- A saturated solution cannot dissolve any more solute at that temperature.
Crystallization
Forming pure crystals from a saturated solution, based on how solubility changes with temperature.
Compound B: 287 g at 60 °C, 241 g at 40 °C, so 46 g crystallizes.
- Filter the hot solution to remove insoluble impurities.
- Slow, undisturbed cooling gives large, well-shaped crystals.
- Salt from seawater: seawater, saturated solution, salt crystals.
Distillation
Heat to vaporise the lower-boiling liquid, then condense the vapour.
Below 25 °C, use fractional distillation, as for crude oil.
- Acetone (56 °C) separates from water (100 °C).
- Also recovers water from salt water; salt stays in the flask.
- Kannauj makes Mitti ka Ittar by the Deg-Bhapka method.
Paper Chromatography
Components move up the paper at different rates and separate.
Chroma = colour, graphein = to write.
- Draw the start line in pencil, 2 cm from the bottom.
- Keep the solvent level below the ink spot.
- Separates inks, food colours, spinach and petal pigments.
Funnel and Sublimation
Immiscible liquids separate by density; a subliming solid separates from one that does not.
Oil at 910 g per litre is less dense than water, so it floats.
- Separating funnel: drain the lower water layer, then collect the oil.
- Camphor or naphthalene sublimes; sand stays behind.
- Dry ice (solid carbon dioxide) also sublimes.
Centrifugation and Coagulation
Spinning or clumping clears mixtures that filtration cannot.
Alum (fitkari) is the coagulant for muddy water.
- Fast spinning sends heavier particles to the bottom of the tube.
- Blood separates into plasma, platelets with white cells, and red cells.
- Lemon juice coagulates milk proteins to make paneer.
Colloids and Tyndall Effect
Solutions, colloids and suspensions differ in particle size; bigger particles scatter light.
Tyndall effect: scattering of light by particles.
- Colloids do not settle and cannot be filtered: milk, blood, ice cream.
- Colloid = dispersed phase + dispersion medium.
- Oil-in-water: milk. Water-in-oil: butter, cold cream.
📝 Practice Exploring Mixtures and their Separation - 10 questions
Concept-check questions · with answers & explanations
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Why A: Gas particles move freely in all directions and mix uniformly with other gases, so gas-gas mixtures such as hydrogen and oxygen are homogeneous; smoke, however, is solid particles suspended in air, so its composition is not uniform and it is heterogeneous, like fog (liquid droplets in air) and dust in air.
Why not B: Gases mix easily and uniformly, which is exactly why most gas mixtures are homogeneous.
Why not C: Smoke is a mixture; its solid particles and the air keep their own identities and can be separated.
Why not D: Colour has nothing to do with mixing; smoke is heterogeneous because of suspended solids, not because of gas colours.
Remember: Gas + gas = homogeneous; solid or liquid particles floating in a gas (smoke, fog, dust) = heterogeneous.
Why C: Step 1 (mass of solution = solute + solvent): A = 100 g, B = 120 g, C = 110 g. Step 2 (% m/m = solute / solution × 100): A = 20 / 100 × 100 = 20%, B = 20 / 120 × 100 = 16.7%, C = 30 / 110 × 100 = 27.3%. Step 3 (compare): 27.3% is the largest, so Student C’s solution is the most concentrated.
Why not A: Student A has a neat 20% but less sugar per gram of solution than Student C.
Why not B: Student B uses the same sugar as A but more water, giving the lowest concentration, 16.7%.
Why not D: The ratios of sugar to solution differ, so the concentrations cannot be equal.
Remember: Concentration compares solute to the whole solution; more solute or less solvent both raise it.
Why A: NCERT explains that slow, undisturbed cooling gives the particles in the solution enough time to come together, resulting in larger, shiny, well-shaped blue crystals; rapid cooling in ice-cold water gives smaller and less well-formed crystals.
Why not B: Some evaporation from an open beaker is harmless; the cooling rate is chosen for crystal quality, not to trap water.
Why not C: Cooling lowers solubility, it does not raise it; that fall is what makes the crystals appear.
Why not D: The solution stays saturated at every temperature during cooling; the excess solute leaves as crystals.
Remember: Slow and still gives big crystals; fast and shaken gives small ones.
Why C: In distillation the water vaporises, passes through the condenser and is collected as pure liquid in a separate vessel, while the salt remains in the distillation flask; NCERT states distillation can separate a liquid from a solution containing dissolved solids and allows recovery of the solvent.
Why not A: Evaporation drives the water off into the air; the salt is recovered but the solvent disappears.
Why not B: Dissolved salt passes straight through filter paper; filtration only removes undissolved particles.
Why not D: Crystallization yields the salt as crystals from a saturated solution but does not collect the water.
Remember: Want the solvent back? Condense it: distil.
Why A: Salt dissolves in water while chalk does not (chalk left a residue on filter paper in the class activity), so adding water gives a salt solution with chalk suspended in it; filtration holds back the chalk as residue, and evaporating the clear filtrate leaves the salt behind.
Why not B: Common salt does not sublime; only substances like camphor and naphthalene change directly to vapour on heating.
Why not C: A separating funnel is for two immiscible liquids; two dry powders form no layers to drain.
Why not D: Distillation collects a vaporised liquid as the distillate; chalk and salt are solids and no liquid is present to distil.
Remember: One soluble, one not: dissolve, filter, evaporate.
Why C: Droplets that stay evenly dispersed without settling yet scatter light show the two signatures of a colloid; clouds, like fog (tiny water droplets in air), are heterogeneous mixtures of this kind, which is why they look bright or hazy in sunlight.
Why not A: A solution is transparent and shows no Tyndall effect because its particles are under 1 nm; scattered sunlight rules this out.
Why not B: Suspension particles are larger than 1000 nm and settle on standing; cloud droplets keep floating.
Why not D: A cloud is water plus air, two substances that keep their identities, so it is a mixture, not a compound.
Remember: Does not settle + scatters light = colloid.
Why D: Muddy water is a suspension, milk and blood are colloids, and all three are heterogeneous because their particles are not truly dissolved and they scatter light; brass is an alloy, a homogeneous mixture of copper and zinc.
Why not A: Smoke is solid particles suspended in air, a heterogeneous mixture, so this set has one wrong entry.
Why not B: Brass and vinegar are homogeneous and muddy water is heterogeneous, so three entries are wrong.
Why not C: Milk looks uniform but is a colloid that scatters light and is classed as heterogeneous.
Remember: Solutions and alloys are homogeneous; suspensions and colloids (milk, blood, smoke, fog) are heterogeneous.
Why D: Step 1 (meaning of 4% m/m): 4 g of zinc oxide is present in every 100 g of powder. Step 2 (scaling to the given mass): 300 g contains 4 × (300 / 100) = 4 × 3 = 12 g of zinc oxide.
Why not A: 4 g is the amount in 100 g only; the sample is three times larger.
Why not B: 1.2 g is 12 g with a decimal-place slip.
Why not C: 75 g comes from 300 / 4, dividing by the percentage instead of taking 4 per hundred.
Remember: Mass of component = (percentage / 100) × total mass.
Why A: Step 1 (solubility at the lower temperature): at 40 °C only 241 g of B can remain dissolved in 100 g of water. Step 2 (excess solute crystallizes): 287 g – 241 g = 46 g can no longer stay in solution and separates out as pure crystals.
Why not B: 241 g is the mass that stays dissolved at 40 °C, not the mass that crystallizes.
Why not C: 287 g is the total solute; most of it remains dissolved because the solution is still saturated at 40 °C.
Why not D: 528 g adds the two solubilities instead of subtracting them.
Remember: Mass deposited = solubility at the hot temperature minus solubility at the cold temperature, per 100 g of water.
Why C: Water and acetone differ by 100 – 56 = 44 °C, comfortably more than about 25 °C, so acetone vaporises first and simple distillation works.
Why not A: Acetone and alcohol differ by 78 – 56 = 22 °C, less than 25 °C, so fractional distillation would be needed.
Why not B: Alcohol and benzene differ by only 2 °C and would vaporise almost together.
Why not D: Alcohol and chloroform differ by 78 – 61 = 17 °C, again below the roughly 25 °C needed.
Remember: Subtract the boiling points; 25 °C or more means simple distillation will do.
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Frequently Asked Questions
A homogeneous mixture has a uniform composition throughout, like sugar in water, which is equally sweet in the first and last sip. It is also called a solution and always remains homogeneous. A heterogeneous mixture is not uniform: sand in water stays visible and settles, and oil and water form separate layers. Alloys such as brass are homogeneous; smoke and fog are heterogeneous.
Mass by mass percentage = (mass of solute / mass of solution) × 100. The mass of solution is solute plus solvent. For 10 g of salt in 90 g of water, the solution is 100 g, so the concentration is 10% m/m. For 20 g of sugar in 100 g of water, the solution is 120 g and the answer is 16.67% m/m, which is lower than 20%.
Mass by volume percentage is used when a weighed solid is dissolved in a liquid whose volume is easier to measure than its mass, as in medicines and laboratories: 5% glucose and 0.9% saline are % m/v. Volume by volume percentage is used when two miscible liquids are mixed, as in perfumes, cosmetics, vinegar and pesticide sprays: 1 mL of pesticide in 100 mL of spray is 1% v/v.
Solubility is the maximum amount of solute that dissolves in a fixed quantity of solvent (100 g or 100 mL) at a given temperature. A solution that cannot dissolve more is saturated. The solubility of solids in liquids generally increases with temperature, while that of gases in liquids generally decreases. In NCERT Table 5.4, potassium nitrate rises from 21 g to 167 g per 100 g of water between 10 °C and 80 °C.
Simple distillation works when the boiling points differ by at least about 25 °C. The lower-boiling liquid vaporises first, its vapour is cooled in a condenser, and the pure liquid is collected. Acetone (about 56 °C) and water (100 °C) differ by 44 °C, so distillation works. For differences below 25 °C, such as the fractions of crude oil, fractional distillation is used.
The mixture is heated gently in a china dish covered by an inverted funnel with a cotton-plugged nozzle. Camphor changes directly from solid to vapour below its melting point, without becoming a liquid. The vapour deposits as solid camphor on the cooler inner wall of the funnel, which is called deposition. Sand does not sublime, so it stays in the dish.
They differ mainly in particle size. Solution particles are less than 1 nm, colloid particles 1 to 1000 nm, and suspension particles more than 1000 nm. Suspension particles are visible, settle and can be filtered. Colloid particles stay evenly dispersed, do not settle and pass through filter paper. Both colloids and suspensions scatter light (Tyndall effect); a true solution does not.
The Tyndall effect is the scattering of light by particles, named after John Tyndall. A beam of light passing through a colloid or suspension becomes visible, while a beam through a transparent solution does not. Examples include a beam of sunlight entering a dark room through a small hole, floodlight beams in a stadium, a laser beam through milk in water, and hazy air in smoky, dusty cities.
