🧪 Important Questions · Class 9 Science — Chemistry
Chapter 5: Exploring Mixtures and their Separation
15 Short + 10 Long Answer Questions | Theory & Application Both
2 & 3 Marks SAQ
5 Marks LAQ
CBSE Pattern
Numericals Included
Separation Methods
Tyndall Effect
5 Marks LAQ
CBSE Pattern
Numericals Included
Separation Methods
Tyndall Effect
📚 Contents
How to Use This Q&A Sheet
- Attempt each question on your own first, then match with the given answer.
- Short Answer Questions (2–3 marks): Write crisp, keyword-rich answers of 3–5 sentences.
- Long Answer Questions (5 marks): Use step-by-step format; include formulas, diagrams (described), and units wherever needed.
- For numericals, always write: Given → Formula → Substitution → Answer with unit.
- Watch for questions asking you to compare two methods — always use a table or side-by-side points.
Exam Strategy
This chapter is one of the most scoring in Class 9 Chemistry. Questions on separation methods, concentration calculations, and Tyndall effect appear almost every year. Know the principle behind each separation technique — examiners love asking “Why is X method used for Y mixture?”
This chapter is one of the most scoring in Class 9 Chemistry. Questions on separation methods, concentration calculations, and Tyndall effect appear almost every year. Know the principle behind each separation technique — examiners love asking “Why is X method used for Y mixture?”
Short Answer Questions — 15 Questions (2–3 Marks Each)
🧪 Classification & Solutions (Q1–Q5)
Q1. [Theoretical] What is a homogeneous mixture? Give two examples. How is it different from a heterogeneous mixture? (2 Marks)
Ans: A homogeneous mixture (also called a solution) is one that has a uniform composition throughout — every part of the mixture looks and tastes the same. Examples: sugar dissolved in water (equally sweet throughout), vinegar (acetic acid in water). In contrast, a heterogeneous mixture does not have uniform composition — different parts look different. Example: sand in water (sand particles are visible and settle down).
Q2. [Theoretical] Define concentration of a solution. Why is maintaining the right concentration important in everyday life? Give one example. (2 Marks)
Ans: Concentration of a solution is the amount of solute dissolved in a given amount of solvent or solution. The right concentration is essential because too little or too much solute changes the effectiveness of the solution. For example, when farmers spray pesticides on crops, too little may not protect the crops, while too much can damage crops, soil and the environment. Similarly, ORS (Oral Rehydration Solution) must have exact concentrations of salt and sugar to be medically effective.
Q3. [Practical/Application] 10 g of salt is dissolved in 90 g of water. Calculate the mass by mass percentage (% m/m) of the solution. (3 Marks)
Ans:
Given: Mass of solute (salt) = 10 g, Mass of solvent (water) = 90 g
Total mass of solution = 10 + 90 = 100 g
Formula: % m/m = (Mass of solute / Mass of solution) × 100
Substituting: % m/m = (10/100) × 100
∴ Mass by mass percentage = 10% m/m
Total mass of solution = 10 + 90 = 100 g
Formula: % m/m = (Mass of solute / Mass of solution) × 100
Substituting: % m/m = (10/100) × 100
∴ Mass by mass percentage = 10% m/m
Q4. [Theoretical] What is solubility? How does temperature affect the solubility of (i) solids in liquids and (ii) gases in liquids? (2 Marks)
Ans: Solubility is the maximum amount of solute that can dissolve in a fixed quantity (100 mL or 100 g) of solvent at a given temperature. A solution that cannot dissolve any more solute at that temperature is called a saturated solution. Effect of temperature: (i) For solids in liquids — solubility generally increases with increase in temperature (e.g., more sugar dissolves in hot water than cold water). (ii) For gases in liquids — solubility generally decreases with increase in temperature (e.g., cold water holds more dissolved oxygen than warm water).
Exam Tip
The word “generally” is important — always write it. Examiners look for this qualifier since there are exceptions.
The word “generally” is important — always write it. Examiners look for this qualifier since there are exceptions.
Q5. [Practical/Application] 5 g of glucose is dissolved in water to make 100 mL of solution. Calculate its concentration in mass by volume percentage (% m/v). Where is this method of expressing concentration commonly used? (3 Marks)
Ans:
Given: Mass of solute (glucose) = 5 g, Volume of solution = 100 mL
Formula: % m/v = (Mass of solute / Volume of solution) × 100
Substituting: % m/v = (5/100) × 100
∴ Mass by volume percentage = 5% m/v
Formula: % m/v = (Mass of solute / Volume of solution) × 100
Substituting: % m/v = (5/100) × 100
∴ Mass by volume percentage = 5% m/v
This method is commonly used in medicines and laboratories where measuring the volume of a liquid is easier than weighing it. For example, glucose intravenous drip (5% w/v) and saline drip (0.9% m/v NaCl) in hospitals.
🌟
India’s Contribution — Salt Crystallization! Ancient Indian coastal communities used solar evaporation to obtain karkatch salt and boiling of sea brines to obtain panga salt. Crystallization of salt has been practised in India for thousands of years — long before modern chemistry was formalized!
🧪 Separation Methods — Homogeneous Mixtures (Q6–Q10)
Q6. [Theoretical] What is crystallization? State the principle on which it works. Give one real-life example. (2 Marks)
Ans: Crystallization is the process of forming pure solid crystals from a saturated solution by cooling it. It is used to separate a solid from a homogeneous mixture (solution) and to purify solids. The principle is based on the difference in the solubility of a substance at different temperatures — when a hot saturated solution is cooled, excess solute that can no longer remain dissolved separates out as pure crystals. Real-life example: Salt crystals obtained by solar evaporation of sea water, or mishri (candy sugar) crystals formed when sugar solution is slowly cooled.
Q7. [Theoretical] What is distillation? State the condition under which it can be used to separate two miscible liquids. (2 Marks)
Ans: Distillation is the process of separating a homogeneous mixture of two miscible liquids by heating the mixture. The liquid with the lower boiling point vaporises first; the vapours are then cooled in a condenser and collected as a pure liquid (called the distillate). The condition for distillation is that the two liquids must differ in their boiling points by at least about 25 °C. For example, acetone (boiling point 56 °C) and water (boiling point 100 °C) can be separated by distillation as their boiling points differ by 44 °C.
Q8. [Practical/Application] How would you separate the ink of a black sketch pen into its component colours? Name the technique and explain the principle behind it. (3 Marks)
Ans: The technique used is paper chromatography. A spot of ink is placed on chromatographic (or filter) paper near one end. The paper is dipped into a solvent (e.g., water) such that the solvent level is below the ink spot. As the solvent rises through the paper, it carries the ink components with it. Different dye components travel at different speeds depending on their interaction with the solvent and the paper. The component that interacts more with the solvent travels farther, while those that interact more with the paper travel less. This separates the ink into different coloured spots (bands) on the paper. The principle is the difference in the rates of movement of components of the mixture on the paper.
Q9. [Theoretical] What is sublimation? How is it different from evaporation? Name a substance that undergoes sublimation. (2 Marks)
Ans: Sublimation is the process in which a solid changes directly into vapour (below its melting point) without passing through the liquid state. When the vapour cools and changes back into a solid without becoming a liquid, it is called deposition. Evaporation, on the other hand, is the conversion of a liquid into vapour. So in sublimation, the solid bypasses the liquid phase entirely, while in evaporation, the liquid phase is already present. Substances that sublime: camphor, naphthalene, and solid carbon dioxide (dry ice). Sublimation is used to separate camphor from sand.
Q10. [Practical/Application] A student accidentally mixed acetone and water in a lab. Suggest a method to separate them. Why can’t they be separated by simple evaporation alone? (3 Marks)
Ans: The mixture can be separated by distillation. The mixture is heated in a distillation flask; acetone (boiling point 56 °C) vaporises first since it has a lower boiling point than water (100 °C). The acetone vapours pass through a condenser where they are cooled and collected as pure liquid acetone in a conical flask. Water remains behind in the distillation flask. They cannot be separated by simple evaporation alone because evaporation would cause the acetone to vaporise and escape into the air — we would recover only water but lose the acetone. Distillation allows us to recover both the liquids separately.
Key Point
Distillation is preferred over evaporation when you want to recover the solvent as well. Evaporation is used only when you want to recover the solute (e.g., salt from salt solution).
Distillation is preferred over evaporation when you want to recover the solvent as well. Evaporation is used only when you want to recover the solute (e.g., salt from salt solution).
🧪 Heterogeneous Mixtures, Colloids & Tyndall Effect (Q11–Q15)
Q11. [Theoretical] What is the Tyndall effect? In which types of mixtures is it observed? Give one example from daily life. (2 Marks)
Ans: The Tyndall effect is the scattering of a beam of light by the particles of a mixture, making the path of the light beam visible. It is named after the scientist John Tyndall who first explained it. The Tyndall effect is observed in colloids and suspensions — where particles are large enough to scatter light — but not in true solutions (where particles are too small to scatter light). Daily life example: When sunlight enters a dark room through a small gap, its path becomes visible due to scattering by dust particles in the air. Floodlights in a sports stadium also demonstrate the Tyndall effect.
Q12. [Theoretical] What is a suspension? How is it different from a solution? Give two examples of suspensions. (2 Marks)
Ans: A suspension is a heterogeneous mixture in which solid particles do not dissolve but remain scattered throughout the medium. The particles are visible to the naked eye (size > 1000 nm), and they settle down when the mixture is left undisturbed. In contrast, a solution is a homogeneous mixture where solute particles (size < 1 nm) are invisible to the naked eye and do not settle down. Examples of suspensions: muddy water (sand in water) and chalk powder in water. Suspensions show the Tyndall effect; solutions do not.
Q13. [Practical/Application] How is centrifugation used to separate blood components? Explain the principle. (3 Marks)
Ans: Centrifugation is the process of spinning a mixture at high speed in a tube. The principle is that when spun rapidly, the centrifugal force (outward force) causes the heavier particles to move outwards and settle at the bottom of the tube, while the lighter liquid stays at the top. When blood is centrifuged: the heavier red blood cells settle at the bottom, while the lighter plasma and platelets/white blood cells remain at the top. This allows doctors to separate blood into its components for medical tests (e.g., detecting malaria, anaemia) and for blood banks to store specific components required by patients.
Q14. [Theoretical] What is a colloid? How is it different from a solution and a suspension in terms of particle size? (2 Marks)
Ans: A colloid is a type of mixture where the particles are larger than in a true solution but smaller than in a suspension. Colloid particles do not settle over time and are uniformly dispersed. The key difference in particle size:
- Solution: Particle size < 1 nm (invisible, do not scatter light)
- Colloid: Particle size 1–1000 nm (scatter light — show Tyndall effect)
- Suspension: Particle size > 1000 nm (visible to naked eye, settle down)
Examples of colloids: milk, blood, tomato sauce, ice cream.
Q15. [Practical/Application] What is coagulation? How is alum (fitkari) used to purify muddy water? (2 Marks)
Ans: Coagulation is the process of adding a substance (called a coagulant) to a suspension to cause the fine particles to clump together, forming larger masses that then settle down by gravity. When powdered alum (fitkari) is added to muddy water, it acts as a coagulant — it causes the fine clay and silt particles to clump together into larger masses. These larger clumps then settle at the bottom of the container (sedimentation). The clear water above can then be separated by decantation or filtration. A similar process is used in water treatment plants to purify drinking water.
Long Answer Questions — 10 Questions (5 Marks Each)
🧪 Solutions & Concentration (Q1–Q3)
Q1. [Theoretical] Explain the three methods of expressing the concentration of a solution using percentages. Give the formula for each and one example where each method is practically used. (5 Marks)
Ans: The concentration of a solution tells us the amount of solute dissolved in a given amount of solvent or solution. It can be expressed in three ways:
- Mass by Mass Percentage (% m/m or % w/w): It tells us how many grams of solute are present in 100 g of the total solution.% m/m = (Mass of solute / Mass of solution) × 100
Used for: labelling packaged foods (milk powder, spice mixtures), heterogeneous mixtures.
- Mass by Volume Percentage (% m/v or % w/v): It tells us how many grams of solute are present in 100 mL of the solution.% m/v = (Mass of solute / Volume of solution) × 100
Used for: medicines and hospital solutions — e.g., 5% glucose drip, 0.9% m/v saline drip.
- Volume by Volume Percentage (% v/v): Used when two miscible liquids are mixed; it tells how many mL of solute are present in 100 mL of solution.% v/v = (Volume of solute / Volume of solution) × 100
Used for: perfumes, cosmetics and vinegar (e.g., 5% v/v acetic acid in vinegar).
- Summary: Use % m/m for solid–solid or solid–liquid; % m/v for medicine and lab solutions; % v/v for liquid–liquid mixtures like perfumes.
Definition to Memorize
Concentration = amount of solute dissolved in a given amount of solvent or solution. Always mention the formula with correct symbols for full marks.
Concentration = amount of solute dissolved in a given amount of solvent or solution. Always mention the formula with correct symbols for full marks.
Q2. [Practical/Application] Three students prepare sugar solutions: Student A dissolves 20 g in 80 g water; Student B dissolves 20 g in 100 g water; Student C dissolves 30 g in 80 g water. Calculate the % m/m concentration for each and identify whose solution is most concentrated. (5 Marks)
Ans:
Formula: % m/m = (Mass of solute / Mass of solution) × 100Student A: Mass of sugar = 20 g, Mass of water = 80 g
Mass of solution = 20 + 80 = 100 g
% m/m = (20/100) × 100 = 20%
Mass of solution = 20 + 80 = 100 g
% m/m = (20/100) × 100 = 20%
Student B: Mass of sugar = 20 g, Mass of water = 100 g
Mass of solution = 20 + 100 = 120 g
% m/m = (20/120) × 100 = 16.67%
Student C: Mass of sugar = 30 g, Mass of water = 80 g
Mass of solution = 30 + 80 = 110 g
% m/m = (30/110) × 100 = 27.27%
∴ Student C has the most concentrated solution (27.27% m/m)
Conclusion: Student C’s solution is most concentrated because it has the highest mass of solute (30 g) dissolved in the smallest total mass of solution (110 g). Concentration depends not just on the amount of solute but on the ratio of solute to total solution mass.
5-Mark Strategy
Write the formula first, then solve each case separately. Compare the three values at the end and clearly state which is highest. Never forget to compute the total mass of solution (solute + solvent) — a very common mistake!
Write the formula first, then solve each case separately. Compare the three values at the end and clearly state which is highest. Never forget to compute the total mass of solution (solute + solvent) — a very common mistake!
Q3. [Theoretical] What is crystallization? Describe the step-by-step procedure for preparing copper sulfate crystals in the laboratory. Why is sulfuric acid added in this process? (5 Marks)
Ans: Crystallization is the process of forming pure, well-shaped crystals from a saturated solution by slow cooling. Its principle is the difference in solubility at different temperatures.
- Prepare saturated solution: Take 1 g of copper sulfate in a 100 mL beaker. Add 25 mL of water and a drop of dilute sulfuric acid. Heat gently in a water bath while stirring. Gradually add more copper sulfate until the solution becomes saturated (no more dissolves).
- Filter the hot solution: Filter the hot saturated solution through a filter paper to remove insoluble impurities. Collect the clear filtrate in a clean beaker.
- Cool slowly: Cover the beaker with a watch glass and allow it to cool slowly at room temperature without disturbing. Slow cooling gives particles enough time to arrange into larger, well-formed crystals.
- Collect and dry crystals: Filter the crystals, rinse with cold water, and allow them to dry on a watch glass.
- Why sulfuric acid? A drop of dilute sulfuric acid is added to prevent unwanted side reactions that could introduce impurities, helping to produce purer, better-shaped crystals.
Observation: Blue, shiny, well-formed crystals of copper sulfate are obtained.
Safety Note
Copper sulfate is toxic — never touch with bare hands. Sulfuric acid is corrosive — handle only under adult supervision.
Copper sulfate is toxic — never touch with bare hands. Sulfuric acid is corrosive — handle only under adult supervision.
🧪 Separation Methods (Q4–Q7)
Q4. [Theoretical] Explain the process of distillation with a labelled description of the apparatus. Why is distillation preferred over evaporation when separating acetone from water? (5 Marks)
Ans: Distillation is the method of separating a homogeneous mixture of two miscible liquids (boiling points differing by ≥ 25 °C) by vaporising the lower-boiling component and condensing it back into liquid.
- Set-up: Pour the mixture of acetone and water into a distillation flask. Place on a tripod stand with wire gauze. Connect the flask neck to a water condenser (Liebig condenser) using a delivery tube. Place a conical flask at the outlet to collect the distillate.
- Heat the flask: Apply gentle heat using a burner. Monitor temperature with a thermometer in the flask neck. Acetone (b.p. 56 °C) begins to vaporise first.
- Condensation: Acetone vapours travel through the condenser (cooled by circulating cold water flowing opposite to vapour direction). They condense back into liquid acetone.
- Collection: Pure liquid acetone collects in the conical flask (distillate). Water remains in the distillation flask.
- Why not evaporation? In evaporation, acetone vapours escape into the air and are lost. Distillation recovers both liquids — acetone as distillate and water in the flask.
India’s Contribution — Deg-Bhapka Method
The ancient distillation technique from Kannauj (UP), called the Deg-Bhapka method, was used to make Mitti ka Ittar (earthy fragrance perfume). This shows that distillation has been practised in India for centuries!
The ancient distillation technique from Kannauj (UP), called the Deg-Bhapka method, was used to make Mitti ka Ittar (earthy fragrance perfume). This shows that distillation has been practised in India for centuries!
Q5. [Practical/Application] Describe the procedure for paper chromatography to separate the components of black ink. What is the principle of this technique? What would happen if the solvent level is above the ink spot? (5 Marks)
Ans:
- Prepare the strip: Cut a 3 cm wide strip of chromatographic (or filter) paper. Draw a horizontal pencil line 2 cm from the bottom. Mark a spot of black ink at the centre of the line.
- Set up the container: Pour a thin layer of water at the bottom of a gas jar, measuring cylinder, or beaker. The water level must be below the ink spot.
- Dip the paper: Place the paper strip vertically in the container so only its lower end dips into the water.
- Observe: As water rises through the paper by capillary action, it carries the ink components with it. Different colours travel at different speeds and separate into distinct coloured bands.
- Principle: Components of the ink differ in their interaction with the solvent (water) and the paper. Components that interact more with the solvent move faster; those that interact more with the paper move slower.
If solvent level is above the spot: The ink spot would dissolve directly into the solvent and wash away before chromatography begins — no separation would occur.
Etymology Fact
The word ‘chromatography’ comes from Greek words chroma (colour) and graphein (to write) — meaning ‘writing with colour’, because it was first used to separate coloured dyes.
The word ‘chromatography’ comes from Greek words chroma (colour) and graphein (to write) — meaning ‘writing with colour’, because it was first used to separate coloured dyes.
Q6. [Practical/Application] You are given a mixture of sand, common salt and naphthalene. Describe step-by-step how you would separate all three components. Name the technique used at each step. (5 Marks)
Ans: This mixture can be separated in two main steps:
- Step 1 — Separate naphthalene by Sublimation:
Place the mixture in a china dish. Invert a glass funnel (with cotton plug) over it. Heat gently. Naphthalene sublimes (changes directly to vapour), travels up, and deposits as white solid on the cooler inner wall of the funnel. Sand and salt remain behind in the china dish. - Step 2 — Separate sand from salt using Filtration + Evaporation/Crystallization:
Add water to the remaining sand + salt mixture. Salt dissolves in water; sand does not. Filter the mixture through filter paper — sand remains as residue on the filter paper; salt solution (filtrate) passes through. - Step 3 — Recover common salt from solution:
Heat the salt solution (filtrate) to evaporate water. White crystals of common salt are obtained. - Summary of techniques:
- Naphthalene from sand + salt → Sublimation
- Sand from salt solution → Filtration
- Salt from water → Evaporation / Crystallization
Exam Tip — Multi-step Separation Questions
Always do sublimation FIRST (before adding water), since water would prevent the sublimation setup from working properly. Sequence: Sublimation → Dissolve in water → Filter → Evaporate.
Always do sublimation FIRST (before adding water), since water would prevent the sublimation setup from working properly. Sequence: Sublimation → Dissolve in water → Filter → Evaporate.
Q7. [Practical/Application] How would you separate mustard oil from water using a separating funnel? Describe the procedure. Why does mustard oil form the upper layer? (5 Marks)
Ans: Mustard oil and water are immiscible liquids — they do not mix. A separating funnel is used to separate them based on their different densities.
- Setup: Pour a mixture of 5 mL mustard oil and 20 mL water into a 50 mL separating funnel. Close the stopcock.
- Allow to settle: Leave the funnel undisturbed. The liquids separate into two distinct layers — yellow mustard oil on top and colourless water at the bottom.
- Drain the lower layer: Slowly open the stopcock and allow the lower layer of water to drain into a clean container. Close the stopcock when water is nearly fully drained.
- Discard the middle portion: Collect and discard the small portion that may contain a mixture of both liquids.
- Collect the oil: Open the stopcock again to collect the mustard oil separately from the separating funnel (or pour it out from the top).
Why does oil form the upper layer? Mustard oil is less dense than water. In any liquid mixture, the less dense liquid floats on top of the denser one. Since water (density ≈ 1 g/mL) is denser than mustard oil (density ≈ 0.91 g/mL), water settles at the bottom.
🧪 Solutions, Suspensions & Colloids (Q8–Q10)
Q8. [Theoretical] Compare solutions, suspensions and colloids on the basis of: (i) particle size, (ii) visibility, (iii) stability (settling), (iv) Tyndall effect, and (v) filtration. (5 Marks)
Ans:
| Property | Solution | Suspension | Colloid |
|---|---|---|---|
| Nature | Homogeneous | Heterogeneous | Appears homogeneous |
| Particle size | < 1 nm | > 1000 nm | 1–1000 nm |
| Visibility | Not visible | Visible (naked eye) | Not visible (naked eye) |
| Settling | Does not settle | Settles on standing | Does not settle |
| Tyndall effect | Not shown | Shown | Shown |
| Filtration | Cannot be filtered | Can be filtered | Cannot be filtered (passes through) |
| Examples | Salt water, vinegar | Muddy water, chalk in water | Milk, blood, fog |
Why does a colloid show Tyndall effect but not settle?
Colloid particles (1–1000 nm) are large enough to scatter light (hence Tyndall effect) but small enough that gravity cannot pull them down to settle. This is the key characteristic that distinguishes colloids from both solutions and suspensions.
Colloid particles (1–1000 nm) are large enough to scatter light (hence Tyndall effect) but small enough that gravity cannot pull them down to settle. This is the key characteristic that distinguishes colloids from both solutions and suspensions.
Q9. [Theoretical] Compare evaporation, crystallization and distillation. In which situation would you prefer each of these over the others? Give one example for each. (5 Marks)
Ans:
Evaporation
The solvent is heated and vaporises, leaving behind the solid solute. The solvent is not recovered. Used when only the solid (solute) is needed and the solvent can be discarded. Example: Getting salt from sea water by solar evaporation.
The solvent is heated and vaporises, leaving behind the solid solute. The solvent is not recovered. Used when only the solid (solute) is needed and the solvent can be discarded. Example: Getting salt from sea water by solar evaporation.
Crystallization
A hot saturated solution is cooled slowly. The excess solute separates out as pure crystals. Used when a pure, well-formed solid is needed from a solution containing impurities. Example: Purifying copper sulfate or common salt crystals.
A hot saturated solution is cooled slowly. The excess solute separates out as pure crystals. Used when a pure, well-formed solid is needed from a solution containing impurities. Example: Purifying copper sulfate or common salt crystals.
Distillation The mixture is heated; the lower-boiling component vaporises first, is condensed, and both components are recovered. Used when two miscible liquids (differing by ≥ 25°C in boiling points) need to be separated. Example: Separating acetone (b.p. 56°C) from water (b.p. 100°C).
- Choose evaporation when only the solute is needed and the solvent is not valuable.
- Choose crystallization when you need the solid in pure, crystal form (free from impurities).
- Choose distillation when you need to recover both the liquid solvent and a pure solute, or to separate two miscible liquids.
Q10. [Practical/Application] Answer the following using the solubility data: Potassium nitrate has solubility 62 g per 100 g water at 40°C and 32 g per 100 g water at 20°C. (i) What mass of KNO₃ is needed for a saturated solution in 50 g of water at 40°C? (ii) If this saturated solution is cooled to 20°C, how many grams of KNO₃ will crystallise out? (iii) What does this principle tell you about crystallization? (5 Marks)
Ans:
Given: Solubility of KNO₃ at 40°C = 62 g per 100 g water; at 20°C = 32 g per 100 g water; Water taken = 50 g(i) Mass of KNO₃ for saturated solution in 50 g water at 40°C:
100 g water dissolves 62 g KNO₃ at 40°C
∴ 50 g water dissolves = (62 / 100) × 50 = 31 g KNO₃
100 g water dissolves 62 g KNO₃ at 40°C
∴ 50 g water dissolves = (62 / 100) × 50 = 31 g KNO₃
(ii) Mass of KNO₃ dissolved in 50 g water at 20°C:
100 g water dissolves 32 g KNO₃ at 20°C
∴ 50 g water dissolves = (32 / 100) × 50 = 16 g KNO₃
KNO₃ that crystallises out = dissolved at 40°C − dissolved at 20°C
= 31 g − 16 g = 15 g
∴ (i) 31 g KNO₃ needed at 40°C; (ii) 15 g KNO₃ crystallises out on cooling to 20°C
(iii) Principle: This illustrates the principle of crystallization — the solubility of most solids increases with temperature. When a hot saturated solution is cooled, the excess solute (which can no longer remain dissolved) separates out as pure crystals. The greater the difference in solubility at the two temperatures, the more crystals are obtained.
Real-World Application
This exact principle is used industrially to purify salt, sugar, and many pharmaceutical compounds. The purity of crystals obtained depends on how slowly the solution is cooled — slow cooling produces larger, purer crystals.
This exact principle is used industrially to purify salt, sugar, and many pharmaceutical compounds. The purity of crystals obtained depends on how slowly the solution is cooled — slow cooling produces larger, purer crystals.
Formula & Key Terms Quick Reference
% m/m (Mass by Mass): % m/m = (Mass of solute / Mass of solution) × 100
% m/v (Mass by Volume): % m/v = (Mass of solute / Volume of solution) × 100
% v/v (Volume by Volume): % v/v = (Volume of solute / Volume of solution) × 100
| Separation Method | Type of Mixture | Principle Used |
|---|---|---|
| Crystallization | Homogeneous (solid in liquid) | Difference in solubility at different temperatures |
| Distillation | Miscible liquids (b.p. diff ≥ 25°C) | Difference in boiling points |
| Paper Chromatography | Coloured components in solution | Difference in rate of movement on paper |
| Separating Funnel | Immiscible liquids | Difference in density |
| Sublimation | Sublimable solid + non-sublimable solid | Sublimation on heating |
| Centrifugation | Solid–liquid (colloidal/suspension) | Centrifugal force — density difference |
| Coagulation + Filtration | Fine suspension (e.g., muddy water) | Coagulant causes particles to clump and settle |
Common Exam Mistakes to Avoid
Mistake 1: Confusing Total Solution Mass in % m/m
Total mass of solution = mass of SOLUTE + mass of SOLVENT. Many students use only the solvent mass. Example: 10 g salt in 90 g water → total = 100 g (not 90 g).
Total mass of solution = mass of SOLUTE + mass of SOLVENT. Many students use only the solvent mass. Example: 10 g salt in 90 g water → total = 100 g (not 90 g).
Mistake 2: Saying “Tyndall Effect is seen in all mixtures”
Tyndall effect is shown ONLY by colloids and suspensions — NOT by true solutions. In a solution, particle size is < 1 nm which is too small to scatter light.
Tyndall effect is shown ONLY by colloids and suspensions — NOT by true solutions. In a solution, particle size is < 1 nm which is too small to scatter light.
Mistake 3: Saying Colloids Settle Like Suspensions
Colloid particles do NOT settle over time — this is a key difference from suspensions. Colloids are stable (particles remain uniformly dispersed) while suspension particles settle if left undisturbed.
Colloid particles do NOT settle over time — this is a key difference from suspensions. Colloids are stable (particles remain uniformly dispersed) while suspension particles settle if left undisturbed.
Mistake 4: Placing Solvent Level Above the Sample in Chromatography
In paper chromatography, the solvent level MUST be below the sample spot. If solvent is above the spot, the sample will dissolve into the solvent directly and no separation will occur.
In paper chromatography, the solvent level MUST be below the sample spot. If solvent is above the spot, the sample will dissolve into the solvent directly and no separation will occur.
Mistake 5: Confusing Sublimation with Evaporation
In sublimation, a SOLID directly becomes a VAPOUR (no liquid phase). In evaporation, a LIQUID becomes vapour. Camphor → vapour (sublimation); Water → vapour (evaporation).
In sublimation, a SOLID directly becomes a VAPOUR (no liquid phase). In evaporation, a LIQUID becomes vapour. Camphor → vapour (sublimation); Water → vapour (evaporation).
Mistake 6: Wrong Order in Multi-step Separation
For sand + salt + naphthalene mixture: ALWAYS sublimation FIRST (to remove naphthalene), THEN add water and filter (to separate sand from salt). Doing it in reverse gives wrong results.
For sand + salt + naphthalene mixture: ALWAYS sublimation FIRST (to remove naphthalene), THEN add water and filter (to separate sand from salt). Doing it in reverse gives wrong results.
Mistake 7: Applying Distillation to Immiscible Liquids
Distillation is used for MISCIBLE liquids. For IMMISCIBLE liquids (like oil and water), use a SEPARATING FUNNEL. Never confuse the two based on just “they are two liquids”.
Distillation is used for MISCIBLE liquids. For IMMISCIBLE liquids (like oil and water), use a SEPARATING FUNNEL. Never confuse the two based on just “they are two liquids”.
Homogeneous vs HeterogeneousHomogeneous = uniform composition (solution). Heterogeneous = non-uniform (suspension, some colloids).
% m/m, % m/v, % v/vThree ways to express concentration; use m/m for solids, m/v for medicine, v/v for liquid–liquid.
CrystallizationCooling a hot saturated solution → pure crystals form. Principle: solubility decreases on cooling.
DistillationSeparates two miscible liquids when b.p. difference ≥ 25°C. Recovers BOTH components.
Paper ChromatographySeparates coloured components by difference in rate of movement on paper with a solvent.
Separating FunnelSeparates immiscible liquids based on density difference. Denser liquid settles at bottom.
SublimationSolid → vapour directly (no liquid stage). Used for camphor + sand. Reverse = deposition.
Solution vs Colloid vs SuspensionParticle size: <1nm / 1-1000nm / >1000nm. Tyndall: No / Yes / Yes. Settling: No / No / Yes.
Final Exam Strategy — Score Full Marks in Mixtures!
For separation method questions: always state (a) the NAME of the technique, (b) the PRINCIPLE behind it, and (c) WHY it works for that specific mixture. For numericals: remember total mass = solute + solvent. For theory: the three key comparisons examiners love are — solution vs suspension, distillation vs evaporation, and sublimation vs evaporation. Prepare short comparison tables for each. Revise the Tyndall effect and colloid examples — these appear as MCQs and assertion-reason questions almost every year!
For separation method questions: always state (a) the NAME of the technique, (b) the PRINCIPLE behind it, and (c) WHY it works for that specific mixture. For numericals: remember total mass = solute + solvent. For theory: the three key comparisons examiners love are — solution vs suspension, distillation vs evaporation, and sublimation vs evaporation. Prepare short comparison tables for each. Revise the Tyndall effect and colloid examples — these appear as MCQs and assertion-reason questions almost every year!
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