Navigating through the newly revised CBSE Class 9 Science curriculum (Exploration) requires a thorough physical and chemical understanding of pure substances versus mixtures, homogeneous and heterogeneous systems, true solutions (solute, solvent, saturation, solubility curves), suspensions (sedimentation, particle size, instability), colloidal systems (dispersed phase, dispersion medium, Tyndall effect, Brownian motion), concentration expressions (mass by mass % m/m, mass by volume % m/v, and volume by volume % v/v), physical separation techniques (filtration, evaporation, crystallization, simple distillation, fractional distillation, separating funnel, sublimation, centrifugation, paper chromatography), and the municipal stages of drinking water purification. Chapter 5 of Class 9 Chemistry, “Exploring Mixtures and their Separation”, establishes the experimental foundation for purifying and analyzing chemical matter. It explores why suspended particles in muddy water settle under gravity while milk colloids remain permanently dispersed; details the quantitative mathematics of solution concentrations; investigates how differences in boiling points, volatilities, solubilities, and densities are harnessed in industrial separation; and explains why crystallization is preferred over simple evaporation for heat-sensitive solutes. To help students master every aspect of this high-weightage chapter, this comprehensive solutions guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE Class 9 evaluation standards.
Every question presented in the official NCERT textbook—ranging from all introductory “Think It Over” sections and in-text “Pause and Ponder” prompts (Pages 72, 75, 76, 79, 82, and 84) to the complete end-of-chapter “Revise, Reflect, Refine” exercises (Questions 1 to 10 on Pages 90–94)—has been solved with exhaustive detail. Numerical concentration problems follow a step-by-step box format with explicit formula statements, mass/volume substitutions, and percentage calculations using clean plain-text symbols without raw LaTeX tags. Key scoring terms, official CBSE exam tags, and comprehensive separation matrices have been highlighted to ensure students secure maximum marks in their examinations.
Master Concept & Comparative Summary Tables
1. Comparison: True Solution vs. Colloid vs. Suspension
| Physical Property | True Solution (e.g., Salt/Sugar in Water) | Colloid (e.g., Milk, Blood, Fog, Starch Sol) | Suspension (e.g., Muddy Water, Chalk in Water) |
|---|---|---|---|
| Type of Mixture | Homogeneous (Uniform composition throughout). | Heterogeneous (Appears homogeneous to naked eye). | Heterogeneous (Non-uniform composition). |
| Particle Size Diameter | Extremely small: less than 1 nm (less than 10⁻⁹ m). | Intermediate: 1 nm to 1000 nm (10⁻⁹ m to 10⁻⁶ m). | Large: greater than 1000 nm (greater than 10⁻⁶ m / 1 µm). |
| Visibility of Particles | Invisible even under high-power electron microscopes. | Invisible to the naked eye; visible under electron microscope. | Visible to the unaided naked eye. |
| Filterability | Passes freely through ordinary filter paper and animal membranes. | Passes through ordinary filter paper; retained by ultra-filters. | Retained on ordinary filter paper as residue. |
| Stability on Standing | Completely Stable; solute never settles down over time. | Stable; dispersed particles do not settle under gravity. | Unstable; suspended particles settle to the bottom over time. |
| Tyndall Effect | Does NOT scatter light (Path of light is invisible). | Scatters light strongly (Path of light beam is brightly illuminated). | Scatters light initially until particles settle down. |
2. Master Separation Techniques Matrix
| Physical Separation Technique | Underlying Physical Principle / Property Exploited | Example Mixture Separated | Key Laboratory / Industrial Application |
|---|---|---|---|
| Filtration | Difference in solubility and particle size (Insoluble solid vs. liquid). | Sand and Water; Chalk powder in Salt solution. | Water treatment plants; separating tea leaves. |
| Evaporation | Difference in volatility (Non-volatile solid solute in volatile liquid solvent). | Common Salt from Seawater; Blue ink dye. | Commercial salt harvesting in coastal pans. |
| Crystallization | Difference in solubility at different temperatures without thermal decomposition. | Pure Copper Sulphate (CuSO₄) from impure sample; Alum (Phitkari). | Sugar purification from cane juice; pharmaceutical drug refining. |
| Sublimation | One component sublimes directly from solid to gas on heating; the other does not. | Ammonium Chloride (NH₄Cl) + Salt; Naphthalene + Sand; Camphor. | Purifying volatile aromatic organic compounds. |
| Separating Funnel | Immiscibility and density difference of two liquid phases. | Kerosene / Mustard Oil and Water; Petrol and Water. | Industrial solvent extraction; oil spill separation. |
| Centrifugation | Rapid spinning causes denser particles to settle at bottom and lighter liquid to rise. | Blood cells from Plasma; Cream from Milk; Urine sediment. | Clinical pathology diagnostic labs; dairies. |
| Simple Distillation | Separation of miscible liquids with a large boiling point difference (difference of 25°C or more). | Acetone (56°C) and Water (100°C); Distilled water. | Desalination of seawater; recovery of solvent. |
| Fractional Distillation | Miscible liquids with close boiling points (difference less than 25°C) using a fractionating column. | Crude Petroleum into petrol, diesel, kerosene; Air into O₂, N₂, Ar. | Petrochemical oil refineries; industrial gas liquefaction. |
| Paper Chromatography | Differences in adsorption and solubility of solutes in a moving mobile solvent. | Colored dyes in black ink; plant pigments (Chlorophyll a, b, Carotene). | Forensic toxicology; drug detection in athletes. |
NCERT In-Text Questions: “Think It Over” (Page No. 72)
Question 1 Why do suspended particles settle in muddy water over time but not in milk? [Exam Favorite]
Answer: The difference in settling behavior is governed by particle size, mass, and kinetic stability:
- Muddy Water is a Suspension: The suspended mud and soil particles are large (greater than 1000 nm) and heavy. The gravitational pull acting on these large particles overcomes the random thermal collisions of water molecules, causing them to settle to the bottom (sedimentation) over time.
- Milk is a Colloid (Emulsion): The dispersed fat droplets and protein micelles are microscopic (1 to 1000 nm). They undergo continuous, chaotic zigzag collisions with surrounding water molecules (Brownian motion). This continuous bombardment counteracts gravity, keeping the fat particles permanently suspended without settling.
Question 2 How is evaporation different from boiling? [Exam Favorite]
Answer: Evaporation and boiling differ fundamentally in three physical aspects:
| Basis of Comparison | Evaporation | Boiling |
|---|---|---|
| Temperature Condition | Occurs spontaneously at any temperature below the boiling point. | Occurs strictly at a fixed constant temperature (Boiling Point). |
| Nature of Phenomenon | It is a surface phenomenon (only molecules at liquid surface escape). | It is a bulk phenomenon (bubbles form throughout the liquid mass). |
| Thermal Effect | Always causes cooling of the surroundings (draws latent heat). | No cooling effect; temperature remains constant during boiling. |
Question 3 Why do you see bright rays of sunlight when it passes through small gaps between the leaves of a dense tree? [Exam Favorite]
Answer: This optical phenomenon is caused by the Tyndall Effect (Scattering of Light):
- The forest mist and atmospheric air contain suspended colloidal particles such as tiny water droplets, dust, and smoke particles.
- As sunlight passes through small gaps in the dense forest canopy, these colloidal particles scatter the light rays in all directions.
- The scattered light enters the observer’s eyes, making the exact illuminated trajectory and path of the sunlight beams clearly visible.
NCERT In-Text Questions: “Pause and Ponder”
Page No. 75/76: Pause and Ponder (Questions 1 to 3)
Question 1 A common talcum powder contains 4% (m/m) zinc oxide, which acts as a skin protectant. How much zinc oxide is present in a 300 g container of this talcum powder? [Exam Favorite]
Answer:
================================================================================
NUMERICAL SOLUTION (MASS BY MASS CONCENTRATION):
--------------------------------------------------------------------------------
GIVEN DATA:
• Concentration of Zinc Oxide = 4% (m/m)
• Total Mass of Talcum Powder = 300 g
CALCULATION:
Formula: Mass % (m/m) = (Mass of Solute / Total Mass of Mixture) × 100
4 = (Mass of Zinc Oxide / 300) × 100
4 = (Mass of Zinc Oxide / 3)
Mass of Zinc Oxide = 4 × 3 = 12 g
FINAL ANSWER:
12 grams (12 g) of zinc oxide is present in the 300 g container.
================================================================================
Question 2 Your mother gives you a bottle of orange juice concentrate. She asks you to dilute 30 mL of concentrate with water to prepare 150 mL of juice. What is the concentration (% v/v) of orange juice concentrate in the prepared drink? [Exam Favorite]
Answer:
================================================================================
NUMERICAL SOLUTION (VOLUME BY VOLUME CONCENTRATION):
--------------------------------------------------------------------------------
GIVEN DATA:
• Volume of Solute (Orange Concentrate) = 30 mL
• Total Volume of Prepared Solution = 150 mL
CALCULATION:
Formula: Concentration % (v/v) = (Volume of Solute / Total Volume of Solution) × 100
Concentration % (v/v) = (30 / 150) × 100
Concentration % (v/v) = (1 / 5) × 100 = 20% (v/v)
FINAL ANSWER:
The concentration of orange juice concentrate in the prepared drink is 20% (v/v).
================================================================================
Question 3 A cake recipe uses dry ingredients: 75 g of sugar, 420 g of all-purpose flour, and 5 g of sodium hydrogen carbonate (baking soda). Express the concentration (% m/m) of each component in the dry mixture. [Exam Favorite]
Answer:
================================================================================ NUMERICAL SOLUTION (MULTI-COMPONENT CONCENTRATION % m/m): -------------------------------------------------------------------------------- STEP 1: Calculating Total Mass of Dry Mixture Total Mass = Mass of Sugar + Mass of Flour + Mass of Baking Soda Total Mass = 75 g + 420 g + 5 g = 500 g STEP 2: Concentration of Sugar (% m/m) % Sugar = (75 / 500) × 100 = 75 / 5 = 15.0% (m/m) STEP 3: Concentration of All-Purpose Flour (% m/m) % Flour = (420 / 500) × 100 = 420 / 5 = 84.0% (m/m) STEP 4: Concentration of Sodium Hydrogen Carbonate (% m/m) % Baking Soda = (5 / 500) × 100 = 5 / 5 = 1.0% (m/m) CHECK: 15% + 84% + 1% = 100% FINAL ANSWER SUMMARY: • Sugar : 15.0% (m/m) • All-purpose Flour : 84.0% (m/m) • Sodium Hydrogen Carbonate : 1.0% (m/m) ================================================================================
Page No. 79: Pause and Ponder (Questions 4 & 5)
Question 4 Refer to the solubility curves of compounds A and B. When saturated solutions of A and B are cooled from 80°C to 60°C, which solution is likely to deposit more solid? [Exam Favorite]
Answer:
- Rule: The amount of solid crystallized out on cooling depends on the difference in solubility (Drop in Solubility = Solubility at 80°C – Solubility at 60°C) between the two temperatures.
- Interpretation:
- The compound whose solubility curve shows a steeper upward slope with temperature undergoes a greater drop in solubility upon cooling.
- If Compound B exhibits a larger reduction in solubility when cooled from 80°C to 60°C, then Compound B will deposit a significantly larger mass of crystallized solid.
Question 5 Does the size of common salt crystals depend upon the rate at which a hot saturated solution is cooled or water is evaporated? Explain. [Exam Favorite]
Answer: Yes, the physical size and geometric perfection of crystals depend critically on the rate of cooling/evaporation:
- Slow Cooling / Slow Evaporation (Forms Large Crystals): When a hot saturated solution is allowed to cool very slowly and undisturbed over several days, solute particles have ample time to migrate and systematically arrange themselves into orderly, well-defined, large crystal lattices.
- Rapid Cooling / Fast Boiling (Forms Small/Fine Powder): Rapid cooling causes sudden supersaturation, triggering instantaneous, disorderly precipitation at thousands of nucleation points simultaneously, resulting in very tiny, irregular microscopic crystals.
Page No. 82 & 84: Pause and Ponder (Questions 6 & 7)
Question 6 State whether the following statements on separation methods are true or false: [Exam Favorite] (i) Salt can be separated from a salt solution by evaporation or distillation. (ii) Distillation can be used for separation of two liquids even when these have the same boiling point. (iii) In paper chromatography, the solvent level should be above the sample spot at the beginning of the experiment. (iv) Evaporation and crystallization are the same processes.
Answer:
- (i) True (Evaporation recovers the solid salt residue; distillation recovers both solid salt and pure liquid water distillate).
- (ii) False (Distillation requires a distinct boiling point difference; liquids with identical boiling points vaporize together and cannot be separated).
- (iii) False (The solvent level must strictly be kept below the ink spot; if submerged, the sample will dissolve directly into the solvent reservoir instead of rising via capillary action).
- (iv) False (Evaporation dries mixtures to dryness often decomposing heat-sensitive solids; crystallization purifies substances by forming pure geometric solid crystals without burning).
Question 7 Why do immiscible liquids form two separate layers in a separating funnel? [Exam Favorite]
Answer: Immiscible liquids (such as oil and water) form two distinct, separate layers due to two physical factors:
- Incompatible Intermolecular Forces (Immiscibility): Water molecules are polar and form strong hydrogen bonds, while oil molecules are non-polar. They do not dissolve in one another and repel mixing.
- Difference in Densities: When left undisturbed in a separating funnel, gravitational forces cause the denser liquid (water, density ~ 1.0 g/cm³) to sink and form the lower layer, while the less dense liquid (oil, density ~ 0.8 g/cm³) floats on top.
NCERT Chapter-End Exercises: “Revise, Reflect, Refine” (Pages 90–94)
Question 1 Classify the following mixtures as Homogeneous (Hm) or Heterogeneous (Ht): [Exam Favorite] (a) Air (b) Milk (c) Sugar solution (d) Muddy water (e) Brass (f) Blood
Answer:
| Mixture Name | Classification (Hm / Ht) | Scientific Justification |
|---|---|---|
| (a) Air | Homogeneous (Hm) | A uniform gaseous solution of nitrogen, oxygen, argon, and carbon dioxide without phase boundaries. |
| (b) Milk | Heterogeneous (Ht) | A colloidal emulsion of liquid fat droplets and proteins dispersed non-uniformly in water. |
| (c) Sugar solution | Homogeneous (Hm) | A true solution where sugar molecules are uniformly distributed at the molecular level. |
| (d) Muddy water | Heterogeneous (Ht) | A coarse suspension of soil and clay particles with distinct visible physical boundaries that settle. |
| (e) Brass | Homogeneous (Hm) | A solid-in-solid metallic alloy of copper (~ 70%) and zinc (~ 30%) with uniform atomic distribution. |
| (f) Blood | Heterogeneous (Ht) | A biological colloid containing cellular elements (RBCs, WBCs, platelets) suspended in plasma. |
Question 2 Match the following mixtures with their appropriate method of separation and state the reason for selection: [Exam Favorite]
| Mixture | Method of Separation | Scientific Reason for Selection |
|---|---|---|
| Mud from muddy water | Sedimentation followed by Filtration | Mud particles are insoluble, coarse (greater than 1000 nm), and are trapped on filter paper. |
| Plasma from blood sample | Centrifugation | High-speed spinning separates dense cellular elements to the bottom while lighter plasma remains on top. |
| Naphthalene and sand | Sublimation | Naphthalene sublimes directly into vapor on gentle heating, leaving non-sublimable sand behind. |
| Chalk powder and common salt | Dissolution in water, Filtration, followed by Evaporation | Salt dissolves in water while chalk is insoluble; filtration removes chalk, and evaporation recovers salt. |
| Common salt and water | Evaporation or Simple Distillation | Water is volatile and vaporizes, while salt is non-volatile and remains as crystalline residue. |
| Oil from water | Separating Funnel | Oil and water are immiscible and have different densities; oil floats on top and can be decanted via stopcock. |
| Pigments of a flower petal | Paper Chromatography | Different pigment solutes have different solubilities in the solvent and travel at different rates up paper. |
Question 3 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. Explain with the principle. [Exam Favorite]
Answer:
- Suggested Method: Simple Distillation.
- Scientific Principle: Distillation is used to separate miscible liquids that boil without decomposition and have a sufficient difference in their boiling points (difference of 25°C or more).
- Separation Process:
- The boiling point difference here is
90°C - 60°C = 30°C(which is greater than 25°C). - When the mixture is heated in a distillation flask, the more volatile liquid A (boiling at 60°C) vaporizes first.
- Its vapors pass through the Liebig condenser, cool, condense into liquid, and collect in the receiver flask. Liquid B (boiling at 90°C) remains behind in the flask.
- The boiling point difference here is
Question 4 You are given a mixture of sand, common salt, and naphthalene. Identify and write down the correct sequence of separation techniques to separate each component. [Exam Favorite]
Answer:
================================================================================ STEP-BY-STEP SEPARATION PROTOCOL (SAND + SALT + NAPHTHALENE): -------------------------------------------------------------------------------- STEP 1: SUBLIMATION (Separates Naphthalene) • Place the mixture in a china dish covered with an inverted glass funnel plugged with cotton. • Heat gently. Naphthalene sublimes directly into vapors and solidifies on the cool inner walls of the funnel. • Scrape off pure solid Naphthalene. The residue contains Sand and Common Salt. STEP 2: DISSOLUTION IN WATER & FILTRATION (Separates Sand) • Add distilled water to the remaining sand-salt residue and stir thoroughly. • Common salt dissolves completely forming a solution; sand remains insoluble. • Filter the mixture through a filter paper. Insoluble Sand is collected as the RESIDUE on the filter paper, washed, and dried. STEP 3: EVAPORATION / CRYSTALLIZATION (Recovers Common Salt) • Heat the clear salt-water FILTRATE in a china dish to evaporate water completely. • Pure Common Salt remains behind as the dry solid residue. ================================================================================
Question 5 Will there be any change in the mass of a solution when 10 g of sugar is dissolved in 100 g of water? What type of mixture is formed? [Exam Favorite]
Answer:
- Change in Mass:No, there is no change in total mass.
- According to the Law of Conservation of Mass, the total mass of the resulting solution will be exactly equal to the sum of the masses of the solute and solvent:
Mass of Solution = Mass of Sugar + Mass of Water = 10 g + 100 g = 110 g.
- According to the Law of Conservation of Mass, the total mass of the resulting solution will be exactly equal to the sum of the masses of the solute and solvent:
- Type of Mixture Formed: A Homogeneous Mixture (True Solution) is formed because sugar molecules dissolve completely and occupy the microscopic intermolecular spaces between water molecules uniformly.
Question 6 Explain why Tyndall effect is observed in a colloidal solution (like starch solution or milk) but not in a true solution (like copper sulphate or salt solution). [Exam Favorite]
Answer: The occurrence of the Tyndall effect depends strictly on particle size relative to the wavelength of visible light:
- In Colloids (Starch Sol / Milk): The dispersed colloidal particles have diameters between 1 nm and 1000 nm, which is comparable to the wavelength of visible light (400 to 700 nm). These particles are large enough to scatter incident light rays in all directions, making the path of the light beam brightly illuminated.
- In True Solutions (Copper Sulphate / Salt Water): Solute ions and molecules are smaller than 1 nm (less than 10⁻⁹ m). These tiny particles cannot interact with or scatter light wavelengths. Light passes through unobstructed, leaving the beam path invisible.
Question 7 A student dissolved 40 g of common salt in 320 g of water at 293 K. Calculate the concentration of the solution in terms of mass by mass percentage (% m/m). [Exam Favorite]
Answer:
================================================================================
NUMERICAL SOLUTION (CONCENTRATION % m/m):
--------------------------------------------------------------------------------
GIVEN DATA:
• Mass of Solute (Common Salt) = 40 g
• Mass of Solvent (Water) = 320 g
STEP 1: Calculating Total Mass of Solution
Mass of Solution = Mass of Solute + Mass of Solvent
Mass of Solution = 40 g + 320 g = 360 g
STEP 2: Calculating Mass by Mass Percentage
Formula: Concentration % (m/m) = (Mass of Solute / Mass of Solution) × 100
Concentration % (m/m) = (40 / 360) × 100
Concentration % (m/m) = (1 / 9) × 100 = 100 / 9 = 11.11% (m/m)
FINAL ANSWER:
The concentration of the salt solution is 11.11% (m/m).
================================================================================
Question 8 How would you distinguish between a true solution, a colloid, and a suspension in the laboratory using simple tests? [Exam Favorite]
Answer: A student can distinguish between them using three sequential laboratory tests:
================================================================================ THREE-STEP IDENTIFICATION PROTOCOL: -------------------------------------------------------------------------------- 1. TORCH-LIGHT TEST (Tyndall Effect): • Pass a flashlight beam through the liquid in a dark room. • True Solution : Beam path remains completely INVISIBLE. • Colloid : Beam path GLOWS brightly with scattered light. • Suspension : Shows scattering initially; path disappears as particles settle. 2. FILTRATION TEST (Ordinary Filter Paper): • Filter the mixture through standard filter paper. • True Solution : No residue on paper; clear filtrate passes through. • Colloid : No residue on paper; passes through completely. • Suspension : Solid particles are RETAINED on filter paper as residue. 3. SETTLING TEST (Stability on Standing): • Leave the sample undisturbed in a test tube for 20 minutes. • True Solution : Particles NEVER settle down (Homogeneous). • Colloid : Particles do NOT settle down (Stable). • Suspension : Heavy particles SETTLE to the bottom (Unstable). ================================================================================
Question 9 Why is crystallization considered a better technique than simple evaporation for purifying solids (like copper sulphate or alum)? [Exam Favorite]
Answer: Crystallization is superior to evaporation for purifying solids for three major reasons:
- Prevents Thermal Decomposition: During simple evaporation to dryness, heat-sensitive compounds (such as sugar or copper sulphate) may char, decompose, or lose their water of crystallization. Crystallization uses gentle heating and slow cooling.
- Eliminates Soluble Impurities: In evaporation, all dissolved soluble impurities remain mixed with the dried solid residue. In crystallization, impurities remain dissolved in the mother liquor while only pure geometric crystals separate out.
- Yields High Purity and Regular Geometric Shapes: Crystallization produces large, pure crystals of definite geometric lattice structures.
Question 10 A community water treatment plant purifies river water to supply potable drinking water to a city. Outline the sequence of separation processes involved from the river source to domestic water taps. [Exam Favorite]
Answer: Municipal water purification involves five consecutive separation stages:
================================================================================
MUNICIPAL DRINKING WATER PURIFICATION FLOWCHART:
--------------------------------------------------------------------------------
1. Reservoir (Raw River Water)
│
▼
2. Sedimentation Tank (Heavier suspended impurities and silt settle by gravity)
│
▼
3. Loading Tank (Alum is added to coagulate and settle fine clay colloids)
│
▼
4. Filtration Tank (Passes through multi-layers of fine sand, coarse sand, & gravel)
│
▼
5. Chlorination Tank (Chlorine gas / bleaching powder added to kill pathogenic bacteria)
│
▼
6. Clean, Safe Potable Water to Domestic Home Taps
================================================================================
Frequently Asked Questions (FAQs) – Class 9 Science Chapter 5
Question 1: Define a pure substance and a mixture. [Exam Favorite] Answer:
- Pure Substance: A form of matter having a constant chemical composition and uniform, definite physical and chemical properties throughout (e.g., pure elements like Gold, and chemical compounds like distilled water, H₂O).
- Mixture: A combination of two or more pure substances physically mixed together in any variable proportion without chemical bonding (e.g., Air, Brass, Seawater).
Question 2: What is an Alloy? Why is it considered a mixture? [Exam Favorite] Answer: An alloy is a homogeneous solid solution of two or more metals (or a metal and a non-metal). It is considered a mixture because: (i) it shows the individual properties of its constituent elements, and (ii) its composition can vary (e.g., Brass is 70% Cu + 30% Zn).
Question 3: What is the Tyndall Effect? [Exam Favorite] Answer: The Tyndall effect is the optical phenomenon of scattering of a visible light beam by colloidal particles in a medium, illuminating the path of the light beam.
Question 4: What is Brownian Motion in colloids? [Exam Favorite] Answer: Brownian motion is the continuous, random, erratic zigzag motion of colloidal particles suspended in a liquid or gas, caused by continuous uneven collisions with molecules of the dispersion medium.
Question 5: Differentiate between Solute and Solvent. [Exam Favorite] Answer:
- Solvent: The component of a solution that is present in the larger quantity and dissolves the other component (e.g., Water).
- Solute: The component present in smaller quantity that gets dissolved in the solvent (e.g., Salt).
Question 6: What is a Saturated Solution and an Unsaturated Solution? [Exam Favorite] Answer:
- Saturated Solution: A solution in which no more solute can be dissolved at that specific given temperature.
- Unsaturated Solution: A solution that contains less than the maximum possible amount of solute, so more solute can be dissolved at that temperature.
Question 7: How does temperature affect the solubility of solids in liquids and gases in liquids? [Exam Favorite] Answer:
- Solids in Liquids: Solubility generally increases with an increase in temperature.
- Gases in Liquids: Solubility decreases with an increase in temperature (which is why boiling water expels dissolved oxygen).
Question 8: What is Fractional Distillation, and when is it used? [Exam Favorite] Answer: Fractional distillation is a separation technique used to separate two or more miscible liquids whose boiling points differ by less than 25°C (or 25 K) using a fractionating column packed with glass beads to provide repetitive condensation-vaporization cycles.
Question 9: What is Sublimation? Name three substances that sublime. [Exam Favorite] Answer: Sublimation is the direct phase transformation of a solid into gas on heating without passing through the intermediate liquid state. Examples: Ammonium Chloride (NH₄Cl), Camphor, Naphthalene, Anthracene, and Dry Ice (solid CO₂).
Question 10: State the principle of Centrifugation. [Exam Favorite] Answer: Centrifugation operates on the principle that when a mixture is spun at very high rotational speeds, denser particles are forced outward to the bottom of the container, while lighter particles remain floating on the top.
Question 11: What is the mobile phase and stationary phase in Paper Chromatography? [Exam Favorite] Answer:
- Stationary Phase: The specialized porous chromatographic filter paper containing adsorbed moisture.
- Mobile Phase: The liquid solvent (water or alcohol) that moves up the paper strip by capillary action.
Question 12: Why is air considered a mixture and not a chemical compound? [Exam Favorite] Answer: Air is considered a mixture because: (i) its constituent gases (N₂, O₂, CO₂, Ar) retain their individual chemical properties, (ii) its composition varies slightly across geographical locations, and (iii) its components can be separated by physical methods (fractional distillation of liquid air).
Question 13: What happens when a saturated solution prepared at 60°C is cooled to 20°C? [Exam Favorite] Answer: Since solubility decreases with decreasing temperature, the excess dissolved solute can no longer remain in solution and crystallizes out as a solid precipitate at the bottom of the beaker.
Question 14: What is the function of the glass beads in a fractionating column? [Exam Favorite] Answer: The glass beads provide a large total surface area for the ascending vapors to cool, condense, and re-vaporize repeatedly, ensuring sharp separation between liquids with close boiling points.
Question 15: Give one everyday example each of: (a) Foam, (b) Aerosol, (c) Gel, and (d) Sol. [Exam Favorite] Answer:
- (a) Foam: Shaving cream, whipped cream (Gas in liquid).
- (b) Aerosol: Fog, mist, cloud, smoke (Liquid/solid in gas).
- (c) Gel: Jelly, butter, cheese (Liquid in solid).
- (d) Sol: Milk of magnesia, paint, ink (Solid in liquid).
Mastering the NCERT Solutions for Class 9 Science Chapter 5 (Exploration), “Exploring Mixtures and their Separation”, equips students with the separation techniques, concentration calculations, and colloidal properties required for high performance in CBSE chemistry evaluations. Review the numerical concentration formulas, the multi-step separation flowcharts, and the 15 high-yield FAQs above to secure full marks in your examinations.