NDA Matter, its changes and the mole previous year questions with solutions
Chemistry begins here, and this chapter is about one General Studies question in thirty. Kinds of change, mixtures and separation methods are asked almost every year; since 2025 colloids return as statement sets and the mole as numericals. Every question UPSC has set on it since 2017, solved. Grouped by the method each needs, newest first; open a question for its answer and worked solution.
8.1 Matter, its states and their interconversion; physical and chemical changes
Answer and solution
Answer: (c) Melting of ice
Method. chemical versus physical change
Burning of coal, rusting and acid-base reaction form new substances. Melting of ice is a physical change (C1-A50).
Answer and solution
Answer: (c) During a physical change, the chemical composition of molecules changes
Method. physical change leaves molecular composition unchanged
In a physical change the chemical composition of molecules does not change, so (c) is wrong. Changes of state are physical, a burning candle has both (wax melts, then burns) and rusting is chemical (C1-A50).
Answer and solution
Answer: (c) Diffusion
Method. diffusion
Intermixing of particles of two different kinds of matter on their own is diffusion. Osmosis is solvent movement across a membrane, Brownian motion the random motion of suspended particles (standard, Class 9 ch 1).
Answer and solution
Answer: (d) Mixing of NaOH and HCl
Method. chemical change forms a new substance
NaOH + HCl -> NaCl + H2O, a neutralisation giving new substances. Sublimation and melting are physical, and heating an iron rod only changes temperature (C1-A36, C1-A50).
Answer and solution
Answer: (d) Burning of magnesium ribbon in air
Method. chemical reaction forms a new substance
Burning of magnesium in air: 2Mg + O2 -> 2MgO, a new compound. Evaporation, dissolving and compressing a gas are physical (C1-A07).
Answer and solution
Answer: (c) deposition
Method. deposition: gas to solid directly
Sulphur vapour cools on the rocks and turns directly into solid sulphur, gas to solid without a liquid stage: deposition (NS5-A64; KC1-A27).
Answer and solution
Answer: (a) sublimation
Method. dry ice sublimes
Solid CO2 (dry ice) passes straight to gas, i.e. sublimation; the cold gas chills air so water vapour condenses as mist. The process named for dry ice is sublimation (NCERT Class 9 ch 1 gives sublimation as the example). Reading the mist alone would give condensation.
Answer and solution
Answer: (d) Milk turning sour
Method. chemical change forms a new substance
Milk turning sour: lactose changes into lactic acid by bacteria, a new substance. Dissolving sugar, melting ice and crystallization are physical changes (C1-A50).
Answer and solution
Answer: (b) Graying of hair naturally
Method. chemical change forms a new substance
Cutting hair or fruit and swelling of resin in water change only shape or size, physical. Natural graying is the loss of melanin pigment in the hair, a chemical change in the hair pigment (standard; compare C1-A50, physical vs chemical change).
Answer and solution
Answer: (c) non-crystalline amorphous solid
Method. glass is an amorphous solid
Glass has no long-range ordered arrangement of particles, so it is a non-crystalline (amorphous) solid, not a crystalline one; it is rigid, not a flowing liquid, and not a colloid (Class 9 Science, states of matter, standard).
8.2 Elements, compounds and mixtures; solutions, colloids and suspensions
I. A solution of iodine in ethanol known as 'tincture of iodine' is a heterogeneous mixture.
II. A suspension is a heterogeneous mixture and its particles scatter a beam of light passing through it, making its path visible.
III. A colloidal solution, although it appears homogeneous, is a heterogeneous mixture.
Which of the statements given above are correct?
Answer and solution
Answer: (b) II and III only
Method. types of mixtures
I is false: tincture of iodine is a solution, homogeneous. II is true: a suspension is heterogeneous and its particles scatter a beam (NS5-A69, NS5-A91). III is true: a colloid looks homogeneous but is heterogeneous (NS5-A86). Correct: II and III only.
How many of the pairs given above are correctly matched?
| (blank) | Example of Colloid | Type of Colloid |
|---|---|---|
| I. | Cloud | Aerosol |
| II. | Mud | Sol |
| III. | Milk | Emulsion |
| IV. | Rubber | Foam |
Answer and solution
Answer: (d) All four
Method. types of colloids
I cloud is liquid in gas, an aerosol, correct. II mud (solid in liquid) is listed as a sol, correct. III milk (liquid in liquid) is an emulsion, correct. IV rubber (foam rubber, gas in solid) is a solid foam, correct. All four (NS5-A92, NS5-A93; standard colloid table).
Pairs II and IV rest on the common textbook table (mud as sol, rubber as foam); strictly mud can be called a suspension.
I. A colloid and a suspension are heterogeneous mixtures.
II. Colloids can be separated from the mixture by the process of centrifugation.
III. Particles of colloids and suspensions can be seen by naked eye.
IV. The aqueous solution of copper sulphate does not show Tyndall effect; however, the mixture of milk and water shows Tyndall effect.
Which of the statements given above is/are correct?
Answer and solution
Answer: (d) I, II and IV
Method. colloids and suspensions
I is true: both are heterogeneous (NS5-A69, NS5-A86). II is true: colloids such as blood are separated by centrifugation (NS5-A72, NS5-A85). III is false: colloid particles are too small to see with the naked eye. IV is true: copper sulphate solution shows no Tyndall effect, milk-water does (NS5-A89, NS5-A91). Correct: I, II and IV.
Answer and solution
Answer: (a) Sulphur in carbon disulphide
Method. solution versus heterogeneous mixture
Sulphur dissolves in carbon disulphide to form a homogeneous solution. The sugar and salt crystals, sandy water and sulphur in water (insoluble) are heterogeneous (NS5-A02, NS5-A03).
Answer and solution
Answer: (c) A compound is an impure substance
Method. properties of a compound
A compound is a pure substance of fixed composition, with properties different from its elements and separable only by chemical or electrochemical means; it is not impure (KC1-A31, KC1-A33, KC1-A34).
Answer and solution
Answer: (b) Colloidal solution is homogenous in nature
Method. properties of colloids
A colloid is heterogeneous, so (b) is not true. Its particles are uniformly spread, scatter light (Tyndall effect) and do not settle on standing (NS5-A86, NS5-A89, NS5-A82).
Answer and solution
No correct option. This question is defective as printed in UPSC's paper.
Method. solution = homogeneous mixture of two or more substances
Milk is a colloid, not a solution (NS5-A82); alloys and air are solutions (NS5-A67, NS5-A02). Sugar is also not a solution: it is a pure compound (a solute).
Two options fit as printed: (b) milk is a colloid and (d) sugar is a pure substance; neither is a solution. The intended reading is milk (b), with sugar taken as a solute.
Answer and solution
Answer: (a) Tin
Method. element versus mixture
Tin is an element, a pure substance. Sea water, soil and air are mixtures (KC1-A28, KC1-A29).
Answer and solution
Answer: (a) Ice
Method. pure substance versus mixture
Ice is solid water, a single compound, so a pure substance. Ice-cream, air and honey are mixtures (NS5-A02, KC1-A28).
Answer and solution
Answer: (c) Milk
Method. heterogeneous mixture
Milk is a colloid of fat droplets in water, a heterogeneous mixture. Hydrochloric acid (HCl in water), vinegar (acetic acid in water) and soda water (CO2 in water) are homogeneous solutions (NS5-A02, NS5-A03, NS5-A82).
8.3 Separation of mixtures
I. Evaporation
II. Crystallization
Select the correct answer using the code given below.
Answer and solution
Answer: (c) Both I and II
Method. recovering a non-volatile solute
Evaporation drives off the solvent and leaves the solute; crystallization separates it as crystals from the concentrated solution. Both work for a non-volatile solid solute (NS5-A31, NS5-A41).
Answer and solution
Answer: (c) Centrifugation
Method. centrifugation
In diagnostic laboratories blood is spun in a centrifuge so the heavier blood cells separate from the plasma (NS5-A72, NS5-A73, NS5-A85).
List I (Process)
A. Separation of acetone and water from their mixture
B. Separation of water and kerosene oil from their mixture
C. Separation of cream from milk
D. Separation of pigments from plant extract
List II (Separation method)
1. Chromatography
2. Centrifugation
3. Distillation
4. Separating Funnel
Code:
A B C D
Answer and solution
Answer: (c) 3 4 2 1
Method. match processes with separation methods
A: acetone and water are miscible liquids with different boiling points, distillation (3). B: water and kerosene are immiscible, separating funnel (4). C: cream from milk, centrifugation (2). D: pigments, chromatography (1). Code 3 4 2 1 (NS5-A45, NS5-A58, NS5-A72, NS5-A53).
Answer and solution
Answer: (c) Fractional distillation
Method. fractional distillation
Kerosene and petrol are miscible liquids with different boiling points, separated by fractional distillation (NS5-A49).
Answer and solution
Answer: (c) Separating funnel
Method. separating funnel for immiscible liquids
Benzene and water do not mix, so they form two layers and are separated with a separating funnel (NS5-A57, NS5-A58). The Dean and Stark trap collects water from a refluxing reaction, not the usual separation.
Answer and solution
Answer: (b) Fractional distillation
Method. fractional distillation
Petroleum is a mixture of liquid fractions of different boiling points, separated in a refinery by fractional distillation (NS5-A48, NS5-A49, NS5-A50).
Answer and solution
Answer: (b) Sublimation
Method. sublimation
Anthracene sublimes on heating while salt does not, so the vapour is condensed and collected; this is the principle of separating a sublimable solid from a non-sublimable one (NS5-A64, NS5-A65).
Answer and solution
Answer: (c) By distillation
Method. distillation separates miscible liquids with different boiling points
Two liquids in one homogeneous mixture cannot be filtered; they are separated by distillation, the one with the lower boiling point vaporising first and being condensed (NS5-A41, NS5-A42, NS5-A45).
8.4 Atoms and molecules; the laws of chemical combination
Answer and solution
Answer: (b) Law of conservation of mass
Method. conservation of mass
Mass before = 1.7 + 0.585 = 2.285 g; mass after = 1.435 + 0.85 = 2.285 g. The totals are equal, so the law of conservation of mass is shown (the water is common to both sides) (C1-A04, KC1-A73).
List I (Element)
A. Phosphorus
B. Nitrogen
C. Neon
D. Sulphur
List II (Atomicity)
1. 2
2. 1
3. 8
4. 4
Code:
A B C D
Answer and solution
Answer: (a) 4 1 2 3
Method. atomicity: P4, N2, Ne, S8
Phosphorus P4 gives 4; nitrogen N2 gives 2; neon is monatomic, 1; sulphur S8 gives 8. Matching to List II numbers (1 = 2, 2 = 1, 3 = 8, 4 = 4): A-4, B-1, C-2, D-3, code 4 1 2 3 (standard, atomicity).
Answer and solution
Answer: (b) Sulphur
Method. atomicity (NCERT Class 9 table)
Aluminium is monatomic and chlorine Cl2 is diatomic. Phosphorus P4 has atomicity 4 and sulphur S8 atomicity 8; NCERT's atomicity table labels phosphorus tetra-atomic and reserves 'polyatomic' for sulphur, so the answer is (b) sulphur.
Strictly P4 (four atoms) is also polyatomic; the question follows NCERT's labelling, under which only sulphur is called polyatomic.
Answer and solution
Answer: (d) [symbol: circle divided into three equal sectors by three radii meeting at the centre (an upside-down Y inside a circle)]
Method. Dalton's symbols
In Dalton's table the letter-in-circle symbols are metals (L lead; P inside a circle is platinum), a circle with a plus (cross) is sulphur, and phosphorus is the circle with three radii from the centre, the upside-down Y (NCERT Class 9 ch 8, Fig. 8.9, Symbols of some elements given by Dalton).
Answer and solution
Answer: (c) Avogadro
Method. Avogadro's law
Equal volumes of all gases at the same temperature and pressure hold equal numbers of particles: Avogadro's law (KC1-A78).
Answer and solution
Answer: (c) Iodine
Method. atomicity of elements
Copper, helium (noble gas) and barium (metals) are monatomic; iodine exists as diatomic I2 molecules (standard, atomicity).
Answer and solution
Answer: (c) Matter can neither be created nor destroyed.
Method. law of conservation of mass
The law of conservation of mass says matter can neither be created nor destroyed in a chemical reaction. (a) is the law of definite proportions, (b) Gay-Lussac's law of gaseous volumes, (d) Avogadro's law (C1-A04, KC1-A73).
Answer and solution
Answer: (a) Avogadro's hypothesis
Method. Avogadro's law
Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules: Avogadro's hypothesis (KC1-A78).
8.5 The mole concept: molar mass, Avogadro's number, molar volume and percentage composition
Answer and solution
Answer: (d) 47% and 7%
Method. percentage composition of urea, CO(NH2)2
Urea CH4N2O = 12+4+28+16 = 60. N = 28/60 x 100 = 46.7%, about 47%. H = 4/60 x 100 = 6.7%, about 7% (KC1-A96, KC1-A97; C 12, H 1, N 14, O 16).
Answer and solution
Answer: (d) 4 gram of hydrogen gas contains 6.022 × 1023 molecules
Method. mole and molar volume (22.4 L at STP, N_A = 6.022x10^23)
(a) 0.5 mol x 22.4 = 11.2 L, correct. (b) NH3 = 14+3 = 17 g/mol, so 17 g = 1 mol = 6.022x10^23 molecules, correct. (c) CO2 = 44 g/mol, 22.4 L = 1 mol = 44 g, correct. (d) H2 = 2 g/mol, 4 g = 2 mol = 2 x 6.022x10^23 molecules, not 6.022x10^23, so incorrect (KC1-A94, KC1-A95).
Answer and solution
Answer: (b) 14 g
Method. mass = moles x molar mass
Molar mass of N2 = 2x14 = 28 g/mol. Mass = 0.5 x 28 = 14 g (KC1-A95).
Answer and solution
Answer: (b) 63
Method. equivalent weight = molar mass / n-factor
H2C2O4.2H2O: molar mass = 2+24+64+36 = 126. Oxalic acid is dibasic (n-factor 2), so equivalent weight = 126/2 = 63 for the hydrate; for the anhydrous acid 90/2 = 45. The question names the hydrated formula, so 63; options (c) 90 and (d) 126 are molar masses (standard; titration chemistry).
Reads as the hydrate's equivalent weight: 63. If the anhydrous acid were meant it would be 45 (option a); the formula printed is the dihydrate.
Answer and solution
Answer: (b) 106 u
Method. formula mass of Na2CO3
2 Na + C + 3 O = 2x23 + 12 + 3x16 = 46 + 12 + 48 = 106 u (KC1-A90).
Answer and solution
Answer: (c) six times the mass percent of C as compared to the mass percent of H
Method. percentage composition of C6H12O4 (C 12, H 1, O 16)
Mass of C = 6x12 = 72, H = 12x1 = 12, O = 4x16 = 64; total 148. C: H = 72 : 12 = 6 : 1, so C has six times the mass percent of H (48.6% vs 8.1%). (a) one molecule has 22 atoms, but a mole has 22 x 6.022x10^23 atoms; (b) H is not twice C; (d) H 12 vs O 64, not thrice (KC1-A88, KC1-A96).
Answer and solution
Answer: (a) 113 kg
Method. mass relation from C + O2 -> CO2
C + O2 -> CO2: 12 g C gives 44 g CO2. 1 kg C gives 44/12 = 11/3 kg CO2 (KC1-A102, KC1-A103; C 12, O 16, said so).
Answer and solution
Answer: (c) 10%
Method. mass percentage = mass of solute / mass of solution x 100
Solution = 20 g + 180 g = 200 g. Mass % = 20/200 x 100 = 10% (NS5-A14, NS5-A16).
Shortcut. Solute is 1/10 of the total mass.
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