Every branch of chemistry you'll study for NEET, JEE, IIT-JAM, BITSAT, GATE, CSIR-NET, or the TGT/PGT exams eventually leans on one idea: the mole. Before you can balance an equation with confidence, work out a limiting reagent, or find the concentration of a titrant, you need this chapter to be second nature. I've built these notes to take you from the very first definitions — matter, atoms, significant figures — all the way through stoichiometry, POAC, concentration terms, and gravimetric analysis, with the exam angle kept front and centre throughout.
- 1. Matter and Its Nature
- 2. Dalton's Atomic Theory
- 3. Atoms, Molecules, Elements, Compounds and Their Properties
- 4. Physical Quantities, SI Units and Scientific Notation
- 5. Significant Figures and Rounding Off
- 6. Mass, Weight, Volume, Density and Temperature
- 7. Dimensional Analysis (Factor-Label Method)
- 8. Laws of Chemical Combination
- 9. Atomic Mass, Molecular Mass and Formula Mass
- 10. The Mole and Avogadro's Constant
- 11. Mole–Mass–Volume Relationships
- 12. Percentage Composition and Empirical/Molecular Formula
- 13. Balancing Equations and Stoichiometry
- 14. Limiting Reagent, Theoretical and Percent Yield
- 15. Principle of Atom Conservation (POAC)
- 16. Concentration Terms
- 17. Gravimetric Analysis
- 18. Common Mistakes and Exam Tips
- 19. Practice MCQs (30+ with Answers)
1. Matter and Its Nature
I like to start exactly where the syllabus starts: matter is anything that occupies space and has mass. That's it — trees, air, glass, your calculator, all of it. What makes the topic interesting for exam purposes isn't the definition but the classification that follows, because questions love to test whether you can correctly slot a given substance into the right box.
Matter exists in three physical states — solid, liquid, gas — and each has a mechanical reason behind its behaviour rather than an arbitrary label. In a solid, particles sit close together in fixed positions and only vibrate, which is why solids hold both a definite shape and a definite volume. Loosen that up a bit, let the particles slide past each other while staying close together, and you get a liquid — definite volume, no fixed shape. Let the particles separate entirely and move independently in every direction, and you get a gas, with neither a fixed shape nor a fixed volume.
Matter splits broadly into pure substances and mixtures. Pure substances have a fixed composition that can't be altered by physical means — only by chemical ones — and come in two flavours: elements (one kind of atom, like the oxygen in air) and compounds (more than one element combined in a fixed ratio, like water). Mixtures, by contrast, have variable composition. Stir sugar into water and you get a homogeneous mixture — uniform throughout, inseparable by filtration, but separable by distillation. Leave sand and pebbles together and you get a heterogeneous mixture, with visibly distinct phases separable by filtration or centrifugation.
Sea water dissolves salts uniformly in water, so it's a solution. Phosphorus and bromine are each a single substance. Air, despite being a mixture of gases, behaves as homogeneous — one phase throughout. Concrete has visibly distinct components (sand, gravel, cement), making it heterogeneous.
Properties of matter fall along two separate axes worth keeping distinct in your head. First, physical vs chemical: a physical property (colour, density, melting point) can be measured without changing what the substance is; a chemical property (iron rusting, gold resisting tarnish) describes how a substance transforms into something else. Second, extensive vs intensive: an extensive property like volume or mass scales with the amount of sample, while an intensive property like colour, density, or boiling point stays the same no matter how much of the substance you have. Intensive properties are the useful ones for identification — two lumps of gold, however different in size, share the same colour and melting point.
2. Dalton's Atomic Theory
John Dalton, working at the start of the 19th century, took the old Greek notion of "atomos" (uncuttable) and gave it scientific teeth by using it to explain the experimentally observed laws of conservation of mass and definite proportions. His theory rests on four pillars: matter is made of indivisible atoms; atoms of a given element are identical in mass and properties while atoms of different elements differ; compounds form when atoms of different elements combine in fixed numerical ratios; and atoms are indestructible — chemical reactions just rearrange them.
Notice how neatly this explains conservation of mass — if a reaction is nothing but atoms swapping partners, and atoms themselves don't gain or lose mass, then the total mass before and after must match. That single insight is why we bother balancing chemical equations at all.
3. Atoms, Molecules, Elements, Compounds and Their Properties
An atom is the smallest particle that still identifies an element; split it further and you lose the identity of that element entirely. A molecule plays the same role for compounds — the smallest unit that still represents the compound's identity. Water is the textbook case: split it down to a single molecule and you still have water (two hydrogens, one oxygen); split that molecule apart and you're left with hydrogen and oxygen gas, neither of which behaves anything like water.
Atomicity — the number of atoms in one molecule of an element — is worth memorising for common elements, since it resurfaces constantly in stoichiometry problems: oxygen and hydrogen are diatomic (O2, H2), ozone is triatomic (O3), and noble gases are monatomic.
4. Physical Quantities, SI Units and Scientific Notation
A measurement without a unit is meaningless — "the distance is 25" tells you nothing until you know whether that's 25 inches or 25 kilometres. The SI system standardises seven base quantities, and I'd suggest memorising the table below cold, since direct "which of these is a base unit" questions show up often.
| Quantity | SI Unit | Symbol |
|---|---|---|
| Length | metre | m |
| Mass | kilogram | kg |
| Time | second | s |
| Electric current | ampere | A |
| Temperature | kelvin | K |
| Amount of substance | mole | mol |
| Luminous intensity | candela | cd |
Everything else — newtons, joules, square metres, dynes — is derived from these seven. A dyne (CGS force unit) and a newton (SI force unit) are both derived; ampere and candela are base units. This distinction shows up as a one-line MCQ often enough that it's worth the two minutes to internalise.
Very large or very small numbers get compacted using scientific notation — a number between 1 and 10 multiplied by a power of ten. Moving the decimal left gives a positive exponent; moving it right gives a negative one. Multiplying two numbers in scientific notation multiplies the coefficients and adds the exponents; dividing subtracts them. Addition and subtraction are fussier — you first have to bring both numbers to the same power of ten before combining the coefficients.
Multiply the coefficients and add the exponents: (2.25 × 3.25) × 10(−4) + (−6) = 7.3125 × 10−10.
The most common table of SI prefixes is worth having handy, especially the ones in bold here, since they're the ones that actually turn up in chemistry problems.
| Prefix | Symbol | Power of 10 |
|---|---|---|
| giga | G | 109 |
| mega | M | 106 |
| kilo | k | 103 |
| deci | d | 10−1 |
| centi | c | 10−2 |
| milli | m | 10−3 |
| micro | µ | 10−6 |
| nano | n | 10−9 |
| pico | p | 10−12 |
5. Significant Figures and Rounding Off
Significant figures tell you how confidently a measurement is known — every digit you're sure of, plus one estimated digit. Getting the counting rules automatic saves you from silly errors in every numerical section of the paper. There are five rules I'd commit to memory:
Rule 1 — all non-zero digits count. 40.23 has four significant figures.
Rule 2 — zeros sandwiched between non-zero digits count. 62.08 has four.
Rule 3 — trailing zeros count only if there's a decimal point. 4.00 has three significant figures, but 40 (no decimal) has only two.
Rule 4 — leading zeros never count; they're just placeholders. 0.0035 has two significant figures.
Rule 5 — exact numbers (things you counted, not measured) have infinite significant figures.
When you combine measurements, the rules differ by operation. For addition and subtraction, match the answer's decimal places to whichever number in the sum has the fewest decimal places. For multiplication and division, match the answer's total significant figures to whichever factor has the fewest.
1. If the first dropped digit is 5 or more, round the retained digit up.
2. If the first dropped digit is 4 or less, leave the retained digit unchanged.
3. If the digit being dropped is exactly 5 (nothing after it), round to the nearest even digit — 1.024568 → 1.024, but 18.03500 → 18.04.
The raw sum is 424.06458. The least precise term, 12.8, has only one decimal place, so the final answer rounds to 424.1 — four significant figures.
6. Mass, Weight, Volume, Density and Temperature
Mass and weight get used interchangeably in daily conversation, but they aren't the same thing in chemistry. Mass is the amount of matter in an object, measured on a balance, and it doesn't change with location. Weight is the force gravity exerts on that mass, measured on a spring scale, and it does change — an astronaut's mass stays constant on the Moon, but their weight drops sharply.
Volume, being derived from (length)3, has the SI unit of cubic metres, though in the lab you'll live in litres and millilitres. Since 1 dm³ = 1000 cm³, and 1 L is defined as exactly 1 dm³, it falls out that 1 mL = 1 cm³ exactly — a conversion that quietly underlies a huge number of density problems.
Density is intensive and temperature-dependent — most substances expand slightly on heating, so density typically drops as temperature rises, though the effect is small for solids and liquids. Watch for questions that give you density data at multiple temperatures and ask you to reason about expansion.
Temperature conversions between Celsius, Fahrenheit, and Kelvin come up constantly, mostly as a setup step buried inside a larger problem rather than as the whole question.
Kelvin and Celsius are exactly the same size of degree — only the zero point differs, with 0 K (absolute zero, nature's coldest possible temperature) sitting 273.15 units below 0°C. That's why negative Kelvin temperatures don't exist, while negative Celsius ones obviously do.
Converting 17°C to Fahrenheit: TF = (9/5)(17) + 32 = 62.6°F. Since this is well below the 148°F melting point, the UF6 remains solid.
7. Dimensional Analysis (Factor-Label Method)
Dimensional analysis treats every numerical conversion as multiplication by one or more conversion factors — fractions built from a true equality between units, so their value is exactly 1 and multiplying by them changes units without changing magnitude. The elegance of the method is that it self-checks: if your units don't cancel down to what you want, you know you've set up the problem wrong before you even look at the number.
20.2 (miles/gallon) × (1.609 km/1 mile) × (1 gallon/3.785 L) = 8.59 km/L. Notice how miles and gallons cancel, leaving exactly the units asked for.
8. Laws of Chemical Combination
Four experimental laws from the 18th and early 19th centuries set the stage for everything that follows in this chapter, and examiners like testing whether you can identify which law a given data set illustrates.
Law of Conservation of Mass
Antoine Lavoisier's 1789 combustion experiments established that no detectable mass is gained or lost in a chemical reaction, provided the system is sealed so nothing escapes or enters.
Law of Definite Proportions
Joseph Proust found that natural and synthetic cupric carbonate both contained identical percentages of copper, carbon, and oxygen — a compound's elemental composition by mass is always fixed, regardless of source or method of preparation.
Law of Multiple Proportions
Dalton's 1803 contribution: when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other stand in a ratio of small whole numbers. Sulphur dioxide and sulphur trioxide both combine with the same mass of sulphur but in an oxygen ratio of exactly 2:3.
Gay–Lussac's Law of Gaseous Volumes
At the same temperature and pressure, gases react in volume ratios that reduce to small whole numbers — hydrogen and oxygen combine 2:1 by volume to form water vapour.
Avogadro's Law
Equal volumes of different gases at the same temperature and pressure contain equal numbers of molecules. This single generalisation explained Gay-Lussac's volume law, revealed the diatomic nature of gases like hydrogen and chlorine, gave chemists a route to molar masses of gases, and eventually fed into kinetic molecular theory.
1 g of phosphorus combines with 91.18/8.82 = 10.33 g of hydrogen (from phosphine), and with 77.43/22.57 = 3.43 g of chlorine (from PCl3). The hydrogen-to-chlorine ratio for a fixed mass of phosphorus is 10.33/3.43 ≈ 3:1. Meanwhile hydrogen chloride itself gives a H:Cl ratio of 97.23/2.77 = 36:1. Since 3:1 and 36:1 are simple multiples of one another, this is the law of reciprocal proportions.
9. Atomic Mass, Molecular Mass and Formula Mass
Atoms are far too small to weigh individually — a barely visible sliver of copper already contains roughly 1017 atoms — so chemistry works with relative masses instead, calibrated against carbon-12, whose single atom is defined as exactly 12 atomic mass units (u). One u equals exactly 1/12th the mass of one ¹²C atom.
Historically, atomic mass could be pinned down through several independent methods, and knowing these by name helps with theory-based MCQs: Dulong and Petit's law (atomic mass ≈ 6.4/specific heat, for solids), the Cannizzaro method (using the smallest mass of an element found across several of its compounds), vapour density of a chloride (molecular mass = 2 × vapour density, from which atomic mass follows once valency is known), and the law of isomorphism (isomorphous compounds — same crystal shape — combine with fixed masses of other elements in the ratio of their atomic masses).
Most elements are mixtures of isotopes, so the number reported on the periodic table — the atomic weight — is a weighted average across natural abundance, not the mass of any single real atom.
(0.198 × 10.0129) + (0.802 × 11.0093) = 10.8 u.
Molecular mass is simply the sum of atomic masses in a molecule; formula mass plays the identical role for ionic compounds, which don't strictly exist as discrete "molecules" in the solid state. Numerically the calculation is the same either way.
10. The Mole and Avogadro's Constant
Chemists needed a bridge between the atomic world (individual atoms, far too numerous and too light to count or weigh directly) and the macroscopic world (grams, litres, things you can actually put on a balance). That bridge is the mole — the amount of substance containing exactly as many elementary entities as there are atoms in 12 g of ¹²C.
Since one ¹²C atom masses exactly 12 u, and one mole of ¹²C atoms masses exactly 12 g, you can work out how many particles sit in a mole by dividing 12 g by the experimentally measured mass of a single ¹²C atom (1.992648 × 10−23 g). Do that division and you land on 6.023 × 1023 — Avogadro's constant, NA.
This number applies to any kind of particle you like — atoms, ions, electrons, molecules, even photons. One mole of Na⁺ ions contains 6.023 × 10²³ ions; one mole of electrons contains 6.023 × 10²³ electrons, a fact historically confirmed independently through Millikan's measurement of a single electron's charge (1.602 × 10−19 C) against the known charge on a mole of electrons (96,485 C, the Faraday constant).
Number of moles = (Number of atoms/ions/molecules) / NA
Converting everything to moles: Pb = 4.55×10²²/6.023×10²³ = 0.0755 mol; C = 8.50/12.0 = 0.708 mol; Zn = 7.14×10²²/6.023×10²³ = 0.118 mol; Ca = 0.280 mol (given directly). Increasing order: Pb < Zn < Ca < C.
11. Mole–Mass–Volume Relationships
Three quantities interconvert once you know molar mass and Avogadro's number, and I find it useful to keep this flow chart in mind whenever a problem hands you one quantity and asks for another:
For gases, there's a fourth route in via volume. NTP (normal temperature and pressure) is 298 K and 1 atm; STP (standard) is 273 K and 1 atm. Applying the ideal gas equation pV = nRT at STP gives the familiar shortcut:
This is a trap for the unwary — since water is a liquid here, not a gas, dividing by 22.4 would be wrong even though the number "18" tempts you toward it. Instead: 18 mL × 1 g/mL = 18 g, and 18 g / 18 g mol⁻¹ = 1 mol.
12. Percentage Composition and Empirical/Molecular Formula
Knowing what fraction of a compound's mass comes from each element lets you go backward from experimental combustion data to a chemical formula — a genuinely important skill, not just an academic exercise.
Finding an empirical formula from percentage composition is a three-step routine: convert percentages to actual masses (assume 100 g of sample so percentages become grams directly), convert those masses to moles using atomic masses, then find the smallest whole-number ratio between the mole values.
In 100 g: 74.9 g C → 74.9/12.011 = 6.24 mol; 25.1 g H → 25.1/1.0079 = 24.9 mol. Ratio H:C = 24.9/6.24 ≈ 3.99 ≈ 4. So for every carbon there are four hydrogens — empirical formula CH4.
The empirical formula only gives the simplest ratio, not necessarily the true molecular formula — C2H8, C3H12, and so on would all reduce to the same CH4 ratio. You need an independent molecular mass measurement to pin down which multiple is correct.
Acetylene and benzene both reduce to the empirical formula CH, yet their molecular formulas — C2H2 and C6H6 — are wildly different compounds. This is exactly the kind of pair examiners like to pull out to test whether "same empirical formula" has quietly been mistaken for "same compound" in your head.
| Substance | Empirical | Molecular |
|---|---|---|
| Acetylene | CH | C₂H₂ |
| Benzene | CH | C₆H₆ |
| Formaldehyde | CH₂O | CH₂O |
| Glucose | CH₂O | C₆H₁₂O₆ |
| Hydrazine | NH₂ | N₂H₄ |
| Diborane | BH₃ | B₂H₆ |
Moles C = (25.56/44) = 0.581 mol → 6.97 g C. Moles H = 2 × (10.46/18) = 1.162 mol → 1.162 g H. Mass of O = 12.78 − 6.97 − 1.16 = 4.65 g → 0.291 mol. Dividing each by the smallest (0.291): C = 2, H = 4, O = 1. Empirical formula: C2H4O.
13. Balancing Equations and Stoichiometry
A balanced equation is just conservation of mass written out atom by atom — the same count of each element must appear on both sides. The routine I'd recommend: write the skeleton equation with correct formulas, balance elements other than H and O first, balance H and O last, then do a final check that no element got knocked out of balance while you were fixing another, and confirm you've used the smallest possible whole-number coefficients.
Stoichiometric coefficients in a balanced equation carry four simultaneous meanings — they represent mole ratios, molecule ratios, mass ratios (via molar mass), and, for gases at the same conditions, volume ratios. For the reaction aA + bB → cC + dD:
Every stoichiometry calculation ultimately follows the same three-step path: convert the mass of your starting substance to moles using its molar mass, use the mole ratio from the balanced equation to convert to moles of the substance you actually want, then convert that back to grams (or whatever unit the question asks for) using the target substance's molar mass.
Moles of P4O10 = 852/284 = 3 mol. Since 1 mol P4O10 needs 6 mol CaO, 3 mol needs 18 mol CaO. Mass of CaO = 18 × 56 = 1008 g.
14. Limiting Reagent, Theoretical and Percent Yield
When reactants aren't mixed in the exact ratio the balanced equation demands, one of them runs out first and caps how much product can form — that's the limiting reagent. The other reactant, present in excess, will have leftover material once the reaction stops.
1. Calculate the amount of product each reactant could theoretically form on its own.
2. Whichever reactant gives the smaller amount of product is limiting.
3. The amount of product actually formed is dictated by the limiting reagent alone.
4. To find leftover excess reactant, calculate how much of it the limiting reagent actually consumed, then subtract from the starting amount.
The FeCl3 needed to consume all the AgNO3: 18.0 g AgNO3 × (1 mol/169.87 g) × (1 mol FeCl3/3 mol AgNO3) × (162.21 g/mol) = 5.73 g. Since 32.4 g FeCl3 is available — far more than 5.73 g needed — FeCl3 is in excess and AgNO3 is limiting. Leftover FeCl3 = 32.4 − 5.73 = 26.7 g.
Real reactions rarely hit the calculated maximum. Side reactions, reversibility, and mechanical losses during transfer between vessels all eat into the yield. The theoretical yield is what the balanced equation predicts; the actual yield is what you measure in the lab; percent yield compares the two.
15. Principle of Atom Conservation (POAC)
POAC is, in my experience, one of the most underused shortcuts students have available — it states simply that moles of atoms of a given element are conserved through a reaction sequence, provided the reactants convert completely to products. What makes it genuinely useful is that you don't need a balanced equation at all, and you don't even need every intermediate step spelled out.
(a) No need to balance the equation.
(b) You don't need the reaction to proceed through a single clean, complete step — multi-step or "several steps" transitions are fine as long as the element in question ends up fully converted.
Apply POAC on sulphur, since every S atom ends up in BaSO4: moles of S before = moles of S after. 2 = 1 × nBaSO₄, so nBaSO₄ = 2 mol.
POAC on carbon: 6 × 1 = 1 × nCO₂ ⟹ nCO₂ = 6. POAC on hydrogen: 12 × 1 = 2 × nH₂O ⟹ nH₂O = 6. POAC on oxygen: 6×1 + 2×nO₂ = 2×nCO₂ + 1×nH₂O ⟹ 6 + 2nO₂ = 12 + 6 ⟹ nO₂ = 6.
16. Concentration Terms
Reactions in the lab almost always happen in solution, so expressing "how much solute is dissolved in how much solvent" precisely matters as much as the stoichiometry itself. A solution has a solute (the thing that dissolves, or — if no state change occurs — whichever component is present in smaller quantity) and a solvent (everything else).
Mass Percent (w/w)
Volume Percent (V/V)
Note the volumes of two liquids aren't strictly additive on mixing, so you can't just assume 70 mL alcohol + 30 mL water gives exactly 100 mL of solution.
Mass by Volume (w/V) Percent
Mole Fraction
Mole fraction is dimensionless, always between 0 and 1, and the mole fractions of all components in a solution sum to exactly 1.
Molarity
Molarity is the workhorse concentration unit in the lab because volumes are so much easier to measure than masses. It's temperature-dependent, though, since solution volume itself changes slightly with temperature.
Molality
Unlike molarity, molality is defined using the mass of solvent, not the volume of solution — which makes it temperature-independent, a distinction examiners test surprisingly often.
Strength of a Solution
Dilution: M1V1 = M2V2
Mixing solutions of the same solute: M1V1 + M2V2 + M3V3 = MTotal(V1+V2+V3)
Molality–molarity link: 1/m = d/M − MB/1000, where d is solution density and MB is the solute's molar mass.
Total solution mass = 1120 g. Volume = 1120/1.15 = 973.9 mL. Moles of urea = 120/60 = 2. Molarity = 2 / 0.9739 L = 2.05 M.
17. Gravimetric Analysis
Gravimetric analysis is a quantitative technique that determines the amount of an analyte by isolating it (or a compound of known composition derived from it) and weighing it in pure form. Two methods dominate this section of the syllabus.
Precipitation method: the substance is dissolved and converted to an insoluble precipitate using a precipitating agent — silver nitrate, potassium dichromate, ammonium molybdate, or organic agents like dimethylglyoxime (DMG) and anthranilic acid are typical choices. The precipitate must be filtered, dried, freed of contaminating impurities, converted to a compound of known stoichiometry, and weighed. A good precipitating agent forms its precipitate quickly, is highly specific to the target compound (so it doesn't drag other impurities down with it), stays chemically stable, and has low solubility in the solvent used. Estimating nickel as a Ni–DMG complex, or lead as PbCrO4, are classic examples.
Volatilization method: the sample is heated or dried to separate a volatile component from a non-volatile residue. The loss in mass measures how much volatile material was originally present. This is the standard approach for determining carbonate content in ores — heat a known mass of ore, drive off CO2, and the mass difference between starting ore and the metal-oxide residue tells you how much carbonate was there.
BaO doesn't decompose on heating; only MCO3 → MO + CO2 does. Mass loss = 4.08 − 3.64 = 0.44 g CO2 = 0.01 mol. So moles of MCO3 = moles of MO = 0.01 mol, i.e. 0.02 gram-equivalents. Milliequivalents of HCl used by the oxides = (1 × 100) − (2.5 × 16) = 60. Setting up the equivalence with x as the atomic weight of M and solving the resulting equation gives x = 40 u — the metal is calcium.
18. Common Mistakes and Exam Tips
- 22.4 L/mol only applies at STP — not NTP, not "room temperature." Check the conditions stated before dividing.
- Molarity vs molality — molarity uses solution volume (temperature-dependent); molality uses solvent mass (temperature-independent). Confusing the two costs easy marks.
- Empirical ≠ molecular formula — always check whether the question wants the simplest ratio or the true formula, and whether a molecular mass has been given to let you calculate n.
- Significant figures in addition vs multiplication — decimal places rule for +/−, total sig figs rule for ×/÷. These are genuinely different rules and mixing them up is one of the most common numerical slips.
- Limiting reagent — always calculate product from both reactants independently; whichever gives less product is limiting. Don't just eyeball which mass is "smaller," since molar masses differ.
- POAC vs stoichiometry — use POAC only when the equation is unbalanced or unspecified; use stoichiometry whenever a balanced equation is actually available.
- Volumes of liquids on mixing are not strictly additive — relevant for volume percent problems.
19. Practice MCQs (30+ with Answers)
(a) gram (b) mole (c) litre (d) kelvin
Answer: (b) mole
(a) 2 (b) 3 (c) 4 (d) 5
Answer: (c) 4 — the leading zeros don't count, but the trailing zero after the decimal does.
(a) 22.4 L (b) 24.46 L (c) 11.2 L (d) 1 L
Answer: (a) 22.4 L
(a) 6.023 × 10²² (b) 6.023 × 10²³ (c) 6.023 × 10²¹ (d) 1.2 × 10²³
Answer: (a) 4/40 = 0.1 mol × 6.023×10²³ = 6.023×10²²
(a) Conservation of mass (b) Definite proportions (c) Multiple proportions (d) Gay-Lussac's law
Answer: (b) Law of definite proportions
(a) CH₂O (b) C₂H₄O₂ (c) CH₃O (d) C₃H₆O₃
Answer: (a) CH₂O — moles C:H:O = 3.33:6.7:3.33 = 1:2:1
(a) Mass (b) Volume (c) Density (d) Heat capacity
Answer: (c) Density
(a) 0°C (b) −100°C (c) −273.15°C (d) 273.15°C
Answer: (c) −273.15°C
(a) 120 (b) 142 (c) 138 (d) 148
Answer: (b) 46 + 32 + 64 = 142 g/mol
(a) 2 g (b) 4 g (c) 8 g (d) 10 g
Answer: (b) 0.2 × 0.5 × 40 = 4 g
(a) N₂ (b) H₂ (c) NH₃ (d) No limiting reagent
Answer: (b) H₂ — needs 6 mol H₂ to fully use 2 mol N₂, only 3 mol available
(a) 60% (b) 70% (c) 80% (d) 90%
Answer: (c) (20/25) × 100 = 80%
(a) 11.2 L of CO₂ (b) 22 g of CO₂ (c) 0.2 mol of CO₂ (d) 6.023 × 10²² molecules of CO₂
Answer: (c) 0.2 mol is the largest — (a) = 0.5 mol, (b) = 0.5 mol, (d) = 0.1 mol
(a) Pressure only (b) Temperature (c) Neither pressure nor temperature (d) Only the identity of solute
Answer: (b) Temperature, since solution volume changes with it
(a) 2 (b) 3 (c) 4 (d) 6
Answer: (b) 96/32 = 3
(a) A balanced equation is given (b) The equation is unbalanced or unspecified (c) Only gases are involved (d) Only ionic reactions are involved
Answer: (b)
(a) 1 L of solution (b) 1 kg of solvent (c) 1000 mL of solution (d) 1 kg of solution
Answer: (b) 1 kg of solvent
(a) CO and CO₂ (b) Acetylene and benzene (c) NaCl and KCl (d) H₂O and H₂O₂
Answer: (b) Both reduce to CH
(a) equal masses (b) equal moles (c) equal number of molecules (d) equal densities
Answer: (c) equal number of molecules (equivalent in practice to equal moles)
(a) negligible/near zero (b) 1 g (c) 96,485 g (d) 6.023 × 10²³ g
Answer: (a) — electron mass is so small it's essentially negligible on the gram scale
(a) 1.22 g/mL (b) 1.45 g/mL (c) 1.64 g/mL (d) 1.88 g/mL
Answer: (a) 1.22 g/mL
(a) Molarity (b) Molality (c) Normality (d) Mass % (w/V)
Answer: (b) Molality
(a) 8 (b) 16 (c) 4 (d) 32
Answer: (b) Molecular mass = 2 × vapour density = 16
(a) 1 (b) 2 (c) 3 (d) 6
Answer: (c) 3, by 1:1 stoichiometry
(a) Brass (b) Air (c) Salt water (d) Muddy water
Answer: (d) Muddy water, with visibly distinct solid and liquid phases
(a) 0.1 (b) 0.2 (c) 0.25 (d) 0.8
Answer: (b) 2/(2+8) = 0.2
(a) 18.03 (b) 18.04 (c) 18.035 (d) 18.05
Answer: (b) 18.04 — since the digit before the dropped 5 is odd (3), it rounds up to even (4)
(a) Newton (b) Joule (c) Candela (d) Pascal
Answer: (c) Candela
(a) 2 M (b) 3 M (c) 4 M (d) 6 M
Answer: (b) (2×0.5 + 4×0.5)/1.0 = 3 M
(a) Metal ion concentration by colour (b) Carbonate content in ores (c) pH of a solution (d) Molecular structure
Answer: (b) Carbonate content in ores
(a) AgNO₃ (b) BaCl₂ (c) Dimethylglyoxime (DMG) (d) K₂Cr₂O₇
Answer: (c) Dimethylglyoxime (DMG)
(a) 1 (b) 2 (c) 3 (d) 6
Answer: (c) 3
(a) Mg (b) I₂ (c) MgI₂ (d) Neither
Answer: (b) I₂ — 4/254 = 0.016 mol vs Mg's 4/24 = 0.167 mol, and the ratio required is 1:1
(a) 1 in = 2.54 cm (b) 1 in = 1 cm (c) 1 in = 0.254 cm (d) 1 in = 25.4 cm
Answer: (a) 1 in = 2.54 cm
(a) 11.1% (b) 88.8% (c) 50% (d) 66.7%
Answer: (b) 16/18 × 100 ≈ 88.8%
