Powered by Vardaan CometSome Basic Concepts of Chemistry
The complete, exam-ready chapter — every concept, key formula, solved example, and a fully worked solution to each practice question. Tap any question to reveal its step-by-step solution.
Importance of Chemistry
Chemistry is the branch of science that studies the composition, structure, properties, and changes of matter. It is often called the central science because it links physics with biology, geology, and medicine.
- Materials & industry: polymers, alloys, ceramics, semiconductors, cement, glass, synthetic fibres.
- Food & agriculture: fertilisers such as urea and ammonium phosphate, pesticides, food preservatives.
- Health & medicine: life-saving drugs — analgesics, antibiotics ($\ce{penicillin}$), antihistamines.
- Energy & environment: fuels, batteries, catalytic converters; understanding CFCs and ozone-layer depletion.
Learn the phrase “central science” plus one named application from two different fields (e.g., a fertiliser and a life-saving drug). CFCs → ozone depletion is the standard example of chemistry solving an environmental problem.
Nature of Matter
Matter is anything that has mass and occupies space (volume). It is classified two independent ways — by physical state and by chemical composition.
2.1 Physical States of Matter
| State | Shape | Volume | Compressibility | Arrangement |
|---|---|---|---|---|
| Solid | Definite | Definite | Negligible | Closely packed, fixed positions |
| Liquid | Not definite | Definite | Very slight | Close but mobile |
| Gas | Not definite | Not definite | High | Far apart, random motion |
These states are interconvertible by changing temperature and pressure: $\text{Solid} \rightleftharpoons \text{Liquid} \rightleftharpoons \text{Gas}$. Such changes are physical — no new substance is formed.
2.2 Chemical Classification
Matter → Pure substances (Elements, Compounds) and Mixtures (Homogeneous, Heterogeneous).
- Element: one kind of atom only; cannot be broken down chemically (e.g., $\ce{Na}$, $\ce{O2}$, $\ce{Fe}$). About 118 are known.
- Compound: two or more elements combined chemically in a fixed ratio by mass; new properties (e.g., $\ce{H2O}$, $\ce{CO2}$, $\ce{NaCl}$).
- Mixture: two or more substances in any ratio; components keep their own properties and are separable physically.
| Feature | Mixture | Compound |
|---|---|---|
| Composition | Variable (any ratio) | Fixed ratio by mass |
| Nature of change | Physical | Chemical |
| Properties | Components retain their own | Entirely new |
| Separation | Physical methods | Chemical methods only |
| Example | Air, brass, sea water | Water, $\ce{CO2}$, salt |
“Homogeneous” ≠ “pure”. A salt solution is homogeneous but is still a mixture. Air is a mixture; water is a compound; diamond and ozone are elements.
Q1Classify each as element, compound or mixture: (a) sea water (b) sodium bicarbonate (c) diamond (d) 22-carat gold (e) air (f) ozone.▾
(a) Sea water — homogeneous mixture (water + dissolved salts, variable amount).
(b) Sodium bicarbonate $\ce{NaHCO3}$ — a compound (fixed ratio of elements).
(c) Diamond — an element (pure carbon).
(d) 22-carat gold — a mixture (gold alloyed with copper/silver; only 24-carat is ~pure).
(e) Air — a homogeneous mixture of gases.
(f) Ozone $\ce{O3}$ — an element (only oxygen atoms; an allotrope of oxygen).
Q2Give two differences between a compound and a mixture with one example each.▾
1. Composition: A compound has a fixed ratio of elements by mass (water is always $2\,\text{H}:16\,\text{O}$); a mixture has a variable ratio (sugar solution can be dilute or concentrated).
2. Separation: A compound can be split only by chemical means (electrolysis of $\ce{H2O}$); a mixture is separated by physical means (evaporating a salt solution).
Q3Which separation technique for: (a) common salt from sea water (b) two miscible liquids of close boiling points (c) iron filings from sulphur (d) camphor from sand?▾
(a) Evaporation / crystallisation — evaporate the water; salt crystals remain.
(b) Fractional distillation — a fractionating column separates liquids whose boiling points are close.
(c) Magnetic separation — iron is attracted by a magnet; sulphur is not.
(d) Sublimation — camphor sublimes on gentle heating and re-solidifies, leaving sand behind.
Q4Why is brass classified as a homogeneous mixture and not a compound?▾
Brass is an alloy of copper and zinc. Its composition is variable (the Cu:Zn ratio can change), the metals retain their metallic properties, and no fixed ratio or new compound with distinct properties is formed. Because the constituents are uniformly blended at the atomic scale it looks uniform (homogeneous), but variable composition makes it a mixture, not a compound.
Properties of Matter & Measurement
Physical properties (colour, density, melting point) are observed without changing the substance; chemical properties (combustibility, reactivity with acids) require a chemical change.
3.1 SI Base Units
| Quantity | Symbol | SI unit | Unit symbol |
|---|---|---|---|
| Length | $l$ | metre | m |
| Mass | $m$ | kilogram | kg |
| Time | $t$ | second | s |
| Electric current | $I$ | ampere | A |
| Temperature | $T$ | kelvin | K |
| Amount of substance | $n$ | mole | mol |
| Luminous intensity | $I_v$ | candela | cd |
Common prefixes: $\text{k}=10^{3}$, $\text{d}=10^{-1}$, $\text{c}=10^{-2}$, $\text{m}=10^{-3}$, $\mu=10^{-6}$, $\text{n}=10^{-9}$, $\text{p}=10^{-12}$, $\text{M}=10^{6}$, $\text{G}=10^{9}$.
3.2 Mass, Volume, Density, Temperature
Mass (amount of matter, constant everywhere) differs from weight ($W = mg$, varies with gravity).
Q. Density of mercury is $13.6\ \text{g cm}^{-3}$. Express in SI units.
$13.6\ \text{g cm}^{-3} = 13.6 \times \dfrac{10^{-3}\ \text{kg}}{10^{-6}\ \text{m}^3} = \class{ans}{1.36\times10^{4}\ \text{kg m}^{-3}}$
The Kelvin scale has no negative values. $-40\,^{\circ}\text{C} = -40\,^{\circ}\text{F}$ (favourite MCQ). Always convert to kelvin before gas-law calculations.
Q1Convert: (a) 5 ft 2 inch to cm (1 inch = 2.54 cm) (b) 2.5 L to m³ (c) 3.2 g cm⁻³ to kg m⁻³.▾
(a) $5\ \text{ft}\,2\ \text{in} = (5\times12)+2 = 62\ \text{in}$. Then $62 \times 2.54 = 157.48\ \text{cm}$.
(b) $1\ \text{L} = 10^{-3}\ \text{m}^3 \Rightarrow 2.5\ \text{L} = 2.5\times10^{-3}\ \text{m}^3$.
(c) $3.2\ \text{g cm}^{-3} \times 1000 = 3200\ \text{kg m}^{-3}$.
Q2The density of a gas is 0.001 g cm⁻³. Express it in kg m⁻³.▾
Multiply by 1000: $0.001 \times 1000 = 1$.
Q3Convert 100°F to °C and K.▾
$^{\circ}C = \tfrac{5}{9}(100-32) = \tfrac{5}{9}\times 68 = 37.78\,^{\circ}\text{C}$.
$K = 37.78 + 273.15 = 310.93\ \text{K}$.
Q4A metal piece has mass 210 g and volume 30 cm³. Find its density in SI units.▾
$d = \dfrac{210}{30} = 7\ \text{g cm}^{-3}$.
In SI: $7 \times 1000 = 7000\ \text{kg m}^{-3}$.
Q5At what temperature are the Celsius and Fahrenheit readings numerically equal?▾
Let the common reading be $x$. Then $x = \tfrac{9}{5}x + 32$.
$x - \tfrac{9}{5}x = 32 \Rightarrow -\tfrac{4}{5}x = 32 \Rightarrow x = -40$.
Uncertainty in Measurement
4.1 Scientific Notation
A number is written as $N \times 10^{n}$ with $1 \le N < 10$. E.g. $232.508 = 2.32508\times10^{2}$ and $0.00016 = 1.6\times10^{-4}$.
For addition/subtraction, first make the exponents equal.
4.2 Precision vs Accuracy
Accuracy = closeness to the true value. Precision = closeness of repeated measurements to one another. A miscalibrated balance can be precise yet inaccurate.
4.3 Significant Figures
- All non-zero digits are significant (285 → 3 s.f.).
- Captive zeros (between non-zeros) count (2005 → 4 s.f.).
- Leading zeros never count (0.0025 → 2 s.f.).
- Trailing zeros count only with a decimal point (0.200 → 3; but 100 → 1 s.f., ambiguous — write $1.00\times10^{2}$).
- Exact numbers (counted objects, defined constants) have infinite s.f.
Add / Subtract: answer keeps the fewest decimal places. Multiply / Divide: answer keeps the fewest significant figures.
Rounding a trailing 5: round the preceding digit to the nearest even number ($6.35\to6.4$, $6.25\to6.2$).
$12.11 + 18.0 + 1.012 = 31.122 \to \class{ans}{31.1}$ (18.0 has 1 decimal place).
$2.5 \times 1.25 = 3.125 \to \class{ans}{3.1}$ (2.5 has 2 s.f.).
4.4 Dimensional Analysis (Factor-Label Method)
Multiply by unit factors (fractions equal to 1) arranged so unwanted units cancel.
Q. Convert 3.0 ft to metres (1 ft = 12 in, 1 in = 2.54 cm).
Q1Count significant figures: (a) 0.0048 (b) 500 (c) 0.05000 (d) 6.023 × 10²³ (e) 2.0034 (f) 8000.0.▾
(a) leading zeros don’t count → 2 (b) trailing zeros, no decimal → 1 (ambiguous) (c) 5, 0, 0, 0 after the leading zeros; decimal point present → 4.
(d) only the mantissa 6.023 counts → 4 (e) captive zeros count → 5 (f) decimal point makes all trailing zeros count → 5.
Q2Round to three significant figures: (a) 12.696 (b) 16.351 (c) 6.545 (d) 0.0025489.▾
(a) 4th digit 9 > 5 → 12.7. (b) 5 followed by 1 (> 5 overall) → 16.4.
(c) exact trailing 5 → round 4 to nearest even → 6.54. (d) 0.00254|89 → 8 > 5 → 0.00255.
Q3Evaluate to correct s.f.: (a) 0.02856 × 298.15 × 0.112 (b) 5.5 + 6.6 + 7.7 (c) 34.0 − 0.212.▾
(a) $0.02856\times298.15 = 8.5152$; $\times\,0.112 = 0.95371$. Least s.f. = 3 (from 0.112) → 0.954.
(b) $5.5+6.6+7.7 = 19.8$ — all have 1 d.p., answer keeps 1 d.p. → 19.8.
(c) $34.0-0.212 = 33.788$; least decimal places = 1 (34.0) → 33.8.
Q4Express in scientific notation with correct s.f.: (a) 0.0000345 (b) 4 500 000 (3 s.f.).▾
(a) Move the decimal 5 places right: $3.45\times10^{-5}$ (3 s.f. preserved).
(b) To show exactly 3 s.f., write $4.50\times10^{6}$ — the trailing zero in the mantissa is now unambiguous.
Q1Convert 25.0 kg into pounds (1 kg = 2.205 lb).▾
Q2The speed of light is 3.0 × 10⁸ m s⁻¹. Express it in km h⁻¹.▾
Q3How many seconds are there in 2 days?▾
Q4Convert 1 atm = 760 mmHg into cm and m of mercury.▾
$760\ \text{mm}\times\dfrac{1\ \text{cm}}{10\ \text{mm}} = 76\ \text{cm}$; and $76\ \text{cm}\times\dfrac{1\ \text{m}}{100\ \text{cm}} = 0.76\ \text{m}$.
Laws of Chemical Combination
Matter can neither be created nor destroyed in a chemical reaction. Total mass of reactants = total mass of products. E.g. $\ce{C + O2 -> CO2}$: $12 + 32 = 44$ g.
A given compound always contains the same elements in the same proportion by mass, whatever its source. Water is always $\text{H}:\text{O} = 1:8$ by mass; $\ce{CO2}$ is always $\text{C}:\text{O} = 3:8$.
When two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in a simple whole-number ratio. In $\ce{CO}$ and $\ce{CO2}$, 12 g C combines with 16 g and 32 g O → $16:32 = 1:2$.
Masses of two elements that separately combine with a fixed mass of a third are in the same (or a simple multiple of the) ratio in which they combine with each other.
Reacting gas volumes bear simple whole-number ratios at the same T and P. $\ce{N2 + 3H2 -> 2NH3}$ reacts as $1:3:2$ by volume.
Equal volumes of all gases at the same T and P contain equal numbers of molecules. This distinguished atoms from molecules and explained Gay-Lussac’s law. Consequence: $M = 2\times\text{vapour density}$.
Q1Using 12 g C, show that the oxygen masses 16 g (in CO) and 32 g (in CO₂) follow the law of multiple proportions.▾
Fix the carbon mass at 12 g in both compounds. Oxygen combining: CO → 16 g; $\ce{CO2}$ → 32 g.
Ratio of oxygen masses $= 16:32 = 1:2$ — a simple whole-number ratio.
Q2Pure water always contains 11.1% hydrogen and 88.9% oxygen by mass. Which law does this illustrate?▾
The composition is fixed regardless of the source (rain, river, lab synthesis). H : O $= 11.1 : 88.9 = 1 : 8$ always.
Q31.0 g of a black copper oxide gives 0.888 g Cu; 1.0 g of a red copper oxide gives 0.798 g Cu. Show the law of multiple proportions is obeyed.▾
Oxide A: Cu = 0.888 g, O = $1.0 - 0.888 = 0.112$ g → O per g Cu $= 0.112/0.888 = 0.126$.
Oxide B: Cu = 0.798 g, O = $0.202$ g → O per g Cu $= 0.202/0.798 = 0.253$.
Ratio of O masses per fixed Cu $= 0.126 : 0.253 \approx 1:2$.
Q42.0 g H combines with 16 g O (water) and with 6 g C (methane). C and O combine as 3 : 8 in CO₂. Verify the law of reciprocal proportions.▾
Fix H = 2.0 g (the third element). Combining masses: O = 16 g, C = 6 g.
Ratio C : O (via hydrogen) $= 6 : 16 = 3 : 8$.
Direct combination in $\ce{CO2}$: C : O $= 12 : 32 = 3 : 8$.
Q5State Avogadro's law and deduce the relation between vapour density and molar mass of a gas.▾
Statement: Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
Vapour density $= \dfrac{\text{mass of }V\text{ of gas}}{\text{mass of }V\text{ of } \ce{H2}}$. Equal volumes hold equal numbers ($n$) of molecules, so this equals $\dfrac{n\times M}{n\times M_{\ce{H2}}} = \dfrac{M}{2}$.
Dalton’s Atomic Theory (1808)
- Matter consists of tiny, indivisible particles called atoms.
- All atoms of an element are identical; atoms of different elements differ in mass and properties.
- Compounds form when atoms combine in a fixed, simple whole-number ratio.
- Chemical reactions only rearrange atoms — none are created or destroyed.
The theory explains the laws: postulate 4 → conservation of mass; postulate 3 → definite proportions; whole-number ratios → multiple proportions.
Limitations
- Atoms are divisible — electrons, protons, neutrons were discovered later.
- Isotopes exist: same element, different mass ($\ce{^1H, ^2H, ^3H}$).
- Isobars exist: different elements, same mass number ($\ce{^{40}Ar}$, $\ce{^{40}Ca}$).
- It cannot explain Gay-Lussac’s volume ratios or the existence of allotropes.
Atomic & Molecular Masses
Modern atomic masses use the carbon-12 standard: one $\ce{^{12}C}$ atom is exactly $12$ u.
7.1 Average Atomic Mass
Elements are isotope mixtures, so we take a weighted average over fractional abundances:
$\ce{^{35}Cl}$ (34.97 u, 75.77%) and $\ce{^{37}Cl}$ (36.97 u, 24.23%):
7.2 Molecular Mass & Formula Mass
Molecular mass = sum of atomic masses in a molecule (covalent species). Formula mass = the same idea for one formula unit of an ionic solid (e.g. $\ce{NaCl}$), which has no discrete molecule.
Glucose: $\ce{C6H12O6} = 6(12)+12(1)+6(16) = \class{ans}{180\ \text{u}}$. $\ce{NaCl} = 23+35.5 = \class{ans}{58.5\ \text{u}}$.
Q1Neon isotopes: ²⁰Ne (19.99 u, 90.9%), ²¹Ne (20.99 u, 0.3%), ²²Ne (21.99 u, 8.8%). Find its average atomic mass.▾
$= \dfrac{1817.1 + 6.3 + 193.5}{100} = \dfrac{2016.9}{100}$.
Q2Copper: ⁶³Cu (62.93 u) and ⁶⁵Cu (64.93 u); average atomic mass 63.55 u. Find the % abundance of each isotope.▾
Let $\ce{^{63}Cu}$ be $x$ fraction, $\ce{^{65}Cu}$ be $(1-x)$.
$64.93 - 2.00x = 63.55 \Rightarrow 2.00x = 1.38 \Rightarrow x = 0.69$.
Q3Calculate the molecular/formula mass of: (a) H₂SO₄ (b) CaCO₃ (c) (NH₄)₂SO₄ (d) CuSO₄·5H₂O.▾
(a) $\ce{H2SO4}$: $2(1)+32+4(16)=2+32+64=98$ u.
(b) $\ce{CaCO3}$: $40+12+48=100$ u.
(c) $\ce{(NH4)2SO4}$: $2(14+4)+32+64 = 36+96 = 132$ u.
(d) $\ce{CuSO4.5H2O}$: $63.5+32+64+5(18) = 159.5+90 = 249.5$ u.
Q4The atomic mass of an element is 35.5 u. Express the mass of one atom in grams.▾
Mass of one atom $= 35.5\ \text{u} \times 1.66\times10^{-24}\ \text{g u}^{-1}$.
Equivalently, $\dfrac{35.5}{6.022\times10^{23}}$ g.
Mole Concept & Molar Mass
One mole is the amount of substance containing as many elementary entities as there are atoms in exactly 12 g of carbon-12 — that number is Avogadro’s number, $N_A = 6.022\times10^{23}\ \text{mol}^{-1}$.
Molar volume at STP: $22.7$ L at $273.15$ K & 1 bar (current NCERT); $22.4$ L at $273.15$ K & 1 atm (older convention, still standard in most JEE numericals). Use whichever the question implies.
Molar mass (g mol⁻¹) is numerically equal to the atomic/molecular/formula mass in u. E.g. $\ce{H2O}$: 18 u → 18 g mol⁻¹.
Q. For 4.4 g of $\ce{CO2}$, find (a) moles (b) molecules (c) total atoms (d) volume at STP (22.4 L).
(a) $n = 4.4/44 = 0.1$ mol (b) $0.1\times6.022\times10^{23} = 6.022\times10^{22}$ molecules
(c) each $\ce{CO2}$ has 3 atoms → $1.807\times10^{23}$ atoms (d) $V = 0.1\times22.4 = \class{ans}{2.24\ \text{L}}$
Q1Calculate the moles in: (a) 11.5 g Na (b) 9.8 g H₂SO₄ (c) 5.6 L N₂ at STP (22.4 L).▾
(a) $n = 11.5/23 = 0.5$ mol. (b) $n = 9.8/98 = 0.1$ mol. (c) $n = 5.6/22.4 = 0.25$ mol.
Q2How many atoms are present in 12 g of carbon? In 0.5 mol of N₂?▾
12 g C = 1 mol of C atoms $= 6.022\times10^{23}$ atoms.
0.5 mol $\ce{N2}$ = $0.5\times6.022\times10^{23} = 3.011\times10^{23}$ molecules; each has 2 N atoms → $6.022\times10^{23}$ atoms.
Q3Calculate the mass of 3.011 × 10²³ molecules of water.▾
$n = \dfrac{3.011\times10^{23}}{6.022\times10^{23}} = 0.5$ mol.
Mass $= 0.5\times18 = 9.0$ g.
Q4What is the mass of one atom of nitrogen in grams?▾
1 mol of N atoms (14 g) contains $6.022\times10^{23}$ atoms.
Q5A cylinder contains 6.4 kg of oxygen. Calculate the moles and molecules of O₂.▾
$6.4\ \text{kg} = 6400$ g; $n = 6400/32 = 200$ mol.
Molecules $= 200\times6.022\times10^{23} = 1.204\times10^{26}$.
Q6Calculate the number of electrons in 1.6 g of CH₄ (each CH₄ molecule has 10 electrons).▾
$n(\ce{CH4}) = 1.6/16 = 0.1$ mol → molecules $= 0.1\,N_A$.
Electrons $= 10\times0.1\,N_A = 1\,N_A = 6.022\times10^{23}$.
Q7How many significant figures are in 6.022 × 10²³, and what is the molar volume of an ideal gas at 273.15 K and 1 bar?▾
Only the mantissa counts: 6, 0, 2, 2 → 4 significant figures.
At 273.15 K and 1 bar (current STP), $V_m = 22.711\ \text{L} \approx$ 22.7 L. (At 1 atm it would be 22.4 L.)
Percentage Composition, Empirical & Molecular Formula
Empirical formula = simplest whole-number atom ratio. Molecular formula = actual atom count.
Method (5 steps)
- Write the mass % of each element (take a 100 g sample).
- Divide by atomic mass → moles of each element.
- Divide all by the smallest → simplest ratio.
- If not whole, multiply all by 2, 3 or 4 (… .5 → ×2; … .33 → ×3; … .25 → ×4).
- Write subscripts; scale to molecular formula using $n$.
Q. 40.0% C, 6.7% H, 53.3% O; molar mass 180. Find both formulas.
Moles: C $=40/12=3.33$; H $=6.7$; O $=53.3/16=3.33$. Divide by 3.33 → $1:2:1$ → EF $=\ce{CH2O}$ (mass 30).
$n = 180/30 = 6$ → MF $= \class{ans}{\ce{C6H12O6}}$.
Q1An oxide of nitrogen contains 30.4% N and 69.6% O. Find its empirical formula.▾
Moles in 100 g: N $= 30.4/14 = 2.17$; O $= 69.6/16 = 4.35$.
Divide by 2.17 → N $=1$, O $=2.0$.
Q2A compound has 24.24% C, 4.04% H, 71.72% Cl; molar mass 99 g mol⁻¹. Find its molecular formula.▾
Moles: C $=24.24/12=2.02$; H $=4.04/1=4.04$; Cl $=71.72/35.5=2.02$.
Divide by 2.02 → $1 : 2 : 1$ → EF $=\ce{CH2Cl}$, mass $= 12+2+35.5 = 49.5$.
$n = 99/49.5 = 2$.
Q3An organic compound gives C 54.5%, H 9.1%, O 36.4%. Its vapour density is 44. Find the molecular formula.▾
Molar mass $= 2\times\text{V.D.} = 2\times44 = 88$ g mol⁻¹.
Moles: C $=54.5/12=4.54$; H $=9.1$; O $=36.4/16=2.28$.
Divide by 2.28 → C $=2.0$, H $=4.0$, O $=1$ → EF $=\ce{C2H4O}$ (mass 44).
$n = 88/44 = 2$.
Q4A hydrocarbon contains 85.7% C and 14.3% H; molar mass 56 g mol⁻¹. Determine its molecular formula.▾
Moles: C $= 85.7/12 = 7.14$; H $= 14.3/1 = 14.3$.
Divide by 7.14 → $1 : 2$ → EF $=\ce{CH2}$ (mass 14).
$n = 56/14 = 4$.
Q5Calculate the % of nitrogen in (a) urea CO(NH₂)₂ (b) NH₄NO₃. Which is the better N-fertiliser by mass %?▾
(a) Urea, M $= 12+16+2(14+2) = 60$; N mass $= 28$. $\%N = \dfrac{28}{60}\times100 = 46.7\%$.
(b) $\ce{NH4NO3}$, M $= 80$; N mass $= 28$. $\%N = \dfrac{28}{80}\times100 = 35\%$.
Stoichiometry & Limiting Reagent
A balanced equation speaks in molecules, moles, masses and gas volumes simultaneously. For $\ce{CH4 + 2O2 -> CO2 + 2H2O}$:
- Moles: 1 mol $\ce{CH4}$ + 2 mol $\ce{O2}$ → 1 mol $\ce{CO2}$ + 2 mol $\ce{H2O}$.
- Mass: 16 g + 64 g → 44 g + 36 g (mass conserved: 80 g each side).
- Volumes at STP: 22.4 L + 44.8 L → 22.4 L + 44.8 L (vapour).
1. Balance the equation → 2. convert given quantity to moles → 3. apply the mole ratio → 4. convert back to the asked quantity.
10.1 Limiting Reagent
The reactant consumed first; it limits the product. Identify it by computing $\dfrac{\text{moles available}}{\text{coefficient}}$ for each reactant — the smallest quotient wins. All product calculations then flow from it.
Q. 50.0 g $\ce{N2}$ + 10.0 g $\ce{H2}$; $\ce{N2 + 3H2 -> 2NH3}$. Find the limiting reagent, mass of $\ce{NH3}$, and leftover reactant.
Moles: $\ce{N2} = 50/28 = 1.79$; $\ce{H2} = 10/2 = 5.0$. Quotients: $1.79/1 = 1.79$ vs $5.0/3 = 1.67$ → $\ce{H2}$ is limiting.
$\ce{NH3} = \tfrac{2}{3}(5.0) = 3.33$ mol $= \class{ans}{56.7\ \text{g}}$. $\ce{N2}$ used $= \tfrac{5}{3} = 1.67$ mol $= 46.7$ g → left $= \class{ans}{3.3\ \text{g}}$.
Never assume the smaller mass is limiting — always compare moles ÷ coefficient. Percentage yield $= \dfrac{\text{actual}}{\text{theoretical}}\times100$, with the theoretical yield computed from the limiting reagent.
Q1Calculate the mass of water produced when 16 g of CH₄ burns completely in oxygen.▾
$\ce{CH4 + 2O2 -> CO2 + 2H2O}$. Moles $\ce{CH4} = 16/16 = 1$ mol.
Ratio $\ce{CH4}:\ce{H2O} = 1:2$ → 2 mol water $= 2\times18$.
Q2100 g of CaCO₃ is heated. What volume of CO₂ is liberated at STP (22.4 L)?▾
$\ce{CaCO3 ->[\Delta] CaO + CO2}$. Moles $\ce{CaCO3} = 100/100 = 1$ mol → 1 mol $\ce{CO2}$.
$V = 1\times22.4$ L.
Q33.0 g H₂ reacts with 29.0 g O₂ to form water. Find the limiting reagent, mass of water formed, and mass of excess reactant left.▾
$\ce{2H2 + O2 -> 2H2O}$. Moles: $\ce{H2} = 3.0/2 = 1.5$; $\ce{O2} = 29.0/32 = 0.906$.
Quotients: $\ce{H2}\to 1.5/2 = 0.75$; $\ce{O2}\to 0.906/1 = 0.906$. Smaller is $\ce{H2}$ → $\ce{H2}$ is limiting.
Water $= 1.5$ mol (1:1 with $\ce{H2}$) $= 1.5\times18 = 27$ g.
$\ce{O2}$ used $= 1.5/2 = 0.75$ mol $= 24$ g → left $= 29.0 - 24.0 = 5.0$ g.
Q42 Al + 3 Cl₂ → 2 AlCl₃. If 27 g Al reacts with 71 g Cl₂, which is limiting and how much AlCl₃ forms? (Al = 27, Cl₂ = 71, AlCl₃ = 133.5)▾
Moles: Al $= 27/27 = 1$; $\ce{Cl2} = 71/71 = 1$.
Quotients: Al $\to 1/2 = 0.50$; $\ce{Cl2} \to 1/3 = 0.33$ → $\ce{Cl2}$ limiting.
$\ce{AlCl3} = \tfrac{2}{3}\times1 = 0.667$ mol $= 0.667\times133.5$.
Q5What volume of 0.5 M H₂SO₄ neutralises 100 mL of 1.0 M NaOH? (H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O)▾
Moles NaOH $= 1.0\times0.100 = 0.10$ mol.
Ratio $\ce{H2SO4}:\ce{NaOH} = 1:2$ → acid needed $= 0.05$ mol.
$V = n/M = 0.05/0.5 = 0.10$ L.
Q66.3 g NaHCO₃ is added to 15 g of acetic-acid solution; the residue weighs 18.0 g. What mass of CO₂ escaped?▾
Law of conservation of mass: total in $= 6.3 + 15.0 = 21.3$ g.
Residue $= 18.0$ g → escaped gas $= 21.3 - 18.0$.
Concentration of Solutions
A solution = solute (less) + solvent (more). The main concentration measures:
Molarity uses volume of solution → temperature-dependent (volume expands on heating). Molality uses mass of solvent → temperature-independent, preferred for colligative properties. Mole fraction, mass % and ppm are also T-independent.
Molarity. 5.85 g NaCl in 500 mL solution: $n = 5.85/58.5 = 0.1$; $M = 0.1/0.5 = \class{ans}{0.2\ \text{M}}$.
Molality. 3.0 g urea (M = 60) in 250 g water: $m = \dfrac{0.05}{0.25} = \class{ans}{0.2\ \text{mol kg}^{-1}}$.
Concentrated acid. 98% $\ce{H2SO4}$, $d = 1.84$: $M = \dfrac{10\times1.84\times98}{98} = \class{ans}{18.4\ \text{M}}$.
Q1Calculate the molarity of a solution containing 4.0 g NaOH in 250 mL of solution. (NaOH = 40)▾
$n = 4.0/40 = 0.1$ mol; $V = 0.250$ L.
Q2Calculate the molality of 6.0 g glucose (M = 180) in 500 g of water.▾
$n = 6.0/180 = 0.0333$ mol; solvent $= 0.500$ kg.
Q320 g NaOH is dissolved to make 500 mL of solution. Calculate molarity and normality.▾
$n = 20/40 = 0.5$ mol; $M = 0.5/0.5 = 1.0$ M.
n-factor of NaOH (acidity) = 1 → $N = M\times1 = 1.0$ N.
Q4Concentrated HNO₃ is 70% by mass with density 1.42 g mL⁻¹. Calculate its molarity. (HNO₃ = 63)▾
Q5Find the mole fraction of ethanol in a solution of 46 g C₂H₅OH and 90 g water.▾
$n_{\text{EtOH}} = 46/46 = 1$ mol; $n_{\text{water}} = 90/18 = 5$ mol.
Q6What volume of water must be added to 200 mL of 0.5 M HCl to make it 0.2 M?▾
$M_1V_1 = M_2V_2$: $0.5\times200 = 0.2\times V_2 \Rightarrow V_2 = 500$ mL.
Water added $= 500 - 200 = 300$ mL (assuming additive volumes).
Q7A commercial H₂SO₄ sample is 18 M with density 1.84 g mL⁻¹. Calculate its mass % of H₂SO₄.▾
Rearrange $M = \dfrac{10\,d\,(\%)}{M_B}$: $\% = \dfrac{M\times M_B}{10\,d}$.
Q810 mL of 0.1 N HCl exactly neutralises V mL of 0.05 N NaOH. Find V.▾
At equivalence: $N_1V_1 = N_2V_2$.
Quick Revision Sheet
$1\ \text{u} = 1.66\times10^{-24}$ g · Molar mass $= 2\times$ vapour density.
$\bar m = \sum m_i f_i$ · $\%\text{element} = \dfrac{\text{mass in 1 mol}}{M}\times100$ · MF $= n\times$EF, $n = \dfrac{M}{\text{EF mass}}$.
$M = \dfrac{n}{V(\text{L})}$ · $m = \dfrac{n}{\text{kg solvent}}$ · $x_A = \dfrac{n_A}{\sum n}$ · $N = M\times n\text{-factor}$ · $M_1V_1 = M_2V_2$ · $M = \dfrac{10\,d\,\%}{M_B}$.
Conservation of mass (Lavoisier) · Definite proportions (Proust) · Multiple proportions (Dalton) · Reciprocal proportions (Richter) · Gaseous volumes (Gay-Lussac) · Equal volumes ↔ equal molecules (Avogadro).
- Molar-i-tee = Litre of solution; Molality = mass (kg) of solvent.
- Limiting reagent = smallest of (moles ÷ coefficient).
- Only molarity & normality change with temperature.
- Leading zeros never count; trailing zeros count only with a decimal point.
- Add/Sub → decimal places; Mult/Div → significant figures.
Final word from Vardaan: master the mole map and the concentration formulas, then drill every practice set until the method is automatic. This chapter is the toolkit for the entire year of physical chemistry — every mark invested here pays back many times over in Boards, NEET and JEE. 🚀