Chapters

Everything NCERT Class 11 Unit 1 (Some Basic Concepts of Chemistry) says that can be asked, plus what JEE Main adds beyond the book (marked Beyond NCERT). The short note stays for the last week; this is for a proper revision pass.

Formula sheet

Every formula in the chapter on one screen, for the last pass. Details and examples are in the sections below.

Quantity	Formula	Watch
Moles	n = m/M = N/Nₐ = V/Vₘ	Vₘ = 22.7 L (1 bar) or 22.4 L (1 atm)
Average atomic mass	Σ (isotope mass × fraction)	Fractions, not %
Mass % of element	(its mass in 1 mol / molar mass) × 100
Molecular formula	n × empirical; n = M / empirical mass
C and H from combustion	C = 12/44 × CO₂; H = 2/18 × H₂O	O by difference
Hydrocarbon burning	CₓHᵧ + (x + y/4) O₂ → x CO₂ + y/2 H₂O	Volumes = moles
Limiting reagent	smallest (moles / coefficient)	Never grams
% yield	actual / theoretical × 100	Theoretical from limiting
Mass % (w/w)	solute / solution × 100
ppm	solute / solution × 10⁶
Mole fraction	xₐ = nₐ / Σn	Σx = 1
Molarity	mol solute / L solution	Changes with T
Molality	mol solute / kg solvent	Solvent, not solution
% and density → M	M = 10 × d × % / Mₘ	d in g/mL
M → m	m = 1000 M / (1000 d − M × Mₘ)
x → m	m = 1000 x₂ / (x₁ × M₁)	M₁ = solvent
Dilution	M₁V₁ = M₂V₂
Mixing	M = ΣMV / ΣV
Neutralisation	MₐVₐ × (H⁺ each) = MᵦVᵦ × (OH⁻ each)
Temperature	K = °C + 273.15; °F = 9/5 °C + 32

0 · What JEE Main asks from this chapter

133 questions in JEE Main 2019–2026, about 1 per paper, nearly all easy. Grouped by type (counts are approximate):

Type	Share	Typical question
Stoichiometry: mass/volume from an equation, limiting reagent, mixtures	~32%	8.7 g MnO₂ + excess HCl → mass of Cl₂
Solutions and titrations: M, m, ppm, dilution, neutralisation	~25%	x mg HCl titrated by 25 mL 0.1 M Ba(OH)₂
Formula: empirical/molecular, combustion analysis, gas volumes	~18%	10 mL hydrocarbon + 55 mL O₂ → 40 mL CO₂
Counting: moles, atoms, molecules, electrons, STP volume	~11%	Order of total atoms in 2 mol C₆H₁₂, 684 g sucrose, 90.8 L H₂
% yield (often an organic reaction)	~7%	3.9 g benzene → 4.92 g nitrobenzene, % yield
Theory: significant figures, Dalton, mass vs weight, precision	~6%	Significant figures in 50000.020 × 10⁻³

Most are numerical-answer questions rounded to the nearest integer, often as "x × 10⁻¹" or "x × 10⁻²". Read the multiplier before you answer.

1 · Matter

• Matter: anything that has mass and occupies space.
• Solid: definite shape and volume; particles close, orderly, little movement.
• Liquid: definite volume, no definite shape; particles close but move around.
• Gas: neither definite volume nor shape; particles far apart, move fast.
• The three states interconvert by changing temperature and pressure.

Classification:

	What it is	Examples
Mixture	Two or more pure substances in any ratio; separable by physical methods (hand-picking, filtration, crystallisation, distillation)
→ Homogeneous	Uniform composition throughout	Sugar solution, air
→ Heterogeneous	Composition not uniform; components sometimes visible	Salt + sugar, grains with stones
Pure substance	Fixed composition
→ Element	One type of atom; particles may be atoms (Na, Cu) or molecules (H₂, N₂, O₂)
→ Compound	Atoms of different elements in a fixed ratio; separable only by chemical methods; properties differ from its elements	H₂O, NH₃, CO₂, sugar

Water shows the last point: H₂ burns with a pop and O₂ supports combustion, yet water puts out fire.

Physical property: measured without changing the substance's identity (colour, odour, m.p., b.p., density). Chemical property: needs a chemical change to observe (composition, combustibility, acidity/basicity, reactivity).

2 · Measurement and units

SI was set up in 1960 by the 11th CGPM (General Conference on Weights and Measures). The CGPM was created by the Metre Convention, signed in Paris in 1875. India's standards are kept by the National Physical Laboratory (NPL), New Delhi.

The 7 SI base quantities:

Quantity	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 (Iᵥ)	candela	cd

Everything else (speed, volume, density) is derived from these. Each base unit is now defined by fixing a constant:

• metre: speed of light, c = 299 792 458 m s⁻¹
• kilogram: Planck constant, h = 6.62607015 × 10⁻³⁴ J s
• second: caesium-133 frequency, 9 192 631 770 Hz
• ampere: elementary charge, e = 1.602176634 × 10⁻¹⁹ C
• kelvin: Boltzmann constant, k = 1.380649 × 10⁻²³ J K⁻¹
• mole: Avogadro constant, Nₐ = 6.02214076 × 10²³ mol⁻¹
• candela: luminous efficacy 683 lm W⁻¹ at 540 × 10¹² Hz

NCERT's "Reference Standard" box still describes the kilogram as the Pt-Ir cylinder at Sèvres (the standard from 1889). That is the old definition. If asked "how is the kilogram defined", use the Planck constant. The box also gives the metre's history: a Pt-Ir bar, then krypton-light wavelengths (1960), then the distance light travels in 1/299 792 458 s (1983).

Prefixes (NCERT lists 10⁻²⁴ to 10²⁴; these are the ones questions use):

Prefix	Symbol	Multiple
femto	f	10⁻¹⁵
pico	p	10⁻¹²
nano	n	10⁻⁹
micro	µ	10⁻⁶
milli	m	10⁻³
centi	c	10⁻²
deci	d	10⁻¹
deca	da	10
kilo	k	10³
mega	M	10⁶
giga	G	10⁹
tera	T	10¹²

The rest: hecto (h) 10², peta (P) 10¹⁵, exa (E) 10¹⁸, zetta (Z) 10²¹, yotta (Y) 10²⁴; atto (a) 10⁻¹⁸, zepto (z) 10⁻²¹, yocto (y) 10⁻²⁴.

Mass vs weight: mass is the amount of matter and is constant; weight is the force of gravity on it and varies from place to place. Mass is measured with an analytical balance. (A 2025 question swapped these two definitions as a trap.)

Volume: SI unit m³. The litre is not an SI unit. Measured with a graduated cylinder, burette or pipette; a volumetric flask makes up a known volume.

Density: higher density means more closely packed particles. SI unit kg m⁻³; chemists use g cm⁻³ (1 g cm⁻³ = 1000 kg m⁻³).

Temperature: water freezes at 0 °C / 32 °F and boils at 100 °C / 212 °F. Negative temperatures exist in Celsius but never in kelvin. °C and °F are equal at −40.

Dimensional analysis (unit-factor method): multiply by a fraction equal to 1, with the unit you want on top.

Example · 3 in × (2.54 cm / 1 in) = 7.62 cm. 2 L = 2000 cm³ × (1 m / 100 cm)³ = 2 × 10⁻³ m³. 2 days × 24 h/day × 60 min/h × 60 s/min = 172 800 s.

3 · Uncertainty in measurement

Scientific notation: N × 10ⁿ with 1 ≤ N < 10. To add or subtract, first bring both numbers to the same power; to multiply or divide, multiply the digit terms and add (or subtract) the exponents.

Example · 232.508 = 2.32508 × 10²; 0.00016 = 1.6 × 10⁻⁴. 6.65 × 10⁴ + 8.95 × 10³ = (6.65 + 0.895) × 10⁴ = 7.545 × 10⁴.

Significant figures are the certain digits plus one uncertain digit. The uncertainty is ±1 in the last digit unless stated. Rules:

1. All non-zero digits count: 285 has 3, 0.25 has 2.
2. Zeros before the first non-zero digit don't count; they only place the decimal point. 0.03 has 1, 0.0052 has 2.
3. Zeros between non-zero digits count: 2.005 has 4.
4. Trailing zeros count only if there is a decimal point: 0.200 has 3, 100 has 1, 100. has 3, 100.0 has 4.
5. Counted objects (2 balls, 20 eggs) and exact numbers have infinite significant figures.
6. In scientific notation every digit of N counts: 4.01 × 10² has 3, 8.256 × 10⁻³ has 4.

The power of 10 never adds or removes significant figures.

Example (2021) · 50000.020 × 10⁻³ has 8. 0.00340 has 3.

In calculations:

• Adding or subtracting: keep as many decimal places as the number with the fewest.
• Multiplying or dividing: keep as many significant figures as the number with the fewest.

Example · 12.11 + 18.0 + 1.012 = 31.122 → 31.1. 2.5 × 1.25 = 3.125 → 3.1.

Rounding off (NCERT's rule):

• Digit to drop > 5: round up (1.386 → 1.39).
• Digit to drop < 5: leave the previous digit alone (4.334 → 4.33).
• Digit to drop = 5: if the previous digit is even, leave it; if odd, add 1 (6.35 → 6.4, but 6.25 → 6.2).

Precision vs accuracy: precision is how close repeated readings are to each other; accuracy is how close they are to the true value. NCERT's example, with true value 2.00 g:

Student	Readings (g)	Verdict
A	1.95, 1.93	precise, not accurate
B	1.94, 2.05	neither
C	2.01, 1.99	precise and accurate

4 · Laws of chemical combination

1. Conservation of mass (Lavoisier, 1789): matter is neither created nor destroyed; no net change in mass in any physical or chemical change. Came from his careful combustion experiments.
2. Definite proportions, also called definite composition (Proust): a compound always contains the same elements in the same proportion by mass, whatever its source.
   • NCERT's evidence: natural and synthetic cupric carbonate both gave Cu 51.35%, C 9.74%, O 38.91%.
3. Multiple proportions (Dalton, 1803): if two elements form more than one compound, the masses of one that combine with a fixed mass of the other are in a small whole-number ratio.
   • With 2 g H, O is 16 g in H₂O and 32 g in H₂O₂ → 1 : 2.
4. Gaseous volumes (Gay-Lussac, 1808): gases combine or form in a simple ratio by volume at the same T and P. It is the law of definite proportions by volume.
   • 100 mL H₂ + 50 mL O₂ → 100 mL water vapour (2 : 1 : 2).
5. Avogadro's law (1811): equal volumes of all gases at the same T and P contain equal numbers of molecules.
   • It explains Gay-Lussac's law if H₂ and O₂ are diatomic. Avogadro was the first to separate atoms from molecules. He was ignored for about 50 years until Cannizzaro revived the idea at the first international chemistry conference, Karlsruhe, 1860.

Example (NCERT exercise 1.21) · To test multiple proportions, fix the mass of one element and compare the other. N₂ and O₂ give, per 14 g N, 16, 32, 16 and 40 g O. That is 2 : 4 : 2 : 5, so the data follow the law of multiple proportions.

Dalton's atomic theory (A New System of Chemical Philosophy, 1808):

1. Matter consists of indivisible atoms.
2. All atoms of an element are identical, including in mass; atoms of different elements differ in mass.
3. Compounds form when atoms of different elements combine in a fixed ratio.
4. Chemical reactions reorganise atoms; atoms are neither created nor destroyed.

What it could not explain: the law of gaseous volumes, and why atoms combine. Later work also showed atoms are divisible (subatomic particles) and atoms of one element can differ in mass (isotopes).

JEE asks (2020, 2024): "Which was not proposed by Dalton?" The answer was Gay-Lussac's law, put to you as if it were a postulate. Know the four above exactly.

5 · Atomic, molecular and formula mass

• Standard since 1961: carbon-12 is assigned exactly 12 u.
• 1 amu = 1/12 of the mass of one ¹²C atom = 1.66056 × 10⁻²⁴ g. "amu" is now written u (unified mass).
• Mass of one ¹²C atom = 12 g / Nₐ = 1.9927 × 10⁻²³ g (NCERT exercise 1.30).
• H atom: 1.6736 × 10⁻²⁴ g = 1.0078 u. ¹⁶O: 15.995 u.

Average atomic mass is what the periodic table lists.

Example · Carbon: 0.98892 × 12 + 0.01108 × 13.00335 + (tiny ¹⁴C) = 12.011 u. Chlorine (exercise 1.9): 0.7577 × 34.9689 + 0.2423 × 36.9659 = 35.45 u.

Example (2020, the reverse) · An average of 35.5 means ³⁵Cl : ³⁷Cl ≈ 3 : 1. Solve 35x + 37(1 − x) = 35.5, giving x = 0.75.

Molecular mass = sum of the atomic masses in the molecule. CH₄ = 12.011 + 4(1.008) = 16.043 u; H₂O = 18.02 u; glucose C₆H₁₂O₆ = 180.162 u.

Formula mass is used for ionic solids, which have no discrete molecules. In NaCl each Na⁺ is surrounded by 6 Cl⁻ and each Cl⁻ by 6 Na⁺. Formula mass of NaCl = 23.0 + 35.5 = 58.5 u.

6 · The mole

• 1 mol contains exactly 6.02214076 × 10²³ elementary entities. That number is the Avogadro constant Nₐ (mol⁻¹), also called the Avogadro number. An entity can be an atom, molecule, ion, electron or a specified group.
• The mole was added as the seventh SI base quantity. One mole of anything has the same number of entities.
• Molar mass is the mass of 1 mol in grams. It is numerically equal to the atomic, molecular or formula mass in u. H₂O is 18.02 g mol⁻¹; NaCl is 58.5 g mol⁻¹.

Arrows towards the centre divide, arrows away from it multiply. Turn whatever is given into moles, use the equation's ratio, then turn moles into what is asked.

Molar volume of an ideal gas:

Conditions	Volume of 1 mol
273.15 K, 1 bar (NCERT's STP)	22.7 L
273.15 K, 1 atm (old STP)	22.4 L
298 K, 1 atm	24.5 L
Any T, P	V = nRT/P

JEE uses both values for STP. Use whichever the question gives. If it gives none, check the options: 22.4 vs 22.7 usually separates them, and recent papers lean to 22.7. With R in L bar, the question means 1 bar.

Atoms vs molecules vs electrons. This is the most common counting trap.

• 1 mol O₂ = Nₐ molecules = 2Nₐ atoms = 16Nₐ electrons.
• 1 mol C₆H₁₂O₆ = 24Nₐ atoms.
• 16 g CH₄ = 1 mol = 5Nₐ atoms = 10Nₐ electrons.
• The same mass of different elements: fewer atoms when the atomic mass is larger. 1 g Li has more atoms than 1 g Na, which has more than 1 g Au.

Example (2026) · Order of total atoms in 2 mol cyclohexane, 684 g sucrose and 90.8 L H₂ at STP. Cyclohexane C₆H₁₂: 2 × 18 = 36 mol atoms. Sucrose C₁₂H₂₂O₁₁ (342 g mol⁻¹): 684 g = 2 mol, × 45 = 90 mol atoms. H₂: 90.8 / 22.7 = 4 mol, × 2 = 8 mol atoms. Order: sucrose > cyclohexane > H₂.

Example (2025) · Among 10⁻⁹ g each of Pb, Po, Pr and Pt, Pr (141) has the most atoms: the smallest atomic mass.

7 · Percentage composition, empirical and molecular formula

A pure sample must match these percentages, so they also check purity.

Example · H₂O: H = 2.016/18.02 = 11.18%, O = 88.79%. Ethanol C₂H₅OH (46.068): C 52.14%, H 13.13%, O 34.73%.

Empirical formula is the simplest whole-number ratio of atoms. Molecular formula gives the actual numbers.

Steps:

1. Take 100 g, so the percentages become grams.
2. Divide each by its atomic mass to get moles.
3. Divide by the smallest.
4. If you get 1.5 or 1.33, multiply all by 2 or 3 to reach whole numbers.
5. Find n from the molar mass.

Example (NCERT Problem 1.2) · H 4.07%, C 24.27%, Cl 71.65%; molar mass 98.96. Moles: H 4.04, C 2.021, Cl 2.021 → H : C : Cl = 2 : 1 : 1 → CH₂Cl (49.48). n = 98.96 / 49.48 = 2 → C₂H₄Cl₂.

Example (exercise 1.3, asked again in 2026) · Fe 69.9%, O 30.1% → Fe 1.25 mol, O 1.88 mol → 1 : 1.5 = 2 : 3 → Fe₂O₃.

Combustion analysis · Beyond NCERT

A compound of C, H (and O) burns to CO₂ and H₂O. Find each element's mass, then proceed as for an empirical formula. O is whatever mass is left.

Example (NCERT exercise 1.34) · Welding gas gives 3.38 g CO₂ and 0.690 g H₂O → C 0.0768 mol, H 0.0767 mol → CH. 10.0 L at STP weighs 11.6 g → M = 26 → C₂H₂.

Gas volumes of a hydrocarbon (eudiometry) · Beyond NCERT

Volumes are in the ratio of moles. x = CO₂ volume / hydrocarbon volume; then x + y/4 = O₂ volume / hydrocarbon volume gives y.

Example (2019) · 10 mL needs 55 mL O₂ and gives 40 mL CO₂. So x = 4 and 4 + y/4 = 5.5, giving y = 6 → C₄H₆.

Hydrates carry their water in the molar mass:

Hydrate	Molar mass (g mol⁻¹)
H₂C₂O₄·2H₂O	126
CuSO₄·5H₂O	249.5
FeSO₄·7H₂O	278
Na₂CO₃·10H₂O	286

8 · Stoichiometry

Stoichiometry (Greek stoicheion, element; metron, measure) is the calculation of the masses and volumes of reactants and products.

What a balanced equation says. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) reads as any of:

Substance	Moles · volume at STP	Mass
CH₄	1 mol · 22.7 L	16 g
O₂	2 mol · 45.4 L	64 g
CO₂	1 mol · 22.7 L	44 g
H₂O	2 mol · 45.4 L	36 g

The coefficients are stoichiometric coefficients; they count molecules and moles alike. (s), (l), (g) and (aq) mark the states.

Balancing: atoms of each element must be equal on both sides. Change coefficients, never subscripts. Balance C, then H, then O: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Another: P₄ + 5O₂ → P₄O₁₀.

The method for every stoichiometry question:

1. Write the balanced equation.
2. Convert every given amount to moles.
3. Use the coefficient ratio.
4. Convert the answer back to the unit asked.

Example (NCERT Problems 1.3, 1.4) · 16 g CH₄ = 1 mol → 2 mol H₂O = 36 g. 22 g CO₂ = 0.5 mol → 0.5 mol CH₄ (8 g). Problem 1.4 came back in 2024, three times.

Limiting reagent: the reactant used up first; it fixes how much product forms. For each reactant divide its moles by its coefficient: the smallest is limiting. Excess left = initial − amount the limiting reagent uses.

Example (NCERT Problem 1.5) · 50.0 kg N₂ = 1786 mol, 10.0 kg H₂ = 4960 mol. N₂ + 3H₂ → 2NH₃: N₂ would need 5358 mol H₂, only 4960 are there, so H₂ limits. NH₃ = 4960 × 2/3 = 3307 mol = 56.1 kg.

Example (2026) · A + 2B → AB₂ with 36 g A (0.6 mol) and 56 g B (0.7 mol). B limits (0.7/2 = 0.35 < 0.6). Forms 0.35 mol AB₂ = 77 g; A left = 0.25 mol = 15 g.

% yield and % purity: theoretical yield comes from the limiting reagent; with an impure reactant, use only the pure part.

Example · 3.9 g benzene (0.05 mol) → theoretical nitrobenzene 0.05 × 123 = 6.15 g; actual 4.92 g → 80%. 150 kg limestone at 75% purity = 112.5 kg CaCO₃ → 63 kg CaO.

Mixtures (two unknowns): set up two equations, one for total mass and one for the product.

Example (2024) · CaCO₃ + MgCO₃ = 2.21 g, heated to CaO + MgO = 1.152 g. With a and b moles: 100a + 84b = 2.21 and 56a + 40b = 1.152.

Multi-step reactions (POAC, the principle of atom conservation) · Beyond NCERT

Atoms of an element are conserved across all the steps, so you can link the first reactant to the last product without the intermediate equations. For example, all the C in a compound ends up in CO₂, so moles of C in the sample = moles of CO₂.

Gas reactions at constant T and V: partial pressures act like moles.

Example (2021) · 2SO₂ + O₂ → 2SO₃ from 250 mbar SO₂ and 750 mbar O₂. SO₂ is used up, consuming 125 mbar O₂ and making 250 mbar SO₃. Total = 0 + 625 + 250 = 875 mbar.

9 · Concentration of solutions

Measure	Definition	Changes with T?
Mass % (w/w)	solute mass / solution mass × 100	No
ppm (by mass)	solute mass / solution mass × 10⁶	No
Mole fraction	xₐ = nₐ / (nₐ + nᵦ); Σx = 1	No
Molarity M	mol solute / litre of solution	Yes (volume changes)
Molality m	mol solute / kg of solvent	No

Example (NCERT Problems 1.6, 1.7) · 2 g A in 18 g water → 2/20 × 100 = 10%. 4 g NaOH (0.1 mol) in 250 mL → 0.4 M.

Example (NCERT Problem 1.8) · 3 M NaCl, density 1.25 g mL⁻¹. 1 L of solution weighs 1250 g and holds 175.5 g NaCl, so water = 1074.5 g and m = 3 / 1.0745 = 2.79 m.

Conversions (Mₘ = molar mass of the solute, M₁ = molar mass of the solvent, d = density in g mL⁻¹):

For a dilute aqueous solution, molarity ≈ molality.

Example · Exercise 1.6: 69% HNO₃, d 1.41 → 10 × 1.41 × 69 / 63 = 15.4 M. Exercise 1.29: x(ethanol) = 0.040 in water → 1000 × 0.04 / (0.96 × 18) = 2.31.

Example (2025, ppm) · 12 ppm Fe in 150 kg wheat = 1.8 g Fe → FeSO₄·7H₂O = 1.8 × 278/56 ≈ 9 g.

Dilution, mixing, reacting:

• Dilution keeps the moles of solute. To make 1 L of 0.2 M from 1 M, take 200 mL and make it up to 1 L. The concentrated starting solution is the stock solution.
• Solutions that react: work in millimoles (M × mL = mmol), find the limiting one, then divide what remains by the total volume.

Neutralisation · Beyond NCERT

Count H⁺ and OH⁻, not formula units. H₂SO₄ and oxalic acid give 2 H⁺; H₃PO₄ gives 3 (H₃PO₃ only 2). Ba(OH)₂ and Ca(OH)₂ give 2 OH⁻; Na₂CO₃ takes 2 HCl.

Example (2026) · 25 mL of 0.1 M Ba(OH)₂ = 2.5 mmol = 5 mmol OH⁻ → 5 mmol HCl = 182.5 mg → answer as x × 10⁻¹ → 1825.

Normality N = M × n-factor. NCERT has dropped it but JEE still uses it now and then: equivalents of acid = equivalents of base. The n-factor in detail is in Redox.

Titration readings · Beyond NCERT · Use the concordant reading, the value that repeats, not the average. For 4.8, 4.9, 5.0, 5.0, 5.0 mL, take 5.0 mL. Asked twice in 2021.

10 · Traps

Trap	What goes wrong	Do this
Molality	Dividing by solution mass	Subtract the solute first; kg of solvent
Limiting reagent	Comparing grams (10 g H₂ vs 10 g O₂)	Moles ÷ coefficient; there O₂ limits
Atoms vs molecules	Counting molecules when atoms are asked	Read what is counted; diatomic gases double it
"x × 10⁻²" answers	Entering 0.5 instead of 50	Write the multiplier beside the question first
STP volume	Using 22.4 when the paper means 22.7	Use the one given; else check the options
Hydrates	Leaving out the water of crystallisation	6.3 g oxalic acid dihydrate = 0.05 mol, not 0.07
Dilution vs mixing	Using M₁V₁ = M₂V₂ for two solutions	Add the moles, divide by the total volume
Reacting solutions	Forgetting one is used up	Millimoles, limiting reagent, then total volume
Trailing zeros	Counting the zeros of 100	100 has 1, 100.0 has 4; the power of 10 changes nothing
Percent by mole	Answering 75% for C in CH₄	By mole it is 1 atom in 5 = 20%
Mole fraction after a reaction (2021)	Dropping the excess reactant	Include every gas left: unused O₂ and the water too

11 · Numbers worth knowing

• Nₐ = 6.022 × 10²³; 1 u = 1.66 × 10⁻²⁴ g; one ¹²C atom = 1.99 × 10⁻²³ g.
• 1 mol of gas: 22.7 L (1 bar, 273 K); 22.4 L (1 atm, 273 K); 24.5 L (1 atm, 298 K).

Compound	g mol⁻¹
CaCO₃	100
MgCO₃	84
NaOH	40
HCl	36.5
H₂SO₄	98
HNO₃	63
Na₂CO₃	106
NaHCO₃	84
NaCl	58.5
glucose	180
sucrose	342
urea	60
KMnO₄	158
K₂Cr₂O₇	294
Ba(OH)₂	171
CH₃COONa	82

12 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• Chemistry in ancient India was called Rasayan Shastra.
• Acharya Kanad (Kashyap, born 600 BCE) proposed indivisible Paramãnu in the Vaiseshika Sutras, about 2500 years before Dalton.
• Nagarjuna wrote Rasratnakar, on mercury compounds and extracting metals.
• Chakrapani discovered mercury sulphide and is credited with inventing soap.
• The Charaka Samhita describes metal bhasmas, which are now known to contain nanoparticles.
• The modern applications NCERT names: cisplatin and taxol (cancer), AZT (AIDS), and CFC substitutes.