NCERT Plus One (Class 11) Chemistry • Chapter 1

Some Basic Concepts of Chemistry

From Ancient Atoms & Matter to SI Units, Chemical Laws, Mole Concept, Empirical Formulas, Limiting Reagents & Solution Concentrations

📄 Official NCERT Textbook PDF

Topic 1.1 – 1.2: Importance & Nature of Matter: Elements, Compounds and Mixtures

Picture your kitchen in the morning. When you boil water for tea, three things happen right in front of you: liquid water turns into rising steam (a gas), sugar dissolves completely into clear tea (a homogeneous mixture), while tea leaves stay suspended and get caught in the strainer (a heterogeneous mixture). Even ancient Indian chemistry (Rasayana Shastra) used alloys, dyes, and medicines crafted from these same building blocks!

Chemistry is the branch of science that studies the composition, structure, properties, and interactions of matter. Everything that has mass and occupies space is matter.

1. Physical States of Matter

  • Solid: Particles are tightly packed in an orderly lattice with minimal freedom of movement. Definite shape and definite volume.
  • Liquid: Particles are close together but free to slide past one another. Definite volume, no definite shape.
  • Gas: Particles are far apart and move rapidly in random directions. Neither definite shape nor definite volume; highly compressible.

2. Chemical Classification of Matter

  • Mixtures: Two or more substances physically mixed in any ratio (separable by filtration, distillation, crystallisation).
    • Homogeneous mixture: Uniform composition throughout (sugar solution, air).
    • Heterogeneous mixture: Non-uniform composition; components visible (salt + sugar, grains + stones).
  • Pure Substances: Fixed composition; cannot be separated by physical methods.
    • Elements: Contain only one type of atom (Na, Cu, H2, O2).
    • Compounds: Two or more atoms of different elements chemically combined in a fixed ratio (H2O, CO2, glucose C6H12O6). Notice how H2 burns with a pop, O2 supports combustion, yet H2O extinguishes fire!
A pure substance does NOT mean "single atom"—O2 and C6H12O6 are pure substances! Also, a homogeneous mixture like sugar-water is still a mixture (not a compound) because its ratio can vary.

Topic 1.3: Properties of Matter, SI Base Units, Density & Temperature Scales

Imagine ordering "2 buckets of milk" in Paris, "5 cups" in Kochi, and "1 gallon" in London—chaos! In 1960, scientists agreed on one universal language: the SI System (maintained in India by NPL, New Delhi) with 7 fundamental base units so a chemist in Kerala and a chemist in Tokyo measure identically.

  • Physical vs Chemical Properties: Physical properties (colour, melting point, boiling point, density) are measured without changing chemical identity. Chemical properties (acidity, combustibility) involve chemical change.
  • 7 SI Base Quantities: Length (m), Mass (kg), Time (s), Electric current (A), Thermodynamic temperature (K), Amount of substance (mol), Luminous intensity (cd).
  • Mass vs Weight: Mass is the amount of matter (constant everywhere), while Weight = m × g (changes with gravity).
  • Volume Relations: 1 L = 1000 mL = 1000 cm3 = 1 dm3.
  • Density: Density = Mass / Volume (SI unit: kg m-3; lab unit: g cm-3).
Temperature Conversions:
K = °C + 273.15   |   °F = (9/5)(°C) + 32
Watch out for negative temperatures: °C and °F can be negative (and meet at -40°!), whereas Kelvin (K) never has negative values. Also, 1 L = 1 dm3, NOT 1 m3 (1 m3 = 1000 L).

Topic 1.4: Uncertainty in Measurement: Scientific Notation, Precision, Accuracy & Significant Figures

Three archery students aim at a bullseye marked 2.00. Student A shoots 1.95 and 1.93—very tight together, but shifted right! Student B shoots wildly at 1.94 and 2.05. Student C hits 2.01 and 1.99—tight AND right on target! That single bullseye story unlocks the difference between Precision and Accuracy.

  • Precision vs Accuracy: Precision = closeness of repeated measurements to each other. Accuracy = closeness of a value to the true value.
  • Scientific Notation (N × 10n): Makes tiny or huge numbers manageable (e.g., 0.00016 = 1.6 × 10-4; Avogadro's number = 6.022 × 1023).

Rules for Counting Significant Figures

  1. All non-zero digits are significant (285 cm → 3 SF).
  2. Zeros preceding the first non-zero digit are NOT significant (0.03 → 1 SF; 0.0052 → 2 SF).
  3. Zeros between two non-zero digits ARE significant (2.005 → 4 SF).
  4. Zeros at the right end are significant ONLY if they lie to the right of a decimal point (0.200 g → 3 SF; 100 → 1 SF, whereas 100.0 → 4 SF).
  5. Exact counted numbers have infinite significant figures (2 balls, 20 eggs).
Calculations & Rounding:
• Addition/Subtraction: Limit to fewest decimal places (12.11 + 18.0 + 1.012 = 31.122 → 31.1).
• Multiplication/Division: Limit to fewest significant figures (2.5 × 1.25 = 3.125 → 3.1).
• Dimensional Analysis (Unit Factor Method): Multiply by conversion factors equal to 1 (e.g., 3 in × 2.54 cm / 1 in = 7.62 cm).
In rounding off when the dropped digit is exactly 5: look at the preceding digit! If it is ODD, add 1 (6.35 → 6.4). If it is EVEN, keep it unchanged (6.25 → 6.2)!

Topic 1.5 – 1.6: Five Laws of Chemical Combination & Dalton's Atomic Theory

In the 1780s, Antoine Lavoisier sealed tin and air inside a glass flask, heated it until ash formed, and placed it on a precision balance. The needle didn't budge by a single speck! That experiment shattered the old myth that fire destroys matter—and birthed the Five Laws of Chemical Combination.

  1. Law of Conservation of Mass (Antoine Lavoisier, 1789): Matter can neither be created nor destroyed in a chemical change (Mass of reactants = Mass of products).
  2. Law of Definite Proportions (Joseph Proust): A given compound always contains exactly the same proportion of elements by weight, regardless of source (Natural vs Synthetic CuCO3: both 51.35% Cu, 9.74% C, 38.91% O).
  3. Law of Multiple Proportions (John Dalton, 1803): If two elements combine to form more than one compound, the masses of one element combining with a fixed mass of the other bear a simple whole-number ratio (In H2O: 2 g H + 16 g O; in H2O2: 2 g H + 32 g O → Ratio of O = 16:32 = 1:2).
  4. Gay-Lussac's Law of Gaseous Volumes (1808): Gases combine or are produced in a simple ratio by volume at constant T and P (100 mL H2 + 50 mL O2 → 100 mL H2O vapour = 2:1:2).
  5. Avogadro's Law (1811): Equal volumes of all gases at the same T and P contain equal number of molecules. (Solved Gay-Lussac's puzzle by realizing H2 and O2 are diatomic molecules!)

Dalton's Atomic Theory (1808)

Matter consists of indivisible atoms; atoms of a given element have identical mass & properties; compounds form in fixed ratios; chemical reactions merely reorganize atoms.

Exams frequently ask: Which pair of compounds illustrates Law of Multiple Proportions? Always pick two compounds made of the SAME two elements (like CO & CO2, or H2O & H2O2, or NO & NO2)—NOT H2O and D2O!

Topic 1.7 – 1.8: Atomic Mass, Molecular Mass, Formula Mass & The Mole Concept

When you buy bananas in a market, you say "one dozen" (12). When a bank counts paper, it says "one ream" (500 sheets). Now imagine counting water molecules in a single teardrop—there are over 1,600,000,000,000,000,000,000 of them! Chemists invented their own mega-dozen: One Mole = 6.022 × 1023 particles.

  • Unified Atomic Mass (u / amu): 1 u = 1/12th the mass of one Carbon-12 (12C) atom = 1.66056 × 10-24 g.
  • Average Atomic Mass: Weighted average of naturally occurring isotopes. E.g., Carbon has 12C (98.892%) and 13C (1.108%), giving 12.011 u.
  • Molecular Mass vs Formula Mass: Sum of atomic masses in a discrete molecule (C6H12O6 = 180.16 u). For ionic lattices like NaCl where no discrete molecule exists, we call it Formula Mass (23.0 + 35.5 = 58.5 u).
  • The Mole & Avogadro Constant (NA): 1 mol = 6.022 × 1023 entities. Mass of 1 mole in grams = Molar Mass (g mol-1), numerically equal to mass in u!
n (moles) = Given Mass (w) / Molar Mass (M) = Number of Particles (N) / NA = Volume at STP (L) / 22.7 L
Watch the difference between 1 mol of N (14 g = 6.022 × 1023 N atoms) and 1 mol of N2 (28 g = 6.022 × 1023 N2 molecules = 2 × 6.022 × 1023 N atoms)! Also, 1 mole of any ideal gas at STP (273.15 K, 1 bar) occupies 22.7 L (22.4 L at old 1 atm STP).

Topic 1.9: Percentage Composition, Empirical Formula & Molecular Formula

Suppose a detective chemist finds an unknown white powder. First, instruments reveal its recipe by weight: 40% carbon, 6.67% hydrogen, 53.33% oxygen. That gives the simplest ratio CH2O (Empirical Formula). Only after measuring its molar mass (180 g/mol) does the detective shout: "Aha! Multiply by 6 — it's Glucose, C6H12O6!"

Mass % of an element = (Mass of that element in 1 mol of compound / Molar mass of compound) × 100

Empirical vs Molecular Formula

  • Empirical Formula (EF): Simplest whole-number ratio of atoms in a compound (e.g., CH for benzene, CH2O for glucose).
  • Molecular Formula (MF): Exact number of atoms in one molecule (C6H6, C6H12O6).
  • Link: Molecular Formula = n × (Empirical Formula), where n = Molar Mass / Empirical Formula Mass.
NCERT Problem 1.2: A compound has 4.07% H, 24.27% C, and 71.65% Cl; Molar mass = 98.96 g.
• Moles in 100 g: C = 24.27/12.01 = 2.021; H = 4.07/1.008 = 4.04; Cl = 71.65/35.453 = 2.021.
• Divide by smallest (2.021): C : H : Cl = 1 : 2 : 1 → Empirical Formula = CH2Cl (Mass = 49.48).
• n = 98.96 / 49.48 = 2 → Molecular Formula = C2H4Cl2!
If dividing by the smallest mole number gives a fraction like 1.5 (as in iron oxide Fe:O = 1 : 1.5), do NOT round 1.5 up to 2! Multiply ALL numbers by 2 to get whole numbers: 2 : 3 → Fe2O3!

Topic 1.10.1: Stoichiometry & Limiting Reagent: Why One Ingredient Stops the Kitchen

Suppose you want to make cheese sandwiches. Each sandwich needs 2 slices of bread and 1 slice of cheese. You open the fridge and find 10 bread slices (enough for 5 sandwiches) and only 3 cheese slices. How many sandwiches can you make? Only 3! Cheese ran out first—in chemistry, cheese is your Limiting Reagent!

Stoichiometry uses balanced chemical equations to relate moles, masses, and volumes of reactants and products. For CH4(g) + 2O2(g) → CO2(g) + 2H2O(g): 1 mol (16 g) CH4 needs 2 mol (64 g) O2 to give 1 mol (44 g) CO2 + 2 mol (36 g) H2O.

Limiting Reagent (LR)

The reactant that gets completely consumed first limits the amount of product formed and is called the Limiting Reagent. The leftover reactant is the excess reagent.

NCERT Problem 1.5: 50.0 kg N2(g) + 10.0 kg H2(g) → NH3(g).
• Moles of N2 = 50,000 / 28.0 = 1786 mol.
• Moles of H2 = 10,000 / 2.016 = 4960 mol.
• From N2 + 3H2 → 2NH3, 1786 mol N2 requires 3 × 1786 = 5358 mol H2.
• Since we have only 4960 mol H2, H2 is the Limiting Reagent!
• NH3 formed = (2/3) × 4960 = 3307 mol = 3307 × 17 g ≈ 56.1 kg NH3.
NEVER assume the reactant with the smaller mass in grams/kg is the limiting reagent! Always convert to moles and divide by the stoichiometric coefficient first—50 kg N2 vs 10 kg H2 still had H2 as limiting reagent only because 1 mol N2 needed 3 mol H2!

Topic 1.10.2: Reactions in Solutions: Mass %, Mole Fraction, Molarity (M) & Molality (m)

Wait—why do high-precision chemists prefer Molality (m) over Molarity (M) when a lab's temperature changes from a cool morning to a hot afternoon? Because water expands slightly when warmed! Volume increases, changing Molarity—while kilogram mass of solvent stays stubbornly rock-solid!

Concentration TermFormulaDepends on Temperature?
1. Mass Per Cent (w/w%)(Mass of solute / Mass of solution) × 100No (mass-based)
2. Mole Fraction (χA)nA / (nA + nB); note χA + χB = 1No (mole-based)
3. Molarity (M)Moles of solute / Volume of solution in LitresYES (Volume changes with T)
4. Molality (m)Moles of solute / Mass of solvent in kgNO (Mass stays constant!)
Dilution Formula: M1V1 = M2V2   |   Density Link: M = (% w/w × d × 10) / Msolute
Notice the denominator difference! Molarity divides by total Volume of SOLUTION (L), whereas Molality divides by Mass of SOLVENT (kg)—NOT total solution mass! Also, 0.50 mol Na2CO3 means fixed moles (53 g), whereas 0.50 M Na2CO3 means 0.50 mol dissolved per 1 litre of solution.

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated Favorite) • 2 Marks): State the Law of Multiple Proportions. Who proposed it? Illustrate with a suitable example.

Statement: Proposed by John Dalton (1803), if two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

Example: Hydrogen (2 g) combines with 16 g of oxygen to form Water (H2O) and with 32 g of oxygen to form Hydrogen Peroxide (H2O2). The ratio of oxygen masses is 16 : 32 = 1 : 2.

Q2 (Kerala HSE SAY / Model (Repeated) • 2 Marks): Differentiate between Molarity and Molality. Why is molality preferred over molarity in expressing concentration when temperature changes?

Molarity (M): Number of moles of solute per litre of solution (mol L-1).
Molality (m): Number of moles of solute per kilogram of solvent (mol kg-1).

Reason: Molality depends on the mass of solvent, which is independent of temperature, whereas molarity depends on volume of solution, which changes with temperature.

Q3 (Kerala HSE March (Standard Numerical) • 3 Marks): A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. Find its empirical and molecular formulas.

In 100 g: Moles of C = 24.27/12.01 = 2.021; H = 4.07/1.008 = 4.04; Cl = 71.65/35.453 = 2.021.
Dividing by smallest (2.021) gives ratio C:H:Cl = 1:2:1 → Empirical Formula = CH2Cl (Mass = 49.48 g).
n = 98.96 / 49.48 = 2 → Molecular Formula = C2H4Cl2.

Q4 (Kerala HSE / NCERT Exemplar • 3 Marks): What is a limiting reagent? In the reaction A + B2 → AB2, identify the limiting reagent when 2 mol of A and 3 mol of B2 are mixed.

Limiting Reagent: The reactant that is completely consumed first in a chemical reaction and limits the amount of product formed.

1 mol of A reacts with 1 mol of B2. So 2 mol of A needs only 2 mol of B2 (3 mol B2 is available in excess). Therefore A is the limiting reagent.

Q5 (Kerala HSE March • 2 Marks): State Avogadro's Law. How does it explain Gay-Lussac's law for the formation of water vapour?

Avogadro's Law: Equal volumes of all gases at the same temperature and pressure contain equal number of molecules.

Since 2 volumes H2 + 1 volume O2 → 2 volumes H2O(g), if 1 volume contains n molecules, then 2n molecules of H2 + n molecules of O2 give 2n molecules of H2O because H2 and O2 are diatomic.

Q6 (Kerala HSE SAY • 1 Marks): How many molecules are present in 18 g of water (H2O)?

Molar mass of H2O = 18 g mol-1. 18 g = 1 mole = 6.022 × 1023 molecules.

Q7 (Kerala HSE Model • 2 Marks): Distinguish between precision and accuracy with an example.

Precision refers to the closeness of various measurements for the same quantity, while accuracy is the agreement of a particular value to the true value. E.g., for true value 2.00 g, readings 1.95 g and 1.93 g are precise but not accurate; 2.01 g and 1.99 g are both precise and accurate.

Q8 (Kerala HSE March • 2 Marks): Calculate the mass percentage of each element in Water (H2O). [Atomic mass H = 1.008 u, O = 16.00 u]

Molar mass of H2O = 18.02 g.
Mass % of H = (2.016 / 18.02) × 100 = 11.19%.
Mass % of O = (16.00 / 18.02) × 100 = 88.81%.

Q9 (Kerala HSE Improvement • 1 Marks): Give the SI unit and symbol for 'amount of substance' and 'thermodynamic temperature'.

Amount of substance: mole (mol). Thermodynamic temperature: kelvin (K).

Q10 (Kerala HSE March • 3 Marks): 50.0 kg of N2(g) and 10.0 kg of H2(g) are mixed to produce NH3(g). Calculate the amount of NH3(g) formed and identify the limiting reagent.

N2 + 3H2 → 2NH3. Moles N2 = 50000/28 = 1785.7 mol; Moles H2 = 10000/2.016 = 4960.3 mol.
Required H2 = 3 × 1785.7 = 5357 mol (more than available 4960.3 mol). Thus H2 is the limiting reagent.
Moles of NH3 = (2/3) × 4960.3 = 3306.9 mol → Mass = 3306.9 × 17 g ≈ 56.1 kg NH3.

Q11 (Kerala HSE / NCERT • 2 Marks): How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?

Molar mass of Na2CO3 = 106 g mol-1. 0.50 mol Na2CO3 represents 0.50 × 106 = 53 g of solid sodium carbonate. 0.50 M Na2CO3 represents a solution containing 0.50 mol (53 g) dissolved in 1 litre of solution.

Q12 (Kerala HSE SAY • 1 Marks): Express 0.00016 and 232.508 in scientific notation, and write the number of significant figures in 0.0048.

0.00016 = 1.6 × 10-4; 232.508 = 2.32508 × 102. Significant figures in 0.0048 = 2.