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Class XI ⚗️ Chemistry ~10 MCQs/year Ch 1 of 9

Some Basic Concepts of Chemistry

CUET unit: Some Basic Concepts of Chemistry

📌 Snapshot

  • Foundation unit of Class XI Chemistry that establishes the language of the subject — SI units, scientific notation, significant figures, and dimensional analysis — together with the quantitative tools (mole concept, stoichiometry, molarity, molality) used in every subsequent chapter.
  • Anchors the five empirical laws of chemical combination — conservation of mass, definite proportions, multiple proportions, Gay-Lussac's law of gaseous volumes, and Avogadro's law — which historically motivated Dalton's atomic theory (1808).
  • Defines the mole as the SI unit of amount of substance and fixes the Avogadro number at exactly 6.02214076 × 10²³ entities per mole (2019 SI redefinition), linking macroscopic mass with microscopic particle count.
  • Lays out the empirical → molecular formula pipeline (percentage composition → moles → simplest ratio → multiplier n) that is tested repeatedly in CUET numericals.
  • Introduces stoichiometric calculations, the limiting reagent concept, and the four standard concentration measures — mass per cent, mole fraction, molarity, molality — with the M₁V₁ = M₂V₂ dilution shortcut.
  • Heavy CUET footprint: numerical MCQs on mole–mass–particle interconversion, limiting reagent, molarity, significant figures, density-percentage-to-molarity conversions, and empirical/molecular formula calculations recur every year; expect 8–10 questions per attempt.

📖 Detailed Notes

2.1 Core concepts

Chemistry is the branch of science that studies the preparation, properties, structure and reactions of material substances. It underpins the manufacture of fertilisers, alkali, acids, salts, dyes, polymers, drugs (cisplatin and taxol in cancer therapy, AZT in HIV control), soaps, detergents and metals, and it is central to weather prediction, ozone-layer protection, and pollution control (NCERT §1.1, p. 4).

Nature and states of matter. Matter is anything that has mass and occupies space. At the macroscopic level it occurs in three physical states — solid (definite volume, definite shape, particles close-packed in fixed positions), liquid (definite volume, no definite shape, particles close but mobile), and gas (no definite volume or shape, particles far apart and moving freely) — and these states are interconvertible by changing temperature or pressure (NCERT §1.2.1, p. 4–5; Fig. 1.1).

Classification of matter. Matter is first classified at the bulk level into pure substances (fixed composition, cannot be separated by physical methods) and mixtures (variable composition, separable by physical methods). Mixtures are homogeneous (sugar in water, air, brass — uniform throughout) or heterogeneous (sugar + salt, grains + pulses + dirt — non-uniform). Pure substances split further into elements (one type of atom — Na, Cu, H₂, O₂) and compounds (two or more elements combined in a fixed ratio — H₂O, CO₂, sugar); a compound's properties differ from those of the elements that make it (NCERT §1.2.2, p. 5–6; Figs. 1.2–1.4).

Properties of matter and SI units. Properties are physical (colour, melting point, boiling point, density — measured without changing chemical identity) or chemical (combustibility, acidity, reactivity — require a chemical change) (NCERT §1.3.1, p. 6). The International System of Units (SI), adopted by the 11th CGPM in 1960 and redefined in 2019, has seven base units — metre (length), kilogram (mass), second (time), ampere (electric current), kelvin (thermodynamic temperature), mole (amount of substance), and candela (luminous intensity) — from which all derived units are built (NCERT §1.3.3, Table 1.1, p. 7). Mass is the amount of matter present and is invariant; weight is the gravitational force on a body and changes with location. Volume has units of (length)³; 1 L = 1000 mL = 1000 cm³ = 1 dm³ (NCERT §1.3.4–1.3.5, p. 9). Density (mass per unit volume) has SI unit kg m⁻³, although chemists routinely use g cm⁻³. The Celsius scale is set by °C = (°F − 32) × 5/9, and the Kelvin scale by K = °C + 273.15; negative readings exist on °C but not on K (NCERT §1.3.6–1.3.7, p. 10; Fig. 1.8).

Uncertainty in measurement. Scientific notation expresses every quantity as N × 10ⁿ with 1 ≤ N < 10, so that very large (6.022 × 10²³) and very small (1.66 × 10⁻²⁴ g) numbers fit comfortably on the page and in calculators (NCERT §1.4.1, p. 11). The rules for significant figures are: all non-zero digits are significant; captive zeros between non-zero digits are significant; leading zeros are not; trailing zeros are significant only if a decimal point is present (so 100 has 1 sig fig but 100. has 3 and 100.0 has 4); counted exact numbers have infinite sig figs (NCERT §1.4.2, p. 12). In addition/subtraction, the answer is limited by the fewest decimal places; in multiplication/division, by the fewest significant figures. Precision is the closeness of repeated readings to one another; accuracy is the closeness of a reading to the true value (Table 1.4, p. 13). Dimensional analysis (factor-label method) uses unit-conversion factors equal to unity (e.g. 2.54 cm/1 in) to interconvert units cleanly (NCERT §1.4.3, p. 13–14).

Laws of chemical combination. (i) Conservation of Mass (Lavoisier, 1789) — matter is neither created nor destroyed in any physical or chemical change. (ii) Definite Proportions (Proust) — a given compound always contains the same elements in the same proportion by mass irrespective of source or method of preparation. (iii) Multiple Proportions (Dalton, 1803) — when two elements form more than one compound, the masses of one combining with a fixed mass of the other bear a small whole-number ratio (H in H₂O vs H₂O₂ → O masses 16 : 32 = 1 : 2). (iv) Gay-Lussac's Law of Gaseous Volumes (1808) — gases combine or are produced in simple volume ratios at the same T and P (H₂ + ½O₂ → H₂O is volumetrically 2 : 1 : 2). (v) Avogadro's Law (1811) — equal volumes of all gases at the same T and P contain equal numbers of molecules (NCERT §1.5, p. 14–16; Fig. 1.9).

Dalton's atomic theory (1808) crystallises these laws: matter is made of indivisible atoms; atoms of an element are identical in mass and properties; compounds form when atoms combine in fixed ratios; chemical reactions only rearrange atoms — never create or destroy them (NCERT §1.6, p. 16).

Atomic, molecular and formula mass. Atomic mass is given relative to ¹²C, which is assigned exactly 12 u. One unified atomic mass unit (u or amu) = 1.66056 × 10⁻²⁴ g. The average atomic mass weights isotopic masses by natural abundance (C = 12.011 u, blending ¹²C, ¹³C and trace ¹⁴C). Molecular mass is the sum of atomic masses (H₂O = 18.02 u, CH₄ = 16.043 u). For ionic substances without discrete molecules (NaCl), the analogous quantity is called formula mass (NaCl = 58.5 u) (NCERT §1.7, p. 16–18; Fig. 1.10).

The mole. One mole is the SI amount of substance containing exactly 6.02214076 × 10²³ elementary entities — atoms, molecules, ions, electrons, or specified groups — and this fixed value defines the Avogadro constant N_A. The molar mass in g mol⁻¹ is numerically equal to the atomic/molecular/formula mass in u (NCERT §1.8, p. 18; Fig. 1.11). Mole↔mass↔particle interconversion uses molar mass and N_A.

Percentage composition, empirical and molecular formulas. Mass per cent of an element = (mass of that element in 1 mol / molar mass) × 100. For ethanol C₂H₅OH: C = 52.14 %, H = 13.13 %, O = 34.73 %. From percentage composition: convert to moles → divide by smallest mole value → multiply to whole numbers → that ratio is the empirical formula. The molecular formula = (empirical formula) × n, where n = (molar mass) / (empirical formula mass) (NCERT §1.9–1.9.1, p. 18–20).

Stoichiometry and limiting reagent. A balanced equation gives molar (and hence mass and volume) ratios. For CH₄ + 2 O₂ → CO₂ + 2 H₂O, combustion of 16 g CH₄ yields 36 g H₂O and 44 g CO₂ (Problem 1.3, p. 20–22). The limiting reagent is the reactant present in the least stoichiometric quantity — it is consumed first and decides the yield. Problem 1.5 (p. 22): mixing 50.0 kg N₂ with 10.0 kg H₂ to make NH₃ — moles needed of H₂ (5358) exceed moles available (4960), so H₂ is limiting; NH₃ formed = 3.30 × 10³ mol = 56.1 kg.

Concentration of solutions. Four common expressions — mass per cent (mass of solute × 100/mass of solution), mole fraction (xₐ = nₐ/(nₐ+n_b)), molarity M = moles of solute / volume of solution in L, and molality m = moles of solute / mass of solvent in kg. The dilution shortcut M₁V₁ = M₂V₂ follows from conservation of moles. Molarity changes with temperature because the solution volume expands; molality does not, because the solvent mass is invariant (NCERT §1.10.2, p. 21–24).

2.2 Definitions to memorise

Term Definition Page
Matter Anything that has mass and occupies space 4
Pure substance Substance with all constituent particles of the same chemical nature and a fixed composition 5
Element Pure substance made of only one kind of atom (Na, H₂, O₂) 6
Compound Pure substance formed when two or more elements combine in a fixed mass ratio (H₂O, CO₂) 6
Mass Amount of matter present in a body — invariant with location 9
Weight Gravitational force acting on a body — varies with location 9
Density Mass per unit volume; SI unit kg m⁻³, commonly g cm⁻³ 10
Significant figures Meaningful digits known with certainty plus one estimated (uncertain) digit 12
Precision Closeness of various measurements of the same quantity to one another 13
Accuracy Closeness of a measured value to the true (accepted) value 13
Law of Conservation of Mass In every physical/chemical change there is no net change in mass — Lavoisier, 1789 14
Law of Definite Proportions A given compound always contains the same elements in the same mass proportion — Proust 15
Law of Multiple Proportions When two elements form more than one compound, the masses of one combining with a fixed mass of the other are in small whole-number ratios — Dalton, 1803 15
Gay-Lussac's Law Gases combine in simple volume ratios at the same T and P — 1808 15
Avogadro's Law Equal volumes of gases at the same T and P contain equal numbers of molecules — 1811 15
Atomic mass unit (u / amu) One-twelfth of the mass of one ¹²C atom; 1 u = 1.66056 × 10⁻²⁴ g 16–17
Average atomic mass Weighted mean of isotopic masses using natural-abundance fractions 17
Molecular mass Sum of atomic masses of all atoms in a molecule 17
Formula mass Analogue of molecular mass for ionic compounds with no discrete molecule (NaCl) 18
Mole SI unit of amount of substance; contains exactly 6.02214076 × 10²³ elementary entities 18
Molar mass Mass of one mole of a substance in grams; numerically equal to atomic/molecular/formula mass in u 18
Empirical formula Simplest whole-number ratio of atoms of different elements in a compound 19
Molecular formula Actual number of each kind of atom in a molecule of the compound 19
Limiting reagent Reactant consumed first that thereby limits the amount of product formed 21
Molarity (M) Number of moles of solute per litre of solution (mol L⁻¹) 23
Molality (m) Number of moles of solute per kilogram of solvent (mol kg⁻¹) 24
Mole fraction (x) Ratio of moles of one component to total moles in the mixture 23

2.3 Diagrams / processes to remember

Make sure you can sketch and label each diagram below. Fig. 1.1 (p. 5) shows the three states with their characteristic particle arrangement: solids close-packed and ordered, liquids close but disordered, gases far apart and rapidly moving. Fig. 1.2 (p. 5) is the master classification flowchart — Matter → Pure substance (Elements/Compounds) or Mixture (Homogeneous/Heterogeneous); CUET frequently tests where a given example (air, milk, brass, sodium chloride solution) lands on this tree. Fig. 1.3 (p. 6) contrasts atoms (Na, Cu) with diatomic molecules (H₂, N₂, O₂), and Fig. 1.4 depicts H₂O and CO₂ molecules — useful for distinguishing elemental versus compound molecules.

Fig. 1.6 (p. 9) is a cube illustrating volume relationships: 1 L = 1000 mL = 1000 cm³ = 1 dm³. Fig. 1.7 (p. 10) shows the four volume-measuring devices a CUET candidate must be able to identify by name and by precision class — graduated cylinder, burette, pipette, volumetric flask. Fig. 1.8 (p. 10) puts °C, °F and K side by side so that you can compare the freezing point of water (0 °C / 32 °F / 273.15 K) and human body temperature (37 °C / 98.6 °F / 310.15 K).

Fig. 1.9 (p. 16) is the central visual of the laws of chemical combination: two volumes of H₂ + one volume of O₂ → two volumes of water vapour, with little boxes showing equal numbers of molecules per equal volumes (Avogadro's bridge from Gay-Lussac). Fig. 1.10 (p. 17) depicts cubic packing of Na⁺ and Cl⁻ in NaCl — each ion surrounded by six counter-ions — explaining why NaCl has a formula mass rather than a molecular mass. Fig. 1.11 (p. 18) lays out one mole of various substances (12 g C, 18 g H₂O, 23 g Na, 32 g S, 56 g Fe, 63.5 g Cu, 200 g Hg) side by side to drive home that "one mole" denotes a count of particles, not a fixed mass.

Three workflow processes are worth memorising. The mole-conversion tree (p. 18) maps mass ↔ moles ↔ number of particles via molar mass and N_A, and mass ↔ volume via density — this is the spine of every CUET numerical. The empirical → molecular formula process (p. 19–20) takes percentage composition → divide by atomic mass to get moles → divide by the smallest mole to get the empirical ratio → multiply by n = M_molar/M_empirical to get the molecular formula. The stoichiometry-with-limiting-reagent process (p. 20–22) is: convert all masses to moles → compare available moles to stoichiometric moles required → identify the limiting reagent → use it to compute product moles → convert back to mass or volume. Practising these three procedural pipelines once each will solve >70 % of CUET kech101 numericals.

2.4 Common confusions / NTA trap points

  • Mass vs weight — students treat them as synonyms; NCERT explicitly states (p. 9) that mass is constant while weight varies with gravitational acceleration. Distractor: "weight is the amount of matter".
  • Significant figures with terminal zeros — 100 has only 1 sig fig, 100. has 3, 100.0 has 4 (p. 12). NTA loves this trap; the decimal point is decisive.
  • Definite vs Multiple Proportions — "Definite" is about ONE compound (always the same composition); "Multiple" is about TWO or more compounds of the same two elements (whole-number mass ratios). Don't confuse the two laws.
  • Gay-Lussac vs Avogadro — Gay-Lussac talks about combining volumes; Avogadro extends it to equal molecules per equal volume. They are different statements, and CUET MCQs deliberately swap the names.
  • Molarity vs molality — molarity depends on solution volume and so varies with temperature; molality uses solvent mass and so is temperature-independent (p. 24).
  • Limiting-reagent shortcut — compare moles available to moles required by stoichiometry; the reactant that yields the fewer moles of product is limiting. Smaller mass alone is NOT the criterion (Problem 1.5).
  • amu definition — exactly 1/12 the mass of one ¹²C atom (p. 16); not "the mass of a hydrogen atom" (that is the older 1803 system).
  • Empirical vs molecular formula — glucose has empirical formula CH₂O but molecular formula C₆H₁₂O₆. CUET asks you to find n = molar mass / empirical formula mass; don't forget to multiply through.
  • Mole fraction has no units — it is a ratio of moles, so its sum over all components must equal 1. Trap: choosing % composition as a "mole fraction".
  • Avogadro number is now exact — since 2019 it is fixed at 6.02214076 × 10²³ mol⁻¹ by definition, not measured (p. 18). NTA may phrase this as "approximate" in a distractor.
  • Counting particles in diatomic molecules — 1 mole of Cl₂ contains 6.022 × 10²³ molecules but 2 × 6.022 × 10²³ Cl atoms. Watch the wording.
  • Density-percentage-to-molarity — M = (10 × density × % w/w) / molar mass is a derived shortcut for problems where concentrated solutions are described by weight percent and density (Exercise 1.38, p. 29).

2.5 Key reactions & formulas

Reaction / Formula Conditions / Notes NCERT page
1 L = 1000 mL = 1000 cm³ = 1 dm³ Volume conversions 9
Density = mass / volume SI kg m⁻³; chemistry g cm⁻³ 10
K = °C + 273.15 Kelvin from Celsius 10
°F = (9/5)°C + 32 Fahrenheit from Celsius 10
1 u = 1.66056 × 10⁻²⁴ g Atomic mass unit, ¹²C reference 17
N_A = 6.02214076 × 10²³ mol⁻¹ Avogadro constant (exact since 2019) 18
n (moles) = mass (g) / molar mass (g mol⁻¹) Mass-to-mole 18
Number of particles = n × N_A Mole-to-particle 18
Mass per cent = (mass of element / molar mass) × 100 Element composition in a compound 18
n (multiplier) = M_molar / M_empirical formula Empirical → molecular formula 19
2 H₂(g) + O₂(g) → 2 H₂O(l) Combustion of hydrogen — illustrates Gay-Lussac/Avogadro 15–16
CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) Stoichiometry worked example (Problem 1.3) 20
N₂(g) + 3 H₂(g) → 2 NH₃(g) Limiting-reagent worked example (Problem 1.5) 22
CaCO₃(s) + 2 HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l) Acid–carbonate stoichiometry (Exercise 1.35) 28
Molarity (M) = n_solute / V_solution (L) Mol L⁻¹; temperature-dependent 23
Molality (m) = n_solute / mass_solvent (kg) Mol kg⁻¹; temperature-independent 24
Mole fraction x_A = n_A / (n_A + n_B) Σ x_i = 1; dimensionless 23
M₁V₁ = M₂V₂ Dilution; moles conserved 24
M = (10 × d × % w/w) / M_molar Density-percentage shortcut for molarity (g cm⁻³, %) 29
Average atomic mass = Σ (fᵢ × mᵢ) Weighted by natural abundance fᵢ 17

🎯 Practice MCQs

First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed

Q1. How many significant figures are present in the measurement 0.00500400?

▸ Show answer & explanation

Answer: C

Leading zeros (0.00) are not significant; the digits 5, 0, 0, 4, 0, 0 are all significant — two captive zeros between non-zero digits and three trailing zeros after the decimal point all count. Total = 6.

Q2. 50.0 kg of N₂(g) is mixed with 10.0 kg of H₂(g) for N₂(g) + 3 H₂(g) → 2 NH₃(g). The limiting reagent and the mass of NH₃ produced are, respectively:

▸ Show answer & explanation

Answer: B

Moles of N₂ = 50000/28 = 1786 mol; moles of H₂ = 10000/2.016 = 4960 mol. For 1786 mol N₂, 3 × 1786 = 5358 mol H₂ are needed, but only 4960 mol are available — so H₂ is limiting. NH₃ formed = (2/3) × 4960 = 3.30 × 10³ mol = 56.1 kg.

Q3. The molarity of a solution prepared by dissolving 4.0 g of NaOH in enough water to give 250 mL of solution is: (M_NaOH = 40 g mol⁻¹)

▸ Show answer & explanation

Answer: C

Moles of NaOH = 4.0/40 = 0.1 mol. Molarity = 0.1 mol / 0.250 L = 0.4 mol L⁻¹.

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