Class 11 Chemistry ⚗️

Chapter 1 & 2 — Complete Notes with Key Points & Short Answer Questions

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⚗️ Chapter 1 — Basic Concepts 🔬 Chapter 2 — Analytical Chemistry 🔢 All Formulas 📝 2 & 3 Mark Questions

Chapter 1 — Some Basic Concepts of Chemistry

Maharashtra HSC Board — Complete Key Points with Explanations

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Chapter 1 — Quick Reference

All Key Points at a Glance

🧪 Classification of Matter
  • Chemistry: Study of matter, its physical & chemical properties and changes
  • Matter: Occupies space and has mass
  • Matter → Pure substances + Mixtures
  • Pure substances → Elements + Compounds
  • Mixtures → Homogeneous + Heterogeneous
  • Pure substance: definite chemical composition, same properties always
  • Mixture: no definite composition, no fixed properties
  • Element: cannot be broken into simpler substances by ordinary chemical changes
  • Compound: can be broken into simpler substances; elements combined in fixed ratio
⚙️ Metals, Non-Metals, Metalloids
  • Metals: lustre, conduct heat & electricity, ductile, malleable. Examples: gold, silver, copper, iron. Mercury is liquid metal.
  • Non-metals: no lustre (exception: diamond, iodine), poor conductors (exception: graphite), brittle. Examples: iodine, nitrogen, carbon
  • Metalloids: properties intermediate between metals and non-metals. Examples: arsenic, silicon, germanium
  • Homogeneous mixture: solution — molecules of solute & solvent uniformly mixed throughout
  • Heterogeneous mixture: molecules not uniformly mixed. Example: suspension of insoluble solid in liquid
📏 SI Units (7 Base Units)
  • 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)
  • 1 kg = 1000 g = 10³ g | 1 nm = 10⁻⁹ m | 1 pm = 10⁻¹² m
  • Volume: 1 L = 1 dm³ = 1000 mL = 1000 cm³
  • Density SI unit: kg/m³ | CGS: g/mL
  • Temperature: K = °C + 273.15 | °F = (9/5)°C + 32
⚖️ Laws of Chemical Combination
  • Law of Conservation of Mass (Lavoisier): mass can neither be created nor destroyed. Total mass of reactants = total mass of products.
  • Law of Definite Proportions (Proust): a compound always contains same proportion of elements by weight
  • Law of Multiple Proportions (Dalton, 1803): when two elements form more than one compound, masses of one element combining with fixed mass of other are in simple whole number ratio
  • Gay Lussac's Law (1808): gases combine or are produced in simple ratio by volume at same T & P
  • Avogadro's Law (1811): equal volumes of all gases at same T & P contain equal number of molecules
⚛️ Dalton's Atomic Theory (1808)
  • Matter consists of tiny, indivisible particles called atoms
  • All atoms of a given element have identical properties including mass
  • Atoms of different elements differ in mass
  • Compounds are formed when atoms combine in a fixed ratio
  • Chemical reactions involve only reorganization of atoms
  • Atoms are neither created nor destroyed in chemical reactions
  • Dalton's theory explained all laws of chemical combination
⚗️ Atomic & Molecular Mass
  • Atomic mass unit (amu / u / Da): 1/12th of mass of Carbon-12. 1 amu = 1.66056 × 10⁻²⁴ g
  • Reference: Carbon-12 assigned mass 12.00000 amu (agreed 1961 by IUPAC)
  • Average atomic mass: weighted average of isotope masses × % abundance
  • Molecular mass: sum of average atomic masses of all atoms in molecule
  • Formula mass: used for ionic compounds (e.g. NaCl) instead of molecular mass
  • Periodic table atomic masses are actually average atomic masses
🔢 Mole Concept & Molar Mass
  • 1 mole: amount of substance containing as many entities as atoms in exactly 12 g of C-12
  • Avogadro's constant Nₐ = 6.0221367 × 10²³ mol⁻¹
  • Molar mass: mass of 1 mole of substance in grams = numerically equal to atomic/molecular mass in u
  • n (moles) = mass (g) / molar mass (g mol⁻¹)
  • Number of particles = n × Nₐ
  • Molar volume: 1 mole of any gas at STP (0°C, 1 atm) = 22.4 dm³
  • n (gas at STP) = volume (dm³) / 22.4 dm³ mol⁻¹
📊 States of Matter & Properties
  • Solids: particles tightly held in order → definite shape & volume
  • Liquids: particles close but can move → fixed volume, no fixed shape
  • Gases: particles far apart → no fixed shape or volume
  • States are interconvertible by changing temperature & pressure
  • Physical properties: observed without changing chemical composition (colour, odour, mp, bp, density)
  • Chemical properties: substance undergoes change in chemical composition (burning, rusting)
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Section 1.1–1.2

Nature of Chemistry & Classification of Matter

📖 Explanation

Chemistry is the study of matter, its physical and chemical properties, and the changes it undergoes. It is called a central science because its knowledge is required in physics, biological sciences, applied sciences, and earth and space sciences.

Chemistry is traditionally classified into five branches: organic (study of carbon compounds), inorganic (all non-organic substances), physical (principles underlying chemistry — atoms, molecules, electrons, energies), bio, and analytical.

Matter is classified on the basis of chemical composition into:
• Pure substances: definite chemical composition, always same properties regardless of origin. Divided into elements and compounds.
• Mixtures: no definite chemical composition, no fixed properties. Can be separated by physical methods. Divided into homogeneous and heterogeneous.

✅ Short Answers
Q. Why is chemistry called a central science? (2 marks)
Chemistry is called a central science because its knowledge is required in studies of physics, biological sciences, applied sciences, and earth and space sciences. Its scope covers every aspect of life — air we breathe, food we eat, clothing, fuel, medicines, engineering, and agriculture.
Q. Distinguish between pure substances and mixtures. (2 marks)
Pure substances: Have a definite chemical composition and always have the same properties regardless of their origin. Examples: pure metal, distilled water.
Mixtures: Have no definite chemical composition and hence no definite properties. Can be separated by physical methods. Examples: paint, concrete, seawater.
Q. What are metalloids? Give examples. (2 marks)
Metalloids (semi-metals) are elements that have properties intermediate between metals and non-metals. They show some metallic and some non-metallic characteristics. Examples: arsenic (As), silicon (Si), and germanium (Ge). Silicon is widely used in semiconductors.
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Section 1.3

Properties of Matter & SI Units

📖 Explanation

Properties are classified as physical (measured without changing composition — colour, odour, melting point, density) and chemical (composition changes — burning, rusting).

Measurement & SI Units: Any quantitative measurement is expressed as a number followed by units. "The arbitrarily decided and universally accepted standards are called units." The SI system (proposed 1960) has 7 base units.

Important physical quantities:
• Mass vs Weight: Mass = quantity of matter (does not vary with position). Weight = mass × gravitational force (varies with distance from Earth's centre). Mass is more fundamental.
• Density = mass / volume. SI unit: kg/m³. CGS unit: g/mL or g cm⁻³.
• Temperature scales: K = °C + 273.15 | °F = (9/5)°C + 32. Kelvin (K) is SI unit.

✅ Short Answers
Q. Distinguish between mass and weight. (2 marks)
Mass: An inherent property of matter — measure of quantity of matter a body contains. It does NOT vary with position. SI unit: kilogram (kg).
Weight: Result of mass and gravitational attraction. It varies because gravitational attraction changes with distance from Earth's centre. Hence mass is more fundamental than weight.
Q. Convert 40°C to Kelvin and Fahrenheit. (2 marks)
K = °C + 273.15 = 40 + 273.15 = 313.15 K
°F = (9/5) × 40 + 32 = 72 + 32 = 104°F
Q. What are physical and chemical properties? Give examples of each. (3 marks)
Physical properties: Properties that can be measured or observed without changing the chemical composition of the substance. Examples: colour, odour, melting point, boiling point, density.

Chemical properties: Properties where substances undergo a chemical change and exhibit change in chemical composition. Examples: coal burns in air to produce carbon dioxide; magnesium wire burns in presence of oxygen to form magnesium oxide; iron rusts in moist air.
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Section 1.4

Laws of Chemical Combination

📖 Explanation

1. Law of Conservation of Mass (Lavoisier, 1743–1794):
After combustion experiments, Lavoisier found weight gained by phosphorus = weight lost by air. "Mass can neither be created nor destroyed." Total mass of reactants = Total mass of products.

2. Law of Definite Proportions (Joseph Proust):
Experiments on natural and synthetic cupric carbonate showed same % composition (Cu: 51.35%, O: 38.91%, C: 9.74%). "A compound always contains exactly the same proportion of elements by weight."

3. Law of Multiple Proportions (Dalton, 1803):
When two elements form more than one compound, masses of element B combining with fixed mass of A are in ratio of small whole numbers. Example: Hydrogen + Oxygen → Water (2g : 16g) and → H₂O₂ (2g : 32g). Ratio of oxygen = 16:32 = 1:2.

4. Gay Lussac's Law of Gaseous Volumes (1808):
When gases combine or are produced they do so in simple ratio by volume, at same T & P. Example: H₂(100mL) + O₂(50mL) → H₂O(100mL) = 2:1:2.

5. Avogadro's Law (1811):
Equal volumes of all gases at same T & P contain equal number of molecules.

✅ Short Answers
Q. State and explain the law of conservation of mass. (3 marks)
Proposed by Lavoisier: "Mass can neither be created nor destroyed."
Total mass of reactants = Total mass of products.

Explanation: Lavoisier performed combustion experiments with phosphorus and mercury in air. He observed that the weight gained by phosphorus was exactly equal to the weight lost by air. Similarly, when hydrogen burns with oxygen to form water, the mass of water = mass of hydrogen + mass of oxygen consumed. This law forms the basis of all chemical calculations.
Q. State the law of definite proportions. Give an example. (2 marks)
Law of Definite Proportions (Proust): "A given compound always contains exactly the same proportion of elements by weight", irrespective of its source.
Example: Cupric carbonate — both natural and synthetic samples contain copper: 51.35%, oxygen: 38.91%, carbon: 9.74%. The proportions are always fixed regardless of origin.
Q. State Gay Lussac's law of gaseous volumes with an example. (3 marks)
Gay Lussac's Law (1808): "When gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume, provided all gases are at the same temperature and pressure."

Example 1: H₂(100mL) + O₂(50mL) → H₂O vapour(100mL) — ratio 2:1:2
Example 2: N₂(1L) + H₂(3L) → NH₃(2L) — ratio 1:3:2
These simple whole number volume ratios confirm Gay Lussac's law.
Q. Explain the law of multiple proportions with an example. (3 marks)
Law of Multiple Proportions (Dalton, 1803): When two elements A and B form more than one compound, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers.

Example: Nitrogen and oxygen form two compounds:
• Nitric oxide: 14g N + 16g O → 30g NO
• Nitrogen dioxide: 14g N + 32g O → 46g NO₂
Masses of oxygen (16g and 32g) combining with fixed 14g nitrogen are in ratio 16:32 = 1:2 (small whole numbers). ✓
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Section 1.5–1.6

Avogadro's Law & Dalton's Atomic Theory

📖 Explanation

Avogadro's Law (1811): Equal volumes of all gases at the same temperature and pressure contain equal number of molecules.

Application: H₂(100mL) + O₂(50mL) → H₂O(100mL). Applying Avogadro's law: if 1 volume = n molecules, then 2n H₂ + n O₂ → 2n H₂O. So 2 molecules H₂ + 1 molecule O₂ → 2 molecules H₂O. Avogadro made a distinction between atoms and molecules, which is clearly understood today.

Dalton's Atomic Theory (1808): Published in "A New System of Chemical Philosophy." Four main postulates:
1. Matter consists of tiny, indivisible particles called atoms.
2. All atoms of a given element have identical properties including mass; atoms of different elements differ in mass.
3. Compounds are formed when atoms of different elements combine in a fixed ratio.
4. Chemical reactions involve only the reorganization of atoms — atoms are neither created nor destroyed.

✅ Short Answers
Q. State and explain Avogadro's law. (3 marks)
Avogadro's Law (1811): "Equal volumes of all gases at the same temperature and pressure contain equal number of molecules."

Explanation: Consider H₂ + O₂ → H₂O. Gay Lussac's law gives 2 vol H₂ + 1 vol O₂ → 2 vol H₂O. By Avogadro's law, 2n molecules H₂ + n molecules O₂ → 2n molecules H₂O. So: 2 molecules H₂ + 1 molecule O₂ → 2 molecules H₂O. This is consistent only if H₂ and O₂ are diatomic molecules, which Avogadro correctly proposed.
Q. State the postulates of Dalton's atomic theory. (3 marks)
1. Matter consists of tiny, indivisible particles called atoms.
2. All atoms of a given element have identical properties including mass. Atoms of different elements differ in mass.
3. Compounds are formed when atoms of different elements combine in a fixed ratio.
4. Chemical reactions involve only the reorganization of atoms — atoms are neither created nor destroyed in a chemical reaction.
Dalton's theory successfully explained all the laws of chemical combination.
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Section 1.7

Atomic Mass, Molecular Mass & Formula Mass

📖 Explanation

Atomic Mass: Mass of an atom relative to mass of Carbon-12. One amu (or u or dalton) = 1/12 × mass of one C-12 atom = 1.66056 × 10⁻²⁴ g.

Average Atomic Mass: Many elements exist as isotopes (different mass numbers). Average atomic mass = weighted average of isotope masses × % abundance.
Example: Carbon isotopes — ¹²C (12.00000u, 98.892%), ¹³C (13.00335u, 1.108%), ¹⁴C (14.00317u, 2×10⁻¹⁰%). Average = 12.011u.

Molecular Mass: Sum of average atomic masses of all atoms in a molecule. Example: H₂O = 2(1u) + 16u = 18u. CO₂ = 12u + 2(16u) = 44u.

Formula Mass: Used for ionic compounds (like NaCl) which don't have discrete molecules. Formula mass of NaCl = 23u + 35.5u = 58.5u.

✅ Short Answers
Q. What is one atomic mass unit (amu)? How is it defined? (2 marks)
One atomic mass unit (amu, u, or dalton) is defined as a mass exactly equal to one-twelfth of the mass of one carbon-12 atom.
1 amu = (1/12) × mass of C-12 = (1/12) × 1.992648 × 10⁻²³ g = 1.66056 × 10⁻²⁴ g.
The mass of an atom of carbon-12 is assigned exactly 12.00000 amu (agreed by IUPAC, 1961).
Q. What is average atomic mass? Why is it needed? (3 marks)
Many naturally occurring elements exist as a mixture of more than one isotope. Isotopes have different atomic masses. The average atomic mass is the weighted average of atomic masses of all isotopes of an element, taking into account their relative abundances (% occurrence).

It is needed because in nature, elements are found as a mixture of isotopes, not pure isotopes. The atomic masses shown in the periodic table are actually average atomic masses.

Example: Carbon has ¹²C (98.892%), ¹³C (1.108%), ¹⁴C (trace). Average atomic mass of C = 12.011u.
Q. Calculate molecular mass of H₂SO₄ and CO₂. (2 marks)
H₂SO₄ = 2(1u) + 32u + 4(16u) = 2 + 32 + 64 = 98 u
CO₂ = 12u + 2(16u) = 12 + 32 = 44 u
Q. What is formula mass? Why is it used for ionic compounds? (2 marks)
Formula mass is the sum of atomic masses of all atoms present in the formula of a compound. It is used for ionic compounds (like NaCl, CuSO₄) instead of molecular mass because these compounds do not contain discrete molecules — they exist as a 3D arrangement of ions. Example: formula mass of NaCl = 23u + 35.5u = 58.5u.
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Section 1.8–1.9

Mole Concept, Molar Mass & Gases at STP

📖 Explanation

Mole: A quantitative adjective like "dozen" or "gross", used to express large numbers of submicroscopic entities (atoms, molecules, ions).
Definition: One mole is the amount of a substance that contains as many entities as there are atoms in exactly 12 g (0.012 kg) of Carbon-12.

From calculation: Number of atoms in 12g C-12 = 12g/1.992648×10⁻²³g = 6.0221367 × 10²³ atoms/mol. This is Avogadro's Constant (Nₐ).

Molar Mass: Mass of one mole of a substance in grams. It is numerically equal to atomic/molecular/formula mass in u. Examples: molar mass of H₂O = 18 g mol⁻¹; NaCl = 58.5 g mol⁻¹.

Calculations:
n (moles) = mass(g) ÷ molar mass(g mol⁻¹)
Number of particles = n × Nₐ

Gases at STP: From Avogadro's law, one mole of any gas occupies 22.4 dm³ at STP (0°C, 1 atm). This is molar volume of gas.
n = Volume at STP (dm³) ÷ 22.4 dm³ mol⁻¹
Note: IUPAC changed standard pressure to 1 bar → molar volume = 22.71 L mol⁻¹.

✅ Short Answers
Q. Define one mole. What is Avogadro's constant? (2 marks)
One mole is the amount of a substance that contains as many entities (atoms, molecules, or ions) as there are atoms in exactly 12g of Carbon-12. Thus, one mole contains 6.0221367 × 10²³ particles.
Avogadro's Constant (Nₐ) = 6.0221367 × 10²³ mol⁻¹. Named in honour of Amedeo Avogadro. In the SI system, mole (symbol: mol) is the seventh base quantity for amount of substance.
Q. What is molar mass? Find molar mass of O₂ and NaCl. (2 marks)
Molar mass is the mass of one mole of a substance (element/compound) in grams. It is numerically equal to the atomic mass or molecular/formula mass in u.
Molar mass of O₂ = 2 × 16 = 32 g mol⁻¹
Molar mass of NaCl = 23 + 35.5 = 58.5 g mol⁻¹
Q. Calculate the number of moles and molecules in 5.6 g of urea (NH₂CONH₂). (3 marks)
Molecular mass of urea = 2(14) + 4(1) + 12 + 16 = 28 + 4 + 12 + 16 = 60u → Molar mass = 60 g mol⁻¹
n = mass / molar mass = 5.6 / 60 = 0.0933 mol
Number of molecules = 0.0933 × 6.022 × 10²³ = 5.618 × 10²² molecules
Q. What is molar volume of a gas? (2 marks)
Molar volume of a gas is the volume occupied by one mole of any gas at STP (Standard Temperature = 0°C, Standard Pressure = 1 atm). Its value is 22.4 dm³ mol⁻¹ (or 22.4 L mol⁻¹).
n (moles) = Volume at STP (dm³) / 22.4 dm³ mol⁻¹
This is deduced from Avogadro's law — equal volumes of all gases at same T & P contain equal number of molecules.

Chapter 2 — Introduction to Analytical Chemistry

Maharashtra HSC Board — Complete Key Points with Explanations

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Chapter 2 — Quick Reference

All Key Points at a Glance

🔬 Analytical Chemistry
  • Analytical chemistry: branch dealing with separation, identification, qualitative and quantitative determination of composition of substances
  • Uses instruments and methods to separate, identify and quantify matter
  • Qualitative analysis: detection of presence/absence of elements
  • Quantitative analysis: determination of relative proportions of elements
  • Analysis carried on a small sample, not on entire bulk
  • Semi-microanalysis: when solid/liquid sample is a few grams
📊 Types of Analysis
  • Chemical analysis: dry method (sample not dissolved) + wet method (sample dissolved first)
  • Dry method = preliminary test in qualitative analysis
  • Classical qualitative: precipitation, extraction, distillation; identification by colour, odour, mp, bp
  • Classical quantitative: volumetric and gravimetric analysis
  • Organic elements: C, H, O, N, S, halogen, P
  • Inorganic qualitative: detection of cationic (basic) and anionic (acidic) radicals
📐 Scientific Notation
  • Format: N × 10ⁿ where 1 ≤ N < 10, n = positive or negative integer
  • 6.022 × 10²³ = Avogadro's number
  • 1.66 × 10⁻²⁴ g = 1 amu
  • Addition/Subtraction: make exponents equal first, then add/subtract coefficients
  • Multiplication: multiply coefficients, add exponents
  • Division: divide coefficients, subtract exponents
🎯 Accuracy & Precision
  • Accuracy: nearness of measured value to true value
  • Precision: reproducibility of readings; if readings match closely → high precision
  • High precision is a prerequisite for high accuracy
  • Absolute error = Observed value − True value
  • Relative error = (Absolute error / True value) × 100%
  • Absolute deviation = |Observed value − Mean|
  • Relative deviation = (Mean absolute deviation / Mean) × 100%
🔢 Significant Figures Rules
  • Sig. figs = all certain digits + one uncertain digit
  • Rule 1: All non-zero digits are significant (127.34 → 5 sig figs)
  • Rule 2: Zeros between non-zero digits are significant (120.007 → 6)
  • Rule 3: Zeros to the left of first non-zero digit are NOT significant (0.025 → 2)
  • Rule 4: Zeros to right of decimal point are significant (0.400 → 3)
  • Rule 5: Terminal zeros without decimal are NOT significant (400 → 1)
  • Rule 6: In scientific notation, all digits are significant
  • Rounding off: if next digit <5 → leave unchanged; if ≥5 → increase by 1
🧬 Molecular Formula
  • Empirical formula: simplest whole number ratio of atoms of constituent elements
  • Molecular formula: actual number of atoms in a molecule = r × empirical formula
  • r = molar mass / empirical formula mass
  • Steps: % → g → moles → divide by smallest → whole number ratio → empirical formula → find r → molecular formula
  • If % total ≠ 100, difference = % oxygen
⚗️ Stoichiometry & Limiting Reagent
  • Stoichiometric calculations: calculations based on balanced chemical equations
  • Balanced equation indicates: moles of reactants & products; relative masses; volume relationships at STP
  • Types: mass-mass, mass-volume, volume-volume problems
  • Limiting reagent: reactant consumed first; limits the amount of product formed
  • Other reactant = excess reagent
  • Identify limiting reagent: calculate product from each reactant; reagent giving less product is limiting
🧪 Concentration of Solution
  • Mass percent (w/w%) = (mass of solute / mass of solution) × 100
  • Mole fraction: χₐ = nₐ / (nₐ + n_B). Sum of all mole fractions = 1
  • Molarity (M) = moles of solute / volume of solution (L). Unit: mol L⁻¹. Changes with temperature.
  • Molality (m) = moles of solute / mass of solvent (kg). Unit: mol kg⁻¹. Does NOT change with temperature.
  • Dilute stock solution: M₁V₁ = M₂V₂
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Section 2.1–2.2

Introduction to Analytical Chemistry & Types of Analysis

📖 Explanation

Analytical chemistry facilitates investigation of chemical composition of substances. It uses instruments and methods to separate, identify and quantify matter. The analysis provides chemical or physical information about a sample.

Importance: Chemical analysis is one of the most important methods of monitoring composition of raw materials, intermediates and finished products. Used in agriculture (soil, fertilizer analysis), medicine (medicinal preparations), forensic science, engineering and industry.

Analysis is carried out on a small sample, not on the entire bulk. When the amount is a few grams, it is called semi-microanalysis.

Chemical methods of qualitative analysis are carried out mainly in two stages:
1. Dry method: sample under test is NOT dissolved (used as preliminary test)
2. Wet method: sample is first dissolved, then analyzed

Classical quantitative methods: volumetric analysis (titrimetric) and gravimetric analysis.

✅ Short Answers
Q. What is analytical chemistry? State its importance. (3 marks)
Analytical chemistry is the branch of chemistry that deals with the study of separation, identification, qualitative and quantitative determination of the compositions of different substances. It uses instruments and methods to separate, identify and quantify matter.

Importance:
1. Monitoring composition of raw materials, intermediates, and finished products
2. In agriculture — analysis of soils and fertilizers
3. In medicine — determination of composition of medicinal preparations
4. Applications in forensic science, engineering and industry
5. Industrial processes and production of new materials are closely associated with analytical chemistry
Q. What is the difference between qualitative and quantitative analysis? (2 marks)
Qualitative analysis: Concerned with the detection of the presence or absence of elements in compounds and mixtures of compounds. It answers "what is present?"

Quantitative analysis: Deals with the determination of the relative proportions of elements in compounds and mixtures. It answers "how much is present?" Methods include volumetric (titration) and gravimetric analysis.
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Section 2.3

Accuracy, Precision & Significant Figures

📖 Explanation

Accuracy: Nearness of the measured value to the true (accepted) value. Larger accuracy → smaller error. Depends on the least count of the instrument.

Precision: If multiple readings of the same quantity match closely, they have high precision. High precision implies reproducibility. High precision is a prerequisite for high accuracy.

Errors:
• Absolute error = Observed value − True value
• Relative error = (Absolute error / True value) × 100%
• Absolute deviation = |Observed value − Mean|
• Relative deviation = (Mean absolute deviation / Mean) × 100%

Significant figures: The number of digits known with certainty plus one uncertain digit. The final result cannot be more accurate than the least accurate measurement used. Smaller least count → more significant figures → more accurate measurement.

✅ Short Answers
Q. Distinguish between accuracy and precision. (2 marks)
Accuracy: Nearness of the measured value to the true value. A measurement is accurate if it is close to the correct (accepted) value. Larger accuracy = smaller error.

Precision: Reproducibility of measurements. If multiple readings of the same quantity closely match each other, they are said to have high precision. High precision is a prerequisite for high accuracy, but highly precise measurements may still be inaccurate if there is a systematic error.
Q. State the rules for deciding significant figures. (3 marks)
1. All non-zero digits are significant. (e.g., 127.34 → 5 sig figs)
2. All zeros between two non-zero digits are significant. (e.g., 120.007 → 6 sig figs)
3. Zeros to the left of first non-zero digit are NOT significant. (e.g., 0.025 → 2 sig figs)
4. Zeros at the end of a number are significant if they are on right side of decimal. (e.g., 0.400 → 3 sig figs)
5. Terminal zeros without decimal point are not significant. (e.g., 400 → 1 sig fig; but 4.00 × 10² → 3 sig figs)
6. In numbers written in scientific notation, all digits are significant. (e.g., 2.035 × 10² → 4 sig figs)
Q. In an experiment, observed value is 3.8g, true value is 3.92g. Find absolute and relative error. (2 marks)
Absolute error = Observed − True = 3.8 − 3.92 = −0.12 g
(Negative sign indicates result is lower than true value)
Relative error = (Absolute error / True value) × 100% = (−0.12 / 3.92) × 100% = −3.06%
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Section 2.4

Empirical Formula & Molecular Formula

📖 Explanation

Molecular formula: indicates the actual number of atoms of constituent elements in a molecule.
Empirical formula: represents the simplest whole number ratio of atoms of the constituent elements in a molecule.

Steps to find empirical formula from % composition:
Step I: Check if % total = 100. If not, difference = % oxygen.
Step II: Convert mass percent to grams (use 100g sample).
Step III: Convert to moles by dividing by atomic mass of each element.
Step IV: Divide all mole values by the smallest mole value to get ratio.
Step V: If ratio is not whole number, multiply by suitable coefficient.
Step VI: Write empirical formula.

To find molecular formula:
r = Molar mass / Empirical formula mass
Molecular formula = r × Empirical formula

✅ Short Answers
Q. What are empirical formula and molecular formula? How are they related? (2 marks)
Empirical formula: The simplest ratio of atoms of the constituent elements in a molecule. Example: CH₂Cl (for CH₂Cl₂).
Molecular formula: The actual number of atoms of constituent elements in one molecule. It can be obtained from empirical formula if molar mass is known.
Relation: Molecular formula = r × Empirical formula, where r = Molar mass / Empirical formula mass. Example: If empirical formula = CH₂Cl (mass 49.48g) and molar mass = 98.96g, then r = 2, so molecular formula = C₂H₄Cl₂.
Q. A compound contains 4.07% H, 24.27% C and 71.65% Cl. Molar mass = 98.96g. Find molecular formula. (3 marks)
Step I: 4.07 + 24.27 + 71.65 = 99.99 ≈ 100. No oxygen.
Step II & III: Moles — H: 4.07/1.008 = 4.04 | C: 24.27/12.01 = 2.02 | Cl: 71.65/35.453 = 2.02
Step IV: Divide by 2.02 → H: 2, C: 1, Cl: 1
Empirical formula = CH₂Cl | Empirical mass = 12 + 2 + 35.5 = 49.5g
r = 98.96 / 49.48 = 2
Molecular formula = C₂H₄Cl₂
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Section 2.5–2.6

Stoichiometric Calculations & Limiting Reagent

📖 Explanation

Stoichiometric calculations: Calculations based on balanced chemical equations. A balanced equation is a symbolic representation of a chemical reaction supplying:
(i) Number of moles of reactants & products
(ii) Relative masses of reactants & products
(iii) Volume relationships of gaseous reactants & products at STP

Types of stoichiometric problems: (a) mass-mass (b) mass-volume (c) volume-volume

Limiting Reagent: When reactants are not in stoichiometric amounts, the reactant consumed completely first is called the limiting reagent. It limits the amount of product formed. The other reactant is the excess reagent.

To identify: Calculate product formed from each reactant separately. The reagent producing less product is the limiting reagent.

✅ Short Answers
Q. What is a limiting reagent? How do you identify it? (3 marks)
In a chemical reaction, when reactants are not present in exact stoichiometric amounts, the reactant which gets consumed completely first is called the limiting reagent. It limits the amount of product formed. The other reactant remaining after the reaction is called the excess reagent.

Identification: Calculate the number of moles of product that would be formed from each reactant separately. The reactant that gives less amount of product is the limiting reagent.
Example: 2NO + O₂ → 2NO₂. With 8 mol NO and 7 mol O₂ — from NO: 8 mol NO₂ produced; from O₂: 14 mol NO₂ possible. Since NO gives less product, NO is the limiting reagent.
Q. How much CaO will be produced by decomposition of 5g CaCO₃? (3 marks)
Balanced equation: CaCO₃ → CaO + CO₂
Molar masses: CaCO₃ = 40 + 12 + 48 = 100 g | CaO = 40 + 16 = 56 g
100 g CaCO₃ produces 56 g CaO
∴ 5 g CaCO₃ produces = (56/100) × 5 = 2.8 g CaO
🧪
Section 2.7

Concentration of Solution (Molarity, Molality, Mole Fraction)

📖 Explanation

Concentration expresses the amount of solute present in a given volume of solution.

1. Mass percent (w/w%) = (Mass of solute / Mass of solution) × 100%

2. Mole fraction (χ): Ratio of moles of a component to total moles of solution.
χ_A = n_A / (n_A + n_B). Sum of all mole fractions = 1.

3. Molarity (M): Number of moles of solute per litre of solution.
M = moles of solute / volume of solution in litres
Unit: mol L⁻¹. Molarity changes with temperature (volume changes).

4. Molality (m): Number of moles of solute per kg of solvent.
m = moles of solute / mass of solvent in kg
Unit: mol kg⁻¹. Molality does NOT change with temperature (mass is unaffected).

✅ Short Answers
Q. Define molarity. Calculate molarity of NaOH solution — 4g dissolved in 250mL solution. (3 marks)
Molarity (M) is defined as the number of moles of solute present in 1 litre of the solution.
M = moles of solute / volume of solution (litres)

Molar mass of NaOH = 23 + 16 + 1 = 40 g mol⁻¹
Moles of NaOH = 4 / 40 = 0.1 mol
Volume = 250 mL = 0.250 L
M = 0.1 / 0.250 = 0.4 mol L⁻¹ = 0.4 M
Q. Define molality. Why does molality not change with temperature? (2 marks)
Molality (m) is defined as the number of moles of solute present in 1 kg of solvent.
m = moles of solute / mass of solvent (kg). Unit: mol kg⁻¹.

Molality does not change with temperature because it is based on mass of solvent, which remains constant regardless of temperature. Molarity changes with temperature because it depends on volume of solution, which expands or contracts with temperature.
Q. What is mole fraction? If 2g substance A is dissolved in 18g water, find mole fraction of A. (Molar mass of A = 32, water = 18) (3 marks)
Mole fraction: Ratio of number of moles of a particular component to the total number of moles of the solution. Sum of all mole fractions in a solution = 1.

Moles of A = 2/32 = 0.0625 mol
Moles of water = 18/18 = 1 mol
χ_A = 0.0625 / (0.0625 + 1) = 0.0625 / 1.0625 = 0.0588

🔢 All Formulas — Quick Revision

⚖️ Laws of Combination Mass reactants = Mass products
Ratio = fixed (definite proportions)
Multiple proportions: small whole numbers
Gay Lussac: simple volume ratio
Avogadro: equal volumes = equal molecules
⚛️ Atomic Mass 1 amu = 1.66056 × 10⁻²⁴ g
Avg. atomic mass = Σ(mass × %)/100
Mol. mass = Σ(atomic masses of atoms)
Nₐ = 6.0221367 × 10²³ mol⁻¹
🔢 Mole Calculations n = mass (g) / molar mass (g/mol)
Particles = n × Nₐ
n (gas) = V(STP) / 22.4 dm³
K = °C + 273.15
°F = (9/5)°C + 32
📏 Density & Volume Density = mass / volume
SI: kg/m³ | CGS: g/mL
1 L = 1 dm³ = 1000 mL
1 nm = 10⁻⁹ m | 1 pm = 10⁻¹² m
🎯 Errors Absolute error = Obs − True
Relative error = (Abs error / True) × 100%
Abs deviation = |Obs − Mean|
Rel deviation = (Mean abs dev / Mean) × 100%
🧬 Molecular Formula % → moles → ratio → empirical formula
r = Molar mass / Emp. formula mass
Mol. formula = r × Emp. formula
If %total ≠ 100 → diff = % oxygen
🧪 Concentration Mass% = (mass solute/mass soln) × 100
χ_A = n_A / (n_A + n_B)
M = mol solute / V(L)
m = mol solute / mass solvent(kg)
⚗️ Stoichiometry Use balanced chemical equation
Find moles → use mole ratio → convert to mass/volume
Limiting reagent → less product
Excess = initial − reacted
🌡️ Temperature K = °C + 273.15
°F = (9/5)(°C) + 32
SI unit of temperature: Kelvin (K)
Absolute zero = 0 K = −273.15°C

📝 2 & 3 Mark Practice Questions

Important short answer questions for HSC Board exams — Chapter 1 & 2

✍️
Short Answer

2 Mark Questions (Chapter 1 & 2)

Q1. Define chemistry. (2 marks)
Chemistry is the study of matter, its physical and chemical properties and the physical and chemical changes it undergoes under different conditions. It is known as a central science as its knowledge is required in physics, biological sciences, applied sciences, and earth and space sciences.
Q2. What are elements? Give examples. (2 marks)
Elements are pure substances which cannot be broken down into simpler substances by ordinary chemical changes. They are further classified as metals (gold, silver, copper), non-metals (iodine, nitrogen), and metalloids (silicon, germanium).
Q3. Give properties of metals. (2 marks)
Metals: (i) have lustre (shiny appearance), (ii) conduct heat and electricity, (iii) can be drawn into wire (ductile), (iv) can be hammered into thin sheets (malleable). Examples: gold, silver, copper, iron. Mercury is a liquid metal at room temperature.
Q4. Distinguish between compounds and mixtures. (2 marks)
Compounds: Pure substances where two or more elements are combined in a fixed proportion. Can be broken down by chemical changes. Example: water (H₂O), salt (NaCl).
Mixtures: Contain two or more substances in no fixed proportions. Can be separated by physical methods. Example: sea water, air, paint.
Q5. What is one amu? What is its value in grams? (2 marks)
One atomic mass unit (amu or u or dalton) is defined as a mass exactly equal to one-twelfth of the mass of one Carbon-12 atom. Its value: 1 amu = (1/12) × 1.992648 × 10⁻²³ g = 1.66056 × 10⁻²⁴ g.
Q6. What is molar volume of a gas? State its value at STP. (2 marks)
Molar volume of a gas is the volume occupied by one mole of any gas at STP (0°C, 1 atm). Its value is 22.4 dm³ mol⁻¹. It is deduced from Avogadro's law. Number of moles of gas = Volume at STP / 22.4 dm³ mol⁻¹.
Q7. What is significant figures? (2 marks)
Significant figures in a measurement or result are the number of digits known with certainty plus one uncertain digit. They indicate the accuracy of a measurement — more significant figures means more accurate measurement. The final result of a calculation cannot be more accurate than the least accurate number used.
Q8. Why does molarity change with temperature but molality does not? (2 marks)
Molarity = moles of solute / volume of solution (L). Volume of solution changes with temperature (expands on heating), so molarity changes with temperature.
Molality = moles of solute / mass of solvent (kg). Mass does not change with temperature, so molality remains constant regardless of temperature change.
Q9. How many significant figures are in 1.50 × 10⁴ g and 0.0025 kg? (2 marks)
1.50 × 10⁴ g: In scientific notation all digits are significant. 1, 5, 0 → 3 significant figures.
0.0025 kg: Zeros to the left of first non-zero digit are not significant. 2 and 5 are significant → 2 significant figures.
Q10. State law of conservation of mass. (2 marks)
Proposed by Lavoisier: "Mass can neither be created nor destroyed." In a chemical reaction, the total mass of the reactants is always equal to the total mass of the products. Example: When hydrogen burns with oxygen, mass of water formed = mass of hydrogen + mass of oxygen consumed.
📋
Long Short Answer

3 Mark Questions (Chapter 1 & 2)

Q1. State and explain Dalton's atomic theory. (3 marks)
Dalton published "A New System of Chemical Philosophy" (1808). His theory has four postulates:
1. Matter consists of tiny, indivisible particles called atoms.
2. All atoms of a given element have identical properties including mass; atoms of different elements differ in mass.
3. Compounds are formed when atoms of different elements combine in a fixed ratio.
4. Chemical reactions involve only reorganization of atoms; atoms are neither created nor destroyed.
Dalton's theory successfully explained all the laws of chemical combination (conservation of mass, definite proportions, multiple proportions).
Q2. Calculate average atomic mass of neon: ²⁰Ne (19.9924u, 90.92%), ²¹Ne (20.9940u, 0.26%), ²²Ne (21.9914u, 8.82%). (3 marks)
Average atomic mass = [Σ (atomic mass × % abundance)] / 100
= [(19.9924 × 90.92) + (20.9940 × 0.26) + (21.9914 × 8.82)] / 100
= [1817.67 + 5.46 + 193.98] / 100
= 2017.11 / 100
= 20.17 u
Q3. Explain the types of matter on the basis of chemical composition with diagram. (3 marks)
Matter is classified based on chemical composition into:
Pure substances: Definite chemical composition, same properties always.
  → Elements: Cannot be broken down further (gold, oxygen, silicon)
    → Metals, Non-metals, Metalloids
  → Compounds: Elements combined in fixed ratio, can be decomposed (water, salt, CO₂)
Mixtures: No fixed composition, separated by physical methods.
  → Homogeneous: Uniform composition throughout (solution, air)
  → Heterogeneous: Non-uniform composition (suspension, soil)
Q4. Calculate no. of moles and molecules in 52g of Helium. (3 marks)
Atomic mass of He = 4u → Molar mass = 4 g mol⁻¹
Number of moles = mass / molar mass = 52 / 4 = 13 mol
Number of atoms = 13 × 6.022 × 10²³ = 78.286 × 10²³ atoms = 7.83 × 10²⁴ atoms
Q5. State and explain the law of multiple proportions with two examples. (3 marks)
Law (Dalton, 1803): When two elements A and B form more than one compound, the masses of B that combine with a fixed mass of A are always in the ratio of small whole numbers.

Example 1: H and O form water (H₂:O = 2g:16g) and H₂O₂ (H₂:O = 2g:32g). Ratio of oxygen = 16:32 = 1:2.

Example 2: N and O form NO (N:O = 14g:16g) and NO₂ (N:O = 14g:32g). Ratio of oxygen = 16:32 = 1:2. (Similar: CO and CO₂ → 1:2 ratio)
Q6. State the rules of significant figures with examples. (3 marks)
1. All non-zero digits significant: 127.34 → 5 sig figs
2. Zeros between non-zero digits significant: 120.007 → 6 sig figs
3. Zeros left of first non-zero NOT significant: 0.025 → 2 sig figs
4. Zeros right of decimal significant: 0.400 → 3 sig figs
5. Terminal zeros without decimal NOT significant: 400 → 1 sig fig
6. In scientific notation, all digits significant: 2.035 × 10² → 4 sig figs
Q7. Explain molarity, molality and mole fraction. (3 marks)
Molarity (M): Number of moles of solute in 1 litre of solution. M = n / V(L). Unit: mol L⁻¹. Changes with temperature.

Molality (m): Number of moles of solute in 1 kg of solvent. m = n / mass(kg). Unit: mol kg⁻¹. Does NOT change with temperature.

Mole fraction (χ): Ratio of moles of one component to total moles of solution. χ_A = n_A / (n_A + n_B). It is dimensionless. Sum of mole fractions of all components = 1.
Q8. What is stoichiometry? List information given by balanced equation. (3 marks)
Stoichiometry: Branch dealing with calculations based on balanced chemical equations. A balanced chemical equation is a symbolic representation of a chemical reaction.

A balanced equation gives the following information:
(i) It indicates the number of moles of reactants involved and the number of moles of products formed.
(ii) It indicates the relative masses of the reactants and products linked with the chemical change.
(iii) It indicates the relationship between the volumes of gaseous reactants and products at STP.
Q9. A compound has molar mass 159 and contains 39.62% Cu and 20.13% S. Find its molecular formula. (Atomic masses: Cu=63, S=32, O=16) (3 marks)
% Cu + % S = 59.75 < 100. So % O = 100 − 59.75 = 40.25%
Moles: Cu = 39.62/63 = 0.629 | S = 20.13/32 = 0.629 | O = 40.25/16 = 2.516
Ratio: Cu:S:O = 0.629:0.629:2.516 = 1:1:4
Empirical formula = CuSO₄
Empirical mass = 63 + 32 + 64 = 159g. Molar mass = 159g.
r = 159/159 = 1. Molecular formula = CuSO₄
Q10. Explain the mole concept. How many atoms are in 1 mole of oxygen atoms? (3 marks)
Expressing large counts is easy with quantitative adjectives like "dozen" or "gross." Even a small amount of substance contains very large numbers of atoms or molecules. A quantitative adjective 'mole' is used to express the large number of submicroscopic entities like atoms, ions, electrons.

Definition: One mole is the amount of a substance that contains as many entities as there are atoms in exactly 12g of Carbon-12.

Number of atoms in 12g C-12 = 12g / (1.992648 × 10⁻²³ g/atom) = 6.0221367 × 10²³ atoms.
So 1 mole of oxygen atoms = 6.0221367 × 10²³ atoms of oxygen. This number is called Avogadro's Constant (Nₐ).

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