Class 11 Notes

Class 11 Chemistry Notes Chapter 1 (Some basic concepts of chemistry) – Chemistry Part-I Book

Philoid

Philoid

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Chemistry Part-I
Alright class, let's begin our focused revision of Chapter 1: 'Some Basic Concepts of Chemistry'. This chapter forms the bedrock for understanding quantitative chemistry, which is crucial for many government exams. Pay close attention to the definitions, laws, and especially the mole concept and stoichiometry.

Chapter 1: Some Basic Concepts of Chemistry - Detailed Notes for Government Exam Preparation

1. Importance of Chemistry:

  • Chemistry deals with the composition, structure, properties, and interaction of matter.
  • It plays a central role in science and is interlinked with other branches like physics, biology, geology, etc.
  • Essential for understanding various natural phenomena and has applications in industries, medicine, agriculture, etc.

2. Nature of Matter:

  • Matter: Anything that has mass and occupies space.
  • Classification:
    • Physical: Solid, Liquid, Gas. (Based on arrangement and movement of particles).
    • Chemical:
      • Pure Substances: Have fixed composition.
        • Elements: Consist of only one type of atom (e.g., Na, O, H). Cannot be broken down further by chemical methods.
        • Compounds: Formed when two or more different elements combine chemically in a fixed ratio by mass (e.g., H₂O, CO₂, NaCl). Properties are different from constituent elements. Can be broken down into elements by chemical methods.
      • Mixtures: Consist of two or more substances (elements or compounds) mixed in any ratio. Components retain their individual properties.
        • Homogeneous Mixture: Uniform composition throughout (e.g., saltwater solution, air, alloys). Also called solutions.
        • Heterogeneous Mixture: Non-uniform composition, components often visible (e.g., sand and water, oil and water, soil).

3. Properties of Matter and their Measurement:

  • Physical Properties: Can be measured or observed without changing the identity or composition of the substance (e.g., colour, odour, melting point, boiling point, density).
  • Chemical Properties: Require a chemical change to occur for observation (e.g., acidity, basicity, combustibility, reactivity).
  • Measurement: Quantitative observation involving a number followed by a unit.
  • SI System (Système Internationale d'Unités): The internationally accepted system of units.
    • Seven Base Units:
      • Length (l): metre (m)
      • Mass (m): kilogram (kg)
      • Time (t): second (s)
      • Electric Current (I): ampere (A)
      • Thermodynamic Temperature (T): kelvin (K)
      • Amount of Substance (n): mole (mol)
      • Luminous Intensity (Iv): candela (cd)
    • Derived Units: Units derived from base units (e.g., Volume (m³), Density (kg/m³), Speed (m/s)).
    • Common Prefixes: Used to indicate multiples or submultiples of units (e.g., kilo (k, 10³), milli (m, 10⁻³), micro (µ, 10⁻⁶), nano (n, 10⁻⁹), pico (p, 10⁻¹²)).

4. Uncertainty in Measurement:

  • Scientific Notation: Expressing numbers as N × 10ⁿ, where N is a number between 1.000... and 9.999... and n is an integer (positive or negative exponent). Useful for very large or small numbers.
  • Significant Figures: Meaningful digits in a measured or calculated quantity. Indicate the precision of measurement.
    • Rules:
      1. All non-zero digits are significant.
      2. Zeros between non-zero digits are significant.
      3. Leading zeros (zeros to the left of the first non-zero digit) are NOT significant. (e.g., 0.003 has 1 significant figure).
      4. Trailing zeros (zeros at the end of a number) are significant ONLY if the number contains a decimal point. (e.g., 1200 has 2; 1200. has 4; 0.01200 has 4).
    • Calculations:
      • Addition/Subtraction: Result should have the same number of decimal places as the number with the fewest decimal places.
      • Multiplication/Division: Result should have the same number of significant figures as the number with the fewest significant figures.
  • Accuracy vs. Precision:
    • Accuracy: Closeness of a measurement to the true value.
    • Precision: Closeness of various measurements for the same quantity (reproducibility).

5. Laws of Chemical Combinations:

  • Law of Conservation of Mass (Lavoisier): Matter can neither be created nor destroyed in a chemical reaction. (Total mass of reactants = Total mass of products).
  • Law of Definite Proportions (Proust): A given compound always contains exactly the same proportion of elements by weight, irrespective of its source or method of preparation. (e.g., H₂O always has H and O in a 1:8 ratio by mass).
  • Law of Multiple Proportions (Dalton): 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 a simple whole-number ratio. (e.g., H₂O and H₂O₂; masses of O combining with fixed mass of H are in 1:2 ratio).
  • Gay Lussac's Law of Gaseous Volumes: 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.
  • Avogadro's Law: Equal volumes of all gases at the same temperature and pressure contain an equal number of molecules. (V ∝ n, where n is the number of moles).

6. Dalton's Atomic Theory:

  • Postulates:
    1. Matter consists of indivisible particles called atoms.
    2. Atoms of a given element are identical in mass and properties.
    3. Atoms of different elements differ in mass and properties.
    4. Compounds are formed when atoms of different elements combine in a fixed, simple whole-number ratio.
    5. Chemical reactions involve only the reorganization of atoms. Atoms are neither created nor destroyed.
  • Limitations: Could not explain isotopes, isobars, reasons for chemical combination, Gay Lussac's Law, internal structure of atom, etc. The idea of atoms being indivisible was later proven wrong.

7. Atomic and Molecular Masses:

  • Atomic Mass Unit (amu or u): Defined as 1/12th the mass of one Carbon-12 atom. (1 amu ≈ 1.66056 × 10⁻²⁴ g).
  • Atomic Mass: Average relative mass of an atom of an element as compared to 1/12th the mass of a C-12 atom. It's usually an average value considering isotopes.
  • Molecular Mass: Sum of atomic masses of all the atoms present in one molecule of a substance.
  • Formula Mass: Sum of atomic masses of all atoms in a formula unit of an ionic compound (used instead of molecular mass for ionic compounds like NaCl).

8. Mole Concept and Molar Masses:

  • Mole (mol): The SI unit for amount of substance. It represents a specific number of entities (atoms, molecules, ions, etc.).
  • Avogadro's Number (N<0xE2><0x82><0x90>): The number of entities in one mole. N<0xE2><0x82><0x90> = 6.022 × 10²³.
  • Definition of Mole: One mole is the amount of substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams (or 0.012 kg) of the Carbon-12 isotope.
  • Molar Mass (M): The mass of one mole of a substance expressed in grams per mole (g/mol). Numerically, it is equal to the atomic/molecular/formula mass expressed in 'u', but the unit is g/mol.
    • Molar mass of an element = Atomic mass in g/mol.
    • Molar mass of a compound = Molecular/Formula mass in g/mol.
  • Key Relationships:
    • 1 mole of any substance contains 6.022 × 10²³ entities.
    • Mass of 1 mole of a substance = Molar Mass (in g).
    • 1 mole of any gas at STP (Standard Temperature and Pressure: 0°C or 273.15 K and 1 atm or 101.325 kPa) occupies 22.4 Litres (or dm³). Note: Some newer standards use 1 bar pressure, where the volume is 22.7 L. Check exam specifications, but 22.4 L is widely used.
  • Calculations:
    • Number of moles (n) = Given Mass (w) / Molar Mass (M)
    • Number of moles (n) = Number of Particles / Avogadro's Number (N<0xE2><0x82><0x90>)
    • Number of moles (n) = Volume of Gas at STP (in L) / 22.4 L/mol

9. Percentage Composition, Empirical and Molecular Formula:

  • Percentage Composition: The percentage by mass of each element present in a compound.
    • Mass % of an element = [(Mass of that element in the compound) / (Molar mass of the compound)] × 100
  • Empirical Formula: Represents the simplest whole-number ratio of various atoms present in a compound.
  • Molecular Formula: Shows the exact number of different types of atoms present in one molecule of a compound.
  • Relationship: Molecular Formula = n × (Empirical Formula), where 'n' is a simple integer (1, 2, 3...).
    • n = Molecular Mass / Empirical Formula Mass
  • Calculation Steps (from % Composition):
    1. Assume 100 g of the compound (so % becomes mass in g).
    2. Convert mass of each element to moles (divide by atomic mass).
    3. Divide the mole value of each element by the smallest mole value obtained.
    4. If ratios are not whole numbers, multiply by a suitable integer to get the simplest whole-number ratio (This gives the Empirical Formula).
    5. Calculate Empirical Formula Mass.
    6. Find 'n' using the given Molecular Mass.
    7. Multiply the subscripts in the Empirical Formula by 'n' to get the Molecular Formula.

10. Stoichiometry and Stoichiometric Calculations:

  • Stoichiometry: Deals with the quantitative relationships between reactants and products in a balanced chemical equation.
  • Balanced Chemical Equation: Provides the mole ratio (stoichiometric coefficients) in which reactants combine and products are formed.
  • Calculations: Based on mole ratios from the balanced equation. Can involve mass-mass, mass-volume, or volume-volume relationships.
    • Steps:
      1. Write a balanced chemical equation.
      2. Convert given amounts (mass, volume, etc.) into moles.
      3. Use mole ratios (stoichiometric coefficients) to calculate the moles of the required substance.
      4. Convert the calculated moles back into the desired quantity (mass, volume, etc.).
  • Limiting Reagent (or Limiting Reactant): The reactant that gets consumed completely first in a reaction and limits the amount of product formed. The other reactant(s) are called excess reagents.
    • Identifying Limiting Reagent: Calculate the moles of product formed based on each reactant (using stoichiometry). The reactant that produces the least amount of product is the limiting reagent. All calculations for product yield must be based on the limiting reagent.

11. Reactions in Solutions - Concentration Terms:

  • Solution: A homogeneous mixture of two or more substances.
  • Solute: Substance present in a smaller amount.
  • Solvent: Substance present in a larger amount.
  • Concentration: Amount of solute dissolved in a known amount of solvent or solution.
    • Mass Percent (w/w): (Mass of Solute / Mass of Solution) × 100
    • Mole Fraction (χ): Ratio of the number of moles of a particular component to the total number of moles of all components in the solution.
      • χ<0xE1><0xB5><0xA0> = n<0xE1><0xB5><0xA0> / (n<0xE1><0xB5><0xA0> + n<0xE1><0xB5><0xA7>) ; χ<0xE1><0xB5><0xA7> = n<0xE1><0xB5><0xA7> / (n<0xE1><0xB5><0xA0> + n<0xE1><0xB5><0xA7>)
      • Sum of mole fractions of all components (χ<0xE1><0xB5><0xA0> + χ<0xE1><0xB5><0xA7>) = 1
    • Molarity (M): Number of moles of solute dissolved in 1 litre (or 1 dm³) of solution.
      • M = Moles of Solute / Volume of Solution (in L)
      • Unit: mol/L or M.
      • Temperature Dependent (as volume changes with temperature).
    • Molality (m): Number of moles of solute dissolved in 1 kg of solvent.
      • m = Moles of Solute / Mass of Solvent (in kg)
      • Unit: mol/kg or m.
      • Temperature Independent (as mass does not change with temperature).

Practice MCQs:

  1. Which law states that a given compound always contains exactly the same proportion of elements by weight?
    (A) Law of Conservation of Mass
    (B) Law of Multiple Proportions
    (C) Law of Definite Proportions
    (D) Avogadro's Law

  2. The number of significant figures in the measurement 0.02040 kg is:
    (A) 3
    (B) 4
    (C) 5
    (D) 6

  3. How many moles are present in 11 g of CO₂? (Atomic mass: C=12 u, O=16 u)
    (A) 0.25 mol
    (B) 0.5 mol
    (C) 1.0 mol
    (D) 4.0 mol

  4. The number of atoms in 0.5 moles of Sodium (Na) is approximately:
    (A) 6.022 × 10²³
    (B) 12.044 × 10²³
    (C) 3.011 × 10²³
    (D) 1.505 × 10²³

  5. What is the Molarity (M) of a solution prepared by dissolving 4 g of NaOH in enough water to form 250 mL of solution? (Molar mass of NaOH = 40 g/mol)
    (A) 0.1 M
    (B) 0.2 M
    (C) 0.4 M
    (D) 0.8 M

  6. Which of the following concentration terms is independent of temperature?
    (A) Molarity
    (B) Molality
    (C) Normality
    (D) Volume percentage

  7. An organic compound contains Carbon, Hydrogen and Oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be: (Atomic mass: C=12, H=1, O=16)
    (A) CHO
    (B) CH₂O
    (C) CH₃O
    (D) CH₄O

  8. In the reaction: N₂(g) + 3H₂(g) → 2NH₃(g), if 1 mole of N₂ reacts completely, how many moles of NH₃ are formed?
    (A) 1 mole
    (B) 2 moles
    (C) 3 moles
    (D) 0.5 moles

  9. If 10 g of H₂ reacts with 64 g of O₂ to form water, which reactant is the limiting reagent? (Reaction: 2H₂ + O₂ → 2H₂O; Atomic mass: H=1, O=16)
    (A) H₂
    (B) O₂
    (C) Both H₂ and O₂
    (D) Cannot be determined

  10. One mole of any gas at Standard Temperature and Pressure (STP) occupies a volume of:
    (A) 22.4 mL
    (B) 2.24 L
    (C) 22.4 L
    (D) 1 L

Answer Key:

  1. (C)
  2. (B) - Leading zero is not significant, zero between non-zeros is significant, trailing zero after decimal is significant.
  3. (A) - Molar mass of CO₂ = 12 + 2*16 = 44 g/mol. Moles = 11g / 44 g/mol = 0.25 mol.
  4. (C) - Atoms = Moles × N<0xE2><0x82><0x90> = 0.5 × 6.022 × 10²³ ≈ 3.011 × 10²³.
  5. (C) - Moles of NaOH = 4g / 40 g/mol = 0.1 mol. Volume = 250 mL = 0.25 L. Molarity = 0.1 mol / 0.25 L = 0.4 M.
  6. (B) - Molality depends on mass of solvent, which is temperature independent. Molarity depends on volume of solution.
  7. (C) - %O = 100 - 38.71 - 9.67 = 51.62%. Moles C = 38.71/12 ≈ 3.23. Moles H = 9.67/1 ≈ 9.67. Moles O = 51.62/16 ≈ 3.23. Ratio C:H:O = (3.23/3.23) : (9.67/3.23) : (3.23/3.23) ≈ 1 : 3 : 1. Empirical formula = CH₃O.
  8. (B) - From the balanced equation, 1 mole of N₂ produces 2 moles of NH₃.
  9. (B) - Moles H₂ = 10g / 2 g/mol = 5 mol. Moles O₂ = 64g / 32 g/mol = 2 mol. From 2H₂ + O₂ → 2H₂O: 5 mol H₂ needs 5/2 = 2.5 mol O₂. We only have 2 mol O₂. 2 mol O₂ needs 2*2 = 4 mol H₂. We have 5 mol H₂ (excess). O₂ is consumed first, so O₂ is the limiting reagent.
  10. (C)

Make sure you understand the concepts behind each question, especially the calculations involving the mole concept and stoichiometry. Good luck with your preparation!

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