NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry
NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry covers the importance of chemistry, nature and classification of matter, laws of chemical combination, Dalton's atomic theory, atomic and molecular masses, the mole concept, percentage composition, empirical and molecular formulas, stoichiometry, limiting reagent, and concentration terms in solutions.
Overview of Topics & Subtopics (Click to Jump)
1. Importance of Chemistry, Matter & Its Classification
Short Note:
• Matter: Anything that occupies space and has mass. Exists in three physical states: solid, liquid, and gas.
• Chemical Classification of Matter:
- Pure Substances: Fixed composition (Elements like Au, O2 and Compounds like H2O, NaCl).
- Mixtures: Variable composition. Homogeneous (uniform composition throughout, e.g., sugar solution) or Heterogeneous (non-uniform composition, e.g., mixture of sand and salt).
• Significant Figures: Meaningful digits in a measured or calculated quantity. Rules include non-zero digits are significant, zeros between non-zeros are significant, and leading zeros are not significant.
2. Laws of Chemical Combinations & Dalton's Atomic Theory
Short Note:
• Law of Conservation of Mass (Lavoisier): Mass 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 chemical compound always contains exactly the same proportion of elements by weight.
• Law of Multiple Proportions (Dalton): When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in small whole number ratios.
• Gay Lussac's Law of Gaseous Volumes: When gases combine in a chemical reaction, they do so in a simple ratio by volume, provided all gases are at same temperature and pressure.
• Avogadro's Law: Equal volumes of all gases under identical conditions of temperature and pressure contain equal number of molecules.
3. Atomic & Molecular Masses, Mole Concept & Molar Mass
Short Note:
• Atomic Mass Unit (amu or u): Mass exactly equal to 1/12th the mass of one carbon-12 atom (1 u = 1.66056 × 10-24 g).
• Average Atomic Mass: Sum of (fractional abundance × atomic mass) of all isotopes of an element.
• The Mole Concept: One mole is the amount of substance that contains as many elementary entities (atoms, molecules, or ions) as there are atoms in exactly 12 g of C-12 isotope.
• Avogadro Constant (NA): 6.022 × 1023 particles/mol.
• Molar Volume: Volume occupied by 1 mole of any ideal gas at STP (273.15 K and 1 bar) = 22.7 L (or 22.4 L at 1 atm).
4. Empirical/Molecular Formula, Stoichiometry & Concentration Terms
Short Note:
• Empirical Formula vs Molecular Formula: Empirical formula gives the simplest whole number ratio of atoms present in a compound, whereas molecular formula gives the exact number of atoms of each element present in a molecule. Molecular Formula = n × Empirical Formula.
• Limiting Reagent: The reactant which gets completely consumed first in a chemical reaction and limits the amount of product formed.
• Concentration Terms:
- Mass Percentage (% w/w): (Mass of solute / Mass of solution) × 100.
- Mole Fraction (x): Moles of component / Total moles in solution.
- Molarity (M): Moles of solute / Volume of solution in liters (unit: mol/L or M). Temperature dependent.
- Molality (m): Moles of solute / Mass of solvent in kilograms (unit: mol/kg or m). Independent of temperature.
Intext & Conceptual Questions
Explanation:
• Molarity (M) is defined as moles of solute per liter of solution. Since volume changes with temperature due to thermal expansion/contraction, molarity changes with temperature.
• Molality (m) is defined as moles of solute per kilogram of solvent. Since mass remains constant and is independent of temperature, molality does not change with temperature.
Hence, molality is preferred when concentration needs to be expressed accurately across varying temperatures.
Textbook Chapter End Exercise Questions
Atomic masses: H = 1.008 u, C = 12.011 u, O = 16.00 u
(i) Molar mass of Water (H2O):
= 2 × (Mass of H) + 1 × (Mass of O)
= 2 × (1.008 u) + 16.00 u = 2.016 + 16.00 = 18.016 g/mol.
(ii) Molar mass of Carbon Dioxide (CO2):
= 1 × (Mass of C) + 2 × (Mass of O)
= 12.011 u + 2 × (16.00 u) = 12.011 + 32.00 = 44.011 g/mol.
(iii) Molar mass of Methane (CH4):
= 1 × (Mass of C) + 4 × (Mass of H)
= 12.011 u + 4 × (1.008 u) = 12.011 + 4.032 = 16.043 g/mol.
Given percentages:
H = 4.07%, C = 24.27%, Cl = 71.65%
Molar mass = 98.96 g/mol
Step 1: Calculate moles of each element (in 100 g compound):
• Moles of H = 4.07 / 1.008 = 4.04
• Moles of C = 24.27 / 12.011 = 2.02
• Moles of Cl = 71.65 / 35.453 = 2.02
Step 2: Determine simplest molar ratio:
Divide each mole value by the smallest number (2.02):
• Ratio for C = 2.02 / 2.02 = 1
• Ratio for H = 4.04 / 2.02 = 2
• Ratio for Cl = 2.02 / 2.02 = 1
Therefore, Empirical Formula = CH2Cl.
Step 3: Calculate Molecular Formula:
Empirical formula mass = 12.011 + (2 × 1.008) + 35.453 = 49.48 g/mol
n = Molar Mass / Empirical Formula Mass = 98.96 / 49.48 = 2
Molecular Formula = 2 × (CH2Cl) = C2H4Cl2 (Dichloroethane).
Read More & Chapter Resources
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