Some Basic Concepts of Chemistry — Previous Year Questions (Class 11 Chemistry)
Fundamental chemistry concepts including atomic mass, molar mass, stoichiometry, and chemical equations. Essential for all chemistry calculations.
TL;DR: Fundamental chemistry concepts including atomic mass, molar mass, stoichiometry, and chemical equations. Essential for all chemistry calculations.
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Fundamental chemistry concepts including atomic mass, molar mass, stoichiometry, and chemical equations. Essential for all chemistry calculations.
Some Basic Concepts of Chemistry — Previous Year Questions with Solutions
Q (2023, 3 marks): Define mole and Avogadro's number. How many atoms are in 0.5 mole of oxygen atoms?
Answer: Mole: SI unit of amount of substance. 1 mole = 6.022 × 10^23 particles (atoms, molecules, ions)
Avogadro's number (NA) = 6.022 × 10^23 mol^-1
Number of atoms in 0.5 mole of oxygen:
N = n × NA = 0.5 × 6.022 × 10^23
N = 3.011 × 10^23 atoms
Q (2022, 2 marks): Calculate molar mass of Ca(OH)2. (Ca = 40, O = 16, H = 1)
Answer: Molar mass of Ca(OH)2 = 40 + 2(16 + 1)
= 40 + 2(17)
= 40 + 34
= 74 g/mol
Q (2023, 2 marks): Balance the chemical equation: Fe + O2 → Fe2O3
Answer: Unbalanced: Fe + O2 → Fe2O3
Fe: 1 on left, 2 on right → multiply Fe by 2
O: 2 on left (from O2), 3 on right → need common multiple
Try: 4Fe + 3O2 → 2Fe2O3
Check:
Fe: 4 = 2 × 2 ✓
O: 3 × 2 = 6 = 2 × 3 ✓
Balanced equation: 4Fe + 3O2 → 2Fe2O3
Q (2021, 3 marks): From the equation 2H2 + O2 → 2H2O, calculate: (a) moles of H2O formed from 4 moles of H2, (b) mass of H2O formed (H = 1, O = 16)
Answer: (a) From equation: 2 moles H2 → 2 moles H2O
Molar ratio = 2:2 = 1:1
From 4 moles H2: 4 moles H2O formed
(b) Molar mass of H2O = 2(1) + 16 = 18 g/mol
Mass = 4 × 18 = 72 g of H2O
Q (2022, 3 marks): What is the empirical formula of a compound containing C = 75%, H = 25%? (C = 12, H = 1)
Answer: Assume 100 g sample:
C: 75 g ÷ 12 = 6.25 mol
H: 25 g ÷ 1 = 25 mol
Simplest ratio: Divide by smallest (6.25)
C: 6.25 ÷ 6.25 = 1
H: 25 ÷ 6.25 = 4
Empirical formula: CH4
Q (2023, 3 marks): In reaction: 2FeCl3 + 3H2S → 2FeCl2 + S + 6HCl, identify oxidation number changes for Fe and S.
Answer: FeCl3: Fe is +3, Cl is -1
FeCl2: Fe is +2, Cl is -1
H2S: H is +1, S is -2
S (elemental): S is 0
HCl: H is +1, Cl is -1
Changes:
Fe: +3 → +2 (reduction, gains 1 electron per Fe)
S: -2 → 0 (oxidation, loses 2 electrons per S)
FeCl3 is oxidizing agent; H2S is reducing agent
Frequently Asked Questions
What is the difference between empirical and molecular formula?
Empirical formula shows the simplest whole-number ratio of atoms (e.g., CH2O). Molecular formula shows the actual number of atoms in a molecule (e.g., C6H12O6). The molecular formula is a whole-number multiple of the empirical formula.
What are oxidation numbers and why are they useful?
Oxidation numbers represent the number of electrons gained/lost by an atom in a compound. Rules: (1) Element in free state = 0, (2) Monoatomic ion = its charge, (3) O usually -2 (except peroxides), (4) H usually +1. They help identify redox reactions and electron transfer.
More Class 11 Chemistry PYQs
- Electrochemistry
- Solutions
- Chemical Kinetics
- Coordination Compounds
- Haloalkanes and Haloarenes
- Aldehydes Ketones and Carboxylic Acids
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