Some Basic Concepts of Chemistry — Previous Year Questions (Class 11 Chemistry)
Master fundamental chemistry concepts including atomic mass, molecular mass, moles, and stoichiometry. These foundations are essential for all higher-level
TL;DR: Master fundamental chemistry concepts including atomic mass, molecular mass, moles, and stoichiometry. These foundations are essential for all higher-…
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Master fundamental chemistry concepts including atomic mass, molecular mass, moles, and stoichiometry. These foundations are essential for all higher-level
Some Basic Concepts of Chemistry — Previous Year Questions with Solutions
Q (2023, 1 mark): Calculate the molecular mass of sulfuric acid (H2SO4). (H = 1, S = 32, O = 16)
Answer: H2SO4
Molecular mass = (2 × H) + (1 × S) + (4 × O)
Molecular mass = (2 × 1) + (1 × 32) + (4 × 16)
Molecular mass = 2 + 32 + 64
Molecular mass = 98 g/mol
Q (2022, 2 marks): How many moles are present in 54 g of water? (H = 1, O = 16, Molecular mass of H2O = 18)
Answer: Given:
Mass of water = 54 g
Molecular mass of H2O = 18 g/mol
Number of moles = Mass/Molecular mass
Number of moles = 54/18
Number of moles = 3 moles
Q (2024, 3 marks): Calculate the number of atoms in 2.8 g of nitrogen (N2). (Atomic mass of N = 14, Avogadro's number = 6.022 × 10^23)
Answer: Given:
Mass of N2 = 2.8 g
Molecular mass of N2 = 2 × 14 = 28 g/mol
Avogadro's number = 6.022 × 10^23
Number of moles = 2.8/28 = 0.1 mol
Number of molecules = 0.1 × 6.022 × 10^23 = 6.022 × 10^22 molecules
Number of atoms = Number of molecules × 2 (since N2 has 2 atoms)
Number of atoms = 6.022 × 10^22 × 2 = 1.204 × 10^23 atoms
Q (2023, 4 marks): A compound contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen. If its molar mass is 180 g/mol, determine its molecular formula. (C = 12, H = 1, O = 16)
Answer: Assume 100 g of compound:
C: 40 g
H: 6.67 g
O: 53.33 g
Convert to moles:
C: 40/12 = 3.33 mol
H: 6.67/1 = 6.67 mol
O: 53.33/16 = 3.33 mol
Simplest mole ratio:
Divide by smallest (3.33):
C: 3.33/3.33 = 1
H: 6.67/3.33 = 2
O: 3.33/3.33 = 1
Empirical formula: CH2O
Empirical formula mass = 12 + 2 + 16 = 30
n = Molar mass/Empirical formula mass = 180/30 = 6
Molecular formula = (CH2O)6 = C6H12O6
Q (2022, 3 marks): In a reaction: 2Na + Cl2 -> 2NaCl, calculate the mass of NaCl produced from 23 g of sodium. (Na = 23, Cl = 35.5, N = 14)
Answer: Given:
Reaction: 2Na + Cl2 -> 2NaCl
Mass of Na = 23 g
Molar mass of Na = 23 g/mol
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Number of moles of Na = 23/23 = 1 mol
From stoichiometry:
2 mol Na produces 2 mol NaCl
1 mol Na produces 1 mol NaCl
Moles of NaCl produced = 1 mol
Mass of NaCl = 1 × 58.5 = 58.5 g
Q (2024, 2 marks): Calculate the percentage composition of carbon in propane (C3H8). (C = 12, H = 1)
Answer: C3H8
Molar mass of C3H8 = (3 × 12) + (8 × 1) = 36 + 8 = 44 g/mol
Mass of carbon = 3 × 12 = 36 g
Percentage of C = (Mass of C/Molar mass of C3H8) × 100
Percentage of C = (36/44) × 100
Percentage of C = 81.82%
Frequently Asked Questions
What is the difference between empirical formula and molecular formula?
Empirical formula shows the simplest whole number ratio of atoms in a compound, while molecular formula shows the actual number of atoms. For example, C2H4 and C4H8 have the same empirical formula (CH2) but different molecular formulas.
Why do we use the mole concept in chemistry?
The mole provides a bridge between the atomic scale (where we count atoms and molecules) and the macroscopic scale (where we measure mass and volume). It allows us to count particles by measuring mass.
More Class 11 Chemistry PYQs
- Electrochemistry
- Solutions
- Chemical Kinetics
- Coordination Compounds
- Haloalkanes and Haloarenes
- Aldehydes Ketones and Carboxylic Acids
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