🧪 CBSE Class 11 Chemistry MCQs — Mole Concept and Chemical Calculations
Q1. Number of atoms in 12 g of carbon-12 is:
A) 6.022 × 10²³
B) 1.204 × 10²⁴
C) 3.011 × 10²³
D) 12 × 6.022 × 10²³
✅ Answer: A
Explanation: 12 g of carbon-12 = 1 mole = Avogadro’s number (6.022 × 10²³) atoms.
Q2. Molar mass of water (H₂O) is:
A) 16 g/mol B) 18 g/mol C) 20 g/mol D) 34 g/mol
✅ Answer: B
Explanation: H = 1 × 2 + O = 16 ⇒ 18 g/mol.
Q3. 0.5 mol of NaCl contains how many formula units?
A) 3.011 × 10²³ B) 6.022 × 10²³ C) 1.204 × 10²⁴ D) 5 × 10²²
✅ Answer: A
Explanation: 0.5 mol × 6.022 × 10²³ = 3.011 × 10²³ formula units.
Q4. Number of moles in 88 g of CO₂ (M = 44 g/mol):
A) 0.5 mol B) 1 mol C) 2 mol D) 4 mol
✅ Answer: C
Explanation: n = mass / molar mass = 88 / 44 = 2 mol.
Q5. 18 g of water corresponds to how many molecules?
A) 3.011 × 10²³ B) 6.022 × 10²³ C) 1.204 × 10²⁴ D) 2 × 6.022 × 10²³
✅ Answer: B
Explanation: 18 g H₂O = 1 mole = 6.022 × 10²³ molecules.
Q6. The empirical formula of a compound with 40% C, 6.7% H, and 53.3% O is:
A) CH₂O B) C₂H₂O₂ C) C₂H₄O D) CHO
✅ Answer: A
Explanation: Atomic ratio C:H:O ≈ 1:2:1 ⇒ CH₂O.
Q7. 2.0 g of H₂ (M = 2 g/mol) contains how many moles?
A) 1 mol B) 0.5 mol C) 2 mol D) 0.25 mol
✅ Answer: A
Explanation: 2 ÷ 2 = 1 mol.
Q8. Which contains the greatest number of atoms?
A) 1 mol of H₂O
B) 1 mol of CO₂
C) 1 mol of NaCl
D) All contain same number of particles
✅ Answer: D
Explanation: Each mole has 6.022 × 10²³ entities (Avogadro’s number).
Q9. Molecular mass and molar mass are related as:
A) Numerically same, units differ
B) Molar mass = molecular mass × Avogadro’s number
C) They are unrelated
D) Molar mass is smaller
✅ Answer: A
Explanation: Molecular mass (u) and molar mass (g/mol) have same numerical value.
Q10. 0.25 mol of glucose (C₆H₁₂O₆) has mass:
A) 45 g B) 90 g C) 180 g D) 22.4 g
✅ Answer: A
Explanation: M = 180 g/mol → 0.25 × 180 = 45 g.
Q11. Percentage of hydrogen in CH₄ is:
A) 25% B) 12.5% C) 5% D) 50%
✅ Answer: A
Explanation: (4/16) × 100 = 25%.
Q12. 0.1 mol of gas at STP occupies volume:
A) 2.24 L B) 22.4 L C) 0.224 L D) 11.2 L
✅ Answer: A
Explanation: 1 mol = 22.4 L ⇒ 0.1 mol = 2.24 L.
Q13. Avogadro’s number has units:
A) mol B) mol⁻¹ C) atoms D) no unit
✅ Answer: B
Explanation: It represents number of entities per mole ⇒ mol⁻¹.
Q14. 1 mole of CO₂ contains:
A) 6.022 × 10²³ atoms
B) 6.022 × 10²³ molecules
C) 12.044 × 10²³ atoms
D) 22.4 atoms
✅ Answer: B
Explanation: 1 mol = 6.022 × 10²³ molecules.
Q15. Empirical formula of C₆H₁₂O₆ is:
A) CH₂O B) C₃H₆O₃ C) CHO D) C₂H₄O₂
✅ Answer: A
Explanation: Simplest ratio → divide subscripts by 6 ⇒ CH₂O.
Q16. 1 mole of electrons carries charge:
A) 1.602 × 10⁻¹⁹ C B) 96485 C C) 6022 C D) 6.022 × 10²³ C
✅ Answer: B
Explanation: Faraday constant = 96485 C/mol e⁻.
Q17. Mass of 3.011 × 10²³ H₂O molecules:
A) 9 g B) 18 g C) 36 g D) 3 g
✅ Answer: A
Explanation: 3.011 × 10²³ = 0.5 mol → 0.5 × 18 = 9 g.
Q18. Molarity of solution with 1 mole solute in 2 L solution:
A) 0.5 M B) 1 M C) 2 M D) 0.25 M
✅ Answer: A
Explanation: M = n/V = 1/2 = 0.5 M.
Q19. Weight percent of oxygen in H₂O:
A) 88.9% B) 16% C) 50% D) 66%
✅ Answer: A
Explanation: (16/18) × 100 = 88.9%.
Q20. Which is true about 1 mole CH₄?
A) Contains 6.022 × 10²³ C atoms
B) Contains 4 × 6.022 × 10²³ H atoms
C) Has mass 16 g
D) All of these
✅ Answer: D
Explanation: 1 mole CH₄ = 16 g = contains 1 mol C & 4 mol H atoms.
Q21. 1 mole of O₂ weighs:
A) 16 g B) 32 g C) 8 g D) 64 g
✅ Answer: B
Explanation: M = 2 × 16 = 32 g.
Q22. Empirical formula for 24.27% C, 4.06% H, 71.67% O:
A) CH₂O B) C₂H₄O₂ C) CO₂H D) CHO
✅ Answer: A
Explanation: Ratio ≈ 1:2:1 ⇒ CH₂O.
Q23. 0.02 mol of electrons corresponds to charge:
A) 1929.7 C B) 1.6 × 10⁻¹⁹ C C) 193 C D) 0.02 C
✅ Answer: A
Explanation: Q = n × F = 0.02 × 96485 = 1929.7 C.
Q24. % composition of Na in Na₂CO₃ (M = 106):
A) 43.4% B) 27.4% C) 68.9% D) 22.8%
✅ Answer: A
Explanation: (46/106) × 100 = 43.4% Na.
Q25. Which of the following is NOT a derived unit?
A) Mole B) Kilogram C) Joule D) Newton
✅ Answer: A
Explanation: Mole is a base SI unit, others are derived.
Q26. Molecular formula from empirical CH and molar mass 78 g/mol:
A) C₆H₆ B) C₂H₂ C) CH D) C₃H₃
✅ Answer: A
Explanation: Empirical mass = 13; 78/13 = 6 ⇒ C₆H₆.
Q27. 1.204 × 10²⁴ molecules correspond to:
A) 2 mol B) 1 mol C) 0.5 mol D) 4 mol
✅ Answer: A
Explanation: (1.204 × 10²⁴) / (6.022 × 10²³) = 2 mol.
Q28. Quantity independent of sample size:
A) Mass B) Volume C) Mole D) Number of particles
✅ Answer: C
Explanation: 1 mole always represents a fixed number of entities.
Q29. Number of moles in 44 g of CO₂:
A) 1 mol B) 2 mol C) 0.5 mol D) 4 mol
✅ Answer: A
Explanation: 44/44 = 1 mol.
Q30. Mass of 6.022 × 10²³ atoms of helium (He = 4):
A) 4 g B) 2 g C) 1 g D) 16 g
✅ Answer: A
Explanation: 1 mol He = 4 g.