Mole Concept (Mole concept, Avogadro Constant, Stoichiometric calculations, Limiting reagent, Limiting yield, Percentage yield): NEET MCQs with Answers

15 free questions on Mole Concept (Mole concept, Avogadro Constant, Stoichiometric calculations, Limiting reagent, Limiting yield, Percentage yield) for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
From the following, identify those that contain an equal number of atoms. [NEET 2025]
A. 212 g of Na₂CO₃(s) [molar mass = 106 g]
B. 248 g of Na₂O(s) [molar mass = 62 g]
C. 240 g of NaOH(s) [molar mass = 40 g]
D. 12 g of H₂(g) [molar mass = 2 g]
E. 220 g of CO₂(g) [molar mass = 44 g]
  1. A, B and C only
  2. A, B and D only
  3. B, C and D only
  4. B, D and E only
Show answer
Answer: (B) A, B and D only
Explanation: Moles of atoms: A: 212/106 = 2 mol × 6 = 12; B: 248/62 = 4 mol × 3 = 12; C: 240/40 = 6 mol × 3 = 18; D: 12/2 = 6 mol × 2 = 12; E: 220/44 = 5 mol × 3 = 15. Hence A, B and D contain equal numbers of atoms (12 Nₐ each).

Question 2

Previous-year question
1.0 g of H₂ contains the same number of molecules as [Re-NEET 2024]
  1. 14 g of N₂
  2. 18 g of H₂O
  3. 16 g of CO
  4. 28 g of N₂
Show answer
Answer: (A) 14 g of N₂
Explanation: 1.0 g H₂ = 1/2 = 0.5 mol. 14 g N₂ = 14/28 = 0.5 mol, so it has the same number of molecules. 18 g H₂O and 28 g N₂ are 1 mol each, and 16 g CO is 0.57 mol.

Question 3

Previous-year question
A 4.74 g sample of an inorganic compound contains 0.39 g of K, 0.27 g of Al, 1.92 g of SO₄ radicals and 2.16 g of water. If the molar mass of the compound is 948 g mol⁻¹, what is its molecular formula? [NEET 2024 (Sawaimadhopur)]
  1. K₂Al₂(SO₄)₄·12H₂O
  2. K₂SO₄·Al₂(SO₄)₃·24H₂O
  3. K₂SO₄·Al₂(SO₄)₃·12H₂O
  4. KAl(SO₄)₂·12H₂O
Show answer
Answer: (B) K₂SO₄·Al₂(SO₄)₃·24H₂O
Explanation: Moles: K = 0.39/39 = 0.01; Al = 0.27/27 = 0.01; SO₄ = 1.92/96 = 0.02; H₂O = 2.16/18 = 0.12. Ratio 1 : 1 : 2 : 12 gives the empirical formula KAl(SO₄)₂·12H₂O (formula mass 474). n = 948/474 = 2, so the molecular formula is K₂Al₂(SO₄)₄·24H₂O = K₂SO₄·Al₂(SO₄)₃·24H₂O.

Question 4

Previous-year question
Which is the correct mass of CO₂ produced on heating 20 g of 20% pure limestone? (Atomic mass of Ca = 40) [NEET 2023]
CaCO₃ —(1200 K)→ CaO + CO₂
  1. 1.76 g
  2. 2.64 g
  3. 1.32 g
  4. 1.12 g
Show answer
Answer: (A) 1.76 g
Explanation: Pure CaCO₃ = 20% of 20 g = 4 g = 4/100 = 0.04 mol. By CaCO₃ → CaO + CO₂, 0.04 mol CO₂ is formed = 0.04 × 44 = 1.76 g.

Question 5

Previous-year question
Match List-I with List-II. (Given: molar volume of a gas at STP = 22.4 L) [NEET 2022 (Overseas)]
List-I:
(a) 4.48 litres of O₂ at STP
(b) 12.022 × 10²² molecules of H₂O
(c) 96 g of O₂
(d) 88 g of CO₂
List-II:
(i) 0.2 moles
(ii) 12.044 × 10²³ molecules
(iii) 6.4 g
(iv) 67.2 litres at STP
  1. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
  2. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
  3. (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
  4. (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
Show answer
Answer: (A) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
Explanation: (a) 4.48 L O₂ at STP = 0.2 mol = 0.2 × 32 = 6.4 g → (iii). (b) 12.022 × 10²² molecules of H₂O = 0.2 mol → (i). (c) 96 g O₂ = 3 mol = 3 × 22.4 = 67.2 L → (iv). (d) 88 g CO₂ = 2 mol = 12.044 × 10²³ molecules → (ii).

Question 6

Previous-year question
A sample of phosphorus trichloride (PCl₃) contains 1.4 moles of the substance. How many atoms does the sample contain? [NEET 2021]
  1. 4
  2. 5.6
  3. 8.431 × 10²³
  4. 3.372 × 10²⁴
Show answer
Answer: (D) 3.372 × 10²⁴
Explanation: Each PCl₃ molecule has 4 atoms. Atoms = 1.4 × 4 × 6.022 × 10²³ = 5.6 × 6.022 × 10²³ = 3.372 × 10²⁴. The value 5.6 is only the moles of atoms, and 8.431 × 10²³ counts molecules, not atoms.

Question 7

Previous-year question
If 1 mL of water contains 20 drops, how many molecules of water are present in one drop? [NEET 2021]
  1. 6.023 × 10²³
  2. 1.673 × 10²¹
  3. 1.344 × 10¹⁸
  4. 4.346 × 10²⁰
Show answer
Answer: (B) 1.673 × 10²¹
Explanation: Volume of one drop = 1/20 = 0.05 mL = 0.05 g of water. Moles = 0.05/18 = 2.78 × 10⁻³ mol. Molecules = 2.78 × 10⁻³ × 6.022 × 10²³ ≈ 1.673 × 10²¹.
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Question 8

Previous-year question
On analysis, an organic compound was found to contain C = 54.5% and H = 9.1% by mass (the rest being oxygen). What is its empirical formula? [NEET 2021]
  1. CHO₂
  2. CH₂O
  3. C₂H₄O
  4. C₃H₄O
Show answer
Answer: (C) C₂H₄O
Explanation: O = 100 − 54.5 − 9.1 = 36.4%. Moles: C = 54.5/12 = 4.54; H = 9.1/1 = 9.1; O = 36.4/16 = 2.275. Dividing by 2.275 gives C : H : O = 2 : 4 : 1, i.e. C₂H₄O.

Question 9

Previous-year question
Which of the following contains the largest number of atoms? [NEET 2020]
  1. 1g of Li(s) [Atomic mass of Li = 7]
  2. 1g of Ag(s) [Atomic mass of Ag = 108]
  3. 1g of Mg(s) [Atomic mass of Mg = 24]
  4. 1g of O₂(g) [Atomic mass of O = 16]
Show answer
Answer: (A) 1g of Li(s) [Atomic mass of Li = 7]
Explanation: Moles of atoms: Li = 1/7 = 0.143; Ag = 1/108 = 0.0093; Mg = 1/24 = 0.042; O atoms in 1 g O₂ = 2 × (1/32) = 0.0625. Lithium has the largest number of moles of atoms, hence the most atoms.

Question 10

Previous-year question
What volume will 1.8 g of water vapour occupy at 374 °C and 1 bar pressure? [Use R = 0.083 bar L K⁻¹ mol⁻¹] [NEET (Odisha) 2019]
  1. 96.66 L
  2. 55.87 L
  3. 3.10 L
  4. 5.37 L
Show answer
Answer: (D) 5.37 L
Explanation: n = 1.8/18 = 0.1 mol; T = 374 + 273 = 647 K. V = nRT/P = (0.1 × 0.083 × 647)/1 = 5.37 L.

Question 11

Previous-year question
Which one of the following contains the maximum number of atoms? [NEET 2020]
  1. 1 g of Mg(s) [atomic mass of Mg = 24]
  2. 1 g of O₂(g) [atomic mass of O = 16]
  3. 1 g of Li(s) [atomic mass of Li = 7]
  4. 1 g of Ag(s) [atomic mass of Ag = 108]
Show answer
Answer: (C) 1 g of Li(s) [atomic mass of Li = 7]
Explanation: For equal masses, the element with the smallest atomic mass gives the most atoms. Moles of atoms: Mg = 1/24 = 0.042; O = 1/16 = 0.063; Li = 1/7 = 0.143; Ag = 1/108 = 0.009. Lithium therefore provides the maximum number of atoms.

Question 12

Previous-year question
The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's process is [NEET 2019]
  1. 10
  2. 20
  3. 30
  4. 40
Show answer
Answer: (C) 30
Explanation: N₂ + 3H₂ → 2NH₃. Two moles of NH₃ require three moles of H₂, so 20 mol NH₃ needs 20 × 3/2 = 30 mol of H₂.

Question 13

Previous-year question
In which case is the number of molecules of water maximum? [NEET 2018]
  1. 18 mL of water
  2. 0.18 g of water
  3. 10⁻³ mol of water
  4. 0.00224 L of water vapour at 1 atm and 273 K
Show answer
Answer: (A) 18 mL of water
Explanation: 18 mL of liquid water weighs 18 g = 1 mol. In contrast 0.18 g = 0.01 mol, the third option is 0.001 mol, and 0.00224 L of vapour at STP = 0.0001 mol. Hence 18 mL of water has the most molecules.

Question 14

Previous-year question
A mixture of 2.3 g of formic acid and 4.5 g of oxalic acid is treated with concentrated H₂SO₄. The evolved gaseous mixture is passed through KOH pellets. The mass (in g) of the remaining product at STP will be [NEET 2018]
  1. 1.4
  2. 3.0
  3. 4.4
  4. 2.8
Show answer
Answer: (D) 2.8
Explanation: n(HCOOH) = 2.3/46 = 0.05 mol gives 0.05 mol CO; n(H₂C₂O₄) = 4.5/90 = 0.05 mol gives 0.05 mol CO and 0.05 mol CO₂. KOH absorbs the CO₂, leaving 0.10 mol of CO. Mass = 0.10 × 28 = 2.8 g.

Question 15

Previous-year question
What is the mass of the precipitate formed when 50 mL of a 16.9% solution of AgNO₃ is mixed with 50 mL of a 5.8% NaCl solution? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5) [AIPMT 2015]
  1. 28 g
  2. 3.5 g
  3. 7 g
  4. 14 g
Show answer
Answer: (C) 7 g
Explanation: AgNO₃ taken = 8.45 g, i.e. 8.45/169.8 = 0.0498 mol; NaCl taken = 2.9 g, i.e. 2.9/58.5 = 0.0496 mol. They react 1 : 1, so about 0.0496 mol of AgCl precipitates. Mass = 0.0496 × 143.3 ≈ 7 g.

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