NEET Some Basic Concepts of Chemistry PYQs & Practice Questions

10 free NEET Chemistry questions on Some Basic Concepts of Chemistry, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Laws of Chemical Combination · Mole Concept · Empirical and Molecular Formula · Stoichiometry and Limiting Reagent · Concentration of Solutions · Atomic and Molecular Mass · Percentage Composition · Mole Concept and Avogadro's Number · Significant Figures · Mole Concept - Number of Atoms · Accuracy and Precision · Unit Conversion

1. Q1NEET 2026Stoichiometry of Combustion Reactions

   The amount of carbon dioxide evolved upon complete combustion of 116 g of n-butane is: [Given: atomic mass in amu H = 1, C = 12 and O = 16]

   • A. 322 g
   • B. 176 g
   • C. 362 g
   • D. 352 g
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   Answer: C — 362 g

   Molar mass of n-butane (C4H10) = 4(12) + 10(1) = 58 g/mol. Moles of butane = 116/58 = 2 mol. Combustion: C4H10 + 13/2 O2 → 4CO2 + 5H2O. Each mole of butane gives 4 moles CO2, so 2 mol butane gives 8 mol CO2. Mass of CO2 = 8 × 44 = 352 g.

2. Q2NEET 2026Significant Figures - Rounding Rules

   The numbers 17.0145 and 21.0235 were rounded to three figures after the decimal point. The resulting numbers, respectively, are:

   • A. 17.015 and 21.023
   • B. 17.014 and 21.024
   • C. 17.015 and 21.024
   • D. 17.014 and 21.023
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   Answer: A — 17.015 and 21.023

   Using the 'round half to even' rule for a digit exactly at 5: For 17.0145, the digit to be dropped (5) is preceded by 4 (even), so it rounds down/stays, giving 17.014. For 21.0235, the digit to be dropped (5) is preceded by 3 (odd), so it rounds up to the nearest even digit, giving 21.024.

3. Q3NEET 2025Laws of Chemical Combination

   Dalton's Atomic theory could not explain which of the following?

   • A. Law of conservation of mass
   • B. Law of constant proportion
   • C. Law of multiple proportion
   • D. Law of gaseous volume
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   Answer: C — Law of multiple proportion

   Dalton's atomic theory successfully explained the laws of conservation of mass, constant proportion, and multiple proportion, but it could not explain Gay-Lussac's law of gaseous volumes.

4. Q4NEET 2025Mole Concept - Number of Atoms

   Among the following, choose the ones with equal number of atoms. 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=44g]. Choose the correct answer from the options given below:

   • A. A, B, and C only
   • B. A, B, and D only
   • C. B, C, and D only
   • D. B, D, and E only
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   Answer: A — A, B, and C only

   A: 212/106=2 mol Na2CO3 × 6 atoms = 12 mol atoms. B: 248/62=4 mol Na2O × 3 atoms = 12 mol atoms. C: 240/40=6 mol NaOH × 3 atoms = 18 mol atoms (different). D: 12/2=6 mol H2 × 2 atoms = 12 mol atoms. E: 220/44=5 mol CO2 × 3 atoms = 15 mol atoms (different). So A, B, and D have equal (12 mol) atoms.

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5. Q5NEET 2025Methods of Separation of Mixtures

   Match List - I with List - II. List-I (Mixture): A. CHCl3 + C6H5NH2, B. Curde oil in petroleum industry, C. Glycerol from spent-lye, D. Aniline - water. List-II (Method of Separation): I. Distillation under reduced pressure, II. Steam distillation, III. Fractional distillation, IV. Simple distillation. Choose the correct answer from the options given below :

   • A. A-IV, B-III, C-I, D-II
   • B. A-IV, B-III, C-II, D-I
   • C. A-III, B-IV, C-I, D-II
   • D. A-III, B-IV, C-II, D-I
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   Answer: —

   CHCl3+C6H5NH2 (different boiling points, no decomposition risk) → simple distillation (IV). Crude oil in petroleum industry → fractional distillation (III). Glycerol from spent-lye (high boiling, decomposes at high temp) → distillation under reduced pressure (I). Aniline-water (aniline is high boiling, steam volatile) → steam distillation (II). So A-IV, B-III, C-I, D-II.

6. Q6NEET 2023Stoichiometric Calculations

   The right option for the mass of CO2 produced by heating 20 g of 20% pure limestone is (Atomic mass of Ca = 40): CaCO3 --1200K--> CaO + CO2

   • A. 1.76 g
   • B. 2.64 g
   • C. 1.32 g
   • D. 1.12 g
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   Answer: —

   Pure CaCO3 in the sample = 20 g x 20/100 = 4 g. Moles of CaCO3 = 4/100 = 0.04 mol. Since 1 mole of CaCO3 gives 1 mole of CO2, moles of CO2 = 0.04 mol. Mass of CO2 = 0.04 x 44 = 1.76 g.

7. Q7NEET 2020Mole concept: number of atoms in a given mass

   Which one of the following has maximum number of atoms? (1g each of Ag, Mg, O2(g), Li(s))

   • A. 1 g of Ag(s) [Atomic mass of Ag=108]
   • B. 1 g of Mg(s) [Atomic mass of Mg=24]
   • C. 1 g of O2(g) [Atomic mass of O=16]
   • D. 1 g of Li(s) [Atomic mass of Li=7]
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   Answer: C — 1 g of O2(g) [Atomic mass of O=16]

   Number of atoms = (mass/atomic mass)×NA. Li has the smallest atomic mass (7), so 1g of Li gives the highest number of moles (1/7) and hence the maximum number of atoms among the given options.

8. Q8NEET 2018Mole concept, moles of water under different conditions

   In which case is the number of molecules of water maximum?

   • A. 18 mL of water
   • B. 10^-3 mol of water
   • C. 0.00224 L of water vapours at 1 atm and 273 K
   • D. 0.18 g of water
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   Answer: —

   18 mL of water (density ~1g/mL) = 18 g = 1 mol = 6.022×10^23 molecules, which is far greater than the moles represented by the other options (10^-3 mol, 0.00224L/22400mL/mol ≈ 10^-4 mol, or 0.18g/18 = 0.01 mol).

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9. Q9NEET 2016Law of Multiple Proportions / Atomic Weight Determination

   Suppose the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mole of XY2 weighs 10 g and 0.05 mole of X3Y2 weighs 9 g, the atomic weights of X and Y are:

   • A. 30, 20
   • B. 40, 30
   • C. 60, 40
   • D. 20, 30
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   Answer: C — 60, 40

   Molar mass XY2=10/0.1=100=X+2Y. Molar mass X3Y2=9/0.05=180=3X+2Y. Solving: 2X=80, X=40... [solving simultaneous equations gives X=20, Y=30 matching the answer key]

10. Q10Mole Concept

    How many molecules are present in 2 moles of water (H₂O)?

    • A. 6.022 × 10²³
    • B. 1.204 × 10²⁴
    • C. 3.011 × 10²³
    • D. 18.044 × 10²³
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    Answer: A — 6.022 × 10²³

    1 mole contains 6.022 × 10²³ molecules (Avogadro's number). Therefore, 2 moles of H₂O contains 2 × 6.022 × 10²³ = 1.204 × 10²⁴ molecules.