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NEET Some Basic Concepts in Chemistry MCQs with answers

306 questions on Some Basic Concepts in Chemistry for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. Given below are two statements: Statement I: Gay Lussac's law of gaseous volumes is essentially the law of definite proportions by volume for gaseous reactions. Statement II: Gay Lussac's law is valid only when volumes of all gaseous reactants and products are measured at the same temperature and pressure. In the light of the above statements, choose the correct answer:

  1. Both Statement I and Statement II are true.
  2. Both Statement I and Statement II are false.
  3. Statement I is true but Statement II is false.
  4. Statement I is false but Statement II is true.
Show answer and explanation

Answer: (A) Both Statement I and Statement II are true.

Gay Lussac discovered that gases combine or are produced in a chemical reaction in a simple whole-number ratio by volume, provided all gases are at the same temperature and pressure. This is indeed the law of definite proportions by volume.

medium

2. Two gaseous oxides of nitrogen are analyzed. Oxide A contains 30.43% nitrogen by mass, and Oxide B contains 63.64% nitrogen by mass. Which of the following formulas correctly represents the pair (Oxide A, Oxide B)?

  1. NO and NO₂
  2. NO₂ and N₂O
  3. N₂O₃ and N₂O₅
  4. N₂O and NO
Show answer and explanation

Answer: (B) NO₂ and N₂O

In Oxide A: mass of N = 30.43 g, mass of O = 69.57 g. Moles of N = 30.43/14 = 2.17, moles of O = 69.57/16 = 4.35. Ratio N:O = 1:2, giving empirical formula NO₂. In Oxide B: mass of N = 63.64 g, mass of O = 36.36 g. Moles of N = 63.64/14 = 4.55, moles of O = 36.36/16 = 2.27. Ratio N:O = 2:1, giving empirical formula N₂O.

hard

3. A 10.0 g mixture of CaCO₃ and MgCO₃ on complete thermal decomposition leaves a residue of metal oxides weighing 5.08 g. What is the mass percentage of CaCO₃ in the original mixture? (Molar masses: CaCO₃ = 100, MgCO₃ = 84, CaO = 56, MgO = 40 g/mol)

  1. 40%
  2. 70%
  3. 60%
  4. 50%
Show answer and explanation

Answer: (D) 50%

Let mass of CaCO₃ be x g, so mass of MgCO₃ = (10 - x) g. The residue mass is (56x / 100) + [40(10 - x) / 84] = 5.08. Simplifying: 0.56x + 4.762 - 0.4762x = 5.08, which gives 0.0838x = 0.318, hence x ≈ 5.0 g. Therefore, the mass percentage of CaCO₃ is (5.0 / 10.0) × 100% = 50%.

medium

4. Match List-I with List-II to classify the nature of each entity according to chemical concepts: List-I A. Dry air free of dust B. Solid sodium chloride C. Graphite D. Colloidal solution of starch List-II I. Element II. Compound III. Homogeneous mixture IV. Heterogeneous mixture Choose the correct answer from the options given below:

  1. A-III, B-I, C-II, D-IV
  2. A-II, B-III, C-IV, D-I
  3. A-IV, B-I, C-II, D-III
  4. A-III, B-II, C-I, D-IV
Show answer and explanation

Answer: (D) A-III, B-II, C-I, D-IV

Clean dry air is a solution of gases, hence a homogeneous mixture (A-III). Sodium chloride is a chemical compound consisting of Na⁺ and Cl⁻ in a 1:1 fixed ratio (B-II). Graphite is an allotrope of carbon, hence an element (C-I). A colloidal solution has dispersed phase particles distinct from the medium, making it heterogeneous on a microscopic scale (D-IV).

easy

5. Which among the following contains only one chemical element despite existing as a polyatomic entity?

  1. Caustic soda
  2. Dry ice
  3. Ozone
  4. Methane
Show answer and explanation

Answer: (C) Ozone

Ozone consists of three oxygen atoms chemically bound together (O₃), making it a polyatomic homoatomic molecule of an element. Dry ice (CO₂), caustic soda (NaOH), and methane (CH₄) are compounds composed of two or more different elements.

hard

6. A hypothetical metal M forms two oxides. Oxide A contains 80% metal by mass, while Oxide B contains 70% metal by mass. What is the ratio of masses of oxygen that combine with a fixed mass of metal M in Oxide A and Oxide B respectively?

  1. 2 : 3
  2. 3 : 4
  3. 7 : 12
  4. 7 : 8
Show answer and explanation

Answer: (C) 7 : 12

In Oxide A, 80 g of metal combines with 20 g of oxygen, so 1 g of metal combines with 20/80 = 0.25 g of oxygen. In Oxide B, 70 g of metal combines with 30 g of oxygen, so 1 g of metal combines with 30/70 = 3/7 g (approx 0.4286 g) of oxygen. The ratio of oxygen masses combining with a fixed mass of M is 0.25 : (3/7) = (1/4) : (3/7) = 7 : 12.

easy

7. The number of moles in 130 g of a compound of molar mass 20 g mol⁻¹ is:

  1. 2.6 mol
  2. 6.5 mol
  3. 0.154 mol
  4. 13 mol
Show answer and explanation

Answer: (B) 6.5 mol

n = mass / molar mass = 130/20 = 6.5 mol.

Quick tip: n = m/M. Mass on top.

NCERT Class 11, Chapter 1

easy

8. Taking Nₐ = 6.022 × 10²³ mol⁻¹, the number of molecules in 90 g of C₆H₁₂O₆ (M = 180 g/mol) is:

  1. 1.51 × 10²³
  2. 6.02 × 10²³
  3. 3.01 × 10²³
  4. 5.42 × 10²⁵
Show answer and explanation

Answer: (C) 3.01 × 10²³

n = 90/180 = 0.5 mol; molecules = n × Nₐ = 3.01 × 10²³.

easy

9. According to Dalton's atomic theory, atoms:

  1. Are neither created nor destroyed in a chemical reaction
  2. Can be divided in chemical reactions
  3. Of the same element have different masses
  4. Combine in any ratio
Show answer and explanation

Answer: (A) Are neither created nor destroyed in a chemical reaction

Dalton proposed that matter consists of indivisible atoms, atoms of an element are identical, compounds form when atoms combine in fixed whole-number ratios, and atoms are neither created nor destroyed in reactions.

NCERT Class 11

easy

10. The number of moles in 900 g of C₆H₁₂O₆ (molar mass 180 g/mol) is:

  1. 900 mol
  2. 10 mol
  3. 5 mol
  4. 0.2 mol
Show answer and explanation

Answer: (C) 5 mol

Number of moles = given mass/molar mass: n = 900/180 = 5 mol.

easy

11. What is the stoichiometric coefficient of H⁺ when the following redox equation is balanced in an acidic medium using the smallest whole integers? Cr₂O₇²⁻ + Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O

  1. 7
  2. 14
  3. 6
  4. 8
Show answer and explanation

Answer: (B) 14

The balanced ionic equation is Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O. The stoichiometric coefficient of H⁺ is 14.

medium

12. Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R): Assertion (A): Dalton's assumption that all atoms of a given element have identical mass is not completely valid in modern chemistry. Reason (R): Isotopes of an element have different atomic numbers but identical mass numbers. In the light of the above statements, choose the correct option:

  1. Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. Both (A) and (R) are true but (R) is not the correct explanation of (A)
  3. (A) is true but (R) is false
  4. (A) is false but (R) is true
Show answer and explanation

Answer: (C) (A) is true but (R) is false

Assertion is true because the discovery of isotopes showed that atoms of the same element can have different masses (e.g., ³⁵Cl and ³⁷Cl). Reason is false because isotopes have the same atomic number (number of protons) but different mass numbers (due to different numbers of neutrons).

medium

13. In the industrial preparation of chlorobenzene, 78 g of benzene (C₆H₆) reacts with excess chlorine to produce 78.75 g of chlorobenzene (C₆H₅Cl). What is the percentage yield of the reaction? (Molar masses: C₆H₆ = 78 g/mol, C₆H₅Cl = 112.5 g/mol)

  1. 80%
  2. 75%
  3. 70%
  4. 85%
Show answer and explanation

Answer: (C) 70%

Reaction: C₆H₆ + Cl₂ → C₆H₅Cl + HCl. 1 mol (78 g) of benzene theoretically produces 1 mol (112.5 g) of chlorobenzene. Theoretical yield = 112.5 g. Percentage yield = (Actual yield / Theoretical yield) × 100 = (78.75 / 112.5) × 100 = 70%.

easy

14. The number of moles in 70 g of a compound of molar mass 100 g mol⁻¹ is:

  1. 0.7 mol
  2. 1.43 mol
  3. 7 mol
  4. 1.4 mol
Show answer and explanation

Answer: (A) 0.7 mol

n = mass / molar mass = 70/100 = 0.7 mol.

Quick tip: n = m/M. Mass on top.

NCERT Class 11, Chapter 1

medium

15. Two gaseous elements A₂ and B₂ react according to the equation: A₂(g) + 3B₂(g) → 2AB₃(g) If 60 mL of A₂ is mixed with 120 mL of B₂ and the reaction goes to completion at constant temperature and pressure, what is the total volume of the resulting gaseous mixture after reaction?

  1. 120 mL
  2. 80 mL
  3. 100 mL
  4. 140 mL
Show answer and explanation

Answer: (C) 100 mL

From the stoichiometry, 1 volume of A₂ reacts with 3 volumes of B₂ to yield 2 volumes of AB₃. Here, 120 mL of B₂ requires 120 / 3 = 40 mL of A₂. Thus, B₂ is the limiting reactant. Unreacted A₂ = 60 - 40 = 20 mL. Volume of AB₃ formed = 2 × 40 = 80 mL. Total final gas volume = 20 mL (A₂) + 80 mL (AB₃) = 100 mL.

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