Some Basic Concepts of Chemistry — Class 11 MCQs with Answers
Class 11 CBSE Chemistry · Chapter 1
90 practice questions · 30 Easy · 30 Medium · 30 Hard · Updated
Practise the most important Class 11 CBSE Chemistry questions from Chapter 1, "Some Basic Concepts of Chemistry". You get 9 timed quizzes made from 90 NCERT-based MCQs, with answers and explanations. The questions are split into 30 Easy, 30 Medium and 30 Hard. Warm up on the basics, then move on to the exam-level questions that set top scorers in CBSE & Maharashtra HSC Board exams, JEE Main, MHT-CET, JEE Advanced and NEET UG apart.
To score well in "Some Basic Concepts of Chemistry", focus on reactions, key concepts and careful numericals. Each MCQ here is timed and uses exam-style marking (+4 correct, −1 wrong, 0 skipped). This trains you to stay accurate under time pressure, as real papers need. Every question has a short explanation, so a wrong answer becomes a quick lesson. It is the fastest way to fix gaps before a test.
Use this chapter for focused revision. Start with the Easy set to check your basics on Some Basic Concepts of Chemistry, then move to Medium and Hard to practise applying them. Your accuracy, streaks and XP save automatically. This chapter also adds to your overall Class 11 Chemistry mastery score. 14 sample questions are solved in full below, with the answer and a worked explanation. Sign in free to start practising.
Key concepts: Some Basic Concepts of Chemistry (Class 11 Chemistry)
Chemistry begins by counting the uncountable: this chapter turns masses you can weigh into numbers of atoms and molecules you cannot. It grounds the laws of chemical combination, the mole and Avogadro's number, empirical and molecular formulae, limiting-reagent stoichiometry, and the concentration terms used all year.
- Law of conservation of mass
- Mass is neither created nor destroyed in a chemical reaction; the total mass of the reactants equals the total mass of the products.
- Law of definite proportions
- A given pure compound always contains the same elements combined in a fixed mass ratio, whatever its source or method of preparation.
- Law of multiple proportions
- When two elements form more than one compound, the masses of one combining with a fixed mass of the other are in small whole-number ratios.
- Gay-Lussac's law of gaseous volumes
- Gases react and form products in volumes that bear simple whole-number ratios to one another, all measured at the same temperature and pressure.
Some Basic Concepts of Chemistry — important questions & MCQs with answers (Class 11 Chemistry)
14 solved questions from this chapter's difficulty levels, each with its answer and explanation. The other 76 are timed and scored when you sign in.
- Q1Easy
Avogadro's number ≈
A.3.14B.6.022×10²³✓ CorrectC.9.81D.1.6×10⁻¹⁹Answer: B. 6.022×10²³
Explanation: Avogadro's number, 6.022×10²³, is the fixed count of entities in one mole, defined via ¹²C — not π (3.14) or an electron charge (1.6×10⁻¹⁹).
- Q2Easy
Law of conservation of mass was given by:
A.ProustB.Lavoisier✓ CorrectC.DaltonD.AvogadroAnswer: B. Lavoisier
Explanation: Lavoisier's law states mass is neither created nor destroyed in a reaction — total reactant mass equals total product mass. Proust and Dalton cover composition and atomic theory instead.
- Q3Easy
Molar mass of H₂O:
A.2 g/molB.20 g/molC.16 g/molD.18 g/mol✓ CorrectAnswer: D. 18 g/mol
Explanation: Add each atom's mass: 2 H at 1 + O at 16 = 18 g/mol. Doubling only one H (1+16=17) or dropping H entirely (16) are the common slips.
- Q4Easy
Law of definite proportions was given by:
A.DaltonB.BoyleC.Gay-LussacD.Proust✓ CorrectAnswer: D. Proust
Explanation: Proust's law of definite proportions says a pure compound always has the same fixed mass ratio of elements. Dalton's atomic theory explains why, but is a separate idea.
- Q5Easy
Molar mass of CO₂:
A.28 g/molB.32 g/molC.44 g/mol✓ CorrectD.22 g/molAnswer: C. 44 g/mol
Explanation: Sum atomic masses: C (12) + 2 O at 16 = 44 g/mol. The frequent error is counting oxygen once (12+16=28) instead of twice.
- Q6Easy
Number of significant figures in 0.00450:
A.2B.3✓ CorrectC.4D.5Answer: B. 3
Explanation: Leading zeros before the first nonzero digit are never significant, but a zero after it counts: in 0.00450, only 4, 5, 0 count — 3 significant figures, not 5.
- Q7Medium
Molecules in 11.2 L of O₂ at STP:
A.3.011×10²³✓ CorrectB.6.022×10²³C.1.2×10²⁴D.6.022×10²²Answer: A. 3.011×10²³
Explanation: Moles = volume ÷ 22.4 L/mol: 11.2/22.4 = 0.5 mol, so molecules = 0.5 × Nₐ ≈ 3.011×10²³. Using the full Nₐ (6.022×10²³) forgets only half a mole is present.
- Q8Medium
Moles in 36 g of water:
A.18B.1C.36D.2✓ CorrectAnswer: D. 2
Explanation: Moles = mass ÷ molar mass: 36 g ÷ 18 g/mol = 2 mol. Forgetting to divide (just reading off 36 or 18) is the usual slip on straightforward mole problems.
- Q9Medium
Atoms in 4.6 g of Na (atomic mass 23):
A.6.022×10²³B.1.2×10²²C.3.011×10²²D.1.2×10²³✓ CorrectAnswer: D. 1.2×10²³
Explanation: n = mass/molar mass = 4.6/23 = 0.2 mol; atoms = 0.2 × Nₐ ≈ 1.2×10²³. Skipping the mole conversion and using Nₐ directly (6.022×10²³) overcounts by 5×.
- Q10Medium
Reactant entirely consumed in a reaction is the:
A.Excess reagentB.Limiting reagent✓ CorrectC.CatalystD.SolventAnswer: B. Limiting reagent
Explanation: The limiting reagent is the reactant fully consumed first, capping how much product forms — the excess reagent is what's left over, not what limits the yield.
- Q11Medium
Volume occupied by 8 g of CH₄ at STP:
A.5.6 LB.2.24 LC.22.4 LD.11.2 L✓ CorrectAnswer: D. 11.2 L
Explanation: n = 8/16 = 0.5 mol; volume at STP = n × 22.4 L = 0.5 × 22.4 = 11.2 L. Using the full 22.4 L and forgetting the 0.5 mol factor is the usual overshoot.
- Q12Hard
Empirical formula of glucose (C₆H₁₂O₆):
A.C₆H₁₂O₆B.C₂H₄OC.CHOD.CH₂O✓ CorrectAnswer: D. CH₂O
Explanation: Divide every subscript by their greatest common factor: C₆H₁₂O₆ ÷ 6 gives CH₂O, the empirical formula. Leaving the formula undivided (C₆H₁₂O₆) confuses molecular with empirical.
- Q13Hard
Molarity is:
A.Moles of solute per litre of solution✓ CorrectB.Moles per kg of solventC.Mass per volumeD.Volume per molesAnswer: A. Moles of solute per litre of solution
Explanation: Molarity (M) is defined as moles of solute per litre of solution, M = n/V — molality instead uses kilograms of solvent, a distinction that trips up unit-heavy questions.
- Q14Hard
Hydrocarbon contains 85.7 % C, 14.3 % H, with molar mass 56 g/mol. Molecular formula:
A.C₂H₄B.C₃H₆C.C₄H₈✓ CorrectD.C₄H₁₀Answer: C. C₄H₈
Explanation: Mole ratio C:H = 85.7/12 : 14.3/1 ≈ 1:2, giving empirical CH₂ (mass 14); n = molar mass/14 = 56/14 = 4, so the molecular formula is C₄H₈, not the empirical CH₂ itself.
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Start this chapter free →Some Basic Concepts of Chemistry — FAQs
What are the key concepts in Class 11 Chemistry Some Basic Concepts of Chemistry?+
Chemistry begins by counting the uncountable: this chapter turns masses you can weigh into numbers of atoms and molecules you cannot. It grounds the laws of chemical combination, the mole and Avogadro's number, empirical and molecular formulae, limiting-reagent stoichiometry, and the concentration terms used all year. Key ideas include Law of conservation of mass, Law of definite proportions, Law of multiple proportions, Gay-Lussac's law of gaseous volumes.
What does Class 11 Chemistry Chapter 1 (Some Basic Concepts of Chemistry) cover on XamBaaz?+
It has 90 NCERT-based MCQs on "Some Basic Concepts of Chemistry": 30 Easy, 30 Medium and 30 Hard. Together they make 9 timed quizzes, and you never get the same set twice. Every question has an instant explanation. They help you prepare for CBSE & Maharashtra HSC Board exams, JEE Main, MHT-CET, JEE Advanced and NEET UG.
Are these "Some Basic Concepts of Chemistry" questions free to practise?+
Yes. Sign in with Google to practise "Some Basic Concepts of Chemistry" free. Full unlimited access is ₹999/year on a launch offer until 1 December 2026. No chapter is charged separately.
How should I revise "Some Basic Concepts of Chemistry" for the exam?+
Start with the Easy quiz to check your basics, then try Medium and Hard to practise applying them. There are 9 timed quizzes on this chapter, so you can come back for a fresh set instead of one you have seen. Read each explanation, retry the questions you miss, and track your accuracy until it stays high.
Are these "Some Basic Concepts of Chemistry" MCQs available with answers?+
Yes. 14 sample questions are shown here in full, each with the correct option and a step-by-step "Why" explanation. Sign in free with Google to start practising, with instant scoring.
Is there negative marking in the "Some Basic Concepts of Chemistry" quizzes?+
Yes. The timed quizzes use exam-style marking: +4 for a right answer, −1 for a wrong one and 0 for a skip, the same negative marking as JEE Main, JEE Advanced and NEET UG. MHT-CET and CBSE board papers have no negative marking. Our mocks for those are scored their way.
What are the important questions from Some Basic Concepts of Chemistry (Class 11 Chemistry)?+
The questions that matter most test Law of conservation of mass, Law of definite proportions, Law of multiple proportions, Gay-Lussac's law of gaseous volumes. This page shows 14 solved important MCQs with answers and explanations; all 90 questions on the chapter are available as timed quizzes once you sign in.
Is there an online quiz for Some Basic Concepts of Chemistry?+
Yes — Class 11 Chemistry Some Basic Concepts of Chemistry has timed online quizzes at Easy, Medium and Hard levels, with instant scoring and a worked explanation on every question. The first quiz on the chapter is free.