Some Basic Concepts of Chemistry — Class 11 MCQs with Answers

Some Basic Concepts of Chemistry — important questions & MCQs with answers (Class 11 Chemistry)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Avogadro's number ≈

   A3.14B6.022×10²³C9.81D1.6×10⁻¹⁹
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   Answer: 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).

2. Q2 of 6Easy

   Law of conservation of mass was given by:

   AProustBLavoisierCDaltonDAvogadro
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   Answer: 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.

3. Q3 of 6Medium

   Molecules in 11.2 L of O₂ at STP:

   A3.011×10²³B6.022×10²³C1.2×10²⁴D6.022×10²²
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   Answer: 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.

4. Q4 of 6Medium

   Moles in 36 g of water:

   A18B1C36D2
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   Answer: 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.

5. Q5 of 6Hard

   Empirical formula of glucose (C₆H₁₂O₆):

   AC₆H₁₂O₆BC₂H₄OCCHODCH₂O
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   Answer: 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.

6. Q6 of 6Hard

   Molarity is:

   AMoles of solute per litre of solutionBMoles per kg of solventCMass per volumeDVolume per moles
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   Answer: 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.

Some Basic Concepts of Chemistry notes: key concepts (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.

Avogadro's law

Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules — the basis of the 22.4 L molar volume at STP.

Mole and Avogadro's number

One mole is 6.022 × 10²³ particles (Avogadro's number, Nₐ) and equals the substance's molar mass expressed in grams.

Molar mass

The mass of one mole in g mol⁻¹, numerically equal to the atomic or molecular mass in u; it converts between grams and moles of a substance.

Empirical and molecular formula

The empirical formula is the simplest whole-number atom ratio; the molecular formula is a whole-number multiple n of it, where n = M ÷ empirical mass.

Percentage composition

Mass percent of an element = (mass of that element in one mole ÷ molar mass) × 100; it is used to derive empirical formulae from experimental data.

Limiting reagent

The reactant consumed completely first; it caps how much product can form, while the other reactant is left over in excess and unreacted.

Stoichiometry

The coefficients of a balanced equation give the mole ratios linking reactants and products, letting you predict masses or gas volumes formed.

Molarity

Moles of solute per litre of solution (mol L⁻¹); it varies with temperature because the solution's volume expands or contracts on heating.

Molality

Moles of solute per kilogram of solvent (mol kg⁻¹); being defined by mass it is independent of temperature, unlike molarity which uses volume.

Mole fraction

The ratio of the moles of one component to the total moles in the mixture; the mole fractions of all components always sum to exactly one.