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NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry covers the nature and importance of chemistry, laws of chemical combination, atomic and molecular masses, mole concept, and methods of expressing concentration.

Overview of Topics & Subtopics (Click to Jump)

1. Laws of Chemical Combination

Short Note:

• Law of Conservation of Mass: matter can neither be created nor destroyed in a chemical reaction; total mass of reactants equals total mass of products.

• Law of Definite Proportions: a pure chemical compound always contains the same elements combined in a fixed ratio by mass, regardless of its source.

• Law of Multiple Proportions: when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.

• Gay Lussac's Law of Gaseous Volumes and Avogadro's Law: equal volumes of all gases under the same conditions of temperature and pressure contain equal numbers of molecules.

2. Atomic and Molecular Masses

Short Note:

• Atomic Mass is expressed in atomic mass units (u), based on the carbon-12 isotope as the standard (1 u = 1/12th the mass of one carbon-12 atom).

• Molecular Mass is the sum of atomic masses of all atoms present in a molecule of a substance.

• Formula Mass is used for ionic compounds that do not exist as discrete molecules, calculated similarly using the formula unit.

3. Mole Concept and Molar Mass

Short Note:

• One mole is the amount of substance that contains exactly as many elementary entities as there are atoms in 12 g of carbon-12, i.e. 6.022 × 10²³ entities (Avogadro's number, NA).

• Molar Mass is the mass of one mole of a substance, expressed in g/mol, and is numerically equal to atomic/molecular mass in u.

4. Percentage Composition & Concentration Terms

Short Note:

• Empirical Formula shows the simplest whole-number ratio of atoms of each element in a compound; Molecular Formula shows the actual number of atoms.

• Molarity (M) = moles of solute / volume of solution in litres. Molality (m) = moles of solute / mass of solvent in kg. Mass percent = (mass of solute/mass of solution) × 100.

NCERT Textbook Exercise Questions

Calculate the molecular mass of glucose (C6H12O6).
Answer: Molecular mass = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 u.
Calculate the amount of water (g) produced by the combustion of 16 g of methane (CH4 + 2O2 → CO2 + 2H2O).
Answer: Molar mass of CH4 = 16 g/mol, so 16 g = 1 mole of CH4. From the equation, 1 mole CH4 produces 2 moles H2O. Mass of 2 moles H2O = 2 × 18 = 36 g.
If the density of methanol is 0.793 kg/L, what is its volume needed for making 2.5 L of its 0.25 M solution?
Answer: Moles required = 0.25 × 2.5 = 0.625 mol. Mass required = 0.625 × 32 (molar mass of CH3OH) = 20 g. Volume = mass/density = 20 g / 793 g/L = 0.0252 L = 25.2 mL.
What is the concentration of sugar (C12H22O11) in mol/L if its 20 g are dissolved in enough water to make a final volume of 2 L?
Answer: Moles of sugar = 20/342 = 0.0585 mol (molar mass of C12H22O11 = 342 g/mol). Molarity = 0.0585/2 = 0.02925 mol/L.
Calculate the mass percent of different elements present in sodium sulphate (Na2SO4).
Answer: Molar mass of Na2SO4 = 2(23)+32+4(16) = 46+32+64 = 142 g/mol. Mass % of Na = 46/142 ×100 = 32.39%; Mass % of S = 32/142 ×100 = 22.54%; Mass % of O = 64/142 ×100 = 45.07%.
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Class 11 › Chemistry › Some Basic Concepts of Chemistry
Chapter notesRead the chapterPractice MCQsCheck what you rememberChapter test10 questions on one pageOnline testTest 1, 2 and 3, one question at a timePYQPrevious year questionsStudy materialPDF notes
Topics in this chapter: Importance of Chemistry, Matter & Its ClassificationLaws of Chemical Combinations & Dalton's Atomic TheoryAtomic & Molecular Masses, Mole Concept & Molar MassEmpirical/Molecular Formula, Stoichiometry & Concentration Terms