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

Class 11 Chemistry NCERT Solutions Chapter 1 provides comprehensive step-by-step solutions to all textbook problems covering fundamental concepts like mole calculations, stoichiometry, and laws of chemical combination essential for CBSE 2025-26 examinations.

The first chapter of CBSE Class 11 Chemistry, titled Some Basic Concepts of Chemistry, establishes the quantitative foundation upon which the entire subject rests. Published by the National Council of Educational Research and Training (NCERT.nic.in), this chapter introduces students to the mathematical language of chemistry—teaching them how matter is measured, classified, and transformed through chemical reactions.

Understanding this chapter thoroughly is crucial because every subsequent topic, from chemical bonding to thermodynamics, relies on the numerical skills developed here. Students who master mole calculations and stoichiometric relationships find themselves confident when tackling complex problems in organic chemistry, equilibrium, and electrochemistry. For a complete overview of all chapters, you can explore the Class 11 Chemistry NCERT Solutions – comprehensive guide.

Class 11 Chemistry NCERT Solutions Chapter 1 – Complete Overview

The NCERT textbook structures Chapter 1 to progressively build understanding from basic definitions to complex numerical applications. Beginning with the classification of matter into elements, compounds, and mixtures, the chapter advances through atomic theory, introduces the mole concept, and culminates in stoichiometric calculations that students will use throughout their chemistry education.

The solutions provided below follow the exact methodology recommended by CBSE marking schemes. Each problem is solved using clearly defined steps, ensuring students understand not just the answer but the reasoning process. This approach develops problem-solving skills that transfer to competitive examinations like JEE Main, NEET, and state-level entrance tests.

Key topics covered in Chapter 1 include:

The Laws of Chemical Combination form the historical and theoretical basis for understanding how elements combine. Lavoisier’s Law of Conservation of Mass states that matter cannot be created or destroyed in a chemical reaction—the total mass of reactants equals the total mass of products. Proust’s Law of Definite Proportions establishes that a compound always contains the same elements in fixed mass ratios, regardless of its source or preparation method.

Students preparing for board examinations should note that these laws frequently appear in one-mark definitional questions and three-mark application problems. The Chemical Kinetics NCERT Solutions – Class chapter later builds upon these foundational principles when discussing reaction rates and mechanisms.

Mole Concept and Avogadro’s Number – Solved Problems

The mole concept represents the central calculation tool in chemistry. One mole of any substance contains exactly 6.022 × 10²³ particles—this value, known as Avogadro’s number (Nₐ), connects the atomic world to measurable laboratory quantities. Understanding how to convert between mass, moles, and number of particles is essential for solving NCERT Chapter 1 problems.

The fundamental relationship for mole calculations is expressed as n = m/M, where n represents the number of moles, m is the given mass in grams, and M is the molar mass in grams per mole. This formula forms the backbone of nearly every numerical problem in Chapter 1.

Worked Example 1: Calculating Moles from Mass

Question: Calculate the number of moles present in 49 g of sulphuric acid (H₂SO₄).

Solution: First, calculate the molar mass of H₂SO₄. Using atomic masses (H = 1, S = 32, O = 16): M = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g/mol. Applying the formula n = m/M: n = 49/98 = 0.5 moles.

Worked Example 2: Converting Moles to Number of Particles

Question: How many molecules are present in 0.25 moles of carbon dioxide?

Solution: The number of particles equals moles multiplied by Avogadro’s number. Number of molecules = n × Nₐ = 0.25 × 6.022 × 10²³ = 1.5055 × 10²³ molecules. Since each CO₂ molecule contains 3 atoms (1 carbon + 2 oxygen), the total number of atoms = 1.5055 × 10²³ × 3 = 4.5165 × 10²³ atoms.

Worked Example 3: Mass from Number of Atoms

Question: Calculate the mass of 3.011 × 10²³ atoms of copper (atomic mass = 63.5 u).

Solution: First, find moles of copper atoms: n = Number of atoms/Nₐ = 3.011 × 10²³/6.022 × 10²³ = 0.5 mol. Then calculate mass using m = n × M: m = 0.5 × 63.5 = 31.75 g.

Stoichiometry and Limiting Reagent Calculations

Stoichiometry involves using balanced chemical equations to calculate quantities of reactants and products. The coefficients in a balanced equation represent molar ratios, enabling predictions about how much product forms from given reactants or how much reactant is needed for a desired product quantity.

Students often find limiting reagent problems challenging, but a systematic approach simplifies them considerably. The limiting reagent is the reactant that gets completely consumed first, thereby determining the maximum amount of product that can form. The other reactant, present in excess, remains partially unreacted.

Worked Example 4: Basic Stoichiometry

Question: How many grams of water are produced when 4 g of hydrogen gas reacts completely with oxygen?

Solution: Write the balanced equation: 2H₂ + O₂ → 2H₂O. Calculate moles of H₂: n = 4/2 = 2 mol. From the equation, 2 mol H₂ produces 2 mol H₂O, so moles of water = 2 mol. Mass of water = 2 × 18 = 36 g.

Worked Example 5: Limiting Reagent Problem

Question: 10 g of calcium carbonate reacts with 10 g of hydrochloric acid. Identify the limiting reagent and calculate the mass of CO₂ produced.

Solution: Balanced equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Moles of CaCO₃ = 10/100 = 0.1 mol. Moles of HCl = 10/36.5 = 0.274 mol. From the equation, 1 mol CaCO₃ requires 2 mol HCl. For 0.1 mol CaCO₃, HCl required = 0.2 mol. Available HCl (0.274 mol) exceeds requirement, so CaCO₃ is the limiting reagent. Moles of CO₂ = moles of CaCO₃ = 0.1 mol. Mass of CO₂ = 0.1 × 44 = 4.4 g.

Understanding stoichiometry thoroughly prepares students for physical chemistry topics. The mathematical relationships established here directly apply when studying NCERT Solutions CBSE Free for Class trigonometric functions, which share similar problem-solving approaches requiring systematic application of formulae.

Empirical and Molecular Formula Determination

Determining empirical and molecular formulae from experimental data represents a critical skill tested in CBSE examinations. The empirical formula shows the simplest whole-number ratio of atoms in a compound, while the molecular formula shows the actual number of atoms present in one molecule.

Worked Example 6: Finding Empirical Formula from Percentage Composition

Question: A compound contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen by mass. Determine its empirical formula.

Solution: Assume 100 g of compound. Mass of C = 40 g, H = 6.67 g, O = 53.33 g. Convert to moles: C = 40/12 = 3.33 mol, H = 6.67/1 = 6.67 mol, O = 53.33/16 = 3.33 mol. Divide by smallest (3.33): C = 1, H = 2, O = 1. Empirical formula: CH₂O.

Worked Example 7: Finding Molecular Formula

Question: The empirical formula of a compound is CH₂O and its molar mass is 180 g/mol. Find the molecular formula.

Solution: Empirical formula mass = 12 + 2(1) + 16 = 30 g/mol. Calculate n = Molar mass/Empirical formula mass