NCERT Solutions for Class 11 Chemistry Chapter 1

NCERT Solutions for Class 11 Chemistry Chapter 1 – Some Basic Concepts of Chemistry are available here for Class 11 students. You can also download these solutions for free in PDF format by clicking the download button below.

Prepared by experienced teachers and subject experts, these NCERT Solutions are designed to meet the academic needs of Class 11 learners. Each solution includes step-by-step explanations to help students confidently tackle similar questions in their board exams.

The content is presented in simple, easy-to-understand language and follows the latest CBSE Syllabus 2025-26. By covering the fundamental concepts clearly, these NCERT Solutions aim to enhance the learning experience and build a strong foundation in Chemistry.

Students can download the Class 11 Chemistry Chapter 1 NCERT Solutions PDF from IL to strengthen their preparation and complete a part of their CBSE syllabus well ahead of the exams.

NCERT Solutions Class 11 Chemistry Chapter 1 - Questions with Answers

Q1. Define the term ‘mole’.

Answer: A mole is the amount of substance that contains the same number of entities (atoms, molecules, or particles) as there are atoms in 12 g of carbon-12 isotope.
Numerically, 1 mole = $6.022 \times 10^{23}$ entities (Avogadro’s number).

Q2. Calculate the number of moles in 9 g of water.

Answer: Molar mass of water (H₂O) = 2(1) + 16 = 18 g/mol
Number of moles = Given mass ÷ Molar mass
= $\frac{9}{18}$ = 0.5 mol

Q3. What is the difference between empirical formula and molecular formula?

Answer: Empirical Formula: Simplest whole-number ratio of atoms of each element in a compound.
Example: Glucose (C₆H₁₂O₆) → CH₂O

Molecular Formula: Actual number of atoms of each element in a molecule.
Example: Glucose → C₆H₁₂O₆

Q4. A sample of CO₂ contains 3.01 × 10²³ molecules. Calculate the mass of CO₂ present.

Answer: Avogadro’s number = $6.022 \times 10^{23}$ molecules/mol
Number of moles = $\frac{3.01 \times 10^{23}}{6.022 \times 10^{23}} = 0.5 \, \text{mol}$
Molar mass of CO₂ = 12 + 32 = 44 g/mol
Mass = 0.5 × 44 = 22 g

Q5. State Dalton’s Atomic Theory in brief.

Answer:

• Matter is made of indivisible atoms.
• Atoms of the same element are identical in mass and properties.
• Atoms combine in fixed whole-number ratios to form compounds.
• Atoms cannot be created or destroyed.

Q6. Define molarity and molality.

Answer:

Molarity (M): Number of moles of solute dissolved in 1 litre of solution.

Molality (m): Number of moles of solute dissolved in 1 kg of solvent.

Q7. A solution contains 5 g of NaOH in 250 mL of solution. Calculate its molarity.

Answer:

• Molar mass of NaOH = 40 g/mol
• Moles of NaOH = 5 ÷ 40 = 0.125 mol
• Volume of solution = 250 mL = 0.25 L
• Molarity = 0.125 ÷ 0.25 = 0.5 M

Q8. Differentiate between atom and molecule.

Answer:

Atom: Smallest indivisible unit of an element that takes part in a chemical reaction. Example: H, O, Na.

Molecule: Combination of two or more atoms held by covalent bonds. Example: H₂, O₂, H₂O.

Q9. Define atomic mass unit (amu).

Answer:
1 atomic mass unit (1 u) is defined as one-twelfth of the mass of a carbon-12 atom.
1 u = $1.66 \times 10^{-27} \, \text{kg}$.

Q10. Calculate the number of atoms in 4.6 g of sodium (Na).

Answer:
Molar mass of Na = 23 g/mol
Number of moles = $\frac{4.6}{23} = 0.2 \, \text{mol}$
Number of atoms = 0.2 × $6.022 \times 10^{23}$ = 1.204 × 10²³ atoms

Q11. What is the law of conservation of mass?

Answer:
The law states that “Matter can neither be created nor destroyed in a chemical reaction.”
Thus, total mass of reactants = total mass of products.

Q12. Calculate the empirical formula of a compound containing 40% carbon, 6.7% hydrogen, and 53.3% oxygen.

Answer:

• Step 1: Assume 100 g of compound → C = 40 g, H = 6.7 g, O = 53.3 g

• C: 40 ÷ 12 = 3.33

Q13. Explain the difference between precision and accuracy.

Answer:

• Precision: How close repeated measurements are to each other (consistency).

Q14. Calculate the mass percent of hydrogen in H₂O.

Answer:
Molar mass of H₂O = 18 g/mol
Mass of H = 2 g
% of H = (2 ÷ 18) × 100 = 11.11%

Q15. State Avogadro’s Law.

Answer:
Avogadro’s Law states that equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.

Q16. A gaseous compound of carbon and hydrogen contains 80% carbon by mass. If its molecular mass is 30, find its molecular formula.

Answer:

• Step 1: Assume 100 g → C = 80 g, H = 20 g

• C: 80 ÷ 12 = 6.67

Q17. Differentiate between molar mass and molecular mass.

Answer:

• Molecular Mass: Sum of atomic masses of all atoms in a molecule. Expressed in amu. Example: H₂O = 18 u.

Q18. What is the difference between molecular weight and formula weight?

Answer:

• Molecular weight: Mass of one molecule of a substance (for covalent compounds). Example: H₂O = 18 u.

Q19. Calculate the number of molecules in 11.2 L of CO₂ at STP.

Answer:
At STP, 1 mole of gas = 22.4 L.
Number of moles = $\frac{11.2}{22.4} = 0.5 \, \text{mol}$
Number of molecules = 0.5 × $6.022 \times 10^{23}$ = 3.011 × 10²³ molecules

Q20. State the law of multiple proportions with an example.

Answer:
The law states that when two elements combine to form more than one compound, the masses of one element combining with a fixed mass of the other bear a simple whole-number ratio.
Example: Carbon + Oxygen → CO (12:16) and CO₂ (12:32). Ratio = 16:32 = 1:2.

Q21. A sample of H₂SO₄ contains 9.8 g of H₂SO₄. Calculate the number of moles.

Answer:
Molar mass of H₂SO₄ = 98 g/mol
Number of moles = $\frac{9.8}{98} = 0.1 \, \text{mol}$

Q22. Distinguish between heterogeneous and homogeneous mixtures.

Answer:

• Homogeneous mixture: Uniform composition throughout. Example: sugar solution.

Q23. Calculate the molality of a solution prepared by dissolving 18 g of glucose (C₆H₁₂O₆) in 90 g of water.

Answer:
Molar mass of glucose = 180 g/mol
Moles of glucose = 18 ÷ 180 = 0.1 mol
Mass of solvent = 90 g = 0.09 kg
Molality = $\frac{0.1}{0.09} = 1.11 \, m$

Q24. Explain the term significant figures with an example.

Answer:
Significant figures are the digits in a number that carry meaningful information about precision.
Example: 0.00540 → 3 significant figures (5, 4, 0).

Q25. A compound contains 70% iron and 30% oxygen by mass. Find its empirical formula.

Answer:

• Step 1: Assume 100 g → Fe = 70 g, O = 30 g

• Fe: 70 ÷ 56 = 1.25

Q26. Explain the term stoichiometry with an example.

Answer:
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction.
Example: $2H₂ + O₂ → 2H₂O$.
It shows 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water.

Q27. Calculate the volume of 0.5 mol of oxygen gas at STP.

Answer:
At STP, 1 mol gas = 22.4 L
Volume = 0.5 × 22.4 = 11.2 L 

Q28. Write any two limitations of Dalton’s Atomic Theory.

Answer:

1. It could not explain isotopes (atoms of the same element with different masses).