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

NCERT Solutions for Class 11 Chemistry Chapter 1: Understanding the basic concepts of chemistry marks the beginning of your Class 11 Chemistry journey. Chapter 1 explains important ideas like the mole concept, laws of chemical combination, chemical formulas, and stoichiometry. NCERT Class 11 Chemistry Ch 1 Exercise Solutions will also help you in Class 12, JEE, NEET, and competitive exams.

Here are the complete NCERT exercise solutions for Class 11 Chemistry Chapter 1 — updated for the 2025 NCERT edition.

Class 11 Chemistry Chapter 1 Exercise Solutions

Here are the complete NCERT exercise solutions for Class 11 Chemistry Chapter 1. These Class 11 Chemistry Chapter 1 NCERT Solutions cover all in-text and back-of-the-book exercise questions in a clear and simple format to help you understand key concepts like mole concept, atomic and molecular masses, and stoichiometry. Some basic concepts of chemistry NCERT solutions are perfect for quick revision and board exam preparation.

Class 11 Chemistry Chapter 1 Intext Questions

Class 11 Chemistry Chapter 1 Question Answer includes concept-based problems from important topics such as laws of chemical combination, scientific notation, significant figures, SI units, mole concept, and percentage composition. These NCERT Solutions for Class 11 Chemistry Chapter 1 short questions help strengthen your understanding of basic chemistry principles and ensure you are well-prepared for upcoming board exams and competitive tests.

Q1. Define chemistry.

Answer:
Chemistry is the branch of science that deals with the study of matter — its composition, properties, structure, and the changes it undergoes during chemical reactions.

Q2. What is matter? Give two examples.

Answer:
Matter is anything that has mass and occupies space.
Examples: Water, Air.

Q3. What are physical and chemical properties?

Answer:
Physical properties are those that can be observed without changing the chemical composition of a substance (e.g., colour, melting point).
Chemical properties describe how a substance reacts with other substances to form new ones (e.g., rusting of iron).

Q4. State the law of conservation of mass.

Answer:
Mass can neither be created nor destroyed during any physical or chemical change; it only changes from one form to another.

Q5. What is the law of definite proportions?

Answer:
A chemical compound always contains the same elements in the same fixed proportion by mass, irrespective of its source or method of preparation.

Q6. Write two postulates of Dalton’s Atomic Theory.

Answer:
• Matter consists of tiny indivisible atoms.
• Atoms of the same element have identical masses and properties.

Q7. What is atomic mass unit (amu)?

Answer:
1 amu is defined as one-twelfth of the mass of one carbon-12 atom.
1 amu = 1.66056 × 10⁻²⁴ g

Q8. Define molar mass with an example.

Answer:
Molar mass is the mass of one mole of a substance, expressed in grams.
Example: Molar mass of water (H₂O) = 18 g/mol.

Q9. What is Avogadro’s number?

Answer:
Avogadro’s number is the number of particles present in 1 mole of a substance, equal to
6.022 × 10²³ particles per mole.

Q10. Define empirical formula.

Answer:
Empirical formula is the simplest formula of a compound containing elements in small whole-number ratio by moles.
Example: Glucose empirical formula = CH₂O.

Q11. What is stoichiometry?

Answer:
Stoichiometry deals with numerical calculations of reactants and products in a chemical reaction based on a balanced equation.

Q12. What is limiting reagent?

Answer:
The reactant that gets consumed first during a reaction and limits the amount of product formed is called the limiting reagent.

Q13. Define significant figures. Give one rule.

Answer:
Significant figures are the meaningful digits in a measurement that indicate its precision.
Rule: All non-zero digits are significant.
Example: 345 → 3 significant figures.

Numerical Questions and Solutions – Some Basic Concepts of Chemistry

Class 11 Chemistry Chapter 1 Back Exercise Questions help students revise the complete chapter thoroughly. These questions cover essential topics like mole concept, atomic and molecular masses, empirical and molecular formulas, laws of chemical combination, and stoichiometry. Also, here we have provided some basic concepts of chemistry class 11 numericals for better practice:

Q1. Calculate the molecular mass of the following:

(i) 2H₂O
(ii) CO₂
(iii) CH₄

 Answer:

(i) H₂O
2 × Atomic mass of H = 2 × 1.008 = 2.016
O = 16
Molecular mass = 18.016 u

(ii) CO₂
C = 12.011
2 × O = 32
Molecular mass = 44.011 u

(iii) CH₄
C = 12.011
4 × H = 4.032
Molecular mass = 16.043 u

Q2. Calculate the mass percent of different elements present in sodium sulphate (Na₂SO₄).

Answer:

Molar mass = (2×23) + (1×32) + (4×16) = 142 g/mol

Q3. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

 Answer:

Fe moles = 69.9 / 55.85 = 1.252
O moles = 30.1 / 16 = 1.881
Ratio = 1.252 : 1.881 ≈ 1 : 1.5
Multiplying by 2 → Fe₂O₃

Empirical formula = Fe₂O₃

Q4. Calculate the amount of carbon dioxide that could be produced when:

i. 1 mole of carbon is burnt in air.
ii. 1 mole of carbon is burnt in 16 g of dioxygen.
iii. 2 moles of carbon are burnt in 16 g of dioxygen.

Answer:

Reaction: C + O₂ → CO₂

Q5. Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution.

Molar mass = 82.0245 g/mol

Answer:

Moles = M × V = 0.375 × 0.5 = 0.1875
Mass = 0.1875 × 82.0245 = 15.38 g

Q6. Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41 g/mL and mass percent 69%.

Answer:

Mass of 1L solution = 1.41×1000 = 1410 g
HNO₃ mass = 69% → 0.69×1410 = 972.9 g
Molar mass = 63 g/mol
Moles = 972.9 / 63 = 15.44 mol
Molarity = 15.44 M

Q7. How much copper can be obtained from 100 g CuSO₄?

Answer:

Mass of Cu = (63.5 / 159.5) × 100 = 39.8 g

Q8. Determine the molecular formula of an oxide of iron… M = 159.69 g/mol

Answer:

Empirical formula = Fe₂O₃
Empirical mass = 159.7 → ratio = 1
Molecular formula = Fe₂O₃

Q9. Calculate the atomic mass (average) of chlorine using the following data:

• ¹³⁵Cl (75.77%, 34.9689 u)

• ¹³⁷Cl (24.23%, 36.9659 u)

 Answer:

= (0.7577×34.9689) + (0.2423×36.9659)
= 35.45 u

Q10. In three moles of ethane (C₂H₆), calculate the following: (i) number of moles of carbon atoms, (ii) number of moles of hydrogen atoms, (iii) number of molecules of ethane.
Answer:
For C₂H₆, 1 molecule has 2 C and 6 H atoms. Given 3 mol C₂H₆:
(i) Moles of C atoms = 3 × 2 = 6 mol
(ii) Moles of H atoms = 3 × 6 = 18 mol
(iii) Number of molecules = 3 × 6.022×10²³ = 1.8066×10²⁴ molecules

Q11. What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L⁻¹ if 20 g are dissolved to make 2 L solution?
Answer:
M(C₁₂H₂₂O₁₁) ≈ 342.30 g mol⁻¹. Moles = 20/342.30 = 0.05843 mol.
Molarity = 0.05843 / 2.00 = 0.0292 M.

Q12. If density of methanol is 0.793 kg L⁻¹, what volume is needed to make 2.5 L of 0.25 M solution?
Answer:
Moles needed = 0.25 × 2.5 = 0.625 mol. M(CH₃OH) ≈ 32.04 g mol⁻¹.
Mass = 0.625 × 32.04 = 20.03 g. Density = 0.793 g mL⁻¹.
Volume = 20.03 / 0.793 = 25.3 mL (≈ 25.2–25.3 mL).

Q13. The pressure is measured as mass of air at sea level = 1034 g cm⁻². Convert the pressure to SI unit (Pa).
Answer:
1 atm ≈ 1033.6 g cm⁻² and 1 atm = 1.013×10⁵ Pa.
Hence pressure ≈ 1.013×10⁵ Pa.

Q14. What is the SI unit of mass? How is it defined?
Answer:
Kilogram (kg). Defined via the fixed numerical value of Planck’s constant (h = 6.62607015×10⁻³⁴ J·s), realized using a Kibble balance.

Q15. Match the following prefixes with their multiples: Micro, Deca, Mega, Giga, Femto.
Answer:
Micro = 10⁻⁶; Deca = 10¹; Mega = 10⁶; Giga = 10⁹; Femto = 10⁻¹⁵.

Q16. What do you mean by significant figures?
Answer:
Digits known with certainty in a measurement plus one estimated (uncertain) digit; they indicate precision and guide rounding during calculations.

Q17. Chloroform (CHCl₃) contamination in water is 15 ppm (w/w). (i) Express as percent by mass. (ii) Calculate molality of CHCl₃ in water.
Answer:
(i) 15 ppm = 15 mg per 1 kg = 0.015 g per 1000 g → 1.5×10⁻³ %.
(ii) M(CHCl₃) ≈ 119.37 g mol⁻¹. Moles = 0.015/119.37 = 1.26×10⁻⁴ mol.
Solvent ≈ 1.000 kg → molality = 1.26×10⁻⁴ m.

Q18. Express the following in scientific notation: 0.00048; 234000; 80008; 500.0; 6.0012.
Answer:
0.00048 = 4.8×10⁻⁴; 234000 = 2.34×10⁵; 80008 = 8.008×10⁴;
500.0 = 5.000×10²; 6.0012 = 6.0012×10⁰.

Q19. How many significant figures are present in the following? 0.0025; 208; 5005; 126000; 500.0; 2.0034.
Answer:
0.0025 → 2; 208 → 3; 5005 → 4; 126000 → 3; 500.0 → 4; 2.0034 → 5.

Q20. Round up the following to three significant figures: 34.216; 10.4107; 0.04597; 2808.
Answer:
34.216 → 34.2; 10.4107 → 10.4; 0.04597 → 0.0460; 2808 → 2810.

Q21. The following data are obtained when dinitrogen and dioxygen react to form different compounds. Masses of N₂ (g): 14, 14, 28, 28; Masses of O₂ (g): 16, 32, 32, 80. (a) Which law is obeyed? Give its statement. (b) Fill in the blanks in the usual metric conversions.
Answer:
(a) Law of Multiple Proportions: masses of one element that combine with a fixed mass of another are in small whole-number ratios.
(b) Common conversions: 1 km = 10⁶ mm = 10¹⁵ pm; 1 mg = 10⁻⁶ kg = 10⁶ ng; 1 mL = 10⁻³ L = 10⁻³ dm³.

Q22. How much distance will light travel in 2.00 ns? (c = 3.00×10⁸ m s⁻¹)
Answer:
t = 2.00×10⁻⁹ s. d = c×t = 3.00×10⁸ × 2.00×10⁻⁹ = 0.600 m.

Q23. Identify the limiting reagent in each mixture (A + B → product, 1:1 stoichiometry): (i) 100 atoms A + 50 molecules B; (ii) 50 atoms A + 100 molecules B; (iii) 100 atoms A + 100 molecules B; (iv) 5 mol A + 2.5 mol B; (v) 2.5 mol A + 5 mol B.
Answer:
(i) B; (ii) A; (iii) none; (iv) B; (v) A.

Q24. Dinitrogen and dihydrogen react to give ammonia: N₂ + 3H₂ → 2NH₃. (i) Calculate mass of NH₃ produced if 2.00×10³ g N₂ reacts with 1.00×10³ g H₂. (ii) Will any reactant remain? (iii) If yes, which one and what mass?
Answer:
Moles N₂ = 2000/28.02 = 71.39 mol; moles H₂ = 1000/2.016 = 496.0 mol.
Needed H₂ = 3×71.39 = 214.2 mol → N₂ limiting.
(i) Moles NH₃ = 2×71.39 = 142.8 mol → mass = 142.8×17.031 ≈ 2.43×10³ g.
(ii) Yes. (iii) H₂ left = (496.0−214.2) mol = 281.8 mol → mass ≈ 281.8×2.016 ≈ 5.68×10² g (≈ 568–571 g with rounding).

Q25. How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
Answer:
0.50 mol Na₂CO₃ is an amount (≈ 53 g).
0.50 M Na₂CO₃ means 0.50 mol in 1.00 L of solution (≈ 53 g made up to 1 L).

Q26. If ten volumes of H₂ react with five volumes of O₂, how many volumes of water vapour are produced?
Answer:
2H₂ + O₂ → 2H₂O; volume ratio 2:1 → 2.
Given 10:5 is stoichiometric; product = 10 volumes H₂O(v).

Q27. Convert the following into basic SI units: (i) 28.7 pm; (ii) 15.15 pm; (iii) 25365 mg.
Answer:
(i) 28.7 pm = 2.87×10⁻¹¹ m; (ii) 15.15 pm = 1.515×10⁻¹¹ m; (iii) 25365 mg = 2.5365×10⁻² kg.

Q28. Which of the following will have the largest number of atoms: 1 g Au(s), 1 g Na(s), 1 g Li(s), 1 g Cl₂(g)?
Answer:
Number of atoms ∝ 1/molar mass. Li has the smallest molar mass (≈ 6.94 g mol⁻¹).
Hence 1 g Li has the largest number of atoms.

Q29. Calculate the molarity of ethanol in water if the mole fraction of ethanol is 0.040.
Answer:
Let total moles = 1 → n(ethanol)=0.040, n(water)=0.960.
Masses: ethanol ≈ 0.040×46.07 = 1.843 g; water ≈ 0.960×18.015 = 17.295 g.
Using the standard textbook approximation for this composition, M ≈ 2.31–2.32 M (often quoted as 2.314 M).

Q30. What is the mass of one atom of carbon-12?
Answer:
Mass of one atom = 12.000 g mol⁻¹ / (6.022×10²³ mol⁻¹) = 1.9926×10⁻²³ g (≈ 1.99×10⁻²³ g).

Q31. State the number of significant figures in the following results (as per the operations shown in the exercise): (i) 0.200 × 15; (ii) 5.00 + 1.234; (iii) 2500 ÷ 3.45.
Answer:
(i) 3 significant figures; (ii) 4 significant figures; (iii) 4 significant figures (after applying the rules for multiplication/division and addition).

Q32. What is the molar mass of argon?
Answer:
39.948 g mol⁻¹ (commonly rounded to 39.95 g mol⁻¹).

Q33. Calculate the number of atoms for each case: (i) 52 mol Ar; (ii) 52 u He; (iii) 52 g He.
Answer:
(i) 52×6.022×10²³ = 3.13×10²⁵ atoms.
(ii) 52 u / 4.003 u per atom ≈ 13 atoms.
(iii) 52/4.003 = 12.99 mol → 12.99×6.022×10²³ = 7.83×10²⁴ atoms.

Q34. The percentage composition of a fuel gas indicates its empirical formula to be CH. If molar mass is 26 g mol⁻¹, find the molecular formula.
Answer:
Empirical formula mass = 13. n = 26/13 = 2.
Molecular formula = C₂H₂.

Q35. What mass of CaCO₃ is required to completely react with 25.0 mL of 0.750 M HCl? (CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O)
Answer:
Moles HCl = 0.750×0.0250 = 0.01875 mol.
Moles CaCO₃ = 0.01875/2 = 0.009375 mol.
M(CaCO₃) ≈ 100.09 g mol⁻¹. Mass = 0.009375×100.09 ≈ 0.94 g.

Q36. How many grams of HCl are needed to react completely with 5.00 g MnO₂? (MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O)
Answer:
M(MnO₂) ≈ 86.94 g mol⁻¹. Moles MnO₂ = 5.00/86.94 = 0.0575 mol.
Moles HCl needed = 4×0.0575 = 0.230 mol.
Mass HCl = 0.230×36.46 = 8.39 g (≈ 8.4 g).

NCERT PDF Download for Chapter 1 – Some Basic Concepts of Chemistry

Below, we have provided the complete NCERT solutions PDF for Class 11 Chemistry Chapter 1 – Some Basic Concepts of Chemistry. 

Students can easily download and use these NCERT solutions for class 11 chemistry chapter 1 PDF download for offline study and quick revision. The updated NCERT Solutions for Class 11 Chemistry Chapter 1 PDF for 2025 edition includes all exercise questions, numerical problems, and detailed step-by-step solutions to help strengthen conceptual understanding and exam preparation.

NCERT PDF Download for Chapter 1 – Some Basic Concepts of Chemistry

Key Formulae and Summary – Chapter 1 Chemistry Class 11

These basic concepts of chemistry class 11 solutions build the foundation of physical chemistry through concepts like the mole, stoichiometry, laws of chemical combination, atomic and molecular masses, and concentration terms. Students should know how to apply these in numerical problems. 

The key idea is to relate mass, moles, particles, and volume to solve chemical equations accurately and predict the quantities of reactants and products.

Important Formulae

• Mole concept:
 Number of moles (n) = Given mass (m) / Molar mass (M)

• Avogadro’s relation:
 Number of particles (N) = n × 6.022×10²³

• Molecular mass:
 Sum of atomic masses of all atoms in a molecule

• Mass–Mole–Particle relation:
 m / M = N / Nₐ

• Laws of chemical combination:
 Law of conservation of mass, law of definite proportions, law of multiple proportions, Gay-Lussac’s law of gaseous volumes, Avogadro’s law

• Molarity (M):
 M = moles of solute / volume of solution (L)

• Molality (m):
 m = moles of solute / mass of solvent (kg)

• Mass percent:
 % mass = (Mass of component / Total mass) × 100

• Empirical formula:
 Simplest whole-number ratio of atoms in a compound

• Stoichiometric calculations:
 Based on balanced chemical equations to find limiting reagent and expected yield

Benefits of Using Class 11 Chemistry NCERT Solutions

Using Class 11 Chemistry NCERT Solutions helps students understand concepts more clearly and solve textbook questions with ease. These NCERT Solutions for Class 11 Chemistry Chapter 1 solutions provide detailed explanations, accurate numerical solving, and exam-focused presentation to improve performance. 

Check below how chapter-wise Class 11 Chemistry NCERT solutions support smooth revision and better exam preparation:

• Chemistry class 11 solutions provide clear, step-by-step answers for every textbook question

• Help in understanding important concepts and solving numerical problems easily

• Follow the latest NCERT patterns for Class 11 and competitive exams

• Improve accuracy and confidence in writing answers during exams

• Useful for quick revision before tests, boards, and entrance exams like JEE/NEET

• Strengthen  skills through structured NCERT Chemistry Class 11 Solutions

• Support self-study with easy language and logical explanations

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