InfinityLearnInfinityLearn
courses
study material
results
more
call.svg
need help? talk to experts
talk to experts
7996668865
call.svg
Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9
Banner 10
Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9
Banner 10
Banner 11
AI Mentor
Book Online Demo
Try Test

Courses

Dropper NEET CourseDropper JEE CourseClass - 12 NEET CourseClass - 12 JEE CourseClass - 11 NEET CourseClass - 11 JEE CourseClass - 10 Foundation NEET CourseClass - 10 Foundation JEE CourseClass - 10 CBSE CourseClass - 9 Foundation NEET CourseClass - 9 Foundation JEE CourseClass -9 CBSE CourseClass - 8 CBSE CourseClass - 7 CBSE CourseClass - 6 CBSE Course
sticky footer img
Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!
  • NCERT Solutions Class 11 Chemistry Chapter 1 - Questions with Answers
  • NCERT Solutions for Class 11 Chemistry Chapter 1 FAQs
NCERT Solutions /
ncertclass11 /
NCERT Solutions for Class 11 Chemistry Chapter 1
NCERT Solutions /
ncertclass11 /
NCERT Solutions for Class 11 Chemistry Chapter 1

NCERT Solutions for Class 11 Chemistry Chapter 1

By Swati Singh

|

Updated on 29 Sep 2025, 14:38 IST

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.

Fill out the form for expert academic guidance
+91
Student
Parent / Guardian
Teacher
submit

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.

Unlock the full solution & master the concept
Get a detailed solution and exclusive access to our masterclass to ensure you never miss a concept

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 = 6.022×10236.022 \times 10^{23}6.022×1023 entities (Avogadro’s number).

Ready to Test Your Skills?
Check Your Performance Today with our Free Mock Tests used by Toppers!
Take Free Test

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
= 918\frac{9}{18}189​ = 0.5 mol

cta3 image
create your own test
YOUR TOPIC, YOUR DIFFICULTY, YOUR PACE
start learning for free

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

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

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.

Ready to Test Your Skills?
Check Your Performance Today with our Free Mock Tests used by Toppers!
Take Free Test

Answer: Avogadro’s number = 6.022×10236.022 \times 10^{23}6.022×1023 molecules/mol
Number of moles = 3.01×10236.022×1023=0.5 mol\frac{3.01 \times 10^{23}}{6.022 \times 10^{23}} = 0.5 \, \text{mol}6.022×10233.01×1023​=0.5mol
Molar mass of CO₂ = 12 + 32 = 44 g/mol
Mass = 0.5 × 44 = 22 g

Q5. State Dalton’s Atomic Theory in brief.

cta3 image
create your own test
YOUR TOPIC, YOUR DIFFICULTY, YOUR PACE
start learning for free

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 = 1.66×10−27 kg1.66 \times 10^{-27} \, \text{kg}1.66×10−27kg.

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

Answer:
Molar mass of Na = 23 g/mol
Number of moles = 4.623=0.2 mol\frac{4.6}{23} = 0.2 \, \text{mol}234.6​=0.2mol
Number of atoms = 0.2 × 6.022×10236.022 \times 10^{23}6.022×1023 = 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

  • Step 2: Convert to moles

    • C: 40 ÷ 12 = 3.33

  • H: 6.7 ÷ 1 = 6.7

  • O: 53.3 ÷ 16 = 3.33

  • Step 3: Simplest whole-number ratio → C:H:O = 3.33:6.7:3.33 ≈ 1:2:1
    Empirical formula = CH₂O

  • Q13. Explain the difference between precision and accuracy.

    Answer:

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

  • Accuracy: How close a measurement is to the true or accepted value.
    Example: Shots clustered together but away from the target = precise but not accurate.

  • 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

  • Step 2: Convert to moles

    • C: 80 ÷ 12 = 6.67

  • H: 20 ÷ 1 = 20

  • Step 3: Ratio = 6.67:20 ≈ 1:3
    → Empirical formula = CH₃

  • Step 4: Empirical formula mass = 15
    Molecular mass ÷ Empirical mass = 30 ÷ 15 = 2
    → Molecular formula = C₂H₆

  • 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.

  • Molar Mass: Mass of 1 mole of molecules (equal to molecular mass in grams). Example: H₂O = 18 g/mol.

  • 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.

  • Formula weight: Mass of one formula unit (for ionic compounds). Example: NaCl = 58.5 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 = 11.222.4=0.5 mol\frac{11.2}{22.4} = 0.5 \, \text{mol}22.411.2​=0.5mol
    Number of molecules = 0.5 × 6.022×10236.022 \times 10^{23}6.022×1023 = 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 = 9.898=0.1 mol\frac{9.8}{98} = 0.1 \, \text{mol}989.8​=0.1mol

    Q22. Distinguish between heterogeneous and homogeneous mixtures.

    Answer:

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

  • Heterogeneous mixture: Non-uniform composition. Example: sand in water.

  • 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 = 0.10.09=1.11 m\frac{0.1}{0.09} = 1.11 \, m0.090.1​=1.11m

    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

  • Step 2: Convert to moles

    • Fe: 70 ÷ 56 = 1.25

  • O: 30 ÷ 16 = 1.875

  • Step 3: Simplest ratio = 1.25 : 1.875 = 2 : 3
    → Empirical formula = Fe₂O₃

  • 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→2H2O2H₂ + O₂ → 2H₂O2H2​+O2​→2H2​O.
    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).

  • It could not explain isobars (different elements with same mass).

  •  

    course

    No courses found

    NCERT Solutions for Class 11 Chemistry Chapter 1 FAQs

    What is Chapter 1 of Class 11 Chemistry about?

    Chapter 1, Some Basic Concepts of Chemistry, introduces the fundamental ideas of Chemistry such as laws of chemical combination, Dalton’s atomic theory, mole concept, stoichiometry, atomic and molecular masses, and concentration terms like molarity and molality.

    Why are NCERT Solutions important for Chapter 1 Chemistry?

    NCERT Solutions provide step-by-step answers to textbook questions. They help students build a strong foundation, understand key concepts, and practice numericals essential for board exams and competitive exams like NEET and JEE.

    Are these NCERT Solutions based on the latest CBSE Syllabus (2025-26)?

    Yes. The solutions are prepared according to the updated CBSE Class 11 Chemistry Syllabus 2025–26, ensuring relevance and accuracy.

    Can I download Class 11 Chemistry Chapter 1 NCERT Solutions in PDF format?

    Yes. Students can download free PDF versions of NCERT Solutions for Chapter 1 from IL and other academic platforms. The PDF format allows offline study and easy access.

    What are the key topics covered in Chapter 1 NCERT Solutions?

    The major topics include:

    • Laws of chemical combination
    • Dalton’s atomic theory
    • Atomic and molecular masses
    • Mole concept and Avogadro’s number
    • Empirical and molecular formulas
    • Stoichiometry and numerical problems
    • Concentration terms (molarity, molality, normality)

    How do NCERT Solutions help in board exam preparation?

    They provide clear explanations, solved examples, and practice questions. By working through them, students improve problem-solving skills and develop confidence to attempt similar board exam questions.

    Are numerical problems in NCERT Solutions sufficient for competitive exams like JEE/NEET?

    NCERT Solutions cover the basics thoroughly. While they are sufficient for board exams, for JEE/NEET aspirants, additional higher-level problem-solving practice from reference books is recommended.

    How many exercises are there in Class 11 Chemistry Chapter 1 NCERT textbook?

    The chapter contains multiple exercise questions covering theory and numericals. NCERT Solutions provide step-by-step answers to all these exercises for easy understanding.

    Who prepares NCERT Solutions for Chapter 1 Chemistry?

    They are prepared by subject experts and experienced teachers to match the CBSE syllabus and exam requirements.

    footerlogos
    call

    1800-419-4247 (customer support)

    call

    7996668865 (sales team)

    mail

    support@infinitylearn.com

    map

    Head Office:
    Infinity Towers, N Convention Rd,
    Surya Enclave, Siddhi Vinayak Nagar,
    Kothaguda, Hyderabad,
    Telangana 500084.

    map

    Corporate Office:
    9th Floor, Shilpitha Tech Park,
    3 & 55/4, Devarabisanahalli, Bellandur,
    Bengaluru, Karnataka 560103

    facebooktwitteryoutubelinkedininstagram
    company
    • About us
    • our team
    • Careers
    • Life at Infinity Learn
    • IL in the news
    • Blogs
    • become a Teacher
    courses
    • JEE Online Course
    • NEET Online Course
    • Foundation Online Course
    • CBSE Online Course
    • HOTS Online Course
    • All India Test Series
    • Book Series
    support
    • Privacy Policy
    • Refund Policy
    • grievances
    • Terms & Conditions
    • Supplier Terms
    • Supplier Code of Conduct
    • Posh
    more
    • AINA - AI Mentor
    • Sri Chaitanya Academy
    • Score scholarships
    • YT Infinity Learn JEE
    • YT - Infinity Learn NEET
    • YT Infinity Learn 9&10
    One Stop Solutions
    • JEE Main One Stop Solutions
    • JEE Advanced One Stop Solutions
    • NEET One Stop Solutions
    • CBSE One Stop Solutions
    Rank Predictor
    • JEE Main Rank College Predictor
    • NEET Rank Predictor
    • JEE Main BITSAT Score Predictor
    State Boards Courses
    • Tamil Nadu Online Course

    Free study material

    NCERT SOLUTIONS
    • NCERT Solutions for Class 12
    • NCERT Solutions for Class 11
    • NCERT Solutions for Class 10
    • NCERT Solutions for Class 9
    • NCERT Solutions for Class 8
    • NCERT Solutions for Class 7
    • NCERT Solutions for Class 6
    CBSE BOARD
    • CBSE Class 12 Board Exam
    • CBSE Class 11 Board Exam
    • CBSE Class 10 Board Exam
    • CBSE Class 9 Board Exam
    • CBSE Class 8 Board Exam
    • CBSE Class 7 Board Exam
    • CBSE Class 6 Board Exam
    MULTIPLE CHOICE QUESTIONS
    • CBSE Class 12 MCQs
    • CBSE Class 11 MCQs
    • CBSE Class 10 MCQs
    • CBSE Class 9 MCQs
    • CBSE Class 8 MCQs
    • CBSE Class 7 MCQs
    • CBSE Class 6 MCQs
    WORKSHEETS
    • CBSE Worksheet for Class 12
    • CBSE Worksheet for Class 11
    • CBSE Worksheet for Class 10
    • CBSE Worksheet for Class 9
    • CBSE Worksheet for Class 8
    • CBSE Worksheet for Class 7
    • CBSE Worksheet for Class 6
    STUDY MATERIALS
    • GK Questions
    • English
    • General Topics
    • Biography
    ACADEMIC ARTICLES
    • Maths
    • Physics
    • Chemistry
    • Biology
    REFERENCE BOOKS
    • RD Sharma Solutions
    • Lakhmir Singh Solutions
    • NCERT Solutions for Class 12
    • NCERT Solutions for Class 11
    • NCERT Solutions for Class 10
    • NCERT Solutions for Class 9
    • NCERT Solutions for Class 8
    • NCERT Solutions for Class 7
    • NCERT Solutions for Class 6
    • CBSE Class 12 Board Exam
    • CBSE Class 11 Board Exam
    • CBSE Class 10 Board Exam
    • CBSE Class 9 Board Exam
    • CBSE Class 8 Board Exam
    • CBSE Class 7 Board Exam
    • CBSE Class 6 Board Exam
    • CBSE Class 12 MCQs
    • CBSE Class 11 MCQs
    • CBSE Class 10 MCQs
    • CBSE Class 9 MCQs
    • CBSE Class 8 MCQs
    • CBSE Class 7 MCQs
    • CBSE Class 6 MCQs
    • CBSE Worksheet for Class 12
    • CBSE Worksheet for Class 11
    • CBSE Worksheet for Class 10
    • CBSE Worksheet for Class 9
    • CBSE Worksheet for Class 8
    • CBSE Worksheet for Class 7
    • CBSE Worksheet for Class 6
    • GK Questions
    • English
    • General Topics
    • Biography
    • Maths
    • Physics
    • Chemistry
    • Biology
    • RD Sharma Solutions
    • Lakhmir Singh Solutions

    © Rankguru Technology Solutions Private Limited. All Rights Reserved

    follow us
    facebooktwitteryoutubelinkedininstagram
    Related Blogs
    NCERT Solutions for Class 11 Chemistry Chapter 2NCERT Solutions for Class 11 English - Hornbill Father to SonNCERT Solutions for Class 11 Snapshots Chapter 2 – The AddressNCERT Solutions for Class 11 English Chapter 4 Snapshots BirthNCERT Solutions For Class 11 English Snapshots The Summer of the Beautiful White HorseNCERT Solutions Class 11 PhysicsNCERT Solution for Class 11 English Woven WordsNCERT Solution for Class 11 English HornbillNCERT Solutions for Class 11 English SnapshotsNCERT Solution for Class 11 English