What is NCERT Exercises and Solutions: Some Basic Concepts of Chemistry in NCERT Class 11 Chemistry?

In NCERT Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry, NCERT Exercises and Solutions: Some Basic Concepts of Chemistry is taught with definitions, examples, and interactive simulations on MyAiSchool aligned with CBSE 2025-26.

What are the most-asked NCERT exercise questions in Some Basic Concepts of Chemistry?

NCERT Class 11 Chemistry Some Basic Concepts of Chemistry exercises cover definitions, derivations, numerical problems, and application-based questions. The MyAiSchool solution set provides full step-by-step worked solutions for every NCERT question, aligned with the CBSE board exam pattern. Students should master core formulas, key concepts, and problem-solving methodology to score full marks.

How should students approach numerical problems in Some Basic Concepts of Chemistry?

For numerical problems in NCERT Class 11 Chemistry Some Basic Concepts of Chemistry: (1) list all given data with proper units, (2) write the relevant formula(e), (3) substitute values carefully checking units, (4) calculate with correct significant figures, (5) state the final answer with the appropriate unit. The MyAiSchool solutions follow this CBSE-aligned step-by-step format.

What CBSE board questions come from Some Basic Concepts of Chemistry?

CBSE Class 11 Chemistry board questions from Some Basic Concepts of Chemistry typically include: (1) 1-mark MCQs on definitions, (2) 2-mark short-answers on concepts, (3) 3-mark numerical/application questions, (4) 5-mark long-answer derivations and problem-solving. The MyAiSchool exercise set tags each question by mark weight and Bloom level (L1-L6) for prioritised exam prep.

How do I check my answers for correctness?

To verify answers in Some Basic Concepts of Chemistry exercises: (1) cross-check units and dimensions on both sides of equations, (2) verify limiting cases (extreme values give sensible results), (3) re-substitute the answer to validate equations, (4) compare with reference answers in the MyAiSchool solution set. Every solution includes a unit check and reasoning summary.

What are common mistakes students make in Some Basic Concepts of Chemistry exercises?

Common mistakes in NCERT Class 11 Chemistry Some Basic Concepts of Chemistry exercises include: (1) unit conversion errors, (2) sign or direction errors, (3) algebra/calculation slips, (4) misreading the question requirements, (5) forgetting to write the conclusion or appropriate units. The MyAiSchool solutions flag these traps with annotations so students avoid losing marks.

How does MyAiSchool solution differ from other NCERT solution sets?

MyAiSchool Class 11 Chemistry Some Basic Concepts of Chemistry solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional/unit checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Verified against NCERT and CBSE marking schemes for accuracy.

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NCERT Exercises and Solutions: Some Basic Concepts of Chemistry

🎓 Class 11 Chemistry CBSE Theory Ch 1 – Some Basic Concepts of Chemistry ⏱ ~8 min

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NCERT Exercises and Solutions: Some Basic Concepts of Chemistry

Chapter Summary — Key Formulas & Ideas

Matter & Measurement

Matter has mass and occupies space; three states — solid, liquid, gas.
Pure substance = element + compound. Mixture = homogeneous + heterogeneous.
SI base units: metre, kilogram, second, ampere, kelvin, mole, candela.
Density \( \rho = m/V \); common unit g cm^-3.
Temperature: \( K = ^\circ C + 273.15;\ ^\circ F = (9/5)^\circ C + 32 \).
Scientific notation: \( N \times 10^n,\ 1 \le N < 10 \).

Laws, Atoms & Moles

Conservation of Mass; Definite Proportions; Multiple Proportions; Gay-Lussac; Avogadro.
\(1\ \text{u} = 1.66056 \times 10^{-24}\) g; \( N_A = 6.022 \times 10^{23} \) mol^-1.
\(n = m/M; \quad N = n N_A; \quad V_{\text{STP}} = 22.4\ L\cdot n\).
Atomic mass = Σ (abundance × isotope mass)/100.

Composition, Formula & Stoichiometry

Mass % = (mass of element / M) × 100.
Empirical formula → Molecular formula: \(MF = n\cdot EF,\ n = M_{\text{mol}}/M_{\text{emp}}\).
Limiting reagent: smallest value of n/stoichiometric coefficient.
\( M = n/V(\text{L});\ m = n/m_{\text{solvent(kg)}};\ x_A = n_A/(n_A+n_B);\ M_1V_1 = M_2V_2 \).

Key Terms & Keywords

Matter

Element

Compound

Mixture (homo/hetero)

SI Base Unit

Significant figure

Scientific notation

Dimensional analysis

Conservation of Mass

Definite Proportions

Multiple Proportions

Gay-Lussac's law

Avogadro's law

Avogadro's number

Atomic mass unit (u)

Average atomic mass

Molecular mass

Formula mass

Mole

Molar mass

Molar volume

Percentage composition

Empirical formula

Molecular formula

Stoichiometry

Limiting reagent

Mass percent

Mole fraction

Molarity (M)

Molality (m)

NCERT Exercises — Step-by-Step Solutions

Use atomic masses: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, S = 32, Cl = 35.5, K = 39, Ca = 40, Fe = 56, Cu = 63.5.

1Calculate the molar mass of (a) H₂O (b) CO₂ (c) CH₄.

(a) M(H₂O) = 2(1) + 16 = 18 g/mol.

(b) M(CO₂) = 12 + 2(16) = 44 g/mol.

2Calculate the mass per cent of each element in sodium sulphate (Na₂SO₄).

M(Na₂SO₄) = 2(23) + 32 + 4(16) = 46 + 32 + 64 = 142 g/mol.

\(\%Na = \dfrac{46}{142}\times 100 = 32.39\%\)

\(\%S = \dfrac{32}{142}\times 100 = 22.54\%\)

\(\%O = \dfrac{64}{142}\times 100 = 45.07\%\)

Sum = 32.39 + 22.54 + 45.07 = 100.00 % ✓

3Determine the empirical formula of an oxide of iron which has 69.9 % iron and 30.1 % dioxygen by mass.

Per 100 g: Fe = 69.9 g, O = 30.1 g.

Moles: n(Fe) = 69.9/56 = 1.248; n(O) = 30.1/16 = 1.881.

Divide by smallest (1.248): Fe = 1; O = 1.881/1.248 = 1.507.

Multiply by 2 → Fe = 2, O = 3. Empirical formula = Fe₂O₃.

4Calculate the amount of carbon dioxide that could be produced when (i) 1 mole of carbon burns in air, (ii) 1 mole of carbon is burnt in 16 g of dioxygen, (iii) 2 moles of carbon burn in 16 g of dioxygen.

Reaction: C(s) + O₂(g) → CO₂(g) (1:1:1).

(i) O₂ is unlimited (air excess). 1 mol C → 1 mol CO₂ = 44 g.

(ii) 16 g O₂ = 0.5 mol. Here O₂ is limiting. 0.5 mol CO₂ = 22 g.

(iii) 0.5 mol O₂ vs 2 mol C. O₂ limiting → 0.5 mol CO₂ = 22 g.

5Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 M aqueous solution. (M = 82 g/mol)

\(n = M \cdot V = 0.375 \times 0.500 = 0.1875\ \text{mol}\)

\(m = n \cdot M_r = 0.1875 \times 82 = 15.375\ \text{g}\)

Answer: 15.375 g.

6Calculate the concentration of HNO₃ in moles per litre in a sample which has a density of 1.41 g mL^-1 and the mass per cent of nitric acid in it being 69 %.

Take 1 L of solution → mass = 1000 × 1.41 = 1410 g.

Mass of HNO₃ = 69 % of 1410 = 972.9 g.

M(HNO₃) = 1 + 14 + 3(16) = 63 g/mol.

\(n = 972.9 / 63 = 15.44\ \text{mol}\)

Molarity = 15.44 mol L^-1.

7How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?

M(CuSO₄) = 63.5 + 32 + 4(16) = 159.5 g/mol.

Fraction of Cu = 63.5/159.5.

\(m(Cu) = 100 \times \dfrac{63.5}{159.5} = 39.81\ \text{g}\)

Answer: 39.81 g of Cu.

8Determine the molecular formula of an oxide of iron whose percentage composition is 69.9 % Fe and 30.1 % O, given its molecular mass is 159.69 g/mol.

From Q3, empirical formula = Fe₂O₃. Empirical mass = 2(56) + 3(16) = 112 + 48 = 160 ≈ 159.69.

n = 159.69/160 ≈ 1.

Molecular formula = Fe₂O₃.

9Calculate the atomic mass (average) of chlorine using the following data: ³⁵Cl (abundance 75.77 %, mass 34.9689 u); ³⁷Cl (abundance 24.23 %, mass 36.9659 u).

\(\overline{m} = \dfrac{75.77 \times 34.9689 + 24.23 \times 36.9659}{100}\)

\(= \dfrac{2649.59 + 895.69}{100} = \dfrac{3545.28}{100} = 35.4528\ u \approx 35.45\ u\)

10In three moles of ethane (C₂H₆), calculate: (i) number of moles of carbon atoms, (ii) number of moles of hydrogen atoms, (iii) number of molecules of ethane.

(i) 1 mol C₂H₆ has 2 mol C → 3 mol × 2 = 6 mol of C atoms.

(ii) 3 mol × 6 = 18 mol of H atoms.

(iii) 3 × 6.022 × 10²³ = 1.807 × 10²⁴ molecules.

11What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L^-1 if its 20 g are dissolved in enough water to make a final volume up to 2 L? (M = 342 g/mol)

\(n = 20/342 = 0.0585\ \text{mol}\)

\(M = 0.0585/2 = 0.02925 \approx 0.029\ \text{mol L}^{-1}\)

12If the density of methanol is 0.793 kg L^-1, what is its volume needed for making 2.5 L of its 0.25 M solution? (M(CH₃OH) = 32 g/mol)

Moles needed: n = 0.25 × 2.5 = 0.625 mol.

Mass: m = 0.625 × 32 = 20 g.

Density = 0.793 g mL^-1 → Volume = 20/0.793 = 25.2 mL.

13Pressure is determined as force per unit area. What is the SI unit of pressure in terms of base units?

P = F/A = (kg·m·s^-2)/m² = kg m^-1 s^-2 (called the pascal, Pa).

14What is the SI unit of mass? How is it defined?

The SI unit of mass is the kilogram (kg). Since 2019, the kilogram is defined by fixing the numerical value of Planck's constant h = 6.62607015 × 10^-34 J·s (equivalently kg·m²·s^-1), together with the defined values of the metre and the second.

15Match the following prefixes with their multiples: micro, deca, mega, giga, femto.

Prefix	Symbol	Multiple
femto	f	10^-15
micro	µ	10^-6
deca	da	10¹
mega	M	10⁶
giga	G	10⁹

16What do you mean by significant figures? How many significant figures are there in (a) 0.0025 (b) 208 (c) 5005 (d) 126,000 (e) 500.0 (f) 2.0034?

Significant figures are meaningful digits in a number representing the precision of a measurement.

(a) 0.0025 → 2. (b) 208 → 3. (c) 5005 → 4. (d) 126,000 → ambiguous (3 to 6); best written in scientific notation. (e) 500.0 → 4. (f) 2.0034 → 5.

17Round each of the following to three significant figures: (a) 34.216 (b) 10.4107 (c) 0.04597 (d) 2808.

(a) 34.216 → 34.2. (b) 10.4107 → 10.4. (c) 0.04597 → 0.0460. (d) 2808 → 2810 (or 2.81 × 10³).

18The following data are obtained when dinitrogen and dioxygen react: (1) 14 g N₂ + 16 g O₂ → 30 g NO; (2) 14 g N₂ + 32 g O₂ → 46 g NO₂; (3) 28 g N₂ + 32 g O₂ → 60 g N₂O; (4) 28 g N₂ + 80 g O₂ → 108 g N₂O₅. Which law of chemical combination is obeyed?

Fix 14 g of N₂. Masses of O₂ combining: 16 g (NO), 32 g (NO₂), 16 g (N₂O → /2 since 28 g), 40 g (N₂O₅ → /2).

Ratio of O₂ masses = 16 : 32 : 16 : 40 = 2 : 4 : 2 : 5 — a simple whole-number ratio. This is the Law of Multiple Proportions.

19If the speed of light is 3.0 × 10⁸ m/s, calculate the distance covered by light in 2.00 ns.

Time = 2.00 × 10^-9 s.

\(d = v \cdot t = 3.0 \times 10^{8} \times 2.00 \times 10^{-9} = 0.60\ \text{m}\)

20In a reaction A + B₂ → AB₂, identify the limiting reagent when (i) 300 atoms of A + 200 molecules of B₂; (ii) 2 mol A + 3 mol B₂; (iii) 100 atoms A + 100 molecules of B₂; (iv) 5 mol A + 2.5 mol B₂; (v) 2.5 mol A + 5 mol B₂.

Stoichiometric ratio A : B₂ = 1 : 1.

(i) 300 A : 200 B₂ → B₂ limiting.

(ii) 2 : 3 → A limiting.

(iii) 100 : 100 → stoichiometric, neither limiting.

(iv) 5 : 2.5 → B₂ limiting.

(v) 2.5 : 5 → A limiting.

21Dinitrogen and dihydrogen react to form ammonia: N₂(g) + 3H₂(g) → 2NH₃(g). (i) Calculate the mass of ammonia produced if 2.00 × 10³ g of N₂ reacts with 1.00 × 10³ g of H₂. (ii) Will any reactant be in excess? How much?

n(N₂) = 2000/28 = 71.43 mol.

n(H₂) = 1000/2 = 500 mol.

Ratio N₂ : H₂ required 1 : 3. Check H₂/3 = 166.67; N₂/1 = 71.43.

Smaller → N₂ is limiting.

NH₃ formed = 2 × 71.43 = 142.86 mol → mass = 142.86 × 17 = 2428.6 g ≈ 2.43 × 10³ g.

H₂ used = 3 × 71.43 = 214.29 mol → 428.57 g.

Excess H₂ = 1000 − 428.57 = 571.43 g.

22How are 0.50 mol of Na₂CO₃ and 0.50 M of Na₂CO₃ different?

0.50 mol Na₂CO₃ = fixed amount = 0.50 × 106 = 53 g of Na₂CO₃ (independent of volume).

0.50 M Na₂CO₃ = concentration of 0.50 mol in every 1 L of solution. The actual amount depends on volume.

Example: 500 mL of 0.50 M contains only 0.25 mol = 26.5 g.

23If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how many volumes of water vapour would be produced?

2H₂(g) + O₂(g) → 2H₂O(g).

Volume ratio 2 : 1 : 2. 10 vol H₂ : 5 vol O₂ matches 2 : 1.

Water vapour produced = 10 volumes.

24Convert the following into basic units: (a) 28.7 pm (b) 15.15 pm (c) 25365 mg.

(a) 28.7 pm = 28.7 × 10^-12 m = 2.87 × 10^-11 m.

(b) 15.15 pm = 1.515 × 10^-11 m.

25A welding fuel gas contains carbon and hydrogen only. Burning a small sample completely gave 3.38 g CO₂, 0.690 g H₂O and no other products. A volume of 10.0 L (measured at STP) of this gas was found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass, (iii) molecular formula.

Find masses of C and H in the sample.

Mass of C = (12/44) × 3.38 = 0.922 g.

Mass of H = (2/18) × 0.690 = 0.0767 g.

Moles: n(C) = 0.922/12 = 0.0768; n(H) = 0.0767/1 = 0.0767.

Ratio C : H = 1 : 1. Empirical formula = CH (emp. mass = 13).

Molar mass: at STP, 22.4 L of gas weighs M grams. So 10.0 L weighs 11.6 g → 22.4 L weighs 11.6 × (22.4/10.0) = 25.98 ≈ 26 g/mol.

n = 26/13 = 2 → Molecular formula = C₂H₂ (acetylene/ethyne).

Tip: When a problem combines combustion data with gas density, first use the combustion data for the empirical formula, then use the STP volume/mass data for molar mass, and finally use n = M/M_emp to get the molecular formula.

Frequently Asked Questions — NCERT Exercises and Solutions: Some Basic Concepts of Chemistry

How are NCERT Class 11 Chemistry Chapter 1 exercises structured?

NCERT Class 11 Chemistry Chapter 1 exercises are organised into numerical problems testing the mole concept, percentage composition, empirical formula determination, stoichiometry and concentration calculations. Each problem follows a logical progression — from simple unit conversions to complex limiting-reagent and solution-stoichiometry problems. The MyAiSchool exercise set provides step-by-step solutions for all NCERT in-text and end-of-chapter questions with worked examples to help students master CBSE board exam patterns.

What are the most important formulas for Chapter 1 problems?

Key formulas for Class 11 Chemistry Chapter 1 exercises include: moles = mass ÷ molar mass, number of particles = moles × 6.022 × 10²³, percentage = (part ÷ whole) × 100, molarity = moles ÷ volume (L), molality = moles ÷ mass of solvent (kg), and mole fraction = moles of component ÷ total moles. For stoichiometry, the mole ratio from the balanced equation drives every calculation. Memorising these formulas and understanding their derivations is essential for solving all Chapter 1 NCERT problems and CBSE board questions.

How do you solve limiting reagent NCERT problems step by step?

To solve limiting reagent problems in NCERT Class 11 Chemistry: (1) write and balance the equation, (2) convert all reactant masses to moles, (3) divide moles by coefficient for each reactant, (4) the smallest ratio is the limiting reagent, (5) use the limiting reagent moles to calculate product moles, (6) convert product moles back to mass. Always show clear unit cancellations. Example problems in this exercise set include hydrogen-oxygen, ammonia synthesis and combustion reactions, all solved using this systematic method.

What is the standard approach for empirical formula problems?

To determine empirical formula in NCERT Class 11 Chemistry exercises: (1) write down the percentage composition, (2) assume 100 g of compound so percentages become grams, (3) divide each mass by the atomic mass to get moles, (4) divide all mole values by the smallest, (5) multiply by a small integer if needed to get whole numbers. The resulting subscripts give the empirical formula. To find molecular formula, multiply by n = molar mass ÷ empirical formula mass. This approach handles all hydrocarbon, oxide and hydrate problems in the chapter.

How do you convert between different concentration units?

To convert between concentration units in NCERT Class 11 Chemistry: from molarity to molality use the density of the solution; from mass percentage to molarity multiply by density and divide by molar mass; from molarity to mole fraction use the moles of solvent calculated from volume and density. The exercise set provides worked examples of each conversion. Always check that mass and volume units match and that you account for the difference between solute mass and total solution mass. These conversions appear frequently in CBSE board exam questions.

What types of questions appear in Chapter 1 board exams?

CBSE Class 11 Chemistry Chapter 1 board exam questions typically include: 1-mark MCQs on definitions and units, 2-mark short-answer questions on laws of chemical combination, 3-mark problems on mole calculations and percentage composition, and 5-mark numerical problems combining stoichiometry, limiting reagent and concentration. Practical applications such as titration calculations and gas-volume problems also appear. The MyAiSchool exercise set covers every question type with model answers, marking schemes and common error warnings to maximise board exam scores.

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