NCERT Solutions for Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry (NCERT 2026–27)

These Class 11 Chemistry Chapter 1 solutions cover Some Basic Concepts of Chemistry — the foundation chapter of the NCERT Chemistry Part I textbook (2026–27). You get every NCERT Exercise (1.1–1.36) reproduced verbatim and solved step by step with correct units, plus the key concepts of the mole concept, laws of chemical combination, significant figures and stoichiometry that you must master before the rest of the syllabus.

Class: 11 Subject: Chemistry Unit: 1 Chapter: Some Basic Concepts of Chemistry Exercises: 1.1 – 1.36 Session: 2026–27
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Class 11 Chemistry Chapter 1 – Overview

Chapter 1, Some Basic Concepts of Chemistry, builds the quantitative language of the subject. It begins with the nature of matter (its three states; classification into elements, compounds and mixtures), then moves to measurement — SI units, scientific notation, significant figures, precision vs. accuracy, and dimensional analysis. The heart of the chapter is the five laws of chemical combination (conservation of mass, definite proportions, multiple proportions, Gay Lussac’s law, Avogadro’s law), Dalton’s atomic theory, atomic and molecular masses, and the all-important mole concept (Avogadro number = 6.022 × 1023). Finally it applies these ideas to percentage composition, empirical & molecular formulae, stoichiometry, limiting reagent and ways of expressing concentration (mass per cent, mole fraction, molarity, molality).

Key Concepts & Definitions

Matter: anything that has mass and occupies space; exists as solid, liquid or gas and is classified as a pure substance (element or compound) or a mixture (homogeneous or heterogeneous).

Mole: the amount of substance that contains exactly 6.02214076 × 1023 elementary entities (Avogadro number, NA).

Molar mass: the mass of one mole of a substance, in g mol−1; numerically equal to its atomic/molecular/formula mass in u.

Average atomic mass: the weighted mean of isotopic masses using their natural fractional abundances.

Empirical formula: the simplest whole-number ratio of atoms in a compound; molecular formula = (empirical formula)n, where n = molar mass ÷ empirical formula mass.

Limiting reagent: the reactant that is completely consumed first and therefore decides the maximum amount of product.

Significant figures: the meaningful digits in a measurement — all certain digits plus one uncertain digit.

Important Formulas (Chapter 1)

Moles: n = mass (g) ÷ molar mass (g mol−1) = number of particles ÷ NA.

Mass per cent: mass % of an element = (mass of that element in 1 mol ÷ molar mass) × 100.

Molarity (M): moles of solute ÷ volume of solution in litres (mol L−1). Dilution: M1V1 = M2V2.

Molality (m): moles of solute ÷ mass of solvent in kg (mol kg−1).

Mole fraction: xA = nA ÷ (nA + nB).

Average atomic mass: Σ(fractional abundance × isotopic mass).

Density: d = mass ÷ volume; common unit g cm−3. Temperature: K = °C + 273.15.

Atomic masses used below (u): H = 1.008, C = 12.011 (≈ 12), N = 14, O = 16, Na = 23, S = 32, Cl = 35.5, Ca = 40, Mn = 55, Fe = 55.85 (≈ 56), Cu = 63.5, Au = 197, Li = 7.

NCERT Exercises (1.1–1.36) – Solutions

1.1 Calculate the molar mass of the following: (i) H2O (ii) CO2 (iii) CH4

ANSWER (i) H2O = 2(1.008) + 16.00 = 2.016 + 16.00 = 18.02 g mol−1. (ii) CO2 = 12.011 + 2(16.00) = 12.011 + 32.00 = 44.01 g mol−1. (iii) CH4 = 12.011 + 4(1.008) = 12.011 + 4.032 = 16.04 g mol−1.

1.2 Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4).

ANSWER Molar mass of Na2SO4 = 2(23) + 32 + 4(16) = 46 + 32 + 64 = 142 g mol−1. Na: (46 ÷ 142) × 100 = 32.39 %. S: (32 ÷ 142) × 100 = 22.54 %. O: (64 ÷ 142) × 100 = 45.07 %. (Check: 32.39 + 22.54 + 45.07 = 100 %.)

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

ANSWER Moles of Fe = 69.9 ÷ 55.85 = 1.251; moles of O = 30.1 ÷ 16.00 = 1.881. Divide by the smaller (1.251): Fe = 1.251/1.251 = 1; O = 1.881/1.251 = 1.504. Multiply by 2 to clear the fraction: Fe = 2, O = 3.01 ≈ 3. Empirical formula = Fe2O3.

1.4 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 (1 mol C + 1 mol O2 → 1 mol CO2 = 44 g). (i) Air has plenty of O2, so all 1 mol C reacts → 1 mol CO2 = 44 g. (ii) 16 g O2 = 16/32 = 0.5 mol O2. Here O2 is the limiting reagent (needs 1 mol for 1 mol C). So 0.5 mol O2 gives 0.5 mol CO2 = 22 g. (iii) 2 mol C with 0.5 mol O2: O2 is again limiting → 0.5 mol CO2 = 22 g (1.5 mol C remains unreacted).

1.5 Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol−1.

ANSWER Moles required = M × V(L) = 0.375 mol L−1 × 0.500 L = 0.1875 mol. Mass = moles × molar mass = 0.1875 × 82.0245 = 15.38 g.

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

ANSWER Mass of 1 L solution = 1000 mL × 1.41 g mL−1 = 1410 g. Mass of HNO3 in it = 69 % of 1410 = 0.69 × 1410 = 972.9 g. Molar mass of HNO3 = 1 + 14 + 48 = 63 g mol−1. Moles = 972.9 ÷ 63 = 15.44 mol. Concentration = 15.44 mol ÷ 1 L = 15.44 mol L−1.

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

ANSWER Molar mass of CuSO4 = 63.5 + 32 + 64 = 159.5 g mol−1. 159.5 g CuSO4 contains 63.5 g Cu. Cu from 100 g = (63.5 ÷ 159.5) × 100 = 39.81 g.

1.8 Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively. (Given molar mass = 159.69 g mol−1.)

ANSWER From Q1.3, the empirical formula is Fe2O3; empirical formula mass = 2(55.85) + 3(16) = 111.7 + 48 = 159.7 g mol−1. n = molar mass ÷ empirical formula mass = 159.69 ÷ 159.7 ≈ 1. Molecular formula = (Fe2O3)1 = Fe2O3.

1.9 Calculate the atomic mass (average) of chlorine using the following data: 35Cl — % natural abundance 75.77, molar mass 34.9689 37Cl — % natural abundance 24.23, molar mass 36.9659

ANSWER Average = Σ(fractional abundance × isotopic mass). = (0.7577 × 34.9689) + (0.2423 × 36.9659) = 26.4959 + 8.9568 = 35.45 u.

1.10 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 (i) 1 mol C2H6 has 2 mol C → 3 × 2 = 6 mol C atoms. (ii) 1 mol C2H6 has 6 mol H → 3 × 6 = 18 mol H atoms. (iii) Molecules = 3 × 6.022 × 1023 = 1.807 × 1024 molecules.

1.11 What 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?

ANSWER Molar mass of C12H22O11 = 12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342 g mol−1. Moles = 20 ÷ 342 = 0.0585 mol. Molarity = 0.0585 mol ÷ 2 L = 0.0293 mol L−1.

1.12 If 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?

ANSWER Moles of methanol needed = 0.25 mol L−1 × 2.5 L = 0.625 mol. Molar mass of CH3OH = 12 + 4(1) + 16 = 32 g mol−1. Mass = 0.625 × 32 = 20 g = 0.020 kg. Volume = mass ÷ density = 0.020 kg ÷ 0.793 kg L−1 = 0.0252 L = 25.2 mL.

1.13 Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal, is 1 Pa = 1 N m−2. If mass of air at sea level is 1034 g cm−2, calculate the pressure in pascal.

ANSWER Pressure = force ÷ area = (mass × g) ÷ area, with g = 9.8 m s−2. Convert: 1034 g cm−2 = 1.034 kg ÷ (10−2 m)2 = 1.034 ÷ 10−4 = 1.034 × 104 kg m−2. Pressure = 1.034 × 104 kg m−2 × 9.8 m s−2 = 1.013 × 105 Pa (i.e. about 1 atm).

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

ANSWER The SI unit of mass is the kilogram (kg). It is defined by fixing the numerical value of the Planck constant h to be 6.62607015 × 10−34 when expressed in J s (= kg m2 s−1), the metre and second being defined through the speed of light c and the caesium frequency νCs.

1.15 Match the following prefixes with their multiples: (i) micro (ii) deca (iii) mega (iv) giga (v) femto, against 106, 109, 10−6, 10−15, 10.

ANSWER (i) micro = 10−6; (ii) deca = 10; (iii) mega = 106; (iv) giga = 109; (v) femto = 10−15.

1.16 What do you mean by significant figures?

ANSWER Significant figures are the meaningful digits in a measured or calculated value — all the digits known with certainty plus one last digit that is uncertain (estimated). They express the precision of a measurement; e.g. 11.2 mL has three significant figures, where 11 is certain and 2 is uncertain.

1.17 A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic. The level of contamination was 15 ppm (by mass). (i) Express this in per cent by mass. (ii) Determine the molality of chloroform in the water sample.

ANSWER (i) 15 ppm = 15 parts per 106 parts → per cent = (15 ÷ 106) × 100 = 1.5 × 10−3 %. (ii) 15 ppm means 15 g CHCl3 in 106 g of water. Molar mass of CHCl3 = 12 + 1 + 3(35.5) = 119.5 g mol−1. Moles = 15 ÷ 119.5 = 0.1255 mol; mass of water = 106 g = 1000 kg. Molality = 0.1255 mol ÷ 1000 kg = 1.25 × 10−4 m.

1.18 Express the following in the scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012

ANSWER (i) 4.8 × 10−3; (ii) 2.34 × 105; (iii) 8.008 × 103; (iv) 5.000 × 102; (v) 6.0012 × 100 (= 6.0012).

1.19 How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,000 (v) 500.0 (vi) 2.0034

ANSWER (i) 2 (leading zeros are not significant); (ii) 3; (iii) 4; (iv) 3 (terminal zeros without a decimal not significant); (v) 4; (vi) 5.

1.20 Round up the following upto three significant figures: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808

ANSWER (i) 34.2; (ii) 10.4; (iii) 0.0460; (iv) 2810 (i.e. 2.81 × 103).

1.21 The following data are obtained when dinitrogen and dioxygen react to form different compounds: (i) 14 g N + 16 g O   (ii) 14 g N + 32 g O   (iii) 28 g N + 32 g O   (iv) 28 g N + 80 g O (a) Which law of chemical combination is obeyed by the above experimental data? Give its statement. (b) Fill in the blanks: (i) 1 km = ___ mm = ___ pm (ii) 1 mg = ___ kg = ___ ng (iii) 1 mL = ___ L = ___ dm3

ANSWER (a) Fixing N at 14 g, the masses of O combining are 16, 32 g (and for 28 g N they are 32, 80 g, i.e. 16, 40 g per 14 g N). For a fixed mass of nitrogen, masses of oxygen are in the ratio 16:32 = 1:2 (and 16:40 = 2:5) — simple whole numbers. So the data obey the Law of Multiple Proportions: if two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. (b)(i) 1 km = 106 mm = 1015 pm. (b)(ii) 1 mg = 10−6 kg = 106 ng. (b)(iii) 1 mL = 10−3 L = 10−3 dm3.

1.22 If the speed of light is 3.0 × 108 m s−1, calculate the distance covered by light in 2.00 ns.

ANSWER 2.00 ns = 2.00 × 10−9 s. Distance = speed × time = (3.0 × 108 m s−1) × (2.00 × 10−9 s) = 6.00 × 10−1 m = 0.600 m.

1.23 In a reaction A + B2 → AB2, identify the limiting reagent, if any, in the following reaction mixtures: (i) 300 atoms of A + 200 molecules of B2 (ii) 2 mol A + 3 mol B2 (iii) 100 atoms of A + 100 molecules of B2 (iv) 5 mol A + 2.5 mol B2 (v) 2.5 mol A + 5 mol B2

ANSWER A and B2 react in a 1:1 ratio. (i) 300 A need 300 B2, only 200 available → B2 is limiting. (ii) 2 A need 2 B2; 3 B2 available → A is limiting. (iii) 100 A + 100 B2 — exact 1:1 → stoichiometric mixture, no limiting reagent. (iv) 5 A need 5 B2, only 2.5 available → B2 is limiting. (v) 2.5 A need 2.5 B2; 5 B2 available → A is limiting.

1.24 Dinitrogen and dihydrogen react to produce ammonia: N2(g) + 3H2(g) → 2NH3(g). (i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass?

ANSWER Moles N2 = 2000 ÷ 28 = 71.43 mol; moles H2 = 1000 ÷ 2 = 500 mol. Required ratio N2:H2 = 1:3. For 71.43 mol N2, H2 needed = 214.3 mol; H2 available (500) is more, so N2 is the limiting reagent. (i) NH3 formed = 2 × moles N2 = 2 × 71.43 = 142.86 mol; mass = 142.86 × 17 = 2.43 × 103 g. (ii) Yes, a reactant remains. (iii) Dihydrogen remains. H2 used = 214.3 mol; H2 left = 500 − 214.3 = 285.7 mol; mass = 285.7 × 2 = 5.71 × 102 g (≈ 572 g).

1.25 How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?

ANSWER 0.50 mol Na2CO3 is a fixed amount of substance = 0.50 × 106 = 53 g (molar mass of Na2CO3 = 106 g mol−1). 0.50 M Na2CO3 is a concentration — 0.50 mol of Na2CO3 dissolved in enough water to make 1 litre of solution. The first is an amount; the second is moles per litre of solution.

1.26 If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

ANSWER 2H2(g) + O2(g) → 2H2O(g); by Gay Lussac’s law, volumes react in the ratio 2:1:2. 10 volumes H2 + 5 volumes O2 (exact 2:1) → 10 volumes (ten volumes) of water vapour.

1.27 Convert the following into basic units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg

ANSWER (i) 28.7 pm = 28.7 × 10−12 m = 2.87 × 10−11 m. (ii) 15.15 pm = 15.15 × 10−12 m = 1.515 × 10−11 m. (iii) 25365 mg = 25365 × 10−3 g = 25.365 g = 2.5365 × 10−2 kg.

1.28 Which one of the following will have the largest number of atoms? (i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl2(g)

ANSWER Atoms = (mass ÷ molar mass) × NA; smaller molar mass → more atoms. (i) Au: 1/197 = 0.0051 mol; (ii) Na: 1/23 = 0.0435 mol; (iii) Li: 1/7 = 0.143 mol of atoms; (iv) Cl2: 1/71 = 0.0141 mol → 0.0282 mol of Cl atoms. Largest is 1 g Li (0.143 mol = 8.6 × 1022 atoms).

1.29 Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

ANSWER Take 1 mol of solution: n(ethanol) = 0.040, n(water) = 0.960. Mass of water = 0.960 × 18 = 17.28 g; with density 1 g mL−1, volume of water ≈ volume of solution = 17.28 mL = 0.01728 L. Molarity = moles of ethanol ÷ volume(L) = 0.040 ÷ 0.01728 = 2.31 mol L−1.

1.30 What will be the mass of one 12C atom in g?

ANSWER 1 mol of 12C atoms = 12 g and contains 6.022 × 1023 atoms. Mass of one atom = 12 ÷ (6.022 × 1023) = 1.99 × 10−23 g.

1.31 How many significant figures should be present in the answer of the following calculations? (i) (0.02856 × 298.15 × 0.112) ÷ 0.5785 (ii) 5 × 5.364 (iii) 0.0125 + 0.7864 + 0.0215

ANSWER (i) Multiplication/division: keep the least number of significant figures among the factors. 0.112 has 3 sig figs (least) → answer has 3 significant figures. (ii) 5 here is an exact number (5.364 has 4 sig figs) → answer has 4 significant figures. (iii) Addition: keep the least number of decimal places; all three terms have 4 decimal places → the sum (0.8204) has 4 decimal places, i.e. 4 significant figures.

1.32 Use the data to calculate the molar mass of naturally occurring argon isotopes: 36Ar — 35.96755 g mol−1, 0.337%   38Ar — 37.96272 g mol−1, 0.063%   40Ar — 39.9624 g mol−1, 99.600%

ANSWER Molar mass = Σ(fractional abundance × isotopic mass). = (0.00337 × 35.96755) + (0.00063 × 37.96272) + (0.99600 × 39.9624) = 0.1212 + 0.0239 + 39.8026 = 39.948 g mol−1.

1.33 Calculate the number of atoms in each of the following: (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.

ANSWER (i) 52 mol Ar = 52 × 6.022 × 1023 = 3.131 × 1025 atoms. (ii) 1 He atom has mass 4 u, so 52 u ÷ 4 u = 13 atoms. (iii) 52 g He = 52/4 = 13 mol; atoms = 13 × 6.022 × 1023 = 7.829 × 1024 atoms.

1.34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (at STP) of this gas weighs 11.6 g. Calculate (i) empirical formula, (ii) molar mass, and (iii) molecular formula.

ANSWER Mass of C = (12/44) × 3.38 = 0.922 g → moles C = 0.922/12 = 0.0768. Mass of H = (2/18) × 0.690 = 0.0767 g → moles H = 0.0767/1 = 0.0767. Ratio C:H = 0.0768 : 0.0767 ≈ 1:1 → empirical formula = CH. (ii) Molar mass: 10.0 L at STP = 10.0/22.4 = 0.4464 mol weighs 11.6 g → molar mass = 11.6/0.4464 = 26.0 g mol−1. (iii) Empirical mass (CH) = 13; n = 26.0/13 = 2 → molecular formula = C2H2 (acetylene/ethyne).

1.35 Calcium carbonate reacts with aqueous HCl: CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l). What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

ANSWER Moles HCl = M × V(L) = 0.75 × 0.025 = 0.01875 mol. From the equation, 2 mol HCl react with 1 mol CaCO3 → moles CaCO3 = 0.01875/2 = 0.009375 mol. Molar mass of CaCO3 = 40 + 12 + 48 = 100 g mol−1. Mass = 0.009375 × 100 = 0.94 g.

1.36 Chlorine is prepared by treating MnO2 with HCl: 4HCl(aq) + MnO2(s) → 2H2O(l) + MnCl2(aq) + Cl2(g). How many grams of HCl react with 5.0 g of manganese dioxide?

ANSWER Molar mass of MnO2 = 55 + 32 = 87 g mol−1; moles = 5.0/87 = 0.0575 mol. From the equation, 1 mol MnO2 needs 4 mol HCl → moles HCl = 4 × 0.0575 = 0.230 mol. Molar mass of HCl = 36.5 g mol−1. Mass = 0.230 × 36.5 = 8.40 g.

Extra Practice Questions

Short Answer Type Questions

Q1. State the law of conservation of mass.

ANSWERIn any physical or chemical change, matter is neither created nor destroyed; the total mass of the reactants equals the total mass of the products. It was put forward by Antoine Lavoisier (1789).

Q2. Differentiate between accuracy and precision.

ANSWERAccuracy is how close a measured value is to the true value, while precision is how close repeated measurements are to one another. Values can be precise but not accurate (consistent yet wrong) or accurate but not precise.

Q3. Calculate the number of moles in 11 g of CO2.

ANSWERMolar mass of CO2 = 44 g mol−1; moles = 11/44 = 0.25 mol.

Q4. Why is molarity temperature-dependent while molality is not?

ANSWERMolarity depends on the volume of solution, which expands or contracts with temperature; molality depends only on the mass of solvent, which does not change with temperature.

Q5. Define empirical and molecular formula with one example.

ANSWEREmpirical formula gives the simplest whole-number ratio of atoms (e.g. CH for benzene), while the molecular formula gives the actual number of atoms in a molecule (C6H6 for benzene). Molecular formula = (empirical formula)n.

Long Answer Type Questions

Q1. State and illustrate the law of multiple proportions.

ANSWERThe law of multiple proportions, proposed by Dalton (1803), states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. For example, carbon and oxygen form CO (12 g C : 16 g O) and CO2 (12 g C : 32 g O). For the same 12 g of carbon, the masses of oxygen are 16 g and 32 g, which are in the simple whole-number ratio 1:2. Similarly, hydrogen and oxygen form water (2 g H : 16 g O) and hydrogen peroxide (2 g H : 32 g O), giving an oxygen ratio of 1:2. Such simple ratios support the idea that compounds are built from whole numbers of atoms.

Q2. Explain the mole concept and the significance of Avogadro’s number with an example.

ANSWERA mole is the SI unit of amount of substance; one mole contains exactly 6.02214076 × 1023 elementary entities (atoms, molecules, ions, etc.), called Avogadro’s number, NA. Because individual atoms are far too small to count or weigh, the mole links the microscopic scale (number of particles) with the macroscopic scale (grams that we can weigh): the molar mass of a substance in grams equals the mass of NA of its entities. For example, 18 g of water is 1 mole and contains 6.022 × 1023 water molecules; 12 g of carbon-12 is 1 mole of carbon atoms. Using moles, chemists convert easily between mass, number of particles and (for gases) volume, which makes stoichiometric calculations possible.

Q3. Outline the steps to determine the molecular formula of a compound from its percentage composition and molar mass.

ANSWERFirst, assume 100 g of the compound so that each element’s percentage becomes its mass in grams. Divide each element’s mass by its atomic mass to get the number of moles of each element. Divide all the mole values by the smallest among them to obtain the simplest mole ratio; if the ratio is not whole, multiply through by a small integer to clear fractions. The resulting whole numbers give the empirical formula, and the sum of the atomic masses gives the empirical formula mass. Next, divide the given molar mass by the empirical formula mass to find n (a whole number). Finally, multiply the subscripts of the empirical formula by n to get the molecular formula = (empirical formula)n. For instance, an oxide of iron that is 69.9% Fe and 30.1% O gives the empirical formula Fe2O3; since its molar mass (159.7 g mol−1) equals the empirical formula mass, n = 1 and the molecular formula is also Fe2O3.

MCQs & Assertion–Reason

1. The number of significant figures in 0.00500 m is:

(a) 1    (b) 2    (c) 3    (d) 5

2. The molar mass of CaCO3 is:

(a) 84 g mol−1    (b) 100 g mol−1    (c) 116 g mol−1    (d) 40 g mol−1

3. The number of atoms in 0.5 mol of oxygen gas (O2) is:

(a) 6.022 × 1023    (b) 3.011 × 1023    (c) 1.204 × 1024    (d) 6.022 × 1022

4. The law that states a compound always contains the same elements in the same fixed proportion by mass is the law of:

(a) conservation of mass    (b) multiple proportions    (c) definite proportions    (d) gaseous volumes

5. The SI base unit for amount of substance is the:

(a) kilogram    (b) mole    (c) candela    (d) kelvin

6. 1 ppm by mass is equal to:

(a) 1 g in 100 g    (b) 1 g in 103 g    (c) 1 g in 106 g    (d) 1 g in 109 g

7. Which sample contains the largest number of atoms?

(a) 1 g Au    (b) 1 g Na    (c) 1 g Li    (d) 1 g Cl2

8. In the reaction N2 + 3H2 → 2NH3, if 1 mol N2 reacts with 1 mol H2, the limiting reagent is:

(a) N2    (b) H2    (c) NH3    (d) none

9. The number 232.508 in scientific notation is:

(a) 2.32508 × 102    (b) 23.2508 × 101    (c) 2.32508 × 103    (d) 0.232508 × 103

10. The mass per cent of nitrogen in NH3 (molar mass 17) is about:

(a) 14%    (b) 47%    (c) 82%    (d) 18%

Answer key: 1-(c), 2-(b), 3-(a), 4-(c), 5-(b), 6-(c), 7-(c), 8-(b), 9-(a), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: One mole of any substance contains 6.022 × 1023 entities.

Reason: This number is the fixed value of the Avogadro constant.

A-R 2. Assertion: Molality is preferred over molarity for experiments done at different temperatures.

Reason: Molality depends on the mass of solvent, which is independent of temperature.

A-R 3. Assertion: The terminal zeros in the number 100 (without a decimal point) are significant.

Reason: All zeros written to the right of a number are always significant.

A-R 4. Assertion: The empirical and molecular formula of a compound are always the same.

Reason: The molecular formula is a whole-number multiple of the empirical formula.

A-R 5. Assertion: Mass of a substance is constant but its weight can vary from place to place.

Reason: Weight is the gravitational force on an object and gravity varies with location.

Answer key: 1-(A), 2-(A), 3-(D), 4-(D), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Confusing molarity (per litre of solution) with molality (per kg of solvent) — they use different denominators.
  • Forgetting to identify the limiting reagent before calculating product mass.
  • Counting leading zeros as significant (they are not) or dropping terminal zeros after a decimal (they are significant).
  • Using atomic mass instead of molar mass of the molecule in mole calculations.
  • Mixing up mass per cent and mole fraction; mole fraction needs moles, not masses.
  • Not converting units to a single system (g vs. kg, mL vs. L) before substituting in a formula.

How to score full marks in this chapter

Always write the balanced equation first, then identify the limiting reagent before any stoichiometry. Show each step with units and cancel them to confirm the final unit is correct. Report the answer to the right number of significant figures (least sig figs in multiplication/division; least decimal places in addition/subtraction). Memorise NA = 6.022 × 1023, molar volume at STP = 22.4 L, and the four concentration formulas — these recur throughout Class 11 and 12 Chemistry.

Frequently Asked Questions

What is Class 11 Chemistry Chapter 1 about?

Chapter 1, Some Basic Concepts of Chemistry, covers the nature and classification of matter, SI units and measurement, scientific notation and significant figures, the five laws of chemical combination, Dalton’s atomic theory, atomic and molecular masses, the mole concept, empirical and molecular formulae, stoichiometry, limiting reagent, and ways of expressing concentration.

How many exercise questions are there in Chapter 1?

The NCERT textbook has 36 numbered exercise questions (1.1 to 1.36). All of them — including every numerical — are solved step by step on this page with units and verified answers.

What is the value of Avogadro’s number?

Avogadro’s number (NA) is 6.02214076 × 1023, usually rounded to 6.022 × 1023. It is the number of elementary entities (atoms, molecules, ions) in one mole of any substance.

Are these Class 11 Chemistry Chapter 1 solutions free?

Yes. All solutions are free and follow the official NCERT Chemistry textbook for session 2026–27, with answers cross-checked against the NCERT answer key.

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