Chapter 1 — Some Basic Concepts of Chemistry
Welcome to HSLC Guru. This English-medium study companion is prepared for ASSEB Class 11 Chemistry students working through Chapter 1 — Some Basic Concepts of Chemistry. The chapter sets the foundation of the subject by introducing matter, measurement, the language of chemistry, and the quantitative reasoning that runs through every later topic. Read the summary, study the textbook-style question and answer set, work the numericals, and then test yourself with the MCQs, fill-in-the-blanks, and true/false items at the end. A glossary and formula table are included for quick revision before the examination.
Summary
Chemistry is the branch of science that deals with the composition, structure, properties, and transformations of matter. Its importance is felt in every aspect of modern life: medicines, fertilisers, plastics, fuels, dyes, soaps, food preservatives, and even the materials of microelectronics all owe their existence to chemical research. Matter is anything that has mass and occupies space. It is classified physically into solids, liquids, and gases, and chemically into pure substances (elements and compounds) and mixtures (homogeneous and heterogeneous). Properties of matter are either physical (observable without changing chemical identity, such as colour, density, melting point) or chemical (observed during a chemical change, such as combustibility or reactivity with acid).
To describe matter quantitatively, chemists use the International System of Units (SI). The seven base units are the metre (m) for length, kilogram (kg) for mass, second (s) for time, kelvin (K) for temperature, ampere (A) for electric current, mole (mol) for amount of substance, and candela (cd) for luminous intensity. Derived quantities such as volume (m³), density (kg m⁻³), and pressure (pascal) follow from these. Measurements are reported with significant figures so that the precision of an experimental result is honestly expressed. Rules for significant figures include: all non-zero digits are significant; zeros between non-zero digits are significant; leading zeros are not significant; trailing zeros after a decimal point are significant. In addition and subtraction the result is rounded to the least number of decimal places, and in multiplication and division to the least number of significant figures.
The laws of chemical combination organise the early experimental facts of chemistry. The law of conservation of mass (Lavoisier) states that mass is neither created nor destroyed in a chemical reaction. The law of definite proportions (Proust) states that a given compound always contains the same elements in the same fixed ratio by mass. The law of multiple proportions (Dalton) states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in small whole-number ratios. Gay-Lussac’s law of gaseous volumes and Avogadro’s law extended these ideas to gases. Dalton’s atomic theory explained these laws through five postulates: matter consists of indivisible atoms; atoms of a given element are identical in mass and properties; compounds are formed by combination of atoms in simple whole-number ratios; chemical reactions involve only rearrangement of atoms; and atoms are neither created nor destroyed.
Atomic mass is expressed in unified atomic mass units (u), where 1 u equals one-twelfth the mass of a carbon-12 atom. Average atomic mass accounts for natural isotopic abundance. Molecular mass is the sum of the atomic masses of the atoms in a molecule, while formula mass is used for ionic compounds. The mole is the SI unit for amount of substance: one mole contains exactly 6.022 × 10²³ elementary entities — atoms, molecules, ions, or formula units. This number is Avogadro’s constant, N_A. The molar mass of any substance, expressed in g mol⁻¹, is numerically equal to its atomic, molecular, or formula mass in u. Concentration of solutions is expressed in several ways: molarity (M) is moles of solute per litre of solution; molality (m) is moles of solute per kilogram of solvent; mole fraction (x) is the ratio of moles of one component to total moles; and mass per cent is the mass of solute per 100 g of solution. Stoichiometry uses balanced equations to relate masses, moles, volumes, and particle numbers of reactants and products. When reactants are not present in stoichiometric proportion, the limiting reagent is the one that is consumed first and therefore decides the maximum yield, while the other reactant is left in excess.
$M = \frac{\text{moles of solute}}{\text{volume of solution in L}}$
$m = \frac{n}{w_{\text{solvent in kg}}}$
$x_A = \frac{n_A}{n_A + n_B}$
Very Short Answer Questions (1 mark)
Q1. Define matter.
Answer: Matter is anything that has mass and occupies space, for example wood, water, and air.
Q2. Write the SI unit of amount of substance.
Answer: The SI unit of amount of substance is the mole (mol).
Q3. State the value of Avogadro’s number.
Answer: Avogadro’s number is 6.022 × 10²³ entities per mole.
Q4. What is meant by 1 u (unified atomic mass unit)?
Answer: One unified atomic mass unit is one-twelfth of the mass of a single carbon-12 atom and equals 1.66 × 10⁻²⁴ g.
Q5. Give one example of a homogeneous mixture.
Answer: A solution of sugar in water is a homogeneous mixture because its composition is uniform throughout.
Q6. Define molarity.
Answer: Molarity is the number of moles of solute dissolved in one litre of solution; its unit is mol L⁻¹ or M.
Q7. What is the molar mass of H₂O?
Answer: The molar mass of water is 2(1) + 16 = 18 g mol⁻¹.
Q8. How many significant figures are present in 0.00450?
Answer: The number 0.00450 contains three significant figures (4, 5, and the trailing 0).
Q9. Name the law that states mass is conserved in a chemical reaction.
Answer: The law of conservation of mass, proposed by Antoine Lavoisier, states that mass is neither created nor destroyed during a chemical change.
Q10. Define limiting reagent.
Answer: The limiting reagent is the reactant that is completely consumed first in a reaction and thus limits the amount of product formed.
Short Answer Questions (2-3 marks)
Q1. State and explain the law of multiple proportions with an example.
Answer: The law of multiple proportions, given by John Dalton, 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 small whole-number ratios. For example, carbon and oxygen form CO and CO₂. With 12 g of carbon, CO contains 16 g of oxygen and CO₂ contains 32 g of oxygen. The ratio 16 : 32 simplifies to 1 : 2, a simple whole-number ratio that confirms the law.
Q2. Distinguish between accuracy and precision.
Answer: Accuracy describes how close a measured value is to the true or accepted value of a quantity, while precision describes how close repeated measurements of the same quantity are to one another. A set of readings can be precise but not accurate if a systematic error is present, and accurate readings on average can still show poor precision if random error is high. Good experimental work aims at both.
Q3. Define mole and explain its significance.
Answer: A mole is the amount of a substance that contains exactly 6.022 × 10²³ elementary entities, where the entities may be atoms, molecules, ions, electrons, or formula units. The mole connects the microscopic world of atoms with the macroscopic world of grams: the mass of one mole of any substance in grams equals its atomic, molecular, or formula mass in u. This makes stoichiometric calculations possible.
Q4. What is meant by molality? Why is it preferred over molarity for studies involving temperature change?
Answer: Molality (m) is the number of moles of solute dissolved per kilogram of solvent. It is preferred over molarity in temperature-dependent studies because it is based on the mass of the solvent, which does not change with temperature. Molarity, in contrast, depends on the volume of solution, and volume expands or contracts with temperature, making molarity temperature-dependent.
Q5. Calculate the number of molecules in 11 g of CO₂.
Answer: Molar mass of CO₂ = 12 + 2(16) = 44 g mol⁻¹. Moles of CO₂ = 11 / 44 = 0.25 mol. Number of molecules = 0.25 × 6.022 × 10²³ = 1.5055 × 10²³ molecules.
Q6. Define mole fraction. What is its sum for a binary mixture?
Answer: The mole fraction of a component is the ratio of the number of moles of that component to the total number of moles of all components in the mixture. For a binary mixture of A and B, x_A + x_B = 1; the sum of all mole fractions in any mixture is always unity, and mole fraction is dimensionless.
Long Answer Questions (5-7 marks)
Q1. State the postulates of Dalton’s atomic theory and discuss its limitations.
Answer: Dalton’s atomic theory (1808) is based on the following postulates: (i) matter is made up of small indivisible particles called atoms; (ii) atoms of a given element are identical in mass and properties, while atoms of different elements differ in mass and properties; (iii) compounds are formed when atoms of different elements combine in simple whole-number ratios; (iv) a chemical reaction involves only the rearrangement of atoms — atoms themselves are neither created nor destroyed; (v) atoms cannot be subdivided, created, or destroyed in any chemical reaction. The theory successfully explained the laws of conservation of mass, definite proportions, and multiple proportions, and gave chemistry its first quantitative atomic foundation. However, it has limitations. Atoms are not indivisible — they are composed of subatomic particles such as protons, neutrons, and electrons. Atoms of the same element can differ in mass because of isotopes (for example ¹H, ²H, ³H), so the postulate of identical mass is incorrect. Some compounds combine in non-integer ratios in non-stoichiometric solids, and atoms of different elements can have the same mass (isobars), such as ⁴⁰Ar and ⁴⁰Ca. In nuclear reactions atoms are converted from one element to another, contrary to Dalton’s view.
Q2. Calculate the molarity of a solution prepared by dissolving 4.0 g of NaOH in enough water to make 250 mL of solution.
Answer: Molar mass of NaOH = 23 + 16 + 1 = 40 g mol⁻¹. Moles of NaOH = 4.0 / 40 = 0.10 mol. Volume of solution = 250 mL = 0.250 L. Molarity, M = moles of solute / volume of solution in L = 0.10 / 0.250 = 0.40 mol L⁻¹. Therefore the prepared solution is 0.40 M NaOH. The same calculation also gives normality for NaOH because it has one replaceable OH⁻ ion per formula unit, so its n-factor is 1, and N = M = 0.40 N. Always remember that molarity changes with temperature, since volume of the solution expands on heating.
Q3. 6.3 g of oxalic acid dihydrate (H₂C₂O₄·2H₂O) is dissolved in water to make 250 mL of solution. Calculate the molarity of the solution.
Answer: Molar mass of H₂C₂O₄·2H₂O = 2(1) + 2(12) + 4(16) + 2(18) = 2 + 24 + 64 + 36 = 126 g mol⁻¹. Moles of oxalic acid dihydrate = 6.3 / 126 = 0.050 mol. Volume = 250 mL = 0.250 L. Molarity = 0.050 / 0.250 = 0.20 mol L⁻¹. Hence the solution is 0.20 M with respect to oxalic acid dihydrate. If a teacher asks for normality of this diprotic acid, n-factor = 2, so N = 2M = 0.40 N. The result also illustrates why you must always include the mass of water of crystallisation in the molar mass.
Q4. In the reaction N₂(g) + 3H₂(g) → 2NH₃(g), if 28 g of N₂ is mixed with 6 g of H₂, identify the limiting reagent and calculate the mass of NH₃ produced.
Answer: Moles of N₂ = 28 / 28 = 1.0 mol. Moles of H₂ = 6 / 2 = 3.0 mol. The balanced equation requires 1 mol N₂ to react with 3 mol H₂. The available ratio is exactly 1 : 3, so neither reactant is in excess in this case — they are present in stoichiometric proportion, and both are fully consumed. From 1 mol N₂ we obtain 2 mol NH₃. Mass of NH₃ produced = 2 × 17 = 34 g. If instead 28 g N₂ were mixed with only 3 g H₂, then moles of H₂ = 1.5, the actual ratio would be 1 : 1.5, far less than the required 1 : 3, so H₂ would be the limiting reagent. From 1.5 mol H₂, moles of NH₃ = (2/3) × 1.5 = 1.0 mol, and mass = 17 g. This contrast highlights how the limiting reagent depends on the supplied amounts, not on the molecular masses alone.
Q5. Calculate the mass per cent and mole fraction of solute in a solution prepared by dissolving 18 g of glucose (C₆H₁₂O₆) in 178.2 g of water.
Answer: Total mass of solution = 18 + 178.2 = 196.2 g. Mass per cent of glucose = (18 / 196.2) × 100 = 9.17 %. Molar mass of glucose = 6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 g mol⁻¹. Moles of glucose = 18 / 180 = 0.10 mol. Moles of water = 178.2 / 18 = 9.90 mol. Mole fraction of glucose, x_glucose = 0.10 / (0.10 + 9.90) = 0.10 / 10.00 = 0.010. Mole fraction of water = 1 − 0.010 = 0.990. Hence the solute is present at about 9.17 % by mass and only 0.010 in mole fraction, illustrating that mass per cent and mole fraction give very different impressions of a dilute solution. Molality = 0.10 / 0.1782 ≈ 0.561 mol kg⁻¹, completing the picture of three different concentration measures for the same solution.
Additional MCQs
Q1. The SI unit of mass is:
(a) gram (b) kilogram (c) pound (d) milligram
Answer: (b) kilogram.
Q2. The number of significant figures in 6.022 × 10²³ is:
(a) 2 (b) 3 (c) 4 (d) 23
Answer: (c) 4.
Q3. The number of moles in 9.0 g of H₂O is:
(a) 0.25 (b) 0.50 (c) 1.0 (d) 2.0
Answer: (b) 0.50.
Q4. Which of the following is a derived SI unit?
(a) metre (b) second (c) newton (d) ampere
Answer: (c) newton.
Q5. Avogadro’s law applies to:
(a) only solids (b) only liquids (c) ideal gases at the same T and P (d) all states of matter
Answer: (c) ideal gases at the same T and P.
Q6. The mass of one atom of carbon-12 is:
(a) 12 g (b) 12 u (c) 1 u (d) 6.022 × 10²³ g
Answer: (b) 12 u.
Q7. Which one of the following is a homogeneous mixture?
(a) Sand and iron filings (b) Oil and water (c) Air (d) Granite
Answer: (c) Air.
Q8. The molality of a solution containing 1 mol solute in 500 g of solvent is:
(a) 1 m (b) 2 m (c) 0.5 m (d) 0.2 m
Answer: (b) 2 m.
Q9. 1 mole of any gas at STP occupies a volume of:
(a) 11.2 L (b) 22.4 L (c) 24.0 L (d) 6.022 L
Answer: (b) 22.4 L.
Q10. The empirical formula of a compound with molecular formula C₆H₁₂O₆ is:
(a) CHO (b) CH₂O (c) C₂H₄O₂ (d) C₃H₆O₃
Answer: (b) CH₂O.
Fill in the Blanks
Q1. One mole of any substance contains __________ particles.
Answer: 6.022 × 10²³.
Q2. The SI base unit of temperature is __________.
Answer: kelvin (K).
Q3. The law of __________ states that a chemical compound always contains the same elements in the same fixed ratio by mass.
Answer: definite proportions.
Q4. Molarity is expressed as moles of solute per __________ of solution.
Answer: litre.
Q5. The reactant that is consumed completely in a reaction is called the __________.
Answer: limiting reagent.
True or False
Q1. The mole fraction of a component is independent of temperature.
Answer: True.
Q2. Mass is conserved in every chemical reaction.
Answer: True.
Q3. Molarity does not change with temperature.
Answer: False — molarity depends on volume, which changes with temperature.
Q4. Atoms of all isotopes of an element have the same mass.
Answer: False — isotopes differ in the number of neutrons and hence in mass.
Q5. The empirical formula always shows the actual number of atoms in a molecule.
Answer: False — that is the molecular formula; the empirical formula gives only the simplest whole-number ratio.
Glossary
| Term | Meaning |
|---|---|
| Matter | Anything that has mass and occupies space. |
| Element | A pure substance that cannot be broken down into simpler substances by chemical means. |
| Compound | A pure substance formed when two or more elements combine in a fixed mass ratio. |
| Mixture | A combination of two or more substances in any ratio, retaining their own properties. |
| SI Unit | An internationally accepted standard unit from the Système International d’Unités. |
| Significant Figure | A digit in a measurement that contributes to its precision. |
| Atomic Mass Unit (u) | 1/12 of the mass of one carbon-12 atom; equals 1.66 × 10⁻²⁴ g. |
| Mole | The amount of substance containing 6.022 × 10²³ particles. |
| Avogadro’s Number | The number of entities in one mole, 6.022 × 10²³ mol⁻¹. |
| Molarity | Moles of solute per litre of solution (mol L⁻¹). |
| Molality | Moles of solute per kilogram of solvent (mol kg⁻¹). |
| Mole Fraction | Ratio of moles of one component to the total moles in a mixture. |
| Mass Per Cent | Mass of solute per 100 g of solution. |
| Empirical Formula | The simplest whole-number ratio of atoms of each element in a compound. |
| Molecular Formula | The actual number of atoms of each element in one molecule of a compound. |
| Limiting Reagent | The reactant consumed first that limits the amount of product formed. |
Formula Table
| Quantity | Formula | Unit |
|---|---|---|
| Number of moles (n) | n = mass / molar mass | mol |
| Number of particles | N = n × N_A | — |
| Molarity (M) | M = moles of solute / volume of solution in L | mol L⁻¹ |
| Molality (m) | m = moles of solute / mass of solvent in kg | mol kg⁻¹ |
| Mole fraction (x_A) | x_A = n_A / (n_A + n_B) | dimensionless |
| Mass per cent | (mass of solute / mass of solution) × 100 | % |
| Density (d) | d = mass / volume | g cm⁻³ or kg m⁻³ |
| Volume at STP | V = n × 22.4 L mol⁻¹ | L |
| Dilution equation | M₁V₁ = M₂V₂ | — |
| °C to K | T(K) = t(°C) + 273.15 | K |
Quick Revision Notes
1. Matter is anything having mass and occupying space; classified as solid, liquid, or gas, and as element, compound, or mixture.
2. The seven SI base units cover length, mass, time, temperature, current, amount of substance, and luminous intensity.
3. Significant figures express the precision of measurement; rules for addition and multiplication differ.
4. The four laws of chemical combination — conservation of mass, definite proportions, multiple proportions, and Gay-Lussac’s law — are the experimental basis for atomic theory.
5. The mole, defined through Avogadro’s number 6.022 × 10²³, links mass to particle count and is the central quantity in stoichiometry.
6. Molarity, molality, mole fraction, and mass per cent are the standard ways of expressing the concentration of a solution; each has different temperature behaviour and use cases.
7. Limiting reagent calculations rely on comparing moles supplied with the stoichiometric ratio in the balanced equation.
This completes the HSLC Guru study set for Class 11 Chemistry Chapter 1 — Some Basic Concepts of Chemistry, prepared in line with the ASSEB syllabus. Practise the numericals on moles, molarity, molality, mole fraction, and limiting reagent until the steps become automatic, because the same skills are needed in every later chapter on stoichiometry, solutions, equilibrium, and electrochemistry. Best wishes for your examinations.