Some Basic
Concepts
of Chemistry
All 36 textbook exercise solutions with theory, step-by-step working, and conceptual clarity. Designed for CBSE board exams and JEE / NEET preparation.
Chapter Overview
Some Basic Concepts of Chemistry is the foundational chapter of Class XI Chemistry, introducing the quantitative language chemists use to describe matter.
Every subsequent topic in physical, organic, and inorganic chemistry builds on these foundations. CBSE boards, JEE, and NEET consistently test this chapter at all difficulty levels.
🔬 Mole Concept
1 mol = 6.022×10²³ entities. Bridges atomic and macroscopic worlds.
⚖️ Stoichiometry
Quantitative relationships in balanced equations; limiting reagent; percent yield.
💧 Concentration Units
Molarity, Molality, Mole fraction, ppm — each for specific conditions.
Essential Formulae & Concepts at a Glance
Molar Mass
Sum of atomic masses (g mol⁻¹); n = m/M
Mass Percent
(mass of element / M_r) × 100
Empirical Formula
mass% → moles → simplest whole-number ratio
Molarity (M)
mol solute / L solution
Molality (m)
mol solute / kg solvent
Limiting Reagent
Reactant consumed first; sets max yield
Avogadro's No.
Nₐ = 6.022 × 10²³ mol⁻¹
Mole Fraction
χ = n_comp / n_total ; Σχ = 1
ppm
mg/kg of solution; 1 ppm = 10⁻⁴ %
Calculate the molar mass of: (i) \(\mathrm{H_2O}\) (ii) \(\mathrm{CO_2}\) (iii) \(\mathrm{CH_4}\)
Molar Mass = sum of atomic masses of all atoms in one formula unit (g mol⁻¹). Key formula: \(n = m/M\). Atomic masses used: H = 1, C = 12, O = 16 g mol⁻¹.
\[\text{Molar mass} = 2(1)+16 = \mathbf{18\,g\,mol^{-1}}\]
\[\text{Molar mass} = 12+2(16) = 12+32 = \mathbf{44\,g\,mol^{-1}}\]
\[\text{Molar mass} = 12+4(1) = \mathbf{16\,g\,mol^{-1}}\]
Calculate mass percent of each element in sodium sulphate \((\mathrm{Na_2SO_4})\).
Mass Percent \(= \dfrac{\text{mass of element in 1 mol}}{\text{molar mass of compound}}\times100\). All percents must sum to 100%. Na=23, S=32, O=16 g mol⁻¹.
\[M_r(\mathrm{Na_2SO_4}) = 2(23)+32+4(16) = 46+32+64 = 142\,\text{g mol}^{-1}\]
\[\%\,\mathrm{Na} = \frac{46}{142}\times100 = \mathbf{32.39\%}\]
\[\%\,\mathrm{S} = \frac{32}{142}\times100 = \mathbf{22.54\%}\]
\[\%\,\mathrm{O} = \frac{64}{142}\times100 = \mathbf{45.07\%}\]
Empirical formula of iron oxide: Fe = 69.9%, O = 30.1% by mass.
Steps: Assume 100 g → mass = %. Divide by atomic mass → moles. Divide by smallest → ratio. If ratio ≈ 1.5, multiply by 2. Fe=56, O=16 g mol⁻¹.
\[\text{mol Fe} = 69.9/56 = 1.248;\quad \text{mol O} = 30.1/16 = 1.881\]
Dividing by 1.248: Fe = 1.00, O = 1.51 ≈ 1.5 → multiply both by 2: Fe = 2, O = 3.
CO₂ produced when: (i) 1 mol C in air (ii) 1 mol C + 16 g O₂ (iii) 2 mol C + 16 g O₂
Reaction: \(\mathrm{C + O_2 \to CO_2}\). Mole ratio 1:1:1. The limiting reagent is consumed first and sets maximum product. M_r(O₂)=32, M_r(CO₂)=44 g mol⁻¹.
O₂ excess in air → C is limiting. 1 mol C → 1 mol CO₂ → \(1\times44 = \mathbf{44\,g}\)
\[\text{mol O}_2 = 16/32 = 0.5\,\text{mol (limiting)}\]
\[0.5\times44 = \mathbf{22\,g\,CO_2}\]
Still only 0.5 mol O₂ (limiting). \(\mathbf{22\,g\,CO_2}\) produced.
Mass of \(\mathrm{CH_3COONa}\) \((M_r=82.0245)\) for 500 mL of 0.375 M solution.
\(M = n/V(\text{L})\). Rearranged: \(n = M\times V\). Then mass \(= n\times M_r\). Convert mL → L first.
\[n = 0.375\times0.500 = 0.1875\,\text{mol}\]
\[\text{Mass} = 0.1875\times82.0245 = \mathbf{15.38\,g}\]
Concentration of HNO₃ (density = 1.41 g mL⁻¹, mass% = 69%) in mol L⁻¹.
Take 1 L solution. Mass = density × 1000. Solute mass = (w%/100) × total mass. Moles = mass/M_r. Formula: \(M = 10\,d\,w\%/M_r\). M_r(HNO₃) = 63 g mol⁻¹.
\[\text{Mass of 1 L solution} = 1.41\times1000 = 1410\,\text{g}\]
\[\text{Mass of HNO}_3 = \frac{69}{100}\times1410 = 972.9\,\text{g}\]
\[\text{Moles} = 972.9/63 = 15.44\,\text{mol}\]
Mass of copper obtainable from 100 g of \(\mathrm{CuSO_4}\).
Mass of element = (mass of element per mole / M_r of compound) × given mass. Cu=63.5, S=32, O=16 g mol⁻¹.
\[M_r(\mathrm{CuSO_4}) = 63.5+32+64 = 159.5\,\text{g mol}^{-1}\]
\[\text{Mass of Cu} = \frac{63.5}{159.5}\times100 = \mathbf{39.8\,g}\]
Molecular formula of iron oxide with Fe = 69.9%, O = 30.1% by mass.
Molecular formula = n × empirical formula. n = actual M_r / empirical formula mass. Without extra M_r data, molecular formula = empirical formula.
Same empirical formula procedure as Q3 gives \(\mathrm{Fe_2O_3}\). No additional molar mass given.
Average atomic mass of Cl: ³⁵Cl (75.77%, 34.9689 u) and ³⁷Cl (24.23%, 36.9659 u).
\(\bar{M} = \sum_i (\text{fraction}_i \times M_i)\). Convert % → fractions by ÷100. Most elements exist as isotope mixtures; periodic table shows this weighted average.
\[\bar{M} = (0.7577\times34.9689)+(0.2423\times36.9659) = 26.4937+8.9565 = \mathbf{35.45\,u}\]
In 3 mol ethane \((\mathrm{C_2H_6})\): (i) mol C atoms (ii) mol H atoms (iii) number of molecules.
1 mol C₂H₆ contains 2 mol C and 6 mol H. Number of molecules = n × Nₐ where Nₐ = 6.022 × 10²³ mol⁻¹.
\[3\times2 = \mathbf{6\,\text{mol C atoms}}\]
\[3\times6 = \mathbf{18\,\text{mol H atoms}}\]
\[3\times6.022\times10^{23} = \mathbf{1.807\times10^{24}\,\text{molecules}}\]
Concentration of sugar \((\mathrm{C_{12}H_{22}O_{11}})\) when 20 g dissolved to make 2 L.
M_r(C₁₂H₂₂O₁₁) = 12(12)+22(1)+11(16) = 342 g mol⁻¹. Steps: n = m/M_r; M = n/V(L).
\[n = 20/342 = 0.0585\,\text{mol};\quad M = 0.0585/2 = \mathbf{2.93\times10^{-2}\,\text{mol L}^{-1}}\]
Volume of methanol (density 0.793 kg L⁻¹) for 2.5 L of 0.25 M solution.
Volume = mass/density. M_r(CH₃OH) = 32 g mol⁻¹. Density = 793 g L⁻¹. Sequence: moles needed → mass → volume.
\[n = 0.25\times2.5 = 0.625\,\text{mol};\quad \text{mass} = 0.625\times32 = 20\,\text{g}\]
\[V = 20/793 = 0.0252\,\text{L} = \mathbf{25.2\,\text{mL}}\]
Pressure in Pascal if mass of air = \(10^3\,\text{g cm}^{-2}\) at sea level.
P = (mass per area) × g. Convert: 10³ g cm⁻² = 10⁴ kg m⁻². g = 9.8 m s⁻². 1 Pa = 1 N m⁻² = 1 kg m⁻¹ s⁻².
\[10^3\,\text{g cm}^{-2} = 10^4\,\text{kg m}^{-2}\]
\[P = 10^4\times9.8 = \mathbf{9.8\times10^4\,\text{Pa}}\]
What is the SI unit of mass? How is it defined?
In 2019, all SI units were redefined using fixed values of fundamental constants, eliminating physical artefacts. The kg is now linked to the Planck constant h.
The SI unit of mass is the kilogram (kg), defined by fixing the Planck constant to exactly:
\[h = 6.62607015\times10^{-34}\,\text{J\,s}\]
Since J = kg m² s⁻², this links the kg to fundamental constants, ensuring universal reproducibility.
Match prefixes with multiples: micro, deca, mega, giga, femto.
Order (large→small): giga(10⁹)→mega(10⁶)→kilo(10³)→deca(10¹)→deci(10⁻¹)→milli(10⁻³)→micro(10⁻⁶)→nano(10⁻⁹)→pico(10⁻¹²)→femto(10⁻¹⁵).
| Prefix | Symbol | Multiple |
|---|---|---|
| micro | μ | 10⁻⁶ |
| deca | da | 10¹ |
| mega | M | 10⁶ |
| giga | G | 10⁹ |
| femto | f | 10⁻¹⁵ |
What do you mean by significant figures?
(1) Non-zero: always significant. (2) Between non-zeros: significant. (3) Leading zeros: NOT significant. (4) Trailing with decimal: significant. (5) Trailing without decimal: NOT counted.
Significant figures are all meaningful digits in a measured quantity — all certain digits plus the first uncertain digit — conveying the precision of measurement.
Example: 2.35 → 3 sig figs; 0.0042 → 2 sig figs (leading zeros not significant). They govern rounding of calculated results: ×/÷ → fewest sig figs; +/− → fewest decimal places.
CHCl₃ contamination at 15 ppm. (i) Express as mass% (ii) Find molality.
1 ppm = 10⁻⁴%. For dilute aqueous solutions: 15 ppm ≈ 15 mg CHCl₃ per kg water. Molality = mol/kg solvent. M_r(CHCl₃) = 12+1+3(35.5) = 119.5 g mol⁻¹.
\[\text{Mass \%} = \frac{15}{10^6}\times100 = \mathbf{1.5\times10^{-3}\%}\]
\[n = 0.015/119.5 = 1.26\times10^{-4}\,\text{mol per kg water}\]
\[\text{Molality} = \mathbf{1.26\times10^{-4}\,m}\]
Scientific notation: (i) 0.0048 (ii) 234,000 (iii) 8008 (iv) 500.0 (v) 6.0012
Form: \(a\times10^n\) where \(1\le a < 10\). Decimal moved left → positive n; right → negative n.
\[0.0048 = \mathbf{4.8\times10^{-3}};\quad 234000 = \mathbf{2.34\times10^5};\quad 8008 = \mathbf{8.008\times10^3}\]
\[500.0 = \mathbf{5.000\times10^2};\quad 6.0012 = \mathbf{6.0012\times10^0}\]
Significant figures in: (i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,000 (v) 500.0 (vi) 2.0034
Apply the rules: leading zeros → not significant; internal zeros → significant; trailing zeros with decimal → significant; trailing zeros without decimal → not counted.
| Number | Sig Figs | Reason |
|---|---|---|
| 0.0025 | 2 | Leading zeros not significant |
| 208 | 3 | Zero between digits is significant |
| 5005 | 4 | Both internal zeros significant |
| 126,000 | 3 | No decimal; trailing zeros not counted |
| 500.0 | 4 | Decimal present; trailing zeros significant |
| 2.0034 | 5 | All digits including internal zeros |
Round to 3 sig figs: (i) 34.216 (ii) 10.4107 (iii) 0.04597 (iv) 2808
Locate 3rd sig fig. Check digit to right: <5 → keep; ≥5 → round up. Use scientific notation if place value changes.
\[34.216\approx\mathbf{34.2};\quad 10.4107\approx\mathbf{10.4};\quad 0.04597\approx\mathbf{0.0460};\quad 2808\approx\mathbf{2.81\times10^3}\]
(a) Law obeyed by N₂ + O₂ data. (b) Unit conversions for km, mg, mL.
Dalton's Law (1803): When two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in simple whole-number ratios.
(a) Law of Multiple Proportions. With 14 g N₂ fixed: O₂ = 16 g and 32 g → ratio 1:2. With 28 g N₂: O₂ = 32 and 80 g → ratio 2:5.
(b)
\[1\,\text{km} = 10^6\,\text{mm} = 10^{15}\,\text{pm}\]
\[1\,\text{mg} = 10^{-6}\,\text{kg} = 10^6\,\text{ng}\]
\[1\,\text{mL} = 10^{-3}\,\text{L} = 10^{-3}\,\text{dm}^3\]
Distance covered by light \((3.0\times10^8\,\text{m s}^{-1})\) in 2.00 ns.
Convert ns to s first: 1 ns = 10⁻⁹ s. Then d = speed × time.
\[d = (3.0\times10^8)\times(2.00\times10^{-9}) = 6.0\times10^{-1}\,\text{m} = \mathbf{0.600\,\text{m}}\]
Reaction A + B₂ → AB₂. Limiting reagent in mixtures (i)–(v).
The reaction requires A:B₂ = 1:1. The component present in smaller stoichiometric amount is limiting. If exactly 1:1, neither is limiting (both fully consumed).
| Case | A | B₂ | Limiting Reagent |
|---|---|---|---|
| (i) | 300 atoms | 200 mol | B₂ |
| (ii) | 2 mol | 3 mol | A |
| (iii) | 100 atoms | 100 mol | Neither |
| (iv) | 5 mol | 2.5 mol | B₂ |
| (v) | 2.5 mol | 5 mol | A |
\(\mathrm{N_2+3H_2\to2NH_3}\). Mass of NH₃ from 2.00×10³ g N₂ + 1.00×10³ g H₂. Which reactant remains? How much?
Find moles of each → identify limiting → calculate product from limiting → find excess. M_r: N₂=28, H₂=2, NH₃=17 g mol⁻¹.
\[n(\mathrm{N_2}) = 2000/28 = 71.43\,\text{mol};\quad n(\mathrm{H_2}) = 1000/2 = 500\,\text{mol}\]
H₂ required = 71.43×3 = 214.29 mol < 500 available → N₂ is limiting.
\[n(\mathrm{NH_3}) = 2\times71.43 = 142.86\,\text{mol};\quad \text{Mass} = 142.86\times17 = \mathbf{2.43\times10^3\,g}\]
Unreacted H₂ = (500−214.29)×2 = \(\mathbf{5.71\times10^2\,g}\)
Difference between 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃.
mol = amount of substance (quantity only; no volume reference). M = mol per litre of solution (specifies both quantity AND volume). M_r(Na₂CO₃) = 106 g mol⁻¹.
0.50 mol Na₂CO₃ = 53 g of substance; no solution volume implied.
0.50 M Na₂CO₃ = 53 g dissolved in enough water to make exactly 1 litre of solution.
10 vol H₂ + 5 vol O₂ → volumes of water vapour produced?
Gaseous reactants and products react in simple whole-number volume ratios at constant T and P. \(\mathrm{2H_2+O_2\to2H_2O}\). Volume ratio = 2:1:2.
Ratio 10:5 = 2:1 ✓ — both fully consumed. 10 vol H₂ → \(\mathbf{10\,\text{volumes of H}_2\text{O vapour.}}\)
Convert to SI base units: (i) 28.7 pm (ii) 15.15 pm (iii) 25365 mg
Length base = metre (m). Mass base = kilogram (kg). 1 pm = 10⁻¹² m; 1 mg = 10⁻⁶ kg.
\[28.7\,\text{pm} = \mathbf{2.87\times10^{-11}\,\text{m}};\quad 15.15\,\text{pm} = \mathbf{1.515\times10^{-11}\,\text{m}}\]
\[25365\,\text{mg} = \mathbf{2.5365\times10^{-2}\,\text{kg}}\]
Largest number of atoms in 1 g of: Au / Na / Li / Cl₂
For same mass: atoms ∝ 1/M_r. Smallest atomic mass → most atoms. Cl₂ is diatomic (atoms = 2 × moles × Nₐ). Masses: Au=197, Na=23, Li=7, Cl₂=71.
| Sample | Moles of atoms in 1 g |
|---|---|
| Au | 1/197 = 5.08×10⁻³ |
| Na | 1/23 = 4.35×10⁻² |
| Li | 1/7 = 1.43×10⁻¹ ← Maximum |
| Cl₂ | 2×(1/71) = 2.82×10⁻² |
Molarity of ethanol in water if mole fraction of ethanol = 0.040 (density water = 1 g mL⁻¹).
Assume 1 total mole. Moles of each component = mole fraction values. Use mass water + density → volume. For dilute ethanol, V_solution ≈ V_water. M = mol ethanol / V(L).
Moles ethanol = 0.040; moles water = 0.960.
\[\text{Mass water} = 0.960\times18 = 17.28\,\text{g} \Rightarrow V = 17.28\,\text{mL} = 0.01728\,\text{L}\]
\[M = 0.040/0.01728 = \mathbf{2.31\,\text{mol L}^{-1}}\]
Mass of one \(^{12}\mathrm{C}\) atom in grams.
Mass of one atom (g) = molar mass / Nₐ. ¹²C has molar mass = 12 g mol⁻¹ by definition. Nₐ = 6.022×10²³ mol⁻¹.
\[\text{Mass} = 12/(6.022\times10^{23}) = \mathbf{1.99\times10^{-23}\,\text{g}}\]
Significant figures in answers of: (i) \(\frac{0.02856\times298.15\times0.112}{0.5785}\) (ii) \(5\times5.364\) (iii) \(0.0125+0.7864+0.0215\)
×/÷: answer has same sig figs as factor with fewest. +/−: answer limited by fewest decimal places. Integer "5" has 1 sig fig as a measurement.
Sig figs: 0.02856→4; 298.15→5; 0.112→3; 0.5785→4. Minimum = 3 significant figures.
5 (1 sig fig) × 5.364 (4 sig figs). Minimum = 1 significant figure.
Sum = 0.0125+0.7864+0.0215 = 0.8204. All have 4 decimal places → 4 significant figures.
Molar mass of natural Ar: ³⁶Ar(0.337%, 35.96755), ³⁸Ar(0.063%, 37.96272), ⁴⁰Ar(99.600%, 39.9624).
Same method as Q9. ⁴⁰Ar dominates at 99.6% → average ≈ 40. IUPAC value = 39.948 g mol⁻¹.
\[\bar{M} = (35.96755\times0.00337)+(37.96272\times0.00063)+(39.9624\times0.99600)\]
\[= 0.12121+0.02392+39.80255 = \mathbf{39.948\,\text{g mol}^{-1}}\]
Number of atoms in: (i) 52 mol Ar (ii) 52 u of He (iii) 52 g of He.
(i) mol → ×Nₐ. (ii) u → ÷ mass per atom (4 u for He). (iii) g → mol → ×Nₐ. M_r(He) = 4 g mol⁻¹.
\[52\times6.022\times10^{23} = \mathbf{3.13\times10^{25}\text{ atoms}}\]
\[52/4 = \mathbf{13\text{ atoms}}\]
\[13\,\text{mol}\times6.022\times10^{23} = \mathbf{7.83\times10^{24}\text{ atoms}}\]
Welding gas (C+H only): 3.38 g CO₂ + 0.690 g H₂O from combustion. 10.0 L at STP = 11.6 g. Find (i) empirical formula (ii) molar mass (iii) molecular formula.
All C → CO₂; all H → H₂O. Mass C = (12/44)×m(CO₂); Mass H = (2/18)×m(H₂O). At STP: 1 mol gas = 22.4 L. Molecular formula: n = M_r / empirical formula mass.
\[\text{C} = (12/44)\times3.38 = 0.923\,\text{g};\quad \text{H} = (2/18)\times0.690 = 0.0767\,\text{g}\]
\[\text{mol C} = 0.923/12 = 0.0769;\quad \text{mol H} = 0.0767/1 = 0.0767\]
Ratio C:H ≈ 1:1 → Empirical formula: CH
\[M_r = (11.6/10.0)\times22.4 = \mathbf{26\,\text{g mol}^{-1}}\]
\[n = 26/13 = 2 \Rightarrow \mathbf{C_2H_2}\text{ (acetylene — the welding gas)}\]
\(\mathrm{CaCO_3+2HCl\to CaCl_2+CO_2+H_2O}\). Mass of CaCO₃ for 25 mL of 0.75 M HCl.
Moles HCl = M×V(L). From equation: 2 mol HCl ≡ 1 mol CaCO₃. M_r(CaCO₃) = 100 g mol⁻¹.
\[n(\mathrm{HCl}) = 0.75\times0.025 = 0.01875\,\text{mol}\]
\[n(\mathrm{CaCO_3}) = 0.01875/2 = 0.009375\,\text{mol}\]
\[\text{Mass} = 0.009375\times100 = \mathbf{0.9375\,g \approx 0.94\,g}\]
\(\mathrm{4HCl+MnO_2\to2H_2O+MnCl_2+Cl_2}\). Grams of HCl reacting with 5.0 g MnO₂.
1 mol MnO₂ reacts with 4 mol HCl. M_r(MnO₂) = 55+32 = 87; M_r(HCl) = 36.5 g mol⁻¹. This is the lab preparation of chlorine gas.
\[n(\mathrm{MnO_2}) = 5.0/87 = 5.75\times10^{-2}\,\text{mol}\]
\[n(\mathrm{HCl}) = 4\times5.75\times10^{-2} = 0.230\,\text{mol}\]
\[\text{Mass HCl} = 0.230\times36.5 = \mathbf{8.40\,g}\]