• Atoms, molecules and ions are so tiny that we can never count them one by one, and we cannot weigh a single one on a laboratory balance.
• Chemists therefore group them into a very large "package", exactly as a shop sells eggs by the dozen or paper by the ream.
• The package used in chemistry is the mole.

• The amount of a chemical substance is expressed in the International System of Units (SI) in the unit mole (symbol: mol).
• One mole of any substance contains the same fixed number of particles (molecules, atoms, ions, formula units or electrons).
• That fixed number is called Avogadro’s number:
• Avogadro’s number = 6.02 × 10²³ particles per mole
• The central relation is therefore:
• Number of particles = number of moles × Avogadro’s number
• or, written with symbols, N = n × 6.02 × 10²³. Rearranged, n = N / (6.02 × 10²³).
• Important notes:

• The word "particle" must always be specified. "1 mol of hydrogen" is ambiguous: it may mean 1 mol of H atoms or 1 mol of H₂ molecules. One mole of H₂ molecules contains 6.02 × 10²³ molecules but twice as many atoms, because each molecule has two atoms.
• The mole of a formula unit (an ionic compound such as NaCl) contains 6.02 × 10²³ formula units, and these release a known number of ions when dissolved or melted.

2.1 Avogadro’s hypothesis

• Solids and liquids of one mole have different volumes, because their particles are packed close together and differ in size.
• Gas particles are very far apart compared to their own size, so the volume of a gas depends almost only on the pressure and temperature, not on what the gas is.
• This is the content of Avogadro’s hypothesis:
• Equal volumes of different gases, under the same conditions of temperature and pressure, contain the same number of molecules.
• Consequently, one mole of any gas occupies the same volume under the same conditions. Measurements showed:

• The molecules differ in mass, but a mole of each takes the same volume.
• Hence 22.4 L of hydrogen weighs 2 g whereas 22.4 L of oxygen weighs 32 g (both at STP).

2.3 The working relations

• M(CO₂) = 44 g/mol; n = 8.8 / 44 = 0.2 mol.
• (a) V = 0.2 × 22.4 = 4.48 L at STP.
• (b) V = 0.2 × 24 = 4.8 L at rtp.
• Worked example 2.2. Mass of 11.2 L of oxygen measured at STP. [O = 16]

1. n = 11.2 / 22.4 = 0.5 mol; M(O₂) = 32 g/mol.
2. Mass = 0.5 × 32 = 16 g.

✍️ Worked examples

• Worked example 2.3. Number of molecules in 5.6 L of nitrogen at STP.
• n = 5.6 / 22.4 = 0.25 mol.
• N = 0.25 × 6.02 × 10²³ = 1.505 × 10²³ molecules (and twice that, 3.01 × 10²³, atoms of nitrogen).
• Worked example 2.4 — using molar volume to find the molar mass of an unknown gas.
• A gas has a density of 1.25 g/L at STP.
• The mass of 22.4 L (one mole) = 1.25 × 22.4 = 28 g, so its molar mass is 28 g/mol (it could be N₂ or CO).
• Worked example 2.5 — density. The density of methane CH₄ at STP = molar mass / molar volume = 16 / 22.4 = 0.714 g/L.

2.4 Notes

• The molar volume 22.4 L applies only to gases. Never use it for liquid water or solid sodium.
• Always check the condition in the question STP gives 22.4, rtp gives 24.
• Volumes of gases that react or form can be compared directly in the ratio of the coefficients, as we see in Lesson 3.

6.1 Molarity

• The concentration of a solution can be expressed in several ways.
• The most important is molar concentration (molarity) the number of moles of solute dissolved in one litre of solution.
• Molarity M (mol/L) = number of moles of solute n (mol) / volume of solution V (L)
• Units mol/L, written M. If the volume is given in mL or cm³, divide by 1000. Also n = M × V (L), mass of solute = M × V (L) × molar mass. Note: V is the volume of the solution, not of the solvent.
• Worked example 6.1. Molarity of a solution made by dissolving 4.9 g of sulphuric acid in water to make 250 mL. [H = 1, S = 32, O = 16]
• M(H₂SO₄) = 98; n = 4.9 / 98 = 0.05 mol; V = 0.25 L.
• M = 0.05 / 0.25 = 0.2 mol/L.
• Worked example 6.2 — mass needed. Mass of Na₂CO₃ needed to prepare 500 mL of 0.2 M solution [Na = 23, C = 12, O = 16]. n = 0.2 × 0.5 = 0.1 mol; mass = 0.1 × 106 = 10.6 g.
• Worked example 6.3 — pattern of the NaOH example. 2 g of NaOH (40) in 500 mL: n = 0.05 mol; M = 0.05 / 0.5 = 0.1 M.

6.2 Concentration of ions

• A strong ionic solute dissociates completely, so ion concentrations are the solute molarity times the number of that ion in the formula.
• Worked example 6.4. 0.3 M Al₂(SO₄)₃ → 2Al³⁺ + 3SO₄²⁻. [Al³⁺] = 2 × 0.3 = 0.6 M; [SO₄²⁻] = 3 × 0.3 = 0.9 M.
• Worked example 6.5 — volume containing a number of moles. What volume of 0.2 M CaCl₂ contains 1 mol of chloride ions? Each CaCl₂ gives 2 Cl⁻, so [Cl⁻] = 0.4 M. V = 1 / 0.4 = 2.5 L.
• Worked example 6.6 — concentration from the ions asked (pattern of the chapter exercise).
• What volume of 0.3 M Na₂SO₄ must be taken to prepare 2 L of a solution with [Na⁺] = 0.4 M? [Na₂SO₄] needed = 0.4/2 = 0.2 M.
• M₁V₁ = M₂V₂: 0.3 × V₁ = 0.2 × 2, V₁ = 1.33 L.

6.3 Dilution

• When water is added to a solution, its volume increases but the number of moles of solute is unchanged, so the concentration decreases.
• Since n = M × V is constant:

M₁ × V₁ = M₂ × V₂

• (1 = concentrated/before dilution; 2 = diluted/after).
• Volume of water added = V₂ − V₁. (V₁ and V₂ may both be in mL or both in L.)
• Worked example 6.7. 20 mL of 5 M HCl are diluted to 250 mL. M₂ = 5 × 20 / 250 = 0.4 M. Water added = 250 − 20 = 230 mL.
• Worked example 6.8. Volume of 2 M NaOH needed to prepare 500 mL of 0.1 M solution: V₁ = 0.1 × 500 / 2 = 25 mL; water to add = 475 mL.
• Worked example 6.9 — volume of water added (pattern of the book’s Na₂CO₃ example).
• To what volume must 250 mL of 0.6 M NaCl be diluted to give 0.2 M?
• V₂ = 0.6 × 250 / 0.2 = 750 mL; water added = 750 − 250 = 500 mL.
• Worked example 6.10 — concentration given as mass per volume. A solution contains 0.4 g of NaOH in each 10 mL. In 1 L: 40 g = 1 mol, so M = 1 M. If 1 L of it is diluted with 3 L of water, V₂ = 4 L and M₂ = 1 × 1 / 4 = 0.25 M.

6.4 Final tips for the chapter

• Always write down the molar masses first.
• Check units (g vs mg, mL vs L).
• Balance the equation before any stoichiometric calculation.
• Limiting reactant compute ratios, then base everything on the smaller one.
• Empirical formula moles → divide by smallest → multiply to get integers.
• Gas volumes 22.4 L at STP, 24 L at rtp.