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Cambridge IGCSE Chemistry · 0620

Chapter 3: Stoichiometry — Part 4

Topic 3.3b · Molar gas volume and concentrations

Molar gas volume

Equal volumes of gases
At the same temperature and pressure, equal volumes of different gases contain the same number of molecules (kinetic theory / Avogadro's law).
Molar gas volume at RTP
At room temperature and pressure (RTP), one mole of any gas occupies approximately 24 dm3 (24 000 cm3).
Volume calculations
Volume (dm3) = moles × 24. Rearrangement: moles = volume ÷ 24. Convert cm3 to dm3 by dividing by 1000 (e.g. 3000 cm3 = 3 dm3).

Exam Traps

  • Do not use 24 000 as dm3 — 24 000 cm3 = 24 dm3 = 1 mol at RTP.
  • Do not assume heavier gases occupy less volume at RTP — equal moles of all gases have the same volume at the same T and P.

Solution concentrations

Concentration in g/dm3
Mass of solute (g) per dm3 of solution. concentration = mass ÷ volume (in dm3).
Concentration in mol/dm3
Moles of solute per dm3 of solution. Use the triangle: n = c × V (c = concentration, V = volume in dm3). Convert cm3 to dm3 before using the formula.
Converting mol/dm3 to g/dm3
Multiply the concentration in mol/dm3 by the Mr of the solute.
Titration data
At the end point, moles of acid = moles of base (in the mole ratio from the equation). Use n = c × V for each solution to find an unknown concentration.

Examiner Report Insights

  • n = c × V requires volume in dm3 — convert cm3 by dividing by 1000 before calculating.

Exam Traps

  • Do not use cm3 directly in n = c × V — 25.0 cm3 = 0.0250 dm3.

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