The volume of a gas varies with changes in pressure and temperature. In order to compare the volumes occupied by different gases, we must adopt a set of standard conditions of temperature and pressure called room temperature and pressure (RTP). Accordingly, 25°C (298.15K) and one atmosphere (760 torr) are used as standard conditions for measurement of volumes of gases at room temperature.
It has been found that one mole of all gases occupies a volume of 24 dm³ at RTP. This is called molar volume. It is worth noting that one mole of each gas contains the same number of molecules, occupies the same volume and at the same time has different mass. In other words the masses and the sizes of the molecules of gases do not change the volume occupied by them. The concept of molar volume is useful because it enables us to calculate the number of moles or masses of the gases if we know their volumes.
📐 Worked examples (molar volume)
Calculate moles N₂ in 2.5 dm³ at RTP.
Solution: (2.5/24) = 0.10 mol
Volume of 0.5 mol H₂ at RTP = 0.5×24 = 12 dm³.
Volume of 200 g CO₂ at RTP: (200/44)×24 = 109 dm³.
⚡ 15.1 Quick Check!
Calculate the amount of CO₂ in 100 dm³ at RTP.
🧪 15.2 Concentration of a solution
Concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solution or solvent. The amount of solute dissolved is expressed either in grams or in moles whereas the quantity of solution is expressed in terms of its volume, dm³ or cm³. The unit used for the mass/volume concentration is then g/dm³ while for the molar concentration it is mol/dm³.
Calculate concentration of a solution in g/dm³ which contains 20 g of sodium chloride dissolved in 0.2 dm³ of water.
Solution: Concentration = mass (g) / volume (dm³) = 20 / 0.2 = 100 g/dm³.
Calculate the mass of copper sulphate in grams dissolved in 2 dm³ of the solution to get a solution with a concentration 15 g/dm³.
Solution: Mass = concentration × volume = 15 × 2 = 30 g.
A solution is prepared by dissolving 10 g of NaOH in 1.2 dm³ of water. Calculate the concentration of the solution in mol/dm³. (Molar mass NaOH = 40)
Solution: No. of moles = 10/40 = 0.25 mol.
Concentration = 0.25 mol / 1.2 dm³ = 0.21 mol/dm³.
Solution: Volume in dm³ = 50/1000 = 0.05 dm³.
Moles = concentration × volume = 0.05 × 0.05 = 0.0025 mol.
Molar mass CuSO₄ ≈ 159.5 g/mol → mass = 0.0025 × 159.5 = 0.399 g ≈ 0.4 g.
• Concentration (g/dm³) = mass(g)/volume(dm³) – “G over V”
• Molar concentration = moles/volume – “moles per liter”
• Molar volume 24 – “RTP 24, like 24/7 gas party”
• Always convert cm³ to dm³ (÷1000) before using formulas.
• 15.3 CuSO₄: 50 cm³ = 0.05 dm³, multiply by 0.05 mol/dm³ gives moles, then ×159.5.
📅 35‑minute professional lesson planner (incl. concentration)
- 00:00–05:00 : Hook – compare gas volumes & drinks concentration.
- 05:00–10:00 : RTP & molar volume – define, 24 dm³/mol, examples 15.1-15.3.
- 10:00–15:00 : Quick check & activity – 100 dm³ CO₂, pair discussion.
- 15:00–22:00 : Concentration definition & units – g/dm³, mol/dm³. Work 15.4, 15.5, 15.7.
- 22:00–28:00 : Advanced example – CuSO₄ (cm³ → dm³) and 15.3 calculation.
- 28:00–35:00 : Quiz practice & exit ticket – 3 MCQs on concentration.
✨ Activity: “Concentration bingo” – students calculate unknown masses or concentrations.
📝 10 MCQs · molar volume & concentration
Click option – correct turns green, others red. Submit to see score.