Concentration and Gas Volumes

Summary: Concentration measured in mol/dm^3 or g/dm^3. Key equation: concentration = moles / volume (in dm^3). 1 dm^3 = 1000 cm^3. One mole of any gas occupies 24 dm^3 at room temperature and pressure (RTP). Gas volume (dm^3) = moles x 24. Titration and gas volume calculations. Tags: igcse chemistry stoichiometry Created: 2026-07-14 Last Updated: 2026-07-16

Concentration

Definition: Concentration is the amount of solute dissolved in a given volume of solvent. It is measured in:

• mol/dm^3 (moles per cubic decimetre) — the standard unit for IGCSE calculations
• g/dm^3 (grams per cubic decimetre) — sometimes used when mass is more convenient

The central equation:

concentration (mol/dm^3) = moles / volume (dm^3)

or:

c = n / V

Where:

• c = concentration (mol/dm^3)
• n = number of moles (mol)
• V = volume (dm^3)

Rearranging the equation:

Memory aid — the triangle: Cover up the quantity needed:

      n
    -----
    c | V

Cover n → c x V; Cover c → n / V; Cover V → n / c

Converting Between cm^3 and dm^3

IGCSE questions often give volumes in cm^3 but the concentration formula requires dm^3.

1 dm^3 = 1000 cm^3

To convert cm^3 to dm^3: divide by 1000
To convert dm^3 to cm^3: multiply by 1000

Always check the units in the question. If volume is given in cm^3, the first step is: volume in dm^3 = volume in cm^3 / 1000.

Example 1: Basic Concentration 4.0 g of sodium hydroxide (NaOH) is dissolved in water to make 250 cm^3 of solution. Calculate the concentration in mol/dm^3. [Ar: Na = 23, O = 16, H = 1]

Step 1: Convert volume to dm^3
V = 250 / 1000 = 0.250 dm^3

Step 2: Calculate moles of NaOH
Mr(NaOH) = 23 + 16 + 1 = 40
n = m / Mr = 4.0 / 40 = 0.10 mol

Step 3: Calculate concentration
c = n / V = 0.10 / 0.250 = 0.40 mol/dm^3

Example 2: Finding Moles from Concentration How many moles of HCl are present in 25.0 cm^3 of 0.50 mol/dm^3 hydrochloric acid?

Step 1: Convert volume to dm^3
V = 25.0 / 1000 = 0.0250 dm^3

Step 2: Calculate moles
n = c x V = 0.50 x 0.0250 = 0.0125 mol

Example 3: Finding Volume from Moles and Concentration What volume (in cm^3) of 0.20 mol/dm^3 H2SO4 contains 0.010 mol of the acid?

Step 1: Calculate volume in dm^3
V = n / c = 0.010 / 0.20 = 0.050 dm^3

Step 2: Convert to cm^3
V = 0.050 x 1000 = 50 cm^3

Converting Between mol/dm^3 and g/dm^3

To convert from mol/dm^3 to g/dm^3, multiply by Mr:

concentration (g/dm^3) = concentration (mol/dm^3) x Mr

Example 4: mol/dm^3 to g/dm^3 A solution of NaCl has concentration 0.50 mol/dm^3. Express this in g/dm^3. [Ar: Na = 23, Cl = 35.5]

Mr(NaCl) = 23 + 35.5 = 58.5
Concentration (g/dm^3) = 0.50 x 58.5 = 29.25 g/dm^3

Titration Calculations

Titration questions combine concentration and reacting masses. The key steps:

1. Calculate moles of the known solution (n = c x V)
2. Use the mole ratio from the balanced equation
3. Calculate concentration or volume of the unknown solution

Example 5: Titration Calculation 25.0 cm^3 of 0.100 mol/dm^3 NaOH is exactly neutralised by 20.0 cm^3 of H2SO4. Calculate the concentration of the H2SO4.

Step 1: Balanced equation
2NaOH + H2SO4 -> Na2SO4 + 2H2O

Step 2: Moles of NaOH
V(NaOH) = 25.0 / 1000 = 0.0250 dm^3
n(NaOH) = c x V = 0.100 x 0.0250 = 0.00250 mol

Step 3: Mole ratio
NaOH : H2SO4 = 2 : 1
n(H2SO4) = 0.00250 / 2 = 0.00125 mol

Step 4: Concentration of H2SO4
V(H2SO4) = 20.0 / 1000 = 0.0200 dm^3
c(H2SO4) = n / V = 0.00125 / 0.0200 = 0.0625 mol/dm^3

Example 6: Titration (1:1 ratio) 25.0 cm^3 of NaOH is neutralised by 30.0 cm^3 of 0.100 mol/dm^3 HCl. Find the concentration of the NaOH.

NaOH + HCl -> NaCl + H2O (1:1 ratio)

n(HCl) = c x V = 0.100 x (30.0/1000) = 0.00300 mol
n(NaOH) = n(HCl) = 0.00300 mol (1:1 ratio)
V(NaOH) = 25.0 / 1000 = 0.0250 dm^3
c(NaOH) = n / V = 0.00300 / 0.0250 = 0.120 mol/dm^3

Molar Gas Volume

Key fact: One mole of ANY gas occupies 24 dm^3 at room temperature and pressure (RTP).

• RTP = approximately 20 degrees C (room temperature) and 1 atm (atmospheric pressure)
• This applies to ALL gases, regardless of their Mr
• This is because the volume of a gas depends on the number of particles, not their mass

The gas volume equation:

volume of gas (dm^3) = moles of gas x 24

or:

V = n x 24

Rearranging:

n = V / 24    (to find moles from volume)

Example 7: Gas Volume from Moles What volume (in dm^3) does 0.50 mol of CO2 occupy at RTP?

V = n x 24 = 0.50 x 24 = 12 dm^3

Example 8: Moles from Gas Volume How many moles of O2 are there in 6.0 dm^3 of oxygen gas at RTP?

n = V / 24 = 6.0 / 24 = 0.25 mol

Example 9: Gas Volume from Mass What volume (in dm^3) does 4.4 g of CO2 occupy at RTP? [Ar: C = 12, O = 16]

Step 1: Moles of CO2
Mr(CO2) = 44
n = m / Mr = 4.4 / 44 = 0.10 mol

Step 2: Volume at RTP
V = n x 24 = 0.10 x 24 = 2.4 dm^3

Combining Gas Volume Calculations with Reacting Masses

Example 10: Reacting Mass + Gas Volume Calculate the volume of CO2 (at RTP) produced when 10 g of CaCO3 is heated strongly. [Ar: Ca = 40, C = 12, O = 16]

Step 1: Balanced equation
CaCO3 -> CaO + CO2

Step 2: Moles of CaCO3
Mr(CaCO3) = 100
n(CaCO3) = 10 / 100 = 0.10 mol

Step 3: Mole ratio
CaCO3 : CO2 = 1 : 1
n(CO2) = 0.10 mol

Step 4: Volume of CO2 at RTP
V = n x 24 = 0.10 x 24 = 2.4 dm^3

Example 11: Gas Volume in a Reaction (Harder) Calculate the volume of hydrogen gas (at RTP) produced when 2.4 g of magnesium reacts with excess hydrochloric acid. [Ar: Mg = 24]

Step 1: Balanced equation
Mg + 2HCl -> MgCl2 + H2

Step 2: Moles of Mg
n(Mg) = 2.4 / 24 = 0.10 mol

Step 3: Mole ratio
Mg : H2 = 1 : 1
n(H2) = 0.10 mol

Step 4: Volume of H2 at RTP
V = n x 24 = 0.10 x 24 = 2.4 dm^3

Example 12: Gas Volume from Reacting Mass (Harder Ratio) What volume of O2 (at RTP) is needed to burn 2.7 g of aluminium completely? [Ar: Al = 27]

4Al + 3O2 -> 2Al2O3

n(Al) = 2.7 / 27 = 0.10 mol
Mole ratio Al : O2 = 4 : 3
n(O2) = 0.10 x (3/4) = 0.075 mol
V(O2) = n x 24 = 0.075 x 24 = 1.8 dm^3

Key Points

• Concentration (mol/dm^3) = moles / volume (dm^3) — c = n / V
• 1 dm^3 = 1000 cm^3 — always convert cm^3 to dm^3 before using the formula
• Concentration (g/dm^3) = concentration (mol/dm^3) x Mr
• 1 mol of gas = 24 dm^3 at RTP — V = n x 24
• RTP = room temperature and pressure (approximately 20 degrees C and 1 atm)
• For titrations: n = c x V for the known, then use the mole ratio, then c = n / V for the unknown

Key Concepts from Past Papers

• Concentration: the amount of solute dissolved in 1 dm^3 of solution (measured in mol/dm^3 or g/dm^3)
• Molar gas volume: the volume occupied by 1 mole of any gas at RTP = 24 dm
• RTP (Room Temperature and Pressure): approximately 20 degrees C and 1 atmosphere pressure
• Titration: a technique used to determine the concentration of a solution by reacting it with a solution of known concentration
• Concentration (mol/dm^3) x Volume (dm^3) = moles
• At RTP, volume of gas (dm^3) = moles x 24
• Divide cm^3 by 1000 to convert to dm

Keywords from Past Papers

molecules, concentration, particles, koh, movement, volume, amount, mix, rate, reaction, lower, spread, move, collide, add

Related Notes

• Relative Masses and Moles
• Empirical and Molecular Formulae
• Reacting Masses

Sources

• OpenStax Chemistry 2e — [Chapter 3: Composition of Substances and Solutions (Molarity)], Rice University (free, CC BY 4.0)
• BBC Bitesize GCSE Chemistry — [Concentration of Solutions], BBC (free educational resource)
• Cambridge IGCSE Chemistry 0620 — Syllabus Section 3: Stoichiometry (Concentration and Gas Volumes), Cambridge Assessment International Education
• CK-12 Chemistry for High School — [Chapter 11: Molarity and Gas Volumes], CK-12 Foundation (free, CC BY-NC 3.0)

Past Paper Sources

• 0620/31 May/June 2015: Q22(b)(iv) (0m), Q44(a)(iii) (0m), Q55(c)(i) (0m) (+3 more)
• 0620/32 Feb/March 2015: Q33(d)(ii) (1m)
• 0620/32 Feb/March 2017: Q33(b)(v) (2m)
• 0620/32 May/June 2018: Q11(b)(i) (1m)
• 0620/32 May/June 2020: Q66(b)(i) (1m)
• 0620/33 May/June 2016: Q66(d)(i) (1m)
• 0620/33 May/June 2017: Q66(b)(iv) (1m)
• 0620/33 May/June 2021: Q22(d)(ii) (3m)
• 0620/33 May/June 2022: Q33(b)(ii) (1m)
• 0620/33 May/June 2023: Q88(c)(v) (0m)
• 0620/33 May/June 2024: Q66(c)(ii) (1m), Q66(c)(ii) (1m)
• 0620/33 October/November 2015: Q77(b)(i) (2m), Q77(b)(ii) (0m), Q77(b)(ii) (0m)

Common Misconceptions