Mini Chemistry
Part of Mini EducationSwitch subject

Molar Volume & Concentration

Molar volume and concentration: use 24 dm³ mol⁻¹ at RTP, c=n/V for solutions, and convert correctly between cm³ and dm³.

  • SEC G3 Pure Chemistry 2027
On this page

Learning objectives

  • calculate stoichiometric reacting masses and volumes of gases (one mole of gas occupies 24 dm3 at room temperature and pressure); calculations involving the idea of limiting reactants may be set (knowledge of the gas laws and the calculations of gaseous volumes at different temperatures and pressures are not required)
  • apply the concept of solution concentration (in mol/dm3 or g/dm3) to process the results of volumetric experiments (e.g. titration) and to solve simple problems (appropriate guidance will be provided where unfamiliar reactions are involved)

This topic depends on unit discipline. Convert cm³ to dm³ before substituting into formulae that require volume in dm³, and state the final unit. If mole-conversion steps are shaky, revise Mole & Molar Mass before combining gas-volume and concentration formulas. When these moles feed into reaction yields, continue with Limiting Reactant to avoid ratio mistakes.

1. Definition

Molar volume is the volume occupied by 1 mole of a gas at a specified temperature and pressure.

Molar concentration (molarity) is the number of moles of solute per dm³ of solution.

2. Key Ideas

  • At RTP, molar volume of gas = 24 dm³ mol⁻¹ (24,000 cm³ mol⁻¹).
  • Concentration: c = n/V where c is in mol dm⁻³ and V is in dm³.
  • Moles from mass: n = m/M where M is molar mass in g mol⁻¹.
  • Keep the volume units consistent:
    • 1000 cm³ = 1 dm³
    • 1 cm³ = 1 mL

3. Detailed Explanations

Quick Recall (units first)
  • Convert volumes: 1000 cm³ = 1 dm³ (divide by 1000).
  • Gas moles at RTP: n = V/24 with V in dm³.
  • Concentration: c = n/V with V in dm³.
  • Also know mass concentration: cₘ = m/V in g dm⁻³.

A. Molar Volume of Gases at RTP

At a fixed temperature and pressure, 1 mole of any gas occupies the same volume (at this level).

  • For the O-Level syllabus, use 24 dm³ mol⁻¹ at room temperature and pressure (r.t.p.) unless the question supplies another value.

Gas mole calculation:

n(gas) = V/(molar volume)
Gas volume at room temperature and pressure and solution concentrationFor gas at room temperature and pressure, moles equal volume in cubic decimetres divided by 24. For solutions, concentration in moles per cubic decimetre equals moles divided by volume in cubic decimetres. Cubic centimetres must be divided by 1000 first.Gas at r.t.p.1 mol occupies 24 dm³Volume, Vdm³÷ 24Amount, nmoln = V/24 · V = 24nIf V is in cm³, divide it by 1000 first.Solution concentrationvolume must be in dm³Amount, nmol÷ VConcentration, cmol dm⁻³c = n/V · n = cVMass concentration: cₘ = m/V in g dm⁻³.1000 cm³ = 1 dm³cm³ → dm³: ÷ 1000 · dm³ → cm³: × 1000
For 6092 calculations, use 24 dm³ mol⁻¹ for gases at r.t.p. and convert every solution volume to dm³ before using c = n/V.

B. Concentration of Solutions

Mass concentration:

concentration (g dm⁻³) = m/V

Molar concentration:

c (mol dm⁻³) = n/V

Because n = m/M, you often combine them:

c = m/MV
Units that must match

If c is in mol dm⁻³, then V must be in dm³. Convert cm³ to dm³ by dividing by 1000.

C. Dilution as a concentration application

If you dilute a solution by adding water, the amount of solute stays the same while the total solution volume increases. Use the familiar relationship n = cV twice:

n(before) = c₁V₁; n(after) = c₂V₂

Because the solute amount is unchanged, these expressions are equal. This gives c₁V₁ = c₂V₂ as a shortcut, not a separate rule to memorise.

What changes in dilution

Dilution changes concentration and volume. It does not change the amount of solute (moles).

4. Common Mistakes

  • Using cm³ directly in c = n/V while keeping units as mol dm⁻³ (unit mismatch).
  • Using a gas molar volume other than 24 dm³ mol⁻¹ at r.t.p. when the question does not supply a different value.
  • Forgetting that 1 cm³ = 1 mL, so 250 cm³ = 250 mL = 0.250 dm³.
  • Using Mᵣ as if it has units. Use molar mass M in g mol⁻¹.
  • Confusing gas volume units (cm³ vs dm³) and forgetting to convert.

5. Exam Tips

Gas volume routine
  1. Convert volume to dm³ 2) Use n = V/24 at r.t.p. 3) Use the mole ratio from the equation if needed
Concentration routine
  1. Convert volume to dm³ 2) Find moles using n = m/M 3) Use c = n/V

6. Worked Examples

Modelled example 1

Gas Moles at RTP

Core

Problem

Calculate the number of moles in 600 cm³ of oxygen gas at RTP.
Study the worked solution

  1. Convert the gas volume

    Method

    Divide the volume in cm³ by 1000.

    Reason

    The RTP molar volume is expressed in dm³ mol⁻¹.

    Working

    V = 600/1000 = 0.600 dm³.

  2. Use molar volume

    Method

    Divide the gas volume by 24 dm³ mol⁻¹.

    Reason

    At RTP, every mole of gas occupies 24 dm³.

    Working

    n = V/24 = 0.600/24 = 0.0250 mol.

  3. Check the scale

    Method

    Compare 0.600 dm³ with 24 dm³.

    Reason

    The sample is 1/40 of the molar volume, so it should contain 1/40 mol.

    Working

    n(O₂) = 0.0250 mol.

Guided practice 2

Molarity From Mass and Volume (KOH)

About 7 min

Problem

5.6 g of potassium hydroxide, KOH, is dissolved to make 250 cm³ of solution. Calculate its molarity. Given Aᵣ(K) = 39, Aᵣ(O) = 16, Aᵣ(H) = 1.

Complete the mass–mole–concentration chain

Hints

Hint 1: find moles first
Use M(KOH) = 39 + 16 + 1, then n = m/M.
Hint 2: make volume compatible
Convert 250 cm³ to 0.250 dm³ before using c = n/V.
View solution step by step

  1. Find molar mass

    Method

    Add the K, O and H relative masses.

    Reason

    Mass must be converted into moles before concentration can be calculated.

    Working

    M(KOH) = 39 + 16 + 1 = 56 g mol⁻¹.

  2. Find amount of solute

    Method

    Divide mass by molar mass.

    Reason

    n = m/M.

    Working

    n = 5.6/56 = 0.100 mol.

  3. Convert solution volume

    Method

    Divide 250 cm³ by 1000.

    Reason

    Concentration in mol dm⁻³ requires volume in dm³.

    Working

    V = 250/1000 = 0.250 dm³.

  4. Calculate concentration

    Method

    Divide moles by solution volume.

    Reason

    c = n/V.

    Working

    c = 0.100/0.250 = 0.400 mol dm⁻³.

Common misconception 3

Check a dilution claim

Find and correct the mistake

Learner response

50.0 cm³ of 0.80 mol dm⁻³ HCl(aq) is diluted to 200 cm³. A student says the concentration becomes four times larger because the volume is four times larger. Explain the error and calculate the new concentration.

Connect volume change to concentration

What remains constant?

View solution step by step

  1. Identify the conserved quantity

    Method

    Keep the moles of HCl unchanged.

    Reason

    Dilution adds solvent, not solute.

    Working

    n(HCl) = c₁V₁ = 0.80 × 0.0500 = 0.0400 mol.

  2. Use compatible volume units

    Method

    Convert the final volume to dm³.

    Reason

    Concentration in mol dm⁻³ requires volume in dm³.

    Working

    V₂ = 200/1000 = 0.200 dm³.

  3. Correct the direction

    Method

    Divide the initial concentration by four.

    Reason

    Spreading unchanged moles through four times the volume makes concentration four times smaller.

    Working

    c₂ = n/V₂ = 0.0400/0.200 = 0.20 mol dm⁻³.

Examiner practice 4

Titration-Style Calculation (Moles Equal)

5 marks

Examination question

25.0 cm³ of NaOH(aq) is neutralised by 0.100 mol dm⁻³ HCl(aq). The average titre is 20.0 cm³. Find the concentration of NaOH(aq). Reaction: H⁺ + OH⁻ → H₂O. [5 marks]

Show conversions, mole ratio and concentration

View solution step by step

  1. Convert acid volume

    1 mark

    Method

    Convert the titre to dm³.

    Reason

    n = cV requires volume in dm³.

    Working

    V(HCl) = 20.0/1000 = 0.0200 dm³.

  2. Find acid moles

    1 mark

    Method

    Multiply acid concentration by acid volume.

    Reason

    n = cV.

    Working

    n(HCl) = 0.100(0.0200) = 0.00200 mol.

  3. Apply mole ratio

    1 mark

    Method

    Transfer the amount through the 1:1 equation.

    Reason

    One mole of H⁺ reacts with one mole of OH⁻.

    Working

    n(NaOH) = 0.00200 mol.

  4. Convert alkali volume

    1 mark

    Method

    Convert 25.0 cm³ to dm³.

    Reason

    The final concentration uses volume in dm³.

    Working

    V(NaOH) = 0.0250 dm³.

  5. Calculate alkali concentration

    1 mark

    Working

    c(NaOH) = 0.00200/0.0250 = 0.0800 mol dm⁻³.

Challenge 5

Convert g/dm3 to mol/dm3

Minimal support

Representation transfer

A solution contains 5.85 g of sodium chloride per dm³. Find its concentration in mol dm⁻³. Given Aᵣ(Na) = 23, Aᵣ(Cl) = 35.5.

Convert the quantity per dm3

Hints

Hint 1: interpret per dm3
Treat the stated quantity as a 1 dm³ sample containing 5.85 g NaCl.
Hint 2: convert grams to moles
Divide the mass in that 1 dm³ by M(NaCl).
View solution step by step

  1. Calculate molar mass

    Method

    Add the Na and Cl relative masses.

    Reason

    Molar mass converts the mass quantity into moles.

    Working

    M(NaCl) = 23 + 35.5 = 58.5 g mol⁻¹.

  2. Convert the per-volume quantity

    Method

    Divide grams per dm³ by grams per mole.

    Reason

    The gram units cancel, leaving moles per dm³.

    Working

    c = 5.85/58.5 = 0.100 mol dm⁻³.

7. Mind Stretchers

Mind stretcher 1: Gas Stoichiometry (Use the Equation)Extension

Question: Magnesium reacts with dilute hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). What volume of H₂(g) at RTP is produced from 0.050 mol of Mg?

Show Answer

Mole ratio Mg:H₂ is 1:1, so n(H₂) = 0.050 mol.

At RTP, V = n × 24:

V = 0.050 × 24 = 1.20 dm³

Mind stretcher 2: Spot the Unit ErrorExtension

Question: A student calculates c = 0.10/250 = 4.0 × 10⁻⁴ mol dm⁻³ for a 250 cm³ solution containing 0.10 mol. Explain what is wrong and give the correct answer.

Show Answer

They used 250 cm³ as if it were 250 dm³.

Correct volume: 250 cm³ = 0.250 dm³.

c = 0.10/0.250 = 0.40 mol dm⁻³

8. Quiz

Quiz time

Ready to check your understanding? Try the interactive quiz, then review any questions you missed.

Go to quiz