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4.6 Expressing Concentration

GCSE Chemistry · AQA · Chapter 4: Chemical Calculations

A free GCSE Chemistry lesson for AQA (8462) from OBA School: concentration in g/dm³ and mol/dm³, converting volumes, and finding the mass or moles of solute. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • Calculate the mass concentration of a solution in g/dm³
  • Convert volumes between cm³ and dm³
  • Describe how to make up a solution of known concentration in a volumetric flask
  • (Higher tier) explain how the mass of solute and the volume of solution affect the concentration
  • (Higher tier) calculate concentration in mol/dm³ and convert between g/dm³ and mol/dm³

Key words

  • Solute: The substance that dissolves (for example, salt).
  • Solvent: The liquid that the solute dissolves in (for example, water).
  • Solution: The mixture formed when a solute dissolves in a solvent.
  • Concentration: The amount of solute in a given volume of solution, in g/dm³ or mol/dm³.
  • dm³ (cubic decimetre): A volume equal to 1000 cm³, the same as 1 litre.

Mass concentration in g/dm³

The concentration of a solution tells you how much solute is dissolved in each cubic decimetre (dm³) of solution.

concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³)

Volumes in the lab are usually measured in cm³. There are 1000 cm³ in 1 dm³, so divide cm³ by 1000 to get dm³. For example, 250 cm³ = 250 ÷ 1000 = 0.250 dm³.

Or keep the volume in cm³ and use: concentration (g/dm³) = mass of solute (g) ÷ volume (cm³) × 1000. Both methods give the same answer – choose the one that matches the units you are given.

Worked example: 12.5 g of sodium chloride dissolved to make 250 cm³ of solution

volume = 250 ÷ 1000 = 0.250 dm³
concentration = 12.5 ÷ 0.250 = 50 g/dm³

Answer: 50 g/dm³

Worked example: mass of solute in 200 cm³ of a 40 g/dm³ solution

mass = concentration × volume = 40 × 0.200 = 8.0 g

Answer: 8.0 g

Making up a solution of known concentration

To make an accurate solution, chemists use a volumetric flask. It is designed to hold one exact volume when filled to its graduation mark.

  • Weigh the solute accurately into a beaker and dissolve it in a little distilled water.
  • Pour the solution into the volumetric flask through a funnel. Rinse the beaker and funnel with distilled water and add the rinsings, so no solute is left behind.
  • Add distilled water until the bottom of the meniscus is on the graduation mark (add the last drops with a dropping pipette, reading at eye level).
  • Stopper the flask and turn it upside down several times so the solution is evenly mixed.

What changes the concentration? (Higher tier)

  • More solute in the same volume → more concentrated (A → B).
  • Less solvent (smaller volume) with the same solute → more concentrated (A → C).
  • Adding water (diluting) spreads the same solute through a bigger volume → less concentrated.

Key fact: When you dilute a solution, the mass (and moles) of solute stays the same. Only the volume changes.

Concentration in mol/dm³ (Higher tier)

Chemists often measure concentration in moles per cubic decimetre, because equations compare moles.

concentration (mol/dm³) = amount of solute (mol) ÷ volume of solution (dm³)

so: moles = concentration × volume (dm³)

Worked example: 4.9 g of sulfuric acid, H₂SO₄, in 250 cm³ of solution

Mr H₂SO₄ = 98, so moles = 4.9 ÷ 98 = 0.050 mol
concentration = 0.050 ÷ 0.250 = 0.20 mol/dm³

Answer: 0.20 mol/dm³

Worked example: moles of hydrochloric acid in 25.0 cm³ of a 0.100 mol/dm³ solution

moles = 0.100 × (25.0 ÷ 1000) = 0.00250 mol

Answer: 0.00250 mol (2.50 × 10⁻³ mol)

Worked example: mass of solute in a solution of known mol/dm³

What mass of sodium carbonate, Na₂CO₃, is dissolved in 50.0 cm³ of a 0.400 mol/dm³ solution?
moles = 0.400 × (50.0 ÷ 1000) = 0.0200 mol
Mr Na₂CO₃ = (2 × 23) + 12 + (3 × 16) = 106
mass = 0.0200 × 106 = 2.12 g

Answer: 2.12 g

Converting between g/dm³ and mol/dm³ (Higher tier)

Worked example: sodium hydroxide solution

A solution of NaOH has a concentration of 20 g/dm³. Mr NaOH = 40.
concentration = 20 ÷ 40 = 0.50 mol/dm³
Going back: 0.50 × 40 = 20 g/dm³

Answer: 0.50 mol/dm³

Exam tip: Always check units. If the volume is in cm³, divide by 1000 before you use it in a concentration equation.

Common mistakes

  • Using cm³ in the equation. 12.5 g in 250 cm³ is not 0.05 g/dm³ – convert to 0.250 dm³ first.
  • Multiplying by 1000 when converting cm³ to dm³. A dm³ is bigger, so the number gets smaller.
  • Thinking dilution changes the amount of solute. Adding water changes the volume only.
  • Confusing g/dm³ with mol/dm³. Read the unit the question asks for and use Mr to convert.

Revision questions

Try these, then check your answers below each one.

Question 1. 15 g of salt is dissolved to make 0.50 dm³ of solution. Calculate the concentration in g/dm³.

Answer: 30 g/dm³

Working: 15 ÷ 0.50 = 30 g/dm³.

Question 2. 2.5 g of copper sulfate is dissolved to make 125 cm³ of solution. Calculate the concentration in g/dm³.

Answer: 20 g/dm³

Working: 125 cm³ = 0.125 dm³; 2.5 ÷ 0.125 = 20 g/dm³.

Question 3. How many cm³ are there in 1 dm³? (Choose from: 10, 100, 1000, 1 000 000.)

Answer: 1000

Working: 1 dm³ = 10 cm × 10 cm × 10 cm = 1000 cm³.

Related lessons

Continue in the full course

This lesson is section 4.6 (AGCCH0406) of our free GCSE Chemistry course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Chemistry course.

 

 

 

 

 

 

 

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