Quantitative chemistry · GCSE Chemistry
Concentration
GCSE Chemistry revision on concentration: mol/dm³ and g/dm³, converting cm³ to dm³, dilution, and using concentration in titration calculations.
Concentration in mol/dm³ = moles ÷ volume in dm³. Divide cm³ by 1000. g/dm³ = mol/dm³ × Mᵣ. Diluting with water lowers concentration; moles of solute stay the same.
The important bits
What you need to know
- 1
Concentration of a solution can be measured in mol/dm³ (molar concentration) or in g/dm³. Always state the unit.
- 2
concentration (mol/dm³) = moles / volume (dm³). Rearrange: moles = concentration × volume; volume = moles / concentration.
- 3
Convert cm³ to dm³ by dividing by 1000. 25.0 cm³ = 0.0250 dm³; 250 cm³ = 0.250 dm³; 1 dm³ = 1000 cm³.
- 4
concentration in g/dm³ = concentration in mol/dm³ × Mᵣ. 0.100 mol/dm³ NaOH (Mᵣ 40) is 4.00 g/dm³.
- 5
A more concentrated solution has more solute particles in the same volume, so collision frequency with another reactant is higher and rate is faster (link to collision theory).
- 6
Dilution: moles of solute stay constant. c₁V₁ = c₂V₂ if you use the same units of volume. Adding water increases volume and decreases concentration.
- 7
Titration uses this formula twice: moles of the known solution from c × V, then concentration of the unknown from moles / V, after applying the equation ratio.
- 8
A standard solution is one of accurately known concentration, made in a volumetric flask to the graduation line, with the meniscus at eye level.
Quotations worth analysing
Short evidence. Real method.
“c = n / V with V in dm³”
The usual trap is leaving V in cm³, which makes the concentration 1000 times too small. Write /1000 in the working every time.
“0.100 mol/dm³ HCl 25.0 cm³ n = 0.100 × 0.0250 = 0.00250 mol”
Three significant figures in the volume, three in the concentration, so three in the moles if the data support it. Then use the mole ratio.
“g/dm³ = mol/dm³ × Mᵣ”
Examiners often ask you to convert. 0.50 mol/dm³ H₂SO₄ (Mᵣ 98) is 49 g/dm³. Do not treat g/dm³ as if it were already moles.
Go deeper
Volume unit first, arithmetic second
Circle every volume in a question and convert it before you touch the calculator. 25 cm³ of 0.200 mol/dm³ is not 5 mol. It is 0.200 × 0.025 = 0.00500 mol. That single conversion is worth more marks across the paper than any other quantitative habit. Measuring cylinders are fine for rate practicals; volumetric flasks and pipettes are for concentration you will calculate from. If you make a solution by dissolving 4.00 g of NaOH in water and making up to 250 cm³, moles = 4.00/40 = 0.100 mol, volume = 0.250 dm³, concentration = 0.400 mol/dm³. “Making up to” means the final volume of the solution, not 250 cm³ of water poured onto solid and ignored.
Go deeper
Dilution keeps moles, changes volume
Pipette 25.0 cm³ of 1.00 mol/dm³ acid into a 250 cm³ volumetric flask and make up to the mark with water. Moles transferred = 1.00 × 0.0250 = 0.0250 mol. New concentration = 0.0250 / 0.250 = 0.100 mol/dm³, a ten-fold dilution. The pH of a strong acid would rise by about 1 (Higher). Students think dilution destroys moles of acid. It only spreads them through more water. In a rate experiment, diluting HCl while keeping the total volume the same is how you change concentration fairly. Invert the flask several times so the solution is uniform before you pipette from it.
Go deeper
Titration is this formula with a ratio in the middle
Mean concordant titre 24.00 cm³ of 0.100 mol/dm³ H₂SO₄ against 25.0 cm³ of NaOH. Moles H₂SO₄ = 0.100 × 0.02400 = 0.00240 mol. Equation H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so moles NaOH = 2 × 0.00240 = 0.00480 mol. Concentration NaOH = 0.00480 / 0.0250 = 0.192 mol/dm³. The 2 from sulfuric acid is the mark people drop. Strength of the acid (strong versus weak) does not change this arithmetic if you titrate to a proper end-point; concentration and ratio do. Convert 24.00 cm³ to 0.02400 dm³ before you multiply.
See the idea in action
A solution contains 5.85 g of NaCl in 250 cm³ of solution. Mᵣ(NaCl) = 58.5. Moles = 5.85 / 58.5 = 0.100 mol. Volume = 250/1000 = 0.250 dm³. Concentration = 0.100 / 0.250 = 0.400 mol/dm³, or 0.400 × 58.5 = 23.4 g/dm³. If 25.0 cm³ of this solution is diluted to 100 cm³, moles in the portion = 0.400 × 0.0250 = 0.0100 mol, new concentration = 0.0100 / 0.100 = 0.100 mol/dm³.
Exam technique
Turn knowledge into marks
Convert cm³ to dm³ in the first line of working. Write moles = c × V, then the equation ratio, then c = n / V for the unknown. Quote mol/dm³ or g/dm³ clearly. For dilutions, moles of solute stay the same.
Common mistakes
Do not give these marks away
- 01
Forgetting to divide cm³ by 1000.
- 02
Treating g/dm³ as mol/dm³, or using the wrong Mᵣ in the conversion.
- 03
Missing the 1:2 ratio when sulfuric acid is titrated against NaOH.
How many moles are in 50.0 cm³ of 0.200 mol/dm³ HCl?
A10.0 mol
B0.0100 mol
C0.200 mol
D4.00 mol
Show the answer
0.0100 mol. Volume = 50.0/1000 = 0.0500 dm³. Moles = 0.200 × 0.0500 = 0.0100 mol. Leaving the volume in cm³ would give a nonsense 10 mol.
Quick questions
If this is the bit you searched
How do you calculate concentration in mol/dm3 GCSE Chemistry?
Divide the number of moles of solute by the volume of solution in dm³. Convert cm³ to dm³ by dividing by 1000.
How do you convert mol/dm3 to g/dm3?
Multiply the concentration in mol/dm³ by the relative formula mass Mᵣ of the solute.
How do you convert cm3 to dm3?
Divide by 1000. 25 cm³ is 0.025 dm³. 1 dm³ is 1000 cm³.
What happens to moles when you dilute a solution?
The number of moles of solute stays the same. Volume increases, so concentration decreases. Use n = cV before and after, or c₁V₁ = c₂V₂.