IB Chemistry HLTopic 1 — Counting Particles by MassPaper 1 & 2Core skill~11 min read
Concentration of Solutions
Most reactions you meet in the lab happen in solution, so you need a way of saying how much stuff is dissolved in how much liquid. The chemistry here is easy. The marks are lost almost entirely on one thing: volume units.
📘 What you need to know
Concentration is the amount of solute dissolved in a given volume of solution.
c = n ÷ V, with V in dm3. Units of c: mol dm−3.
1 dm3 = 1000 cm3 = 1 litre. To convert cm3 to dm3, divide by 1000.
Square brackets mean concentration in mol dm−3: [NaCl] = 0.25 mol dm−3.
Mass concentration is g dm−3. Multiply by molar mass to go from mol dm−3 to g dm−3; divide to come back.
Parts per million (ppm) is used for very dilute samples. In water, 1 ppm = 1 mg per dm3.
On dilution the amount of solute does not change, so c1V1 = c2V2.
What concentration really means
Concentration is a ratio, not an amount. Two spoons of sugar in a mug is sweet; two spoons in a bucket is not. Same solute, very different concentration.
Because it is a ratio, you can take a small sample from a bottle and its concentration is identical to the concentration of the whole bottle. That fact is what makes titrations work.
Concentrated and dilute describe how much is dissolved. Strong and weak describe how far an acid ionises — completely different ideas that share a lot of exam questions.
Keep “concentrated / dilute” and “strong / weak” in separate boxes in your head. You can have a very dilute solution of a strong acid, and a very concentrated solution of a weak one. Mixing these up is a classic Paper 1 trap.
The units trap
Nearly every mark lost in this topic comes from the same place. Volumes are measured in the lab in cm3, because that is what pipettes and burettes are marked in. Concentration is defined per dm3. So a conversion is nearly always needed, and it is easy to forget.
Do the conversion as the very first line of your working, not halfway through. It is far easier to spot a missing factor of 1000 that way.
The three ways of writing concentration
Molar concentrationc (mol dm−3) = moles of solute (mol) ÷ volume of solution (dm3)
Mass concentrationρ (g dm−3) = mass of solute (g) ÷ volume of solution (dm3)
The two are linked by the molar mass, exactly as mass and moles were on the last page.
Unit
Means
Typical use
Convert to mol dm−3 by
mol dm−3
moles of solute per dm3
everything quantitative
already there
g dm−3
grams of solute per dm3
bottle labels, solubility
divide by molar mass
ppm
1 mg per dm3 of water
pollutants, drinking water
convert mg to g, then divide by molar mass
Why 1 ppm = 1 mg dm−3 in water: 1 dm3 of water has a mass of about 1 kg, which is 1 000 000 mg. So 1 mg in that dm3 is literally one part per million. The shortcut only holds for dilute aqueous solutions.
Dilution
When you add water to a solution, you add nothing to the solute. The number of moles stays exactly the same — it is just spread through a bigger volume. That single sentence gives you the equation.
Dilutionc1V1 = c2V2
Both sides are simply “moles of solute”. You can use cm3 on both sides here, as long as you are consistent, because the volume units cancel.
Worked examples
WORKED EXAMPLE
Making up a standard solution
Calculate the mass of potassium manganate(VII), KMnO4, needed to make 250 cm3 of a 0.0200 mol dm−3 solution.
Step 1: convert the volume first250 ÷ 1000 = 0.250 dm³Step 2: moles needed = c × Vn = 0.0200 × 0.250 = 5.00 × 10⁻³ molStep 3: molar mass of KMnO₄39.10 + 54.94 + (4 × 16.00) = 158.04Step 4: mass = moles × molar massm = 5.00 × 10⁻³ × 158.04 = 0.7902m = 0.790 gskip step 1 and you get 790 g — a thousand times too much
WORKED EXAMPLE
Converting g dm⁻³ to mol dm⁻³
A saline solution is labelled 8.50 g dm−3 sodium chloride. Calculate its concentration in mol dm−3.
Step 1: molar mass of NaCl22.99 + 35.45 = 58.44 g mol⁻¹Step 2: divide the mass concentration by it8.50 ÷ 58.44 = 0.14545…0.145 mol dm⁻³the volume is already 1 dm³ on both sides, so it simply cancels
WORKED EXAMPLE
Dilution
25.0 cm3 of 2.00 mol dm−3 hydrochloric acid is transferred to a volumetric flask and made up to 500 cm3 with water. Calculate the concentration of the diluted acid.
Step 1: the moles of HCl do not changen = 2.00 × 0.0250 = 0.0500 molStep 2: same moles, new volumec = 0.0500 ÷ 0.500 = 0.1000.100 mol dm⁻³Check with c₁V₁ = c₂V₂2.00 × 25.0 = c₂ × 500 → c₂ = 0.100the volume went up 20 times, so the concentration fell 20 times
WORKED EXAMPLE
Parts per million
A 2.0 dm3 sample of river water is found to contain 1.4 mg of nitrate ions. Calculate the concentration in ppm.
Step 1: ppm in water means mg per dm³1.4 mg ÷ 2.0 dm³ = 0.70 mg dm⁻³0.70 ppmno molar mass needed — ppm here is a mass ratio, not a mole ratio
💡 Exam tip
Convert cm3 to dm3 on the first line of every solution question, before anything else. Make it a reflex.
Read the units in the answer line. mol dm−3 and g dm−3 are not interchangeable, and questions sometimes switch between them deliberately.
For dilutions, the volume in V2 is the final total volume, not the volume of water you added.
“Made up to 250 cm3” means the final volume is 250 cm3. “250 cm3 of water was added” does not — read carefully.
Square brackets always mean mol dm−3, so [HCl] = 0.10 needs no unit conversion.
Sanity check: typical school solutions are between about 0.01 and 2 mol dm−3. An answer of 400 usually means a missing 1000.
⚠ Common mix-up
Leaving the volume in cm3. By far the most common error in the whole topic, and it makes every answer 1000 times out.
Confusing concentrated with strong. Concentration is about how much is dissolved; strength is about how far it ionises.
Multiplying by 1000 the wrong way. There are fewer dm3 than cm3, so the number must get smaller when you convert to dm3.
Using the volume of solvent instead of the volume of solution. Concentration is defined per dm3 of the final solution.
Forgetting the molar mass when converting g dm−3 to mol dm−3. The two numbers are never equal unless M happens to be 1.
Treating ppm as a percentage. 1 ppm is 0.0001 %, not 1 %.
Up next: Avogadro’s Law — the same counting idea again, but for gases, where volume alone is enough to tell you the ratio of particles.
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