Cambridge IGCSE Chemistry · Syllabus 0620 · Stoichiometry
Concentration
What is Concentration?
Concentration states how much solute is dissolved in a given volume of solution, and it can be measured in either of two units: grams of solute per cubic decimetre of solution, written g per dm3, or moles of solute per cubic decimetre of solution, written mol per dm3. A larger number in either unit means a more concentrated solution. Concentration is a statement about the dissolved solute only, which is what distinguishes it from density.
This definition is part of the Stoichiometry chapter in Cambridge IGCSE Chemistry.
Common mistakes with Concentration
- "Concentration is measured in \(\mathrm{g/cm^3}\)." Fix That is the usual unit of density. The two units named for concentration are \(\mathrm{g/dm^3}\) and \(\mathrm{mol/dm^3}\), and both are per cubic decimetre.
- "A concentrated solution is one with a large mass." Fix Concentration says nothing about how much solution you have. A drop of \(2\ \mathrm{mol/dm^3}\) acid is more concentrated than a bucket of \(0.1\ \mathrm{mol/dm^3}\) acid, and the bucket contains far more solute in total.
- Using the volume of water added rather than the volume of solution. Fix Concentration is per \(\mathrm{dm^3}\) of finished solution. "Made up to \(250\ \mathrm{cm^3}\)" gives you the volume you need; "dissolved in \(250\ \mathrm{cm^3}\) of water" is a different statement, and a question will say which it means.
- Dividing by the titre in step 5. Fix Divide by the volume of the solution whose concentration you are finding. Label both volumes with their solution's name as you write them down.
Examiner tips on Concentration
- Working the selector on a question you have not seen "\(50.0\ \mathrm{cm^3}\) of \(0.200\ \mathrm{mol\,dm^{-3}}\) hydrochloric acid is added to excess magnesium. Calculate the volume of hydrogen produced at r.t.p." I have: a concentration and a volume, of hydrochloric acid. I want: a gas volume in \(\mathrm{dm^3}\), of hydrogen. Different substances, so a ratio is needed. Route: \(n = cV\) (converting \(50.0\ \mathrm{cm^3}\) to \(0.0500\ \mathrm{dm^3}\)) → ratio from \(\mathrm{Mg} + 2\mathrm{HCl} \rightarrow \mathrm{MgCl_2} + \mathrm{H_2}\), which is \(2\ \mathrm{HCl} : 1\ \mathrm{H_2}\) → \(V = n \times 24\). Answer: \(n(\mathrm{HCl}) = 0.200 \times 0.0500 = 0.0100\ \mathrm{mol}\); \(n(\mathrm{H_2}) = 0.00500\ \mathrm{mol}\); \(V = 0.00500 \times 24 = 0.120\ \mathrm{dm^3}\), that is \(120\ \mathrm{cm^3}\) at r.t.p. The word "excess" told you the magnesium never needed checking.