What does the concentration of a solution measured in mol/dm³ tell you?

It tells you how many moles of solute are dissolved in each cubic decimetre of solution.

4.0 g of sodium hydroxide, relative formula mass 40, is dissolved to make 500 cm³ of solution. Calculate the concentration in mol/dm³.

4.0 ÷ 40 = 0.10 mol of solute in 0.500 dm³, so the concentration is 0.10 ÷ 0.500 = 0.20 mol/dm³, given to 2 significant figures.

Calculate the mass of sodium hydroxide, relative formula mass 40, in 100 cm³ of a solution of concentration 2.00 mol/dm³.

2.00 × 0.100 = 0.200 mol, and 0.200 × 40 = 8.00 g, given to 3 significant figures.

OCR GCSE Chemistry A J248

Chemical Reactions

Topic overview Writing formulae of elements and compounds Balanced equations, and deducing formulae from ions Ionic equations, and state symbols The mole and the Avogadro constant Conservation of mass, and changes of mass in open systems Stoichiometry, and calculating reacting masses Exothermic and endothermic reactions, and reaction profiles Activation energy, and bond energy calculations Oxidation and reduction, by oxygen and by electrons Acids, alkalis and neutralisation The ionic equation for neutralisation, and reactions of acids Dilute and concentrated, weak and strong The pH scale, and hydrogen ion concentration pH as a factor of ten, and measuring pH Products at the electrodes, and molten ionic compounds Electrolysis of aqueous solutions Inert and non-inert electrodes

Elements, Compounds and Mixtures

Topic overview Purity, and melting point data Relative formula mass and empirical formulae Formulations, and separating mixtures Paper and thin layer chromatography Interpreting chromatograms and Rf values Choosing a purification method Metals and non-metals in the Periodic Table Electron arrangement and position in the Periodic Table Ionic, covalent and metallic bonding compared Dot and cross diagrams, and the limits of models Reactions, electron arrangement and Mendeleev's table Carbon's four bonds, and the variety of organic compounds Diamond, graphite, fullerenes and graphene Bond strength, intermolecular forces and predicting state Bulk properties of ionic, molecular and giant structures Nanoparticles: size and surface area to volume Uses and risks of nanoparticulate materials

Global Challenges

Topic overview Extracting metals by reduction and by electrolysis Biological methods of metal extraction Rate against yield in industry, and reaction conditions The Haber process and fertiliser production Nitrogen, phosphorus and potassium in agriculture Life cycle assessment Recycling, and the decisions behind it Alloys, and the corrosion of metals Comparing glass, ceramics, polymers, composites and metals Functional groups and homologous series Predicting the products of organic reactions Addition polymerisation Condensation polymerisation, and making a polymer DNA and other natural polymers Fractional distillation of crude oil The fractions, and crude oil as a feedstock Dependence on hydrocarbons, and cracking Chemical cells and fuel cells The Earth's early atmosphere, and how oxygen developed The greenhouse effect, and evidence for climate change Effects of carbon dioxide and methane, and air pollutants Producing potable water

Monitoring and Controlling Chemical Reactions

Topic overview Concentration in mol/dm³, and titration Titration calculations, and the mass of a solute Molar gas volume and the volumes of gases Theoretical mass and percentage yield Atom economy, and choosing a reaction pathway Measuring the rate of a reaction, and rate graphs Factors affecting rate, and collision theory Surface area and rate Catalysts and how they work Reversible reactions and dynamic equilibrium Changing the conditions of an equilibrium Activation energy, and enzymes as catalysts

Particles

Topic overview The particle model and changes of state Limitations of the particle model How the atomic model changed, and the structure of the atom The size of atoms, and the subatomic particles Calculating protons, neutrons and electrons

Predicting and Identifying Reactions

Topic overview Groups 1, 7 and 0, and why they behave as they do Transition metals Predicting reactivity from the Periodic Table Reactivity of metals with water and acids Tests for oxygen, hydrogen, carbon dioxide and chlorine Tests for aqueous cations and anions Flame tests, and identifying metal ions Identifying a species from test results Instrumental methods of analysis

Monitoring and Controlling Chemical Reactions

Concentration in mol/dm³, and titration

3 things to know
Paper 4

Overview

What concentration in mol/dm³ means and how it relates to moles and volume, converting cm³ to dm³, calculating concentrations and masses and converting from g/dm³, the effect of dilution, the pipette, burette, white tile and indicators used in a titration, what the end point is, reading a burette, rough titrations and concordant titres.

What you need to know

Answer each one before you reveal it.

Your recallNot started

Key term

State the relationship between concentration, number of moles of solute and volume of solution.

Numbers

How do you convert a volume measured in cm³ into dm³?

Numbers

A solution contains 0.50 mol of solute in 250 cm³ of solution. Calculate its concentration in mol/dm³.

Questions from this subtopic

Each one with the model answer underneath.

1
What does the concentration of a solution measured in mol/dm³ tell you?
Model answer
It tells you how many moles of solute are dissolved in each cubic decimetre of solution.
2
State the relationship between concentration, number of moles of solute and volume of solution.
Model answer
Concentration in mol/dm³ equals the number of moles of solute divided by the volume of the solution in dm³.
3
How do you convert a volume measured in cm³ into dm³?
Model answer
Divide the volume in cm³ by 1000.
4
A solution contains 0.50 mol of solute in 250 cm³ of solution. Calculate its concentration in mol/dm³.
Model answer
250 cm³ is 0.250 dm³, so the concentration is 0.50 ÷ 0.250 = 2.0 mol/dm³, given to 2 significant figures.
5
4.0 g of sodium hydroxide, relative formula mass 40, is dissolved to make 500 cm³ of solution. Calculate the concentration in mol/dm³.
Model answer
4.0 ÷ 40 = 0.10 mol of solute in 0.500 dm³, so the concentration is 0.10 ÷ 0.500 = 0.20 mol/dm³, given to 2 significant figures.
6
Calculate the mass of sodium hydroxide, relative formula mass 40, in 100 cm³ of a solution of concentration 2.00 mol/dm³.
Model answer
2.00 × 0.100 = 0.200 mol, and 0.200 × 40 = 8.00 g, given to 3 significant figures.

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Chemical Reactions

Elements, Compounds and Mixtures

Global Challenges

Monitoring and Controlling Chemical Reactions

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