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AQA GCSE Chemistry Smart practice A mixed round aimed at your weak spots
Unit 1
Atomic structure and the periodic table
A simple model of the atom, symbols, relative atomic mass, electronic charge and isotopes
Atoms, elements and compounds
0%
Mixtures
0%
The development of the model of the atom
0%
Relative electrical charges of subatomic particles
0%
Size and mass of atoms
0%
Relative atomic mass
0%
Electronic structure
0%
The periodic table
The periodic table
0%
Development of the periodic table
0%
Metals and non-metals
0%
Group 0
0%
Group 1
0%
Group 7
0%
Properties of transition metals (chemistry only)
Comparison with Group 1 elements (chemistry only)
0%
Typical properties (chemistry only)
0%
Unit 2
Bonding, structure, and the properties of matter
Chemical bonds, ionic, covalent and metallic
Chemical bonds
0%
Ionic bonding
0%
Ionic compounds
0%
Covalent bonding
0%
Metallic bonding
0%
How bonding and structure are related to the properties of substances
The three states of matter
0%
State symbols
0%
Properties of ionic compounds
0%
Properties of small molecules
0%
Polymers
0%
Giant covalent structures
0%
Properties of metals and alloys
0%
Metals as conductors
0%
Structure and bonding of carbon
Diamond
0%
Graphite
0%
Graphene and fullerenes
0%
Bulk and surface properties of matter including nanoparticles (chemistry only)
Sizes of particles and their properties (chemistry only)
0%
Uses of nanoparticles (chemistry only)
0%
Unit 3
Quantitative chemistry
Chemical measurements, conservation of mass and the quantitative interpretation of chemical equations
Conservation of mass and balanced chemical equations
0%
Relative formula mass
0%
Mass changes when a reactant or product is a gas
0%
Chemical measurements
0%
Use of amount of substance in relation to masses of pure substances
Moles (HT only)
0%
Amounts of substances in equations (HT only)
0%
Using moles to balance equations (HT only)
0%
Limiting reactants (HT only)
0%
Concentration of solutions
0%
Yield and atom economy of chemical reactions (chemistry only)
Percentage yield (chemistry only)
0%
Atom economy (chemistry only)
0%
Using concentrations of solutions in mol/dm³ (chemistry only) (HT only)
0%
Use of amount of substance in relation to volumes of gases (chemistry only) (HT only)
0%
Unit 4
Chemical changes
Reactivity of metals
Metal oxides
0%
The reactivity series
0%
Extraction of metals and reduction
0%
Oxidation and reduction in terms of electrons (HT only)
0%
Reactions of acids
Reactions of acids with metals
0%
Neutralisation of acids and salt production
0%
Soluble salts
0%
The pH scale and neutralisation
0%
Titrations (chemistry only)
0%
Strong and weak acids (HT only)
0%
Electrolysis
The process of electrolysis
0%
Electrolysis of molten ionic compounds
0%
Using electrolysis to extract metals
0%
Electrolysis of aqueous solutions
0%
Representation of reactions at electrodes as half equations (HT only)
0%
Unit 5
Energy changes
Exothermic and endothermic reactions
Energy transfer during exothermic and endothermic reactions
0%
Reaction profiles
0%
The energy change of reactions (HT only)
0%
Chemical cells and fuel cells (chemistry only)
Cells and batteries (chemistry only)
0%
Fuel cells (chemistry only)
0%
Unit 6
The rate and extent of chemical change
Rate of reaction
Calculating rates of reactions
0%
Factors which affect the rates of chemical reactions
0%
Collision theory and activation energy
0%
Catalysts
0%
Reversible reactions and dynamic equilibrium
Reversible reactions
0%
Energy changes and reversible reactions
0%
Equilibrium
0%
The effect of changing conditions on equilibrium (HT only)
0%
The effect of changing concentration (HT only)
0%
The effect of temperature changes on equilibrium (HT only)
0%
The effect of pressure changes on equilibrium (HT only)
0%
Unit 7
Organic chemistry
Carbon compounds as fuels and feedstock
Crude oil, hydrocarbons and alkanes
0%
Fractional distillation and petrochemicals
0%
Properties of hydrocarbons
0%
Cracking and alkenes
0%
Reactions of alkenes and alcohols (chemistry only)
Structure and formulae of alkenes (chemistry only)
0%
Reactions of alkenes (chemistry only)
0%
Alcohols (chemistry only)
0%
Carboxylic acids (chemistry only)
0%
Synthetic and naturally occurring polymers (chemistry only)
Addition polymerisation (chemistry only)
0%
Condensation polymerisation (HT only) (chemistry only)
0%
Amino acids (HT only) (chemistry only)
0%
DNA (deoxyribonucleic acid) and other naturally occurring polymers (chemistry only)
0%
Unit 8
Chemical analysis
Purity, formulations and chromatography
Pure substances
0%
Formulations
0%
Chromatography
0%
Identification of common gases
Test for hydrogen
0%
Test for oxygen
0%
Test for carbon dioxide
0%
Test for chlorine
0%
Identification of ions by chemical and spectroscopic means (chemistry only)
Flame tests (chemistry only)
0%
Metal hydroxides (chemistry only)
0%
Carbonates (chemistry only)
0%
Halides (chemistry only)
0%
Sulfates (chemistry only)
0%
Instrumental methods (chemistry only)
0%
Flame emission spectroscopy (chemistry only)
0%
Unit 9
Chemistry of the atmosphere
The composition and evolution of the Earth's atmosphere
The proportions of different gases in the atmosphere
0%
The Earth's early atmosphere
0%
How oxygen increased
0%
How carbon dioxide decreased
0%
Carbon dioxide and methane as greenhouse gases
Greenhouse gases
0%
Human activities which contribute to an increase in greenhouse gases in the atmosphere
0%
Global climate change
0%
The carbon footprint and its reduction
0%
Common atmospheric pollutants and their sources
Atmospheric pollutants from fuels
0%
Properties and effects of atmospheric pollutants
0%
Unit 10
Using resources
Using the Earth's resources and obtaining potable water
Using the Earth's resources and sustainable development
0%
Potable water
0%
Waste water treatment
0%
Alternative methods of extracting metals (HT only)
0%
Life cycle assessment and recycling
Life cycle assessment
0%
Ways of reducing the use of resources
0%
Using materials (chemistry only)
Corrosion and its prevention (chemistry only)
0%
Alloys as useful materials (chemistry only)
0%
Ceramics, polymers and composites (chemistry only)
0%
The Haber process and the use of NPK fertilisers (chemistry only)
The Haber process (chemistry only)
0%
Production and uses of NPK fertilisers (chemistry only)
0%
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Using concentrations of solutions in mol/dm\\\\u00b3 (chemistry only) (HT only)

Practise 20 exam-style questions on Using concentrations of solutions in mol/dm\\\\u00b3 (chemistry only) (HT only) for AQA GCSE Chemistry. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Calculate the amount, in moles, of hydrochloric acid in 25.0 cm³ of 0.200 mol/dm³ hydrochloric acid.

Model answer

volume = 25.0 ÷ 1000 = 0.0250 dm³
amount = concentration × volume = 0.200 × 0.0250 = 0.00500 mol

Explanation

Amount in moles = concentration in mol/dm³ × volume in dm³. Always convert cm³ to dm³ by dividing by 1000 first; using 25.0 directly gives an answer 1000 times too big.

Mark scheme
  1. 1 mark

    volume = 25.0 ÷ 1000 = 0.0250 dm³

    allow 0.200 × 25.0 ÷ 1000

  2. 1 mark

    0.00500 (mol)

    allow 5 × 10⁻³; allow 0.005; correct answer with no working scores 2 marks

Question 2

2 marks

Solution A is 50.0 cm³ of 0.400 mol/dm³ copper(II) sulfate.
Solution B is 200 cm³ of 0.0800 mol/dm³ copper(II) sulfate.

Show by calculation which solution contains the greater amount of copper(II) sulfate in moles.

Model answer

A: moles = 0.400 × (50.0 ÷ 1000) = 0.0200 mol
B: moles = 0.0800 × (200 ÷ 1000) = 0.0160 mol
Solution A contains more copper(II) sulfate.

Explanation

The amount of solute depends on both the concentration and the volume. Solution B has the larger volume, but A is five times more concentrated, so A contains more moles. Do not judge by concentration or volume alone.

Mark scheme
  1. 1 mark

    A: 0.400 × 0.0500 = 0.0200 mol

  2. 1 mark

    B: 0.0800 × 0.200 = 0.0160 mol, so A contains more

    conclusion must be consistent with the calculated values

Question 3

3 marks

Explain how the concentration of a solution in mol/dm³ is related to the mass of solute dissolved and the volume of the solution.

Model answer

The mass of solute is converted into moles by dividing by its relative formula mass. The concentration is then the number of moles divided by the volume of the solution in dm³. So dissolving a greater mass of solute in the same volume gives a higher concentration, while dissolving the same mass in a larger volume gives a lower concentration.

Explanation

Concentration in mol/dm³ = (mass ÷ Mr) ÷ volume in dm³. Doubling the mass doubles the concentration; doubling the volume halves it. This relationship lets you calculate any one of mass, volume or concentration if you know the other two and the Mr.

Mark scheme
  1. 1 mark

    the mass of solute is converted to an amount in moles by dividing by its relative formula mass (Mr)

    allow moles = mass ÷ Mr

  2. 1 mark

    concentration = amount (moles) ÷ volume of solution in dm³

    allow concentration = mass ÷ (Mr × volume)

  3. 1 mark

    a greater mass of solute gives a higher concentration, and a larger volume of solution (for the same mass) gives a lower concentration

    allow concentration is proportional to mass and inversely proportional to volume

Question 4

4 marks

Magnesium reacts with hydrochloric acid:
Mg + 2HCl → MgCl₂ + H₂

Calculate the volume of 2.0 mol/dm³ hydrochloric acid, in cm³, needed to react exactly with 1.2 g of magnesium.
Relative atomic mass (Ar): Mg = 24

Model answer

moles of Mg = 1.2 ÷ 24 = 0.050 mol
From the equation, 1 mol Mg reacts with 2 mol HCl, so moles of HCl = 0.10 mol
volume = moles ÷ concentration = 0.10 ÷ 2.0 = 0.050 dm³
volume = 0.050 × 1000 = 50 cm³

Explanation

Rearranging concentration = moles ÷ volume gives volume = moles ÷ concentration. The answer comes out in dm³, so multiply by 1000 to give cm³, and remember the 1 : 2 ratio from the balanced equation.

Mark scheme
  1. 1 mark

    moles of Mg = 1.2 ÷ 24 = 0.050

  2. 1 mark

    moles of HCl = 2 × 0.050 = 0.10

    allow ecf

  3. 1 mark

    volume = 0.10 ÷ 2.0 = 0.050 dm³

    allow ecf

  4. 1 mark

    50 (cm³)

    allow ecf; 25 cm³ (ratio not used) scores 3 marks; correct answer with no working scores 4 marks

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