AQA GCSE Combined Science (8464), Higher tier · Chemistry › Quantitative chemistry › Use of amount of substance in relation to masses of pure
Practise Concentration of solutions. 14 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Concentration tells you how much solute is dissolved in a given volume of solution. Here it is measured in grams per dm3 (g/dm3). You must calculate concentrations and masses of solute, convert cm3 to dm3 and, on the Higher tier, explain how the mass of solute and the volume of solution affect the concentration.
Grade by grade
What you need to be able to do, from the first marks up to the top grade.
3
Recall the unit of concentrationMass of solute per volume of solution, usually grams per dm3 (g/dm3).
4
Convert cm3 to dm3Divide by 1000, e.g. 250 cm3 = 0.250 dm3.
5
Calculate concentration in g/dm3Concentration = mass of solute (g) ÷ volume of solution (dm3).
6
Calculate the mass of solute in a solutionMass = concentration (g/dm3) × volume (dm3).
7
Explain how mass and volume affect concentration (HT)More solute in the same volume, or the same solute in less volume, gives a higher concentration.
8
Solve multi-step concentration problemsRearrange the equation, e.g. find the volume of solution that contains a given mass.
Notes
What concentration means
Many reactions take place in solutions. A solution is a solute dissolved in a solvent.
The concentration of a solution can be measured as the mass of solute per volume of solution, in g/dm3.
1 dm3 = 1000 cm3 (1 litre). To convert cm3 to dm3, divide by 1000.
Calculations
concentration (g/dm3) = mass of solute (g) ÷ volume of solution (dm3)
mass of solute (g) = concentration (g/dm3) × volume (dm3)
Example: 5.0 g of sodium chloride is dissolved to make 250 cm3 of solution. Volume = 250 ÷ 1000 = 0.250 dm3. Concentration = 5.0 ÷ 0.250 = 20 g/dm3.
Example: the mass of solute in 50 cm3 of a 40 g/dm3 solution = 40 × 0.050 = 2.0 g.
How mass and volume affect concentration (Higher tier) grade 7+
If the volume stays the same, the concentration is proportional to the mass of solute: double the mass, double the concentration.
If the mass of solute stays the same, the concentration is inversely proportional to the volume of solution: double the volume (e.g. by adding water), halve the concentration.
A concentrated solution has a lot of solute in each dm3; a dilute solution has only a little.
Cheatsheet
Concentration (g/dm3) = mass (g) ÷ volume (dm3)
Mass (g) = concentration × volume (dm3)
Volume (dm3) = mass ÷ concentration
1 dm3 = 1000 cm3
cm3 → dm3: divide by 1000
Same volume, more solute → higher concentration; same solute, more volume → lower concentration (HT) grade 7+
How to answer each type of question
Calculate the concentration
2 to 3 marks5
Convert the volume to dm3 (÷ 1000).
Divide the mass in grams by the volume in dm3.
Give the unit, g/dm3.
Example. A student dissolved 3.0 g of copper sulfate in water to make 200 cm3 of solution. Calculate the concentration of the solution in g/dm3. [3 marks]
Multiply the concentration by the volume and give the unit, g.
Example. Calculate the mass of sodium hydroxide in 25.0 cm3 of a solution with a concentration of 80.0 g/dm3. [2 marks]
Show the model answer
25.0 cm3 = 0.0250 dm3 (1) mass = 80.0 × 0.0250 = 2.00 g (1)
Explain how the concentration changes (HT)
2 marks7
Say what stays the same (the mass of solute or the volume).
Say what changes and by how much, then give the effect on the concentration.
Example. A student has 100 cm3 of sugar solution with a concentration of 50 g/dm3. She adds water until the volume of the solution is 200 cm3. Explain what happens to the concentration. [2 marks]
Show the model answer
The mass of sugar stays the same, but the volume of solution doubles (1). So the concentration halves, to 25 g/dm3 (1).
Calculate the volume of solution needed
3 marks8
Rearrange: volume (dm3) = mass ÷ concentration.
Calculate the volume in dm3.
Convert to cm3 if asked (× 1000).
Example. A plant food solution contains 12 g/dm3 of a nitrogen compound. A gardener needs 0.60 g of the compound for one plant. Calculate the volume of solution needed, in cm3. [3 marks]
Formula triangle: mass on top, concentration × volume underneath.
Unit check: g ÷ dm3 = g/dm3. If the units do not come out right, the formula is upside down.
A dm3 is 1000 times bigger than a cm3, so the number in dm3 is always 1000 times smaller.
Sense check: if the volume is less than 1 dm3, the concentration is a bigger number than the mass.
Where marks are lost
Using the volume in cm3 without dividing by 1000, which makes the concentration 1000 times too small.
Multiplying by 1000 instead of dividing when converting cm3 to dm3.
Using the volume of water added instead of the final volume of the solution.
Leaving out the unit, or writing g/cm3 instead of g/dm3.
Exam technique
Write the conversion to dm3 as its own line of working: it is often a mark.
Give the unit g/dm3 with every concentration.
In Higher tier 'explain' questions, say what stays the same (mass or volume) and what changes.
Quick recall
Cover the answers and test yourself. The app has these as flashcards that come back just before you'd forget them.
concentration in g/dm3 = mass of solute in g ÷ volume of solution in dm3 How many cm3 are there in 0.250 dm3?
250 cm3
Name the piece of apparatus used to weigh the solid solute accurately.
(electronic) balance
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy5 marks
The concentration of a solution can be calculated using the equation: concentration in g/dm3 = mass of solute in g ÷ volume of solution in dm3
(a) A student dissolved 12 g of sodium chloride in water to make 0.50 dm3 of solution. Calculate the concentration of the solution.[2]
(b) How many cm3 are there in 1 dm3?[1]
(c) Calculate the mass of solute in 2.0 dm3 of a solution with a concentration of 15 g/dm3.[2]
Show the answer and mark scheme
(a)Answer: 24 g/dm3
12 ÷ 0.50
24 (g/dm3)
(b)Answer: 1000
1000
(c)Answer: 30 g
15 × 2.0
30 (g)
Question 2Medium6 marks
Many chemical reactions take place in solutions. The concentration of a solution can be measured in g/dm3.
(a) A student dissolved 5.0 g of copper sulfate in water to make 250 cm3 of solution. Calculate the concentration of the solution in g/dm3.[2]
(b) Calculate the mass of sodium hydroxide in 25.0 cm3 of a solution with a concentration of 40 g/dm3.[2]
(c) A student needs 3.0 g of potassium nitrate. The student has a solution of potassium nitrate with a concentration of 60 g/dm3. Calculate the volume of solution, in cm3, that contains 3.0 g of potassium nitrate.[2]
Show the answer and mark scheme
(a)Answer: 20 g/dm3
250 cm3 = 0.25 dm3
5.0 ÷ 0.25 = 20 (g/dm3)
(b)Answer: 1.0 g
40 × 0.0250
1.0 (g)
(c)Answer: 50 cm3
3.0 ÷ 60 = 0.05 (dm3)
50 (cm3)
Question 3Hard8 marks
A student needs to make 250 cm3 of sodium carbonate solution with a concentration of 10.6 g/dm3.
(a) Calculate the mass of sodium carbonate the student needs.[2]
(b) Describe how the student should make the solution so that its concentration is accurate. Include the names of the apparatus used.[6]
Show the answer and mark scheme
(a)Answer: 2.65 g
10.6 × 0.250
2.65 (g)
(b)
weigh 2.65 g of sodium carbonate on a balance (reading to at least 0.01 g), e.g. in a weighing boat
reweigh the empty weighing boat / weigh by difference to find the mass actually transferred
dissolve the solid in a small volume of distilled / deionised water in a beaker, stirring with a glass rod
pour the solution into a 250 cm3 volumetric flask through a funnel
rinse the beaker, rod and funnel with distilled water and add the rinsings to the flask (so no solute is lost)
add distilled water until the bottom of the meniscus is on the graduation line, adding the last few drops with a dropping pipette
stopper the flask and invert it several times to mix the solution
use the mass actually transferred to calculate the exact concentration
Marked with levels of response: the full level descriptors are in the app.