Moles in Solution — Concentration (g/dm³)

This lesson covers how to calculate concentration in g/dm³, as required by the AQA GCSE Combined Science Trilogy specification (8464). You will learn how to convert between mass, volume and concentration, and how to solve problems involving solutions.

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What Is Concentration?

Concentration tells you how much solute is dissolved in a given volume of solution.

The simplest unit is grams per cubic decimetre (g/dm³):

$\text{concentration (g/dm}^3) = \frac{\text{mass of solute (g)}}{\text{volume of solution (dm}^3)}$concentration (g/dm3)=volume of solution (dm3)mass of solute (g)​

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Key Unit Conversions

From	To	Conversion
cm³	dm³	Divide by 1000
dm³	cm³	Multiply by 1000

$1 \text{ dm}^3 = 1000 \text{ cm}^3 = 1 \text{ litre}$1 dm3=1000 cm3=1 litre

Exam Tip (AQA 8464): The most common mistake is forgetting to convert cm³ to dm³. Always check the units in the question.

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The Concentration Triangle

graph TD
    A["<b>mass of solute (g)</b>"] --- B["<b>concentration (g/dm³)</b>"]
    A --- C["<b>volume (dm³)</b>"]
    B --- C
    style A fill:#f59e0b,color:#000,stroke:#d97706
    style B fill:#3b82f6,color:#fff,stroke:#2563eb
    style C fill:#3b82f6,color:#fff,stroke:#2563eb

To Find	Formula
Concentration	$c = \dfrac{m}{V}$c=Vm​
Mass of solute	$m = c \times V$m=c×V
Volume	$V = \dfrac{m}{c}$V=cm​

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Worked Examples

Example 1 — Finding Concentration

Question: 5.0 g of sodium chloride is dissolved in 250 cm³ of water. What is the concentration?

Step 1: Convert volume: $250 \text{ cm}^3 = \dfrac{250}{1000} = 0.25 \text{ dm}^3$250 cm3=1000250​=0.25 dm3

Step 2: Calculate concentration: $c = \frac{5.0}{0.25} = 20 \text{ g/dm}^3$c=0.255.0​=20 g/dm3

Example 2 — Finding Mass of Solute

Question: What mass of potassium chloride is needed to make 500 cm³ of a 30 g/dm³ solution?

Step 1: Convert volume: $500 \text{ cm}^3 = 0.5 \text{ dm}^3$500 cm3=0.5 dm3

Step 2: Calculate mass: $m = c \times V = 30 \times 0.5 = 15 \text{ g}$m=c×V=30×0.5=15 g

Example 3 — Finding Volume

Question: What volume of a 40 g/dm³ solution contains 10 g of solute?

$V = \frac{m}{c} = \frac{10}{40} = 0.25 \text{ dm}^3 = 250 \text{ cm}^3$V=cm​=4010​=0.25 dm3=250 cm3

Example 4 — Larger Volume

Question: 80 g of copper sulfate is dissolved in 2.0 dm³ of solution. What is the concentration?

$c = \frac{80}{2.0} = 40 \text{ g/dm}^3$c=2.080​=40 g/dm3

Example 5 — Small Volume

Question: What is the concentration if 2.0 g of NaOH is dissolved in 50 cm³?

$V = \frac{50}{1000} = 0.05 \text{ dm}^3$V=100050​=0.05 dm3

$c = \frac{2.0}{0.05} = 40 \text{ g/dm}^3$c=0.052.0​=40 g/dm3

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Dilution

When you add more water (solvent) to a solution:

• The mass of solute stays the same (you haven't added or removed any solute).
• The volume increases.
• Therefore the concentration decreases.

Worked Example — Dilution

Question: 100 cm³ of a 20 g/dm³ NaCl solution is diluted to 400 cm³. What is the new concentration?

Step 1: Find the mass of solute in the original solution: $m = c \times V = 20 \times 0.1 = 2.0 \text{ g}$m=c×V=20×0.1=2.0 g

Step 2: Calculate the new concentration with the new volume: $c = \frac{2.0}{0.4} = 5.0 \text{ g/dm}^3$c=0.42.0​=5.0 g/dm3

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Comparing Solutions

Solution	Concentration	More or Less Concentrated?
5 g in 100 cm³	50 g/dm³	More concentrated
5 g in 500 cm³	10 g/dm³	Less concentrated
5 g in 1000 cm³	5 g/dm³	Least concentrated

The same mass of solute in a smaller volume gives a higher concentration.

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Common Mistakes to Avoid

Mistake	Correction
Not converting cm³ to dm³	Always divide by 1000: $250 \text{ cm}^3 = 0.25 \text{ dm}^3$250 cm3=0.25 dm3
Confusing dm³ and cm³	1 dm³ = 1000 cm³ = 1 litre
Using the wrong triangle formula	Use the concentration triangle: $c = m \div V$c=m÷V
Forgetting units in the answer	Always state g/dm³ (or the unit asked for)

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Summary

• Concentration (g/dm³) = mass of solute (g) ÷ volume of solution (dm³).
• Always convert cm³ to dm³ by dividing by 1000.
• Use the concentration triangle to rearrange for mass or volume.
• Diluting a solution decreases the concentration but does not change the mass of solute.
• This topic is tested on both Foundation and Higher tier AQA papers.

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Extended Worked Examples — g/dm³ in Depth

Concentration in g/dm³ underpins every solutions question in the Trilogy specification. These extended examples tie g/dm³ together with $M_r$Mr​, conservation of mass, and reacting mass logic.

Worked Example 6 — Preparing a Standard Solution

Question: A technician must prepare 250 cm³ of a 20 g/dm³ sodium chloride solution. What mass of NaCl must be weighed?

Step 1 — volume in dm³. $V = 250 / 1000 = 0.25$V=250/1000=0.25 dm³.

Step 2 — mass. $m = c \times V = 20 \times 0.25 = 5.0$m=c×V=20×0.25=5.0 g.

Worked Example 7 — Finding the Volume Needed

Question: What volume of a 50 g/dm³ sugar solution contains 12.5 g of sugar?

$V = \frac{m}{c} = \frac{12.5}{50} = 0.25 \text{ dm}^3 = 250 \text{ cm}^3$V=cm​=5012.5​=0.25 dm3=250 cm3

Worked Example 8 — Small Volume Laboratory Problem

Question: A pipette delivers 25.0 cm³ of a 4.0 g/dm³ copper sulfate solution. What mass of copper sulfate is in the pipette?

$V = 25.0 / 1000 = 0.025$V=25.0/1000=0.025 dm³.

$m = c \times V = 4.0 \times 0.025 = 0.10 \text{ g}$m=c×V=4.0×0.025=0.10 g

Worked Example 9 — Dilution Factor

Question: 50 cm³ of a 60 g/dm³ solution is diluted to 300 cm³. What is the new concentration?

Step 1 — mass of solute in original solution. $m = 60 \times 0.050 = 3.0$m=60×0.050=3.0 g.

Step 2 — new concentration. $c = 3.0 / 0.300 = 10$c=3.0/0.300=10 g/dm³.

Shortcut: the volume increases by a factor of 6, so the concentration falls by a factor of 6 (60 → 10 g/dm³). Dilution does not change the total mass of solute.

Worked Example 10 — Comparing Two Solutions

Solution	Mass of solute	Volume	Concentration
A	2.0 g	250 cm³	$2.0 / 0.25 = 8$2.0/0.25=8 g/dm³
B	4.5 g	500 cm³	$4.5 / 0.50 = 9$4.5/0.50=9 g/dm³

Solution B is slightly more concentrated, even though it contains more water in absolute terms, because it has proportionally more solute per dm³.