Concentration in mol/dm³ (Higher Tier)

This lesson covers how to calculate concentration in mol/dm³ and convert between g/dm³ and mol/dm³, as required by the AQA GCSE Combined Science Trilogy specification (8464) at Higher tier. This builds on the g/dm³ concentration work from the previous lesson.

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What Is Concentration in mol/dm³?

While g/dm³ tells you the mass of solute per unit volume, mol/dm³ tells you the number of moles of solute per unit volume. This is more useful for chemists because reactions depend on the number of particles, not their mass.

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

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The Two Concentration Triangles

Triangle 1: mol/dm³

graph TD
    A["<b>moles of solute</b>"] --- B["<b>concentration (mol/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 (mol/dm³)	$c = \dfrac{n}{V}$c=Vn​
Moles	$n = c \times V$n=c×V
Volume (dm³)	$V = \dfrac{n}{c}$V=cn​

Triangle 2: g/dm³ (reminder)

To Find	Formula
Concentration (g/dm³)	$c = \dfrac{m}{V}$c=Vm​
Mass (g)	$m = c \times V$m=c×V
Volume (dm³)	$V = \dfrac{m}{c}$V=cm​

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Converting Between g/dm³ and mol/dm³

$\text{concentration (g/dm}^3) = \text{concentration (mol/dm}^3) \times M_r$concentration (g/dm3)=concentration (mol/dm3)×Mr​

$\text{concentration (mol/dm}^3) = \frac{\text{concentration (g/dm}^3)}{M_r}$concentration (mol/dm3)=Mr​concentration (g/dm3)​

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Worked Examples

Example 1 — Finding Concentration in mol/dm³

Question: 0.4 mol of NaOH is dissolved in 2.0 dm³ of solution. What is the concentration in mol/dm³?

$c = \frac{n}{V} = \frac{0.4}{2.0} = 0.2 \text{ mol/dm}^3$c=Vn​=2.00.4​=0.2 mol/dm3

Example 2 — Finding Moles from Concentration

Question: How many moles of HCl are in 250 cm³ of 2.0 mol/dm³ solution?

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

$n = c \times V = 2.0 \times 0.25 = 0.5 \text{ mol}$n=c×V=2.0×0.25=0.5 mol

Example 3 — Converting g/dm³ to mol/dm³

Question: A NaOH solution has a concentration of 8.0 g/dm³. What is this in mol/dm³?

$M_r$Mr​ of NaOH = 23 + 16 + 1 = 40

$c \text{ (mol/dm}^3) = \frac{8.0}{40} = 0.2 \text{ mol/dm}^3$c (mol/dm3)=408.0​=0.2 mol/dm3

Example 4 — Converting mol/dm³ to g/dm³

Question: A solution of $\text{H}_2\text{SO}_4$H2​SO4​ has a concentration of 0.5 mol/dm³. What is this in g/dm³?

$M_r$Mr​ of $\text{H}_2\text{SO}_4 = (2 \times 1) + 32 + (4 \times 16) = 98$H2​SO4​=(2×1)+32+(4×16)=98

$c \text{ (g/dm}^3) = 0.5 \times 98 = 49 \text{ g/dm}^3$c (g/dm3)=0.5×98=49 g/dm3

Example 5 — Finding Mass from mol/dm³ and Volume

Question: What mass of NaCl is dissolved in 200 cm³ of a 0.5 mol/dm³ solution?

$M_r$Mr​ of NaCl = 23 + 35.5 = 58.5

Step 1: Find moles: $n = c \times V = 0.5 \times 0.2 = 0.1 \text{ mol}$n=c×V=0.5×0.2=0.1 mol

Step 2: Find mass: $m = n \times M_r = 0.1 \times 58.5 = 5.85 \text{ g}$m=n×Mr​=0.1×58.5=5.85 g

Example 6 — Finding Volume

Question: What volume of 0.1 mol/dm³ HCl contains 0.025 mol?

$V = \frac{n}{c} = \frac{0.025}{0.1} = 0.25 \text{ dm}^3 = 250 \text{ cm}^3$V=cn​=0.10.025​=0.25 dm3=250 cm3

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Linking to Titrations and Reacting Masses

In many exam questions, you will combine concentration calculations with reacting mass calculations:

1. Use $n = c \times V$n=c×V to find moles of one reactant.
2. Use the mole ratio from the balanced equation.
3. Use $m = n \times M_r$m=n×Mr​ to find the mass or $c = n \div V$c=n÷V to find a concentration.

Worked Example — Combining Skills

Question: 25.0 cm³ of 0.1 mol/dm³ NaOH reacts with excess $\text{HCl}$HCl. What mass of NaCl is produced?

$\text{NaOH}(aq) + \text{HCl}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$NaOH(aq)+HCl(aq)→NaCl(aq)+H2​O(l)

Step 1: Moles of NaOH: $n = 0.1 \times 0.025 = 0.0025 \text{ mol}$n=0.1×0.025=0.0025 mol

Step 2: Ratio NaOH : NaCl = 1 : 1, so moles of NaCl = 0.0025 mol.

Step 3: $M_r$Mr​ of NaCl = 58.5 $m = 0.0025 \times 58.5 = 0.146 \text{ g}$m=0.0025×58.5=0.146 g

Exam Tip (AQA 8464): Higher tier questions often combine concentration in mol/dm³ with reacting mass calculations. Practise linking these skills together.

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Common Mistakes to Avoid

Mistake	Correction
Confusing g/dm³ and mol/dm³	g/dm³ uses mass; mol/dm³ uses moles — check which is asked for
Forgetting to convert cm³ to dm³	Divide by 1000 every time
Multiplying instead of dividing (or vice versa) when converting	g/dm³ → mol/dm³: divide by $M_r$Mr​; mol/dm³ → g/dm³: multiply by $M_r$Mr​
Forgetting to calculate $M_r$Mr​ first	Always find $M_r$Mr​ before converting between concentration units

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Summary

• Concentration in mol/dm³ = moles ÷ volume (dm³).
• Convert g/dm³ to mol/dm³ by dividing by $M_r$Mr​.
• Convert mol/dm³ to g/dm³ by multiplying by $M_r$Mr​.
• Always convert cm³ to dm³ first.
• This topic is Higher tier only on the AQA Trilogy specification (8464).

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Extended Worked Examples — mol/dm³ in Context

The mol/dm³ unit is more fundamental to chemistry than g/dm³ because reactions occur in mole ratios. The examples below combine concentration, $M_r$Mr​, moles, and balanced equations to handle the kinds of questions seen on AQA Higher Paper 1.

Worked Example 7 — Standard Solution Preparation

Question: A chemist prepares 500 cm³ of a 0.2 mol/dm³ sodium carbonate solution. What mass of $\text{Na}_2\text{CO}_3$Na2​CO3​ must be weighed?

$M_r$Mr​ of $\text{Na}_2\text{CO}_3 = 106$Na2​CO3​=106; $V = 0.500$V=0.500 dm³.

Step 1 — moles required. $n = c \times V = 0.2 \times 0.5 = 0.1$n=c×V=0.2×0.5=0.1 mol.

Step 2 — mass. $m = n \times M_r = 0.1 \times 106 = 10.6$m=n×Mr​=0.1×106=10.6 g.

Worked Example 8 — Moles in a Sample

Question: How many moles of HCl are in 50 cm³ of a 0.5 mol/dm³ solution?

$V = 0.050$V=0.050 dm³. $n = 0.5 \times 0.050 = 0.025$n=0.5×0.050=0.025 mol.

Worked Example 9 — Mass from mol/dm³

Question: What mass of NaOH is in 250 cm³ of 0.4 mol/dm³ solution?

$M_r$Mr​ of NaOH = 40; $V = 0.250$V=0.250 dm³.

$n = 0.4 \times 0.250 = 0.1$n=0.4×0.250=0.1 mol. $m = 0.1 \times 40 = 4.0$m=0.1×40=4.0 g.

Worked Example 10 — Converting g/dm³ → mol/dm³