GCSE · Chemistry · AQA · Spec 8462

Concentration in mol/dm3 (chem HT)

A glass of squash poured from the jug holds far less drink, yet tastes exactly the same. Add water and it tastes weaker. That difference is concentration.

Chemistry · Concentration

Pour out as much as you like — the label never changes
0 cm³200 cm³400 cm³600 cm³800 cm³1000 cm³drag to take more or less solution →

Volume taken: 5/20. Volume taken 250 cm³. Same volume in dm³ 0.25 dm³. Moles of solute taken 0.50 mol. Concentration of the solution 2.0 mol/dm³

Volume taken250 cm³Same volume in dm³0.25 dm³Moles of solute taken0.50 molConcentration of the solution2.0 mol/dm³

This bottle holds a solution labelled 2.0 mol/dm³. The bar is the full 1 dm³ (1000 cm³, the same as 1 litre). Drag to take more or less of it and watch what changes — and what doesn't.

Exam line: moles of solute = concentration (mol/dm³) × volume (dm³). At exactly 1 dm³, the moles you hold ARE the concentration.
Watch out: A smaller sample holds fewer moles, but it is not a weaker solution. It still came out of the same 2.0 mol/dm³ bottle.

Predict, then check

Careful — this is not the same as pouring some out.

You have a beaker of solution. First you (a) dissolve more solute in it without changing the volume. Then, separately, you (b) add water to a fresh beaker of the same solution, making the volume bigger. What happens to the concentration in (a) and in (b)?

One relationship, three jobs

Tap the quantity you want to find to cover it. What's left tells you whether to multiply or divide.

Tap the quantity you want to find. The triangle shows you the formula.

÷

Cover moles of solute, concentration or volume of solution to reveal its rearranged formula, then plug in numbers to solve.

Two ways to write a concentration

g/dm³vsmol/dm³

Same idea — solute per 1 dm³ — but one weighs the solute and the other counts it.

Focus

What it measures

g/dm³

The mass of solute, in grams, in each 1 dm³ of solution

mol/dm³

The amount of solute, in moles, in each 1 dm³ of solution

The insight

g/dm³ weighs the solute; mol/dm³ counts it. Counting is what chemists need, because reactions happen particle by particle.

Getting from one to the other

g/dm³

Divide by the mass of one mole (the Mr in grams) to get mol/dm³

mol/dm³

Multiply by the mass of one mole (the Mr in grams) to get g/dm³

Example: glucose (Mr 180) and sucrose (Mr 342), both 0.10 mol/dm³

g/dm³

Glucose 18 g/dm³, sucrose 34.2 g/dm³ — different numbers

mol/dm³

Both 0.10 mol/dm³ — the same number

Using it in a reaction calculation

g/dm³

Not directly: turn grams into moles first, using the Mr

mol/dm³

Directly: moles = concentration × volume in dm³

Spot the mistake

Where does this answer go wrong?

What mass of sodium chloride (Mr = 58.5) is dissolved in 200 cm³ of a 0.50 mol/dm³ sodium chloride solution?

A student's answer — which line goes wrong?

Exam line: Before you multiply or divide by a volume, check it is in dm³.

Titration calculation, step by step

Problem

A titration is used to find the concentration of some sulfuric acid. Each time, 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide solution is exactly neutralised by the acid. The volumes of acid needed (the titres) were: rough 21.20 cm³, then 20.15 cm³, 20.05 cm³ and 20.10 cm³. The equation is 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O. Calculate the concentration of the sulfuric acid in mol/dm³.

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

How many moles of solute sit in each 1 dm³ of solution — and how to use that to count moles, find masses and crack titration results.

What you need to know

  • Concentration in mol/dm³ is the number of moles of solute in each 1 dm³ of solution.
  • 1 dm³ = 1000 cm³, so divide a volume in cm³ by 1000 to get dm³.
  • moles of solute = concentration × volume (dm³), and concentration = moles ÷ volume (dm³).
  • The mass of one mole of a substance is its relative formula mass (Mr) in grams, so mass = moles × Mr.
  • Concentration can also be given in g/dm³: the mass of solute in each 1 dm³. Divide by Mr to turn g/dm³ into mol/dm³.
  • In a titration, take the mean of the concordant titres, use the balanced equation's mole ratio, then find the unknown concentration.

The big picture

Concentration in mol/dm³ tells you how many moles of solute are dissolved in each 1 dm³ of solution. Because it is a ratio, pouring out a smaller sample gives you fewer moles but the same concentration, while adding solute or adding water changes it. Moles = concentration × volume in dm³ (so cm³ must be divided by 1000 first), the Mr in grams converts moles to mass, and in a titration you use the concordant mean titre and the balanced equation's mole ratio to find an unknown concentration.

Key points

1A concentration depends on both the amount of solute and the volume of solution it is dissolved in.
2Taking a smaller sample of a solution does not change its concentration; it only reduces the moles you have.
3Adding solute to the same volume raises the concentration; adding water to the same solute lowers it.
4Equal volumes of solutions with the same concentration in mol/dm³ contain equal numbers of solute particles.
5The coefficients in a balanced equation give the mole ratio in which substances react.
6Concordant titres are results that agree closely; only these are averaged for the mean titre.

Worked example

Problem

4.0 g of sodium hydroxide (Mr = 40) is dissolved in water and the solution is made up to 250 cm³. Calculate the concentration of the solution in mol/dm³.

⚠ Watch out

Putting a volume in cm³ straight into moles = concentration × volume. The volume must be in dm³ (divide cm³ by 1000), or every answer comes out 1000 times too big or too small.

🧠

Memory hook

The label is 'per 1 dm³'. Pour out less and you get fewer moles, but the label stays the same. Only adding solute or adding water rewrites the label.

✓

Check yourself

Try it: 0.15 mol of solute is dissolved to make 600 cm³ of solution. What is the concentration? (0.25 mol/dm³: 600 cm³ = 0.600 dm³, and 0.15 ÷ 0.600 = 0.25.)

Flashcards

(13)
What does a concentration of 1.5 mol/dm³ mean?
There are 1.5 moles of solute dissolved in every 1 dm³ of the solution.
What does a concentration in g/dm³ measure?
The mass of solute, in grams, dissolved in each 1 dm³ of solution.
How many cm³ are in 1 dm³?
1000 cm³. So divide a volume in cm³ by 1000 to get dm³ (for example, 25.0 cm³ = 0.0250 dm³).
How do you work out the moles of solute in a solution?
moles = concentration (mol/dm³) × volume (dm³).
What is the mass of one mole of a substance?
Its relative formula mass (Mr) in grams. For example, one mole of water (Mr 18) has a mass of 18 g.
You know the moles of solute. How do you find its mass?
mass (g) = moles × Mr.
How do you turn a concentration in g/dm³ into mol/dm³?
Divide by the Mr of the solute.
You pour 10 cm³ out of a bottle of 0.5 mol/dm³ solution. What is the concentration of your 10 cm³?
Still 0.5 mol/dm³. You have fewer moles, but the ratio of solute to solution is unchanged.
What happens to the concentration if you add water to a solution? Why?
It goes down. The moles of solute stay the same but they are spread through a larger volume.
Two solutions have the same volume and the same concentration in mol/dm³. What else do they have in common?
They contain equal numbers of dissolved solute particles, even if the solutes have different masses.
What are concordant titres?
Titration results that agree closely with each other. Only these are used to calculate the mean titre.
Where does the mole ratio for a titration calculation come from?
The coefficients (the big numbers in front of each formula) in the balanced equation.
What is the last step in finding an unknown concentration from a titration?
Divide the unknown solution's moles by its volume in dm³.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

Learning with Lightbulb is opening soon

You can use this lesson now. Join the waitlist and we'll let you know when the full Lightbulb experience is ready.

Keep me posted

More AQA GCSE Chemistry topics

See the full AQA Chemistry curriculum →

How this lesson was checked. This AQA GCSE Chemistry (specification 8462)lesson was published through Lightbulb Learning's human-designed editorial process — the educational standards, accuracy rules and publication checks it must pass were authored and approved by Philip Halpin. It passed subject-specific assessment, automated educational checks and technical publication verification before going live (publication checks completed 30 September 2026). Published pages are monitored, human spot-checking is ongoing across the lesson library, and anything found wrong is corrected or withdrawn. How our lessons are made and checked. Spotted a mistake? Email hello@lightbulblearning.co and we'll review it.