How to calculate concentration in g/dm³ and mol/dm³

Concentration questions run right through the acids, bases and salts chapter and every titration that follows. SPM uses two units for concentration, g dm⁻³ and mol dm⁻³, and the single most common error is not the chemistry at all, it is forgetting that volumes are usually given in cm³ but concentration is measured per dm³. This guide sets out both units, the conversion between them, and the volume trap, with each type worked through.

The two units, side by side

Concentration is amount of solute divided by volume of solution. The two units differ only in how they measure “amount”:

• Concentration in g dm⁻³ measures the solute in grams: concentration = mass of solute (g) ÷ volume of solution (dm³).
• Concentration in mol dm⁻³, also called molarity, measures the solute in moles: concentration = moles of solute ÷ volume of solution (dm³).

Both divide by the volume in dm³, which is where students slip.

The volume trap: cm³ to dm³

A litre and a cubic decimetre are the same thing, and 1 dm³ = 1000 cm³. Laboratory volumes are usually quoted in cm³ (25.0 cm³ of acid, 500 cm³ of solution), so before you divide you must convert:

volume in dm³ = volume in cm³ ÷ 1000

Miss this and every concentration comes out a thousand times too small or too large. Make converting the volume your automatic first move.

Worked example one: concentration in g dm⁻³

20 g of sodium hydroxide is dissolved in water to make 500 cm³ of solution. Find the concentration in g dm⁻³.

• Convert the volume: 500 cm³ ÷ 1000 = 0.5 dm³.
• Concentration = mass ÷ volume = 20 ÷ 0.5 = 40 g dm⁻³.

Worked example two: concentration in mol dm⁻³

Find the concentration of the same solution in mol dm⁻³. (Molar mass of NaOH = 40 g mol⁻¹)

• Moles of NaOH = mass ÷ molar mass = 20 ÷ 40 = 0.5 mol.
• Concentration = moles ÷ volume = 0.5 ÷ 0.5 = 1.0 mol dm⁻³.

Notice both answers describe the same solution, 40 g dm⁻³ and 1.0 mol dm⁻³ are two ways of saying the same thing. That is exactly why you can convert between them.

Converting directly between the two units

You do not always need to go back to mass and moles. The two concentrations are linked by the molar mass:

concentration (g dm⁻³) = concentration (mol dm⁻³) × molar mass

and, rearranged,

concentration (mol dm⁻³) = concentration (g dm⁻³) ÷ molar mass

Check it against the NaOH result: 1.0 mol dm⁻³ × 40 g mol⁻¹ = 40 g dm⁻³. Consistent. Practise this switch at converting concentration units.

Worked example three: mol dm⁻³ to g dm⁻³

A sulfuric acid solution has a concentration of 0.1 mol dm⁻³. Express this in g dm⁻³. (Molar mass of H₂SO₄ = 2 + 32 + 64 = 98 g mol⁻¹)

• Concentration (g dm⁻³) = 0.1 × 98 = 9.8 g dm⁻³.

Worked example four: g dm⁻³ to mol dm⁻³

A sodium chloride solution has a concentration of 11.7 g dm⁻³. Express this in mol dm⁻³. (Molar mass of NaCl = 23 + 35.5 = 58.5 g mol⁻¹)

• Concentration (mol dm⁻³) = 11.7 ÷ 58.5 = 0.2 mol dm⁻³.

These two examples are mirror images: multiply by the molar mass in one direction, divide by it in the other. If you can see which way the molar mass goes, you never need to memorise both formulae separately.

A word on dilution

Changing a solution’s concentration by adding water is a related but separate skill, using the relationship that moles of solute stay constant as volume increases. It is its own topic, and you can rehearse it with our molarity and dilution calculator once the two units here feel solid.

Protect your marks

Three habits keep concentration work clean. First, convert the volume to dm³ before anything else, write ”÷ 1000” as your first line so it is never skipped. Second, decide which unit the question wants and pick the matching formula; a g dm⁻³ answer needs mass, a mol dm⁻³ answer needs moles. Third, carry the unit through every line so an answer in the wrong unit is caught before you box it. Lay the working out clearly and method marks follow even if one number slips.

Concentration is the foundation of titration calculations, so time spent here pays off across the whole quantitative side of the syllabus. Drill a mixed set at molarity and concentration. And if you would like a teacher to watch your volume conversions and molar masses live and correct the exact step you get wrong, our online one-to-one lessons with our experienced SPM Chemistry teachers are built for that, from RM50 an hour with a paid one-hour trial lesson.