To convert molarity to concentration, multiply by the molar mass: concentration (g dm−3) = molarity (mol dm−3) x molar mass (g mol−1). To convert the other way, divide the concentration in g dm−3 by the molar mass.
Concentration is quoted two ways in SPM Chemistry, molarity in mol dm−3 and concentration in g dm−3, and questions often give you one but ask for the other. Our online one-to-one teachers make this a single, reliable step so you never lose marks simply because the data and the answer are in different units.
When you use it
Use this whenever a question mixes the two concentration units: a bottle labelled in g dm−3 but a calculation that needs mol dm−3, or a molarity you must express as a mass per cubic decimetre. It is the small conversion that connects the “molarity” and “concentration” ideas.
The formula
concentration (g dm−3) = molarity (mol dm−3) x molar mass (g mol⁻¹)
Rearranged the other way:
molarity (mol dm−3) = concentration (g dm−3) / molar mass (g mol⁻¹)
The only extra information you ever need is the molar mass of the solute, which you build from the formula.
Units
Molarity is in mol dm−3, concentration in g dm−3, and molar mass in g mol⁻¹. Notice how the units cancel: (mol dm−3) x (g mol⁻¹) = g dm−3, which is a quick way to check you have multiplied rather than divided.
Why the molar mass is the bridge
The two units describe the same solution, so the only thing that changes is whether you count the solute in moles or in grams. The molar mass is the exchange rate between moles and grams, which is why it is the single number that carries you across. If you ever feel unsure of the direction, write the units under the numbers and see which arrangement leaves you with the unit the question wants. That habit turns a step people often reverse into one you can trust, and it costs only a few seconds in the exam.
Worked example 1 (easy), sodium hydroxide
Convert 0.5 mol dm−3 sodium hydroxide, NaOH, to g dm−3. (M of NaOH = 40 g mol⁻¹.)
- Step 1, concentration = molarity x molar mass = 0.5 x 40.
- Step 2, concentration = 20 g dm−3.
Worked example 2 (medium), sulfuric acid
Convert 9.8 g dm−3 sulfuric acid, H2SO4, to mol dm−3. (M of H2SO4 = 98 g mol⁻¹.)
- Step 1, molarity = concentration / molar mass = 9.8 / 98.
- Step 2, molarity = 0.1 mol dm−3.
Worked example 3 (SPM level), calcium hydroxide with a bracket
Convert 2 mol dm−3 calcium hydroxide, Ca(OH)2, to g dm−3. (Ar: Ca = 40, O = 16, H = 1.)
- Step 1, molar mass: M(Ca(OH)2) = 40 + 2(16 + 1) = 40 + 34 = 74 g mol⁻¹.
- Step 2, concentration = molarity x molar mass = 2 x 74 = 148 g dm−3.
The bracket is the trap here: there are two oxygen and two hydrogen atoms, giving 74, not 57.
Common traps
- Multiplying when you should divide, or vice versa. Use the unit check: the units must come out as g dm−3 or mol dm−3.
- Getting the molar mass wrong, especially with brackets such as Ca(OH)2.
- Forgetting that this conversion does not involve volume at all, it is per cubic decimetre on both sides.
- Dropping the units from the final answer.
- Rounding the molar mass and shifting the answer.
How we help
Our teachers pair this conversion with the molarity and dilution formulae, and with bracketed formulae like Ca(OH)2, so the molar-mass step never catches you out. Lessons are one-to-one, taught in English, from RM50 per hour, with a paid one-hour trial so you can test the approach first. Once you can switch units confidently, titration and standard-solution questions read much more smoothly, because you can accept the data in whichever unit the examiner chooses.
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