Molarity and concentration

Concentration questions run right through the Acids, Bases and Salts chapter and into titration, so they appear reliably in SPM Chemistry. Our online one-to-one teachers make sure you can move between the two ways of stating concentration, molarity and grams per cubic decimetre, and never trip over the cm-cubed-to-dm-cubed conversion.

When you use it

Use molarity whenever a question involves a solution: preparing it, diluting it, or reacting it in a titration. Concentration in g dm⁻³ is the everyday-mass version of the same idea, and you will often convert from one to the other.

The formulae

Molarity (mol dm⁻³): molarity = n / V

• n = moles of solute, in mol
• V = volume of solution, in dm cubed (dm³)

Concentration (g dm⁻³): concentration = mass of solute (g) / V (dm³)

The two are linked because n = mass / molar mass, so molarity x molar mass = concentration in g dm⁻³.

Units

Molarity is in mol dm⁻³; concentration is in g dm⁻³. Volume must be in dm cubed. Convert cm cubed to dm cubed by dividing by 1000 (250 cm³ = 0.25 dm³).

Reading the words carefully

A concentrated solution has more solute in the same volume; a dilute solution has less. Neither word tells you a number on its own, so the exam always gives you either the moles and the volume, or the mass and the volume, and expects you to combine them. Watch for questions that describe a solute dissolved and “made up to” a stated volume: that stated volume is the volume of solution, and it is the one you use, not the volume of water you started with. The phrase “made up to 250 cm³ with distilled water” is a signal that the final volume is 0.25 dm³.

Worked example 1 (easy), sodium chloride

0.5 mol of sodium chloride is dissolved to make 2 dm³ of solution. Find the molarity.

• Step 1, molarity = n / V = 0.5 / 2.
• Step 2, molarity = 0.25 mol dm⁻³.

Worked example 2 (medium), sodium hydroxide from a mass

4 g of sodium hydroxide, NaOH, is dissolved to make 500 cm³ of solution. Find the molarity. (M of NaOH = 40 g mol⁻¹.)

• Step 1, moles: n = m / M = 4 / 40 = 0.1 mol.
• Step 2, convert volume: 500 cm³ = 0.5 dm³.
• Step 3, molarity = n / V = 0.1 / 0.5 = 0.2 mol dm⁻³.

Worked example 3 (SPM level), mass needed for a target solution

What mass of sodium carbonate, Na₂CO₃, is needed to make 250 cm³ of a 0.1 mol dm⁻³ solution? (M of Na₂CO₃ = 106 g mol⁻¹.)

• Step 1, convert volume: 250 cm³ = 0.25 dm³.
• Step 2, moles needed: n = molarity x V = 0.1 x 0.25 = 0.025 mol.
• Step 3, mass: m = n x M = 0.025 x 106 = 2.65 g.

So you weigh out 2.65 g. This “work back from a target concentration” style is exactly what a standard-solution question asks.

Common traps

• Leaving the volume in cm cubed. Always convert to dm cubed first (divide by 1000).
• Mixing up molarity (mol dm⁻³) and concentration (g dm⁻³); check which unit the answer needs.
• Using the volume of water added rather than the total volume of solution.
• Forgetting the mass-to-moles step when the question gives grams.
• Dropping the units, which the marking scheme awards.

How we help

Our teachers give you the full ladder, molarity from moles, molarity from a mass, and mass from a target molarity, so you are ready for the standard-solution and titration questions that build on it. Lessons are one-to-one, taught in English, from RM50 per hour, with a paid one-hour trial so you can see the method first. Secure concentration now and dilution, standard solutions and titration all follow the same clear logic.