Concentration and molarity of solutions

This page covers one Form 4 content standard: concentration and molarity of solutions. It turns the qualitative idea of a “strong” or “dilute” solution into a precise number, which you then use in every titration and dilution calculation later in the chapter. Two units are examined, grams per cubic decimetre and moles per cubic decimetre, so the skill is knowing both and converting cleanly between them.

What concentration means

The concentration of a solution is the quantity of solute dissolved in a given volume of solution. There are two ways to state it in SPM Chemistry (code 4541):

• Concentration in g dm⁻³, the mass of solute (in grams) dissolved in 1 dm³ of solution.
• Molarity (molar concentration) in mol dm⁻³, the number of moles of solute dissolved in 1 dm³ of solution.

The unit dm³ (cubic decimetre) is the same volume as one litre, and 1 dm³ = 1000 cm³. Because most laboratory volumes are measured in cm³, converting cm³ to dm³ (divide by 1000) is the step students most often forget.

The two key formulae

Concentration in grams per cubic decimetre is:

Concentration (g dm⁻³) = mass of solute (g) ÷ volume of solution (dm³)

Molarity is the same idea but counted in moles:

Molarity (mol dm⁻³) = number of moles of solute (mol) ÷ volume of solution (dm³)

Both use the volume of the solution, not the volume of water added, a small but important distinction.

Linking the two units

Because moles and mass are connected by the molar mass, the two concentration units are connected the same way:

Concentration (g dm⁻³) = molarity (mol dm⁻³) × molar mass (g mol⁻¹)

and, rearranged,

Molarity (mol dm⁻³) = concentration (g dm⁻³) ÷ molar mass (g mol⁻¹)

So the recipe for any conversion is: find the molar mass, then multiply (mol → g) or divide (g → mol). This is exactly the mass–mole link from the Mole Concept chapter, applied per cubic decimetre.

Worked example

Question. 4.0 g of sodium hydroxide, NaOH, is dissolved in water to make 250 cm³ of solution. Calculate (a) the concentration in g dm⁻³, (b) the molarity in mol dm⁻³. [Ar: Na = 23, O = 16, H = 1]

Step 1, Convert the volume to dm³. Volume = 250 cm³ ÷ 1000 = 0.25 dm³.

Step 2, Concentration in g dm⁻³. Concentration = mass ÷ volume = 4.0 g ÷ 0.25 dm³ = 16 g dm⁻³.

Step 3, Find the molar mass of NaOH. M = 23 + 16 + 1 = 40 g mol⁻¹.

Step 4, Molarity in mol dm⁻³. Molarity = concentration ÷ molar mass = 16 ÷ 40 = 0.4 mol dm⁻³. (Check: moles = 4.0 ÷ 40 = 0.1 mol; molarity = 0.1 ÷ 0.25 = 0.4 mol dm⁻³.)

Answer. (a) 16 g dm⁻³. (b) 0.4 mol dm⁻³.

Practice question

Question. A solution of copper(II) sulfate, CuSO₄, has a molarity of 0.2 mol dm⁻³. (a) Calculate its concentration in g dm⁻³. (b) What mass of CuSO₄ is present in 500 cm³ of this solution? [Ar: Cu = 64, S = 32, O = 16]

Answer. Molar mass of CuSO₄ = 64 + 32 + (4 × 16) = 160 g mol⁻¹. (a) Concentration = molarity × molar mass = 0.2 × 160 = 32 g dm⁻³. (b) Volume = 500 cm³ = 0.5 dm³; mass = concentration × volume = 32 × 0.5 = 16 g (or: moles = 0.2 × 0.5 = 0.1 mol; mass = 0.1 × 160 = 16 g).

Exam tip

Convert the volume to dm³ before you divide, dividing by a volume in cm³ is the single biggest source of lost marks on this standard, and it makes the answer 1000 times too small. Write the molar mass with its unit, g mol⁻¹, and check that your final unit matches what the question asked (g dm⁻³ or mol dm⁻³). If a question mixes the two units, decide which one it wants first, then convert once at the end rather than switching back and forth mid-calculation.

Choosing the right route

Every question on this standard is a short chain built from three quantities, mass, moles and volume, plus the molar mass. Underline what you are given and what is required, then pick the shortest path. If you are given a mass and a volume and asked for molarity, go mass → moles (divide by molar mass) → molarity (divide by volume in dm³). If you are given a molarity and asked for a mass in a certain volume, go molarity × volume → moles → mass (× molar mass). Laying the chain out in labelled steps means an examiner can award method marks even if one arithmetic step slips.

A common variation gives you a concentration and asks how to make a more dilute or more concentrated solution, which is the bridge to the next standard. For now, fix the definitions: concentration is amount per unit volume, molarity counts that amount in moles, and the molar mass is the only thing you need to move between grams and moles. Everything in titration and dilution is built on these three sentences.

Where this fits

This is content standard 6.3 of the Acids, Bases and Salts chapter. It supplies the units used in standard solutions, dilution and titration, and it reuses the mass–mole skill from the Mole Concept chapter. Practise the conversions in the worked examples and test yourself with the practice questions. In our online 1-to-1 SPM Chemistry lessons, taught in English, from RM50/hr, our teachers drill the cm³-to-dm³ conversion until it is automatic, because that one habit protects the marks in every later calculation in the chapter.

Quick recap

• Concentration = amount of solute per volume of solution.
• Concentration (g dm⁻³) = mass ÷ volume (dm³); molarity (mol dm⁻³) = moles ÷ volume (dm³).
• Convert between units with the molar mass: g dm⁻³ = mol dm⁻³ × molar mass.
• Always change cm³ to dm³ (÷ 1000) before dividing.