The concentration of a solution can be measured in two different ways – in grams per dm³ (g/dm³) or in moles per dm³ (mol/dm³). Concentrations in mol/dm³ tell us the amount of solute (in moles) dissolved in every dm³ of solvent. For example, a sodium chloride solution with a concentration of 5 mol/dm³ means that in every dm³ of water there are five moles of dissolved sodium chloride. We can calculate the concentration of a solution by dividing the moles by the volume. For example, if I have 2 moles of magnesium sulfate and dissolve it in 500 dm³ of water, the concentration of the solution formed would be 2/500 = 0.004 mol/dm³.

100 grams of potassium chloride (KCl) is dissolved in 200 dm³ of water. Give the concentration of the resulting solution in mol/dm³.
Mr of potassium = 39, Mr of chlorine = 35.5

• First work out the moles of potassium chloride by dividing the mass by the Mr
• 100 / 74.5 = 1.34 mol
• Then find out the concentration of the solution by dividing the moles by the volume
• 1.34 / 200 = 0.0067 mol/dm³

A 1 mol/dm³ sodium chloride solution with a volume of 500 cm³ completely reacts with 350 cm³ of silver nitrate solution of unknown concentration. A precipitate of silver chloride and aqueous sodium nitrate are formed. Calculate the concentration of the silver nitrate solution.

• Concentration x volume of sodium chloride solution = concentration x volume of silver nitrate solution
• 1 x 500 = ? x 350
• 500 / 350 = 1.43
• Concentration of silver nitrate solution = 1.43 mol/dm³

One mole of a gas always occupies a volume of 24 dm³ (or 24,000 cm³). This is true of all gases when they are at room temperature (20^oC) and standard pressure (1 atm). Make sure you memorise this molar gas volume because they won’t give you it in the exam!

This means that whenever we know the amount of gas (i.e. the moles), or we can calculate the moles from its mass, then we also know its volume. To find the volume in dm³, all we need to do is multiply the number of moles by 24. Remember that to convert between dm³ to cm³, multiply your answer by 1000.

Calculate the volume of 7.52 g of ammonia (NH₃) gas at room temperature and pressure.

• Moles = mass / Mr
• 7.52 / 17 = 0.44 mol
• Volume = moles x 24
• 0.44 x 24 = 10.56 dm³

3.25 g of zinc is reacted with an excess of hydrochloric acid to form zinc chloride and hydrogen gas, as shown in the equation below. Calculate the volume of hydrogen gas (in cm³) formed in this reaction.

• We don’t know how many moles of hydrogen we have, but we do have the mass of zinc so we can work out the moles of that and use our balanced equation to work out the moles of hydrogen.
• Moles of zinc = 3.25 / 65 = 0.05 mol
• Using our balanced equation, we can see that there is a 1 : 1 ratio between zinc and hydrogen. This means that if there are 0.05 moles of zinc reacting, 0.05 moles of hydrogen are formed.
• Volume of hydrogen = moles x 24
• 0.05 x 24 = 1.2 dm³
• 1.2 x 1000 = 1200 cm³