1. Moles and Solutions g n gfm To calculate the number of moles in a solution we use the following n CV n = number of moles C = concentatration (mol/l) V = volume in litres

2. Moles and Solutions Calculate the no. of moles of sodium hydroxide present in 2 litres of 1 mol/l solution. From previous triangle we see that if we cover the letter we want that n = C × V therefore n = 1 × 2 = 2 moles Ex.1 Ex.2 Calculate the no. of moles of lithium nitrate present in 250cm 3 of 2 mol/l solution. n = C × V therefore n = 2 × (250/1000) = 0.5 moles

3. Moles and Solutions Calculate the concentration of a solution with 3 moles of hydrochloric acid dissolved in 2 litres. C = n / V therefore C = 3 / 2 = 1.5 mol/l Ex.3 Ex.4Calculate the concentration of a solution with 222g of calcium chloride dissolved in 4 litres. First we need to find out how many moles of calcium chloride we have. G.F.M of CaCl 2 = 111g n = 222 / 111 = 2 moles therefore C = n / V therefore C = 2 / 4 = 0.5 mol/l n = g / G.F.M

4. Moles and Solutions Calculate the volume of solution that can be produced when 6 moles of sulphuric acid is used to prepare a 2 mol/l solution. V = n / C therefore V = 6 / 2 = 3 litres Ex.5 Ex.6Calculate the volume of solution that can be produced when 80g of sodium hydroxide is used to prepare a 4 mol/l solution. First we need to find out how many moles of sodium hydroxide we have. G.F.M of NaOH = 40g n = 80 / 40 = 2 moles therefore n = g / G.F.M V = n / C therefore V = 2 / 4 = 0.5 litres