1. Moles and solutions By the end of section you should be able to… Calculate the amount of substance in mol, using solution volume and concentration Describe a solution’s concentration using the terms concentration and dilute.

2. The concentration of a solution is the amount of solute, in mol, dissolve per 1 dm 3 (1000 cm 3 ) of solution. The amount, in moles, in a solution is: n = c x V(in dm 3 ) or n = c x V(in cm 3 ) 1000

3. Worked examples 1. What is the amount in moles of sodium hydroxide in 50 cm 3 of a solution of concentration 2 mol/ dm 3 Answer n = c x V (in cm 3 ) 1000 n = 2 x 50 1000 = 0.01 mol

4. 2. If 0.4 mol of sodium carbonate is dissolved to make 200cm 3 of solution, what is its concentration? Answer Re-arrange the equation n = c x V(dm 3 ) c = n V(dm 3 ) c = 0.4 mol 0.20 dm 3 c = 2 mol/dm 3

5. 3. What volume of a solution of hydrochloric acid of concentration 0.5 mol/dm 3 contains 0.15 mol? Give the answer in cm 3 Answer Re-arrange the equation n = c x V(dm 3 ) V(dm 3 ) = n / c = 0.15 / 0.5 = 0.3 dm 3 To convert 0.3dm 3 to cm 3, multiply 1000 0.3 x 1000 = 300 cm 3

6. Standard solution A standard solution is a solution of known concentration. Standard solutions are normally used in titration to determine unknown information about another substance.

7. Mass concentration( g/dm 3 ) Find the amount, in mol,required for 1 dm 3 of the solution.. convert moles to grams, i.e from mass = n x M.

8. Find the mass concentration, in g/dm 3 for the following solutions (a) 0.02 moles of HCl dissolved in 500cm 3 of solution. ( b) 0.25 moles of HNO 3 in 2 dm 3 of solution. Answers (a) n = mole x 1 dm 3 V n = 0.02 x 1 dm3 0.5 dm 3 n = 0.04 mol/dm 3

9. From mass = n x M M(HCl) = 1 + 35.5 = 36.5 mass conc. (g/dm 3 ) = 0.04 x 36.5 = 1.46 g/dm 3

10. (b) 0.25 moles of HNO 3 in 2 dm 3 n = mole x 1 dm 3 V n = 0.25 mol x 1 dm 3 2 n = 0.125 mol/dm 3

11. from mass conc. = n x M M(HNO3) = 1 + 14+ (16x3) = 63 i.e. mass conc. = 0.125 x 63 = 7.88 g/dm3

12. NOW TRY QUESTIONS 1 -3 PAGE 17

13. Dilute solution A dilute solution is a solution with a small amount of solute per dm 3 Normal bench solutions of acids usually have concentrations of 1 mol/dm 3 or 2 mol/dm 3. These are dilute solutions.

14. Preparing dilute solution The simple formula usually used is as follows: M 1 V 1 = M 2 V 2 Where M 1 = original concentration V 1 = original volume M 2 = new concentration V 2 = new volume

15. How can 500 cm 3 of a 0.4 mol/dm 3 solution of sulphuric acid be prepared from a solution of concentration 2 mol/dm 3. Answer 2 x V = 500 x 0.4 volume = 500 x 0.4 2 = 100 cm 3 therefore, 100 cm 3 of the original solution be further diluted by adding 400 cm 3 of water to make 500 cm 3 of 0.4 mol/dm 3

16. What is the concentration in mol/dm 3 of 375 cm3 of dilute sodium hydroxide solution prepared by diluting 75 cm 3 of sodium hydroxide solution of concentration 0.6 mol/dm 3 with water? Answer From M 1 V 1 = M 2 V 2 75 x 0.6 = 375 x M 2 M 2 = 75 x 0.6 375 concentration, M 2 = 0.12 mol/dm 3