1. Moles and gas volumes At the end of this section you should be able to calculate the amount of substance in moles, using gas volume.

2. Avogadro’s hypothesis Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

3. At room temperature and pressure (RTP): One mole of gas molecules occupies approximately 24.0dm 3 (24000cm 3 ) The volume per mole of gas molecules is 24.0 dm 3 mol -1

4. To work out amount, in moles, in a solution, use the equations: n = V(in dm 3 ) 24.0 n = V(in dm 3 ) 24000 Hence, V= n x 24000cm 3

5. Worked examples What amount, in mol, of gas molecules are in the following gas volumes at RTP? (i) 36 dm 3 (ii) 250cm 3

6. (i)n = 36 24.0 = 1.5mol (ii) n = 250 24000 = 0.01mol

7. What is the volume of the following at RTP? (i) 2 mol SO 2 (ii) 0.15 mol H 2

8. (i) V = 2 x 24000 cm 3 = 48000 cm 3 (ii) V = 0.15 x 24000 cm 3 = 3600 cm 3

9. What is the mass of the following at RTP? (i) 0.6 dm 3 N 2 (ii) 1950 cm 3 C 2 H 4 NB: mass = mole x molecular mass

10. (i)Mole = 0.6 24.0 = 0.025 mol from, mass = mole x M M(N 2 ) = 14 x 2 = 28 mass = 0.025 x 28 = 0.7g

11. (ii) Mole = 1950 24000 = 0.0813mol mass = mole x M M(C 2 H 4 ) = (12 x 2) + 1x4 = 28 hence, mass = 0.0813 x 28 = 2.275g

12. What is the volume of the following at RTP? (i) 1.282 g SO 2 (ii) 2.5 g CO 2 NB: For this question, you need to calculate the moles and then follow it up with volume

13. (i)Mole = mass/molecular mass M(SO 2 ) = 32 + 16x2 = 64 mole = 1.282 / 64 = 0.02 mol from, V = n x 24000 cm 3 = 0.02 x 24000 = 480.75cm 3

14. Now try question 1 – 3 Page 15 Question 8 (a) and (b) Page 35