Question 18M.3.sl.TZ1.b.ii
| Date | May 2018 | Marks available | [Maximum mark: 2] | Reference code | 18M.3.sl.TZ1.b.ii |
| Level | sl | Paper | 3 | Time zone | TZ1 |
| Command term | Calculate | Question number | b.ii | Adapted from | N/A |
Palmitic acid has a molar mass of 256.5 g mol−1.
The apparatus in the diagram measures the surface pressure created by palmitic acid molecules on the surface of water. This pressure is caused by palmitic acid molecules colliding with the fixed barrier. The pressure increases as the area, A, available to the palmitic acid is reduced by the movable barrier.
When a drop of a solution of palmitic acid in a volatile solvent is placed between the barriers, the solvent evaporates leaving a surface layer. The graph of pressure against area was obtained as the area A was reduced.
The solution of palmitic acid had a concentration of 0.0034 mol dm−3. Calculate the number of molecules of palmitic acid present in the 0.050 cm3 drop, using section 2 of the data booklet.
[2]
amount of acid = «5.0 × 10–5 dm3 × 0.0034 mol dm–3» = 1.7 × 10–7 «mol»
number of molecules = «1.7 × 10–7 mol × 6.02 × 1023 mol–1 =» 1.0 × 1017
Award [2] for correct final answer.
Award [1] for “1.0 × 1020”.
[2 marks]
