DP Chemistry (last assessment 2024)
Question 21M.2.hl.TZ2.c(ii)
| Date | May 2021 | Marks available | [Maximum mark: 2] | Reference code | 21M.2.hl.TZ2.c(ii) |
| Level | hl | Paper | 2 | Time zone | TZ2 |
| Command term | Determine | Question number | c(ii) | Adapted from | N/A |
c(ii).
[Maximum mark: 2]
21M.2.hl.TZ2.c(ii)
Limestone can be converted into a variety of useful commercial products through the lime cycle. Limestone contains high percentages of calcium carbonate, CaCO3.
The second step of the lime cycle produces calcium hydroxide, Ca(OH)2.
Determine the volume, in dm3, of 0.015 mol dm−3 calcium hydroxide solution needed to neutralize 35.0 cm3 of 0.025 mol dm−3 HCl (aq).
[2]
Markscheme
«nHCl = 0.0350 dm3 × 0.025 mol dm−3 =» 0.00088 «mol»
OR
nCa(OH)2 = nHCl/0.00044 «mol» ✓
«V = =» 0.029 «dm3» ✓
Award [2] for correct final answer.
Award [1 max] for 0.058 «dm3».
