DP Chemistry (last assessment 2024)
Question 23M.3.SL.TZ2.8
| Date | May 2023 | Marks available | [Maximum mark: 4] | Reference code | 23M.3.SL.TZ2.8 |
| Level | SL | Paper | 3 | Time zone | TZ2 |
| Command term | Calculate, Identify | Question number | 8 | Adapted from | N/A |
8.
[Maximum mark: 4]
23M.3.SL.TZ2.8
(a)
Identify the type of bond and by-product when monosaccharides combine.
Bond: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
By-product: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
[2]
Markscheme
Bond:
glycosidic ✓
By-product:
water/H2O ✓
Accept “ether/C-O-C”
OR
“covalent/polar covalent” for M1.
(b)
Calculate the energy produced from the combustion of 15.00 g of sucrose, C12H22O11.
ΔHc = −5640 kJ mol−1
[2]
Markscheme
«15.00 g / 342.34 g kJ mol−1 =» 0.04382 «mol» ✓
«0.04382 mol × 5640 kJ mol−1 = » 247.1 «kJ» ✓
Award [2] for correct final answer.
Accept “−247.1 «kJ»” for M2.
