| Date | November 2017 | Marks available | 3 | Reference code | 17N.2.hl.TZ0.2 |
| Level | HL | Paper | 2 | Time zone | TZ0 |
| Command term | Determine | Question number | 2 | Adapted from | N/A |
Question
Analytical chemistry uses instruments to separate, identify, and quantify matter.
Menthol is an organic compound containing carbon, hydrogen and oxygen.
Nitric oxide reacts with chlorine.
2NO (g) + Cl2 (g) → 2NOCl (g)
The following experimental data were obtained at 101.3 kPa and 263 K.
Outline how this spectrum is related to the energy levels in the hydrogen atom.
A sample of magnesium has the following isotopic composition.
Calculate the relative atomic mass of magnesium based on this data, giving your answer to two decimal places.
Complete combustion of 0.1595 g of menthol produces 0.4490 g of carbon dioxide and 0.1840 g of water. Determine the empirical formula of the compound showing your working.
0.150 g sample of menthol, when vaporized, had a volume of 0.0337 dm3 at 150 °C and 100.2 kPa. Calculate its molar mass showing your working.
Determine the molecular formula of menthol using your answers from parts (d)(i) and (ii).
Deduce the order of reaction with respect to Cl2 and NO.
State the rate expression for the reaction.
Calculate the value of the rate constant at 263 K.
Markscheme
electron transfer/transition between high«er» energy level to low«er» energy level
OR
electron transitions into first energy level causes UV series
OR
transition into second energy level causes visible series
OR
transition into third energy level causes infrared series
Accept any of the points shown on a diagram.
24 x 0.786 + 25 x 0.101 + 26 x 0.113
24.33
Award [2] for correct final answer.
Award [0] for 24.31 with no working (data booklet value).
carbon: «\(\frac{{0.4490\,{\text{g}}}}{{44.01\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}\) =» 0.01020 «mol» / 0.1225 «g»
OR
hydrogen: «\(\frac{{0.1840 \times 2}}{{18.02\,g\,mo{l^{ - 1}}}}\) =» 0.02042 «mol» / 0.0206 «g»
oxygen: «0.1595 – (0.1225 + 0.0206)» = 0.0164 «g» / 0.001025 «mol»
empirical formula: C10H20O
Award [3] for correct final answer.
Do not award M3 for a hydrocarbon.
«temperature =» 423 K
OR
M \( = \frac{{mRT}}{{pV}}\)
«M \( = \frac{{0.150\,{\text{g}} \times 8.31\,{\text{J}}{{\text{K}}^{ - 1}}\,{\text{mol}}{}^{ - 1} \times 423\,{\text{K}}}}{{100.2\,{\text{kPa}} \times 0.0337\,{\text{d}}{{\text{m}}^3}}} = \)» 156 «g mol–1»
Award [1] for correct answer with no working shown.
Accept “pV = nRT AND n = \(\frac{m}{M}\)” for M1.
C10H20O
[1 Mark]
Cl2: first
NO: second
rate = k [NO]2 [Cl2]
180 / 1.80 x 102 «dm6 mol–2 min–1»
