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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.

Nitric oxide reacts with chlorine.

2NO (g) + Cl₂ (g) → 2NOCl (g)

The following experimental data were obtained at 101.3 kPa and 263 K.

Menthol is an organic compound containing carbon, hydrogen and oxygen.

Outline how this spectrum is related to the energy levels in the hydrogen atom.

[1]

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.

[2]

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.

[3]

d.i.

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.

[2]

d.ii.

Determine the molecular formula of menthol using your answers from parts (d)(i) and (ii).

[1]

d.iii.

Deduce the order of reaction with respect to Cl₂ and NO.

[2]

e.i.

State the rate expression for the reaction.

[1]

e.ii.

Calculate the value of the rate constant at 263 K.

[1]

e.iii.

Markscheme

electron transfer/transition between high«er» energy level to low«er» energy level

electron transitions into first energy level causes UV series

transition into second energy level causes visible series

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.

d.i.

«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.

d.ii.

C₁₀H₂₀O

[1 Mark]

d.iii.

Cl₂: first
NO: second

e.i.

rate = k [NO]² [Cl₂]

e.ii.

180 / 1.80 x 10² «dm⁶ mol^–2 min^–1»

e.iii.

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

d.iii.

[N/A]

e.i.

[N/A]

e.ii.

[N/A]

e.iii.

Syllabus sections

Core » Topic 1: Stoichiometric relationships » 1.2 The mole concept

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