DP Chemistry (last assessment 2024)

Question 22M.2.hl.TZ1.b

Date	May 2022	Marks available	[Maximum mark: 2]	Reference code	22M.2.hl.TZ1.b
Level	hl	Paper	2	Time zone	TZ1
Command term	Determine	Question number	b	Adapted from	N/A

[Maximum mark: 2]

22M.2.hl.TZ1.b

Ammonia is soluble in water and forms an alkaline solution:

NH₃(g) + H₂O (l) $⇌$ NH₄⁺(aq) + HO^–(aq)

Determine the concentration, in mol dm^–3, of the solution formed when 900.0 dm³ of NH₃(g) at 300.0 K and 100.0 kPa, is dissolved in water to form 2.00 dm³ of solution. Use sections 1 and 2 of the data booklet.

[2]

Markscheme

amount of ammonia $⟨ ⟨ = \frac{P . V}{R . T} = \frac{100.0 k P a \times 900.0 d m^{3}}{8.31 J K^{- 1} m o l^{- 1} \times 300.0 K} ⟩ ⟩ = 36.1 « mol »$ ✔

concentration $⟨ ⟨ = \frac{n}{V} = \frac{36.1}{2.00} ⟩ ⟩ = 18.1 « mol {dm}^{- 3} »$ ✔

Award [2] for correct final answer.

Examiners report

Good performance. Some teachers suggested the question had an error in units, but this was not the case. The question had to be solved, first by using the data provided for application of gas law to determine the number of moles of gas. Next, given volume of solution, V = 2.00 dm³, determine its concentration.

Syllabus sections

Core

Core » Topic 1: Stoichiometric relationships » 1.3 Reacting masses and volumes

Core » Topic 1: Stoichiometric relationships

Question 22M.2.hl.TZ1.b

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