DP Chemistry (last assessment 2024)

Question 22N.2.sl.TZ0.1

Date	November 2022	Marks available	[Maximum mark: 18]	Reference code	22N.2.sl.TZ0.1
Level	sl	Paper	2	Time zone	TZ0
Command term	Calculate, Deduce, Determine, Justify, State, Write	Question number	1	Adapted from	N/A

[Maximum mark: 18]

22N.2.sl.TZ0.1

Ammonium nitrate, NH₄NO₃, is used as a high nitrogen fertilizer.

(a)

Calculate the percentage by mass of nitrogen in ammonium nitrate. Use section 6 of the data booklet.

[1]

Markscheme

« $% N = \frac{2 \times 14.01 g m o l^{- 1}}{(2 \times 14.01 g m o l^{- 1} + 4 \times 1.01 g m o l^{- 1} + 3 \times 16.00 g m o l^{- 1})} \times 100 % =$ » $35.00$ « $%$ » ✔

Examiners report

Most candidates were able to calculate the percentage by mass of nitrogen.

(b)

State, with a reason, whether the ammonium ion is a Brønsted-Lowry acid or base.

[1]

Markscheme

«Brønsted-Lowry» acid AND can donate a proton/H⁺

«Brønsted-Lowry» acid AND cannot accept proton/H⁺ ✔

Examiners report

40% of candidates identified the ammonium ion as a Bronsted-Lowry acid with a valid reason. The most common mistake was considering "ammonia" instead of the "ammonium ion".

(c)

Deduce the Lewis (electron dot) structure for the nitrate anion.

[1]

Markscheme

✔

Negative charge must be included on square bracket or singly-bonded oxygen atom.

Accept .

Accept any combination of dots/crosses or lines to represent electron pairs.

Examiners report

Only 20% of the candidates scored the mark for the Lewis structure of the nitrate ion. Some candidates had the correct bonds but did not include the charge while a big proportion of candidates made mistakes in drawing the bonds in the ion.

(d)

Calculate the pH of an ammonium nitrate solution with [H₃O⁺] = 1.07× 10⁻⁵ mol dm⁻³. Use section 1 of the data booklet.

[1]

Markscheme

«pH = –log (1.07× 10⁻⁵) =» 4.97 ✔

Examiners report

A very well answered question. The majority of candidates calculated the pH using the given [H₃O⁺].

Cold packs contain ammonium nitrate and water separated by a membrane.

(e.i)

The mass of the contents of the cold pack is 25.32 g and its initial temperature is 25.2 °C. Once the contents are mixed, the temperature drops to 0.8 °C.

Calculate the energy, in J, absorbed by the dissolution of ammonium nitrate in water within the cold pack. Assume the specific heat capacity of the solution is 4.18 J g⁻¹ K⁻¹. Use section 1 of the data booklet.

[1]

Markscheme

«q = mcΔT = 25.32 g × 4.18 J g⁻¹ K⁻¹ × (25.2 °C – 0.8 °C) =» 2580 «J» ✔

Do not accept a negative value.

Examiners report

A well answered question. 60% of the candidates calculated the energy absorbed in the cold pack.

(e.ii)

The change in enthalpy when ammonium nitrate dissolves in water is 25.69 kJ mol⁻¹. Determine the mass of ammonium nitrate in the cold pack using your answer obtained in (e)(i) and section 6 of the data booklet.

If you did not obtain an answer in (e)(i), use 3.11 × 10³ J, although this is not the correct answer.

[2]

Markscheme

« $2.58 \times 10^{3} J \times \frac{1 k J}{1000 J} \times \frac{1 m o l}{25.69 k J} =$ » $0.100$ «mol» ✔

« $0.100$ mol $\times 80.06$ g mol⁻¹ =» 8.01 «g» ✔

Award [2] for the correct final answer.

Accept range of $8.00 - 8.10$ «g».

If $3.11 \times 10^{3}$ J used then answer is $9.69$ «g».

Examiners report

The question was less familiar, however, almost half of the candidates obtained the mass of ammonium nitrate in the pack. This question discriminated well between high achieving and low achieving candidates.

(e.iii)

The absolute uncertainty in mass of the contents of the cold pack is ±0.01 g and in each temperature reading is ±0.2 °C. Using your answer in (e)(ii), calculate the absolute uncertainty in the mass of ammonium nitrate in the cold pack.

If you did not obtain an answer in (e)(ii), use 6.55 g, although this is not the correct answer.

[3]

Markscheme

«fractional / % uncertainty in ΔT =» $\frac{0.4}{24.4}$ / 0.02 / 2«%» ✔

«fractional / % uncertainty in $m$ =» $\frac{0.01}{25.32}$ / 0.0004 / 0.04«%»

fractional / % uncertainty in $m$ is much smaller than uncertainty in ΔT ✔

«2% x 8.01 g =» 0.2 «g» ✔

Award [3] for correct final answer.

Accept range of 0.1 g – 0.2 «g».

If 6.55 g used then the answer is 0.1 «g».

Examiners report

This question discriminated well between high achieving and low achieving candidates. Only about a fifth of the candidates propagated the uncertainties correctly throughout the calculation. Some candidates only obtained the first mark for the fractional/percentage uncertainty in the change in temperature.

(e.iv)

The cold pack contains 9.50 g of ammonium nitrate. Calculate the percentage error in the experimentally determined mass of ammonium nitrate obtained in (e)(ii).

If you did not obtain an answer in (e)(ii), use 6.55 g, although this is not the correct answer.

[1]

Markscheme

«% error = $|\frac{9.50 g - 8.01 g}{9.50 g}| \times 100 % =$ » 15.7«%» ✔

Accept range 14.7 – 15.8«%».

If 6.55 g used then answer is 31.1«%».

Examiners report

Half of the candidates calculated the percentage error correctly. Some candidates left this question blank, some divided by the actual yield and some calculated the percentage yield instead.

Solid ammonium nitrate can decompose to gaseous dinitrogen monoxide and liquid water.

(f.i)

Write the chemical equation for this decomposition.

[1]

Markscheme

NH₄NO₃ (s) → N₂O (g) + 2H₂O (l) ✔

Examiners report

A very well answered question. The majority of candidates wrote a balanced equation given the names of the reactant and products.

(f.ii)

Calculate the volume of dinitrogen monoxide produced at STP when a 5.00 g sample of ammonium nitrate decomposes. Use section 2 of the data booklet.

[2]

Markscheme

«5.00 g ÷ 80.06 g mol⁻¹ =» 0.0625 mol «NH₄NO₃» ✔

«1:1 mol ratio»

« $0.0625 m o l N_{2} O \times \frac{22.7 d m^{3}}{m o l} =$ » 1.42 «dm³» ✔

Award [2] for correct final answer.

Accept range 1.36 – 1.43 «dm³».

Accept calculations based on PV=nRT.

Examiners report

This question was well answered though some candidates made silly mistakes. 60% of the candidates calculated the volume of gaseous product correctly. The majority used PV = nRT instead of the molar volume of 22.7 dm³. The question discriminated well between high achieving and low achieving candidates.

(f.iii)

Calculate the standard enthalpy change, $Δ H^{⦵}$ , of the reaction. Use section 12 of the data booklet.

$Δ H_{f}^{⦵}$ ammonium nitrate = −366 kJ mol⁻¹

$Δ H_{f}^{⦵}$ dinitrogen monoxide = 82 kJ mol⁻¹

[2]

Markscheme

2 x –285.8 «kJ mol⁻¹» ✔

«1 mol (82 kJ mol⁻¹) + 2 mol (– 285.8 kJ mol⁻¹) – 1 mol (–366 kJ mol⁻¹) =» –124 «kJ» ✔

Award [2] for correct final answer.

Examiners report

Less than half of the candidates were able to calculate the standard enthalpy change correctly. Errors included not including DH_f of water, not multiplying DH_f of water by two, subtracting incorrectly (reactants − products), and arithmetic errors. This question also had a high discrimination index.

(f.iv)

Deduce the Lewis (electron dot) structure and shape for dinitrogen monoxide showing nitrogen as the central atom.

[2]

Markscheme

Lewis structure:

✔

Shape: linear ✔

Accept

Only award M2 if the shape corresponds to that expected for the Lewis structure given.

Examiners report

Only a small proportion of candidates were able to deduce the Lewis structure of dinitrogen monoxide. Some candidates achieved the mark for the shape as error carried forward. A quarter of the candidates did not attempt this question.