OCR A-Level Chemistry AS Depth in chemistry (02), June 2019: Question 2

11 marks · Medium difficulty · Structured Questions

Describe the preparation of a magnesium nitrate solution, calculate the volume of nitric acid needed to react with magnesium carbonate, and determine oxidation numbers in thermal decomposition.

Practise this question

Question

A three-part chemistry question about magnesium compounds. Part (a) asks students to describe how to prepare 250.0 cm3 of a 0.4000 mol dm-3 solution of magnesium nitrate starting from crystals, worth 6 marks. Part (b) provides an equation for the reaction between magnesium carbonate and nitric acid, asking to calculate the minimum volume of 1.75 mol dm-3 HNO3 needed to prepare a solution containing 5.00 g of Mg(NO3)2, worth 3 marks. Part (c) gives a thermal decomposition equation for magnesium nitrate and asks to identify the elements oxidised and reduced along with their oxidation number changes, worth 2 marks.
Question text

2 Magnesium nitrate is used in fertilisers as a source of nitrogen.

(a)* A student plans to prepare 250.0 cm3 of a 0.4000 mol dm–3 solution of magnesium nitrate,

starting from magnesium nitrate crystals, Mg(NO3)2•6H2O.

Describe how the student would prepare the solution, giving full details of quantities,

apparatus and method. [6]

Additional answer space if required

(b) A solution of magnesium nitrate can be prepared by reacting magnesium carbonate, MgCO3,

with nitric acid, HNO3(aq).

The equation is shown below.

MgCO3(s) + 2HNO3(aq) Mg(NO3)2(aq) + H2O(l) + CO2(g)

Calculate the minimum volume, in cm3, of 1.75 mol dm–3 HNO that is needed to prepare a

solution containing 5.00 g of Mg(NO3)2.

Give your answer to 3 significant figures.

volume = … cm3 [3]

(c) Magnesium nitrate decomposes when heated, as shown in the equation.

2Mg(NO3)2(s) 2MgO(s) + 4NO2(g) + O2(g)

Using oxidation numbers, show which element has been oxidised and which has been

reduced when magnesium nitrate decomposes.

State the changes in oxidation numbers, including all signs.

Element oxidised …

Oxidation number change: from … to …

Element reduced …

Oxidation number change: from … to …

[2]

Mark scheme

Show the mark scheme The mark scheme provides a level-based response rubric for part (a) detailing calculation of mass (25.63 g or 14.83 g) and specific preparation steps such as weighing, dissolving, transferring to a volumetric flask, and making up to the mark. For part (b), it shows working for calculating moles of magnesium nitrate (0.0337 mol), moles of nitric acid (0.0674 mol), and the final volume (38.5 cm3 to 3 significant figures). For part (c), it lists oxygen being oxidised from -2 to 0, and nitrogen being reduced from +5 to +4.

x = SrCl2 : H2O = ALLOW ECF from n(SrCl2) and/or n(H2O)

0.0102……

AO3.2 Answer must be to TWO significant figures

= 5.8

ALLOW 2 marks for 5.83 (answer must be to 2 SF)

11 AO

Question Answer Marks element Guidance

(ii) To make sure all the water had been removed 1 AO3.4 IGNORE just ‘to weigh to constant mass’

(iii) Use balance that weighs to 3/more decimal places 2 AO3.4 ALLOW more precise/more accurate/ more

×2 sensitive/higher resolution/smaller division/weigh to

0.001

IGNORE ‘less error/smaller interval balance’

Use a larger mass (of hydrated strontium chloride)

IGNORE any reference to lid on crucible

(water can’t escape)

IGNORE ‘weigh straight after heating’

IGNORE idea of repeating the experiment/ taking

an average/ getting concordant results /larger

sample size, etc.

Total 18

AO

Question Answer Marks Guidance

element

2 (a) Please refer to the marking instructions on page 4 of this 6 AO2.8 Indicative scientific points may include:

mark scheme for guidance on how to mark this question. ×2

Calculation:

Level 3 (5–6 marks) 250.0

n = × 0.4000 = 0.1(000) (mol)

Calculates the correct mass of Mg(NO3)2•6H2O or 1000

Mg(NO3)2.

AND M(Mg(NO3)2•6H2O) = 256.3

Explains the preparation steps, with most fine detail. Mass = 0.1000 × 256.3 = 25.63 g

There is a well-developed line of reasoning which is clear and AO2.3 OR M(Mg(NO3)2) = 148.3

logically structured. The information presented is relevant and ×2 Mass = 14.83 g

substantiated.

12 ALLOW small slip/rounding errors such as errors

Level 2 (3–4 marks)

on Mr (e.g. use of 24 instead of 24.3 for Mg Ar)

Attempts a calculation which is partly correct.

AND

Preparation steps (apparatus and method):

Outlines the preparation steps, with some fine detail.

• Weigh mass of crystals

There is a line of reasoning presented with some structure. • Dissolve in (distilled/deionised) water

The information presented is relevant and supported by some • Transfer to 250 cm3 volumetric flask

evidence.

• Make up to the mark with more water so that

Level 1 (1–2 marks) AO2.7 bottom of meniscus is on the mark

Attempts the calculation but makes little progress or ×2

makes errors. IGNORE removing the water of crystallisation

OR

Briefly outlines the preparation steps, which may be Fine detail:

incomplete • 2 or more decimal place balance

• Rinse beaker and transfer washings to flask

There is an attempt at a logical structure with a line of • Use of dropping pipette when filling to mark

reasoning. The information is in the most part relevant. • Stopper, invert several times to mix

0 marks

H032/02 No response or no response worthy of credit. Mark Scheme June 2019

AO

Question Answer 13 Marks Guidance

element

2 (b) FIRST CHECK THE ANSWER ON ANSWER LINE 3 AO2.8

If answer = 38.5 cm3 award 3 marks ×3

---------------------------------------------------------------------------

5.00

n(Mg(NO3)2) = = 0.0337 … (mol) Calculator: 0.03371544167

148.3

n(HNO3) = 2 × 0.0337 … = 0.0674 … (mol) ALLOW ECF from n(Mg(NO3)2)

Calculator: 0.06743088334

1000 3

volume = 0.0674 … × = 38.5 (cm )

1.75 ALLOW ECF from n(HNO3)

3 SF required

(c) 2 AO2.2 MAX 1 mark if no ‘+’ sign for oxidation number

×2

Element oxidised: Oxygen/O ALLOW 2–

Change from: –2 to 0

ALLOW 5+ AND 4+

Element reduced: Nitrogen/N:

Change form + 5 to +4 ALLOW O2 for oxygen

ALLOW 1 mark for all oxidation numbers correct,

but oxidised and reduced the wrong way around

IGNORE numbers around equation

i.e. treat as rough working

Total 11

How to answer it

Magnesium Nitrate: Solutions, Stoichiometry & Redox

What this question tests

This question assesses core Practical Skills (making up a standard solution from a solid, including specific glassware and fine practical details), Stoichiometric Calculations (moles, concentrations, volumes, and reacting ratios), and Redox understanding (assigning oxidation numbers and identifying species oxidized and reduced).

Question Part (a) - 6 Marks

Preparing a Standard Solution

💡 Key Knowledge

  • To make 250.0 cm³ of 0.4000 mol dm⁻³ solution, you first calculate required moles: 0.2500 × 0.4000 = 0.100 mol .
  • Using hydrated crystals Mg(NO₃)₂·6H₂O (Mᵣ = 256.3), mass required = 0.100 × 256.3 = 25.63 g . (Note: anhydrous mass = 14.83 g is also acceptable if specified).

🧠 Exam Technique (Method Steps)

  • 1. Weighing: Weigh the accurately calculated mass of solid using a 2 decimal place (or more) balance.
  • 2. Dissolving: Dissolve the solid in a beaker using distilled/deionised water (less than 250 cm³).
  • 3. Transferring: Transfer solution quantitatively into a 250 cm³ volumetric flask, rinsing beaker and washings into the flask.
  • 4. Toing up: Add water dropwise until the bottom of the meniscus rests exactly on the graduation mark.
  • 5. Mixing: Stopper and invert the flask several times to ensure thorough mixing.

❌ Common Errors

  • Forgetting to account for the water of crystallization in Mg(NO₃)₂·6H₂O when calculating mass.
  • Omitting crucial fine details like rinsing the beaker/glass rod or using a dropping flask for the final drops to the meniscus.
  • Failing to mention inversion/mixing at the end.
Mark Allocation: 6 marks total (Level of Response based on calculation correctness, precision of method steps, and fine laboratory details included).
Question Part (b) - 3 Marks

Stoichiometry and Solution Volume Calculation

📐 Step-by-Step Calculation

  1. Find Molar Mass of Mg(NO₃)₂ :
    Mg = 24.3, N = 14.0 × 2, O = 16.0 × 6 → Mᵣ = 148.3 g mol⁻¹
  2. Calculate moles of Mg(NO₃)₂ :
    n = mass / Mᵣ = 5.00 / 148.3 = 0.033715 mol
  3. Use reacting ratios from the equation:
    Equation ratio MgCO₃ + 2HNO₃ → Mg(NO₃)₂ + H₂O + CO₂ shows 1 mol Mg(NO₃)₂ : 2 mol HNO₃ .
    Moles of HNO₃ needed = 0.033715 × 2 = 0.06743 mol
  4. Calculate volume of HNO₃ :
    Volume = (moles / concentration) × 1000
    Volume = (0.06743 / 1.75) × 1000 = 38.53 cm³

✅ Correct Final Answer

38.5 cm³ (strictly required to 3 significant figures as requested by the prompt).

❌ Common Errors & Traps

  • Omitting the multiplying factor of 2 for the acid-to-salt ratio.
  • Rounding intermediate numbers too early, leading to slight inaccuracies in the final decimal place.
  • Failing to give the final answer to the requested 3 significant figures.
Mark Allocation: 3 marks (1 mark for correct moles of salt, 1 mark for correct moles of acid using 1:2 ratio, 1 mark for final volume to 3 SF).
Question Part (c) - 2 Marks

Redox and Oxidation Number Changes

💡 Key Knowledge & Rules

  • Oxidation numbers must always include the sign ( + or - ).
  • Oxygen in compounds is almost always -2 , and uncombined elements ( O₂ ) have an oxidation number of 0 .
  • Oxidation is an increase in oxidation number (loss of electrons). Reduction is a decrease in oxidation number (gain of electrons).

✅ Correct Answers

  • Element Oxidised: Oxygen (or O / O²⁻ )
  • Oxidation number change: From -2 to 0
  • Element Reduced: Nitrogen (or N / NO₃⁻ )
  • Oxidation number change: From +5 to +4

❌ Common Errors

  • Forgetting to include the + sign for positive oxidation states (e.g., writing 5 instead of +5 costs a mark).
  • Confusing oxidation and reduction directions.
  • Swapping the 'from' and 'to' states in the description.
Mark Allocation: 2 marks (1 mark for correct identification and change for oxygen; 1 mark for correct identification and change for nitrogen. Must include signs).

Topics

Module 2: Foundations in chemistry · Practical Activity Groups · PAG 1: Moles determination · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.