AQA GCSE Chemistry Chemistry Paper 2 (Higher), 2024: Question 1

10 marks · Standard Demand difficulty · Short Answer

Identify sodium and chloride ions using chemical tests, explain the purpose of heating to constant mass, identify the correct calculation for mass of solute, and calculate the mean concentration of sodium ions in a solution given the percentage by mass.

Practise this question

Question

The image shows a multi-part chemistry question. Part 01.1 and 01.2 ask for tests and results for sodium and chloride ions. Part 01.3 and 01.4 involve a method for determining the concentration of sodium chloride by evaporation, with 01.4 being a multiple-choice question about calculating the mass of solid. Part 01.5 provides a table of four trial results for the concentration of sodium chloride and asks the student to calculate the mean concentration of sodium ions given that sodium ions make up 39.3% of the mass.
Question text

01 A student investigated an aqueous solution of a salt.

The student identified that the salt solution contained only sodium ions and chloride

ions.

01.1 Describe a test to identify sodium ions.

Give the result of the test.

[2 marks]

Test for sodium ions

Result

01.2 Describe a test to identify chloride ions.

Give the result of the test.

[2 marks]

Test for chloride ions

Result

The student determined the concentration of sodium chloride in the salt solution.

This is the method used.

1. Weigh an empty evaporating dish.

2. Add 25.0 cm3 of the salt solution into the evaporating dish.

3. Heat the evaporating dish and contents.

4. Weigh the evaporating dish and contents.

5. Repeat steps 3 to 4 until there is no further change in mass.

6. Repeat steps 1 to 5 three more times.

01.3 Why did the student heat the evaporating dish and contents until the mass did not

change?

[1 mark]

01.4 How did the student calculate the mass of solid sodium chloride remaining after steps

1 to 5?

[1 mark]

Tick ( ) one box.

Mass of 25 cm3 of salt solution + mass of empty evaporating dish

Mass of 25 cm3 of salt solution − mass of empty evaporating dish

Mass of evaporating dish and dry contents + mass of empty

evaporating dish

Mass of evaporating dish and dry contents − mass of empty

evaporating dish 4

01.5 The student calculated the concentration of sodium chloride in the salt solution.

Table 1 shows the results.

Table 1

Concentration of sodium chloride in g/dm3

Trial 1 Trial 2 Trial 3 Trial 4

*03* 35.2 34.6 36.4 33.8

The percentage by mass of sodium ions in sodium chloride is 39.3%.

Calculate the mean concentration of sodium ions in the salt solution.

[4 marks]

Mean concentration = g/dm3

Mark scheme

Show the mark scheme The mark scheme provides the answers for parts 01.1 through 01.5. It lists the required chemical tests (flame test for sodium, acidified silver nitrate for chloride) and the corresponding results. It also shows the step-by-step mathematical calculations for the mean concentration of sodium ions, including alternative acceptable methods.

Question 1

AO /

Question Answers Extra information Mark

Spec. Ref.

01.1 (test) 1 AO1

flame test 4.8.3.1

RPA7

(result) 1

yellow (flame)

OR

(test)

flame emission spectroscopy (1) allow FES

(result)

lines match sodium spectrum (1)

AO /

Spec. Ref.

01.2 (test) AO1

(add acidified) silver nitrate 1 4.8.3.4

(solution) RPA7

(result) 1

white precipitate MP2 is dependent upon the

award of MP1

AO /

Spec. Ref.

01.3 to ensure that all the water has 1 AO3

evaporated 4.10.1.2

– – –

RPA8

AO /

Spec. Ref.

01.4 mass of evaporating dish and 1 AO1

dry contents − mass of empty – – – 4.10.1.2

evaporating dish RPA8

AO /

Spec. Ref.

01.5 (mean concentration of NaCl =) AO2

35.2 + 34.6 + 36.4 + 33.8

1 4.10.1.2

RPA8

or

allow 1 mark for

140 35.2 + 34.6 + 33.8

4 = 34.5

= 35.0 (g/dm3) 1

(mean concentration of Na+ =)

39.3 allow correct use of an 1

35.0 × incorrectly determined mean

concentration of sodium chloride

= 13.8 (g/dm3) allow 13.755 correctly rounded 1

to at least 3 significant figures

alternative approach 1:

(total concentration of NaCl =

35.2 + 34.6 + 36.4 + 33.8 = 140

total concentration of Na+ =) allow 1 mark for

39.3 (35.2 + 34.6 + 33.8 = 103.6)

140 × (1) 39.3

100 103.6 × = 40.71

= 55.02 (g/dm3) (1)

(mean concentration of Na+ =)

55.02 allow correct use of an

(1)

4 incorrectly determined total

concentration of Na+

8 =13.8 (g/dm3) (1) allow 13.755 correctly rounded

to at least 3 significant figures

– – –

alternative approach 2:

(concentrations of Na+ =)

39.3

35.2 ×

39.3

34.6 ×

allow 1 mark if a

39.3 concentration of 36.4 is

36.4 × treated as an anomaly and

not used

39.3

33.8 × (1)

= 13.83 13.60 14.31 13.28 (1)

(mean concentration of Na+ =)

13.83 + 13.60 + 14.31 + 13.28 allow correct use of incorrectly

4 determined concentration(s) of

(1) Na+

=13.8 (g/dm3) (1) allow 13.755 correctly rounded

to at least 3 significant figures

Total Question 1 10

How to answer it

Analyzing Sodium Chloride Solutions

What this question tests

This question assesses your ability to identify ions using chemical tests (Required Practical 7), understand experimental procedures for evaporation (Required Practical 8), and perform multi-step concentration calculations involving means and percentages.

Part (01.1) & (01.2): Ion Identification

Testing for Cations and Anions

💡 Key Knowledge

  • Sodium (Na⁺): Identified using a flame test.
  • Chloride (Cl⁻): Identified using silver nitrate solution (acidified with nitric acid).

✅ Correct Answers

01.1: Test: Flame test. Result: Yellow flame.

01.2: Test: Add silver nitrate solution. Result: White precipitate.

🧠 Exam Technique

When describing the chloride test, you should ideally mention adding nitric acid first to remove any carbonate ions that might give a false positive. However, the mark scheme focuses on the reagent (silver nitrate) and the specific color of the precipitate (white).

Part (01.3) & (01.4): Experimental Methods

Evaporation to Dryness

🧠 Heating to Constant Mass

In 01.3, the student repeats heating and weighing. This is a standard technique to ensure all the water has evaporated. If the mass is still changing, there is still water being lost!

❌ Common Errors

In 01.4, students often get confused about which mass to subtract. To find the mass of the salt only, you must take the total mass (dish + salt) and subtract the mass of the empty dish.

✅ Correct Answer (01.4)

Tick the box: Mass of evaporating dish and dry contents − mass of empty evaporating dish

Part (01.5): Multi-Step Calculation

Calculating Mean Concentration of Sodium Ions

📐 Step-by-Step Calculation

This is a 4-mark question. You must show your working clearly!

  1. Calculate the mean concentration of Sodium Chloride (NaCl):
    (35.2 + 34.6 + 36.4 + 33.8) ÷ 4 = 35.0 g/dm³
  2. Identify the percentage of Sodium (Na⁺) in the salt:
    The question states sodium is 39.3% of the total mass.
  3. Calculate the concentration of Sodium ions:
    35.0 × (39.3 ÷ 100) = 13.755 g/dm³
  4. Final Answer (Rounding):
    Round to a sensible number of significant figures (usually 3).
    Result: 13.8 g/dm³

🧠 Examiner Insight

If you made a mistake in the mean (Step 1) but used that "wrong" number correctly in Step 3, you can still get "Error Carried Forward" (ECF) marks. Never leave a calculation blank!

❌ Calculation Traps

  • Dividing by 3: Some students ignore one trial thinking it's an anomaly. Only exclude a result if it is significantly different from the others. Here, they are all close.
  • Units: Ensure your final answer is in g/dm³ as requested.
Mark Breakdown:
1 Mark: Correct mean of NaCl (35.0)
1 Mark: Correct method for percentage (35.0 × 0.393)
1 Mark: Correct calculation result (13.755)
1 Mark: Correct rounding to 13.8

Topics

Chemistry · Required Practicals · C8: Chemical Analysis · C10: Using Resources · Required Practicals

Question and mark scheme from the AQA GCSE Chemistry examination, Chemistry Paper 2 (Higher), 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.