AQA A-Level Chemistry Paper 2, 2023: Question 7

10 marks · Medium difficulty · State/Explain/Numerical

Gas Laws and Volume Calculations This question involves calculating the volume of propanone gas using the ideal gas equation, determining volume changes with temperature, evaluating uncertainty caused by balance precision, and calculating the total gas volume after the combustion of propanone.

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Question

AQA A-Level Chemistry Paper 2, 2023: Question 7
Question text

07 A gas syringe that does not have any graduations is calibrated using a known mass of

propanone (boiling point = 56.2 °C).

The sealed gas syringe contains 0.146 g of propanone (Mr = 58.0) at a temperature of

95 °C and a pressure of 103 kPa

07.1 Calculate the volume, in cm3, of propanone in the gas syringe.

The gas constant, R = 8.31 J K−1 mol−1

[4 marks]

Volume of propanone cm3

07.2 The gas syringe is then cooled to 75 °C, without changing the pressure.

Calculate the decrease in volume.

(If you were unable to calculate the volume in Question 07.1, you should use the

volume 89 cm3. This is not the correct answer.)

[2 marks]

21 3

Decrease in volume cm

07.3 The total uncertainty in using the balance to measure the mass of propanone in

Question 07.1 is ±0.001 g

*20* 3

Calculate the uncertainty that this causes in the volume, in cm , of propanone

calculated in Question 07.1.

(If you were unable to calculate the volume in Question 07.1, you should use the

volume 89 cm3. This is not the correct answer.)

[2 marks]

Uncertainty cm3

07.4 A 600 cm3 sample of propanone is mixed with 2800 cm3 of oxygen in a container at

60 °C and 100 kPa. The mixture is ignited.

When the reaction is complete, the remaining mixture of gases is cooled to 60 °C at

100 kPa

CH3COCH3(g) + 4O2(g) → 3CO2(g) + 3H2O(l)

Calculate the total volume of the remaining gas mixture.

[2 marks]

Volume cm3

Mark scheme

Show the mark scheme Mark scheme for AQA A-Level Chemistry Paper 2, 2023: Question 7

Question Answers Additional Comments/Guidelines Mark

0.146 -3

M1 n(propanone) = (= 2.52 × 10 )

M2 Conversion of T and P (T = 368K and P = 103000Pa)

07.1 M1 × 8.31 × 368

nRT V = scores M2 and M3 (4 x AO2)

M3 V = rearranged for V as subject (in algebraic or numbers) 103000

P

63 Allow 74-75

M4 their evaluated M3 × 1 × 10 = 75 cm

Marked with Q7.1

Using alternate answer

348 3 348 3

M1 V = × M4 = 71 cm M1 V = × 89 = 84 cm

368 368

07.2 M2 Decrease = M4 – M1 = 4 cm3 M2 89 – 84= 5 cm3

(2 x AO2)

Allow answer for M1 calculated as 70.8 cm3 after

substitution of values into pV = nRT. Could then

lead to a difference of 18.2 cm3 if compared to the

alternate value for M4 of 89 cm3

0.001 Marked with Q7.1

M1 % uncertainty = × 100 = 0.685%

0.146

07.3 Allow 0.6 cm3 if 89 cm3 used

M1 (2 x AO2)

M2 Vol uncertainty = × M4 = 0.5 cm3

M1 Vol CO formed = 3 × 600 = 1800 cm3 If PV=nRT method used

07.4 M1 n(CO2) = 0.0651

3 (2 x AO2)

M2 Total Vol left = 1800 + 400 = 2200 cm

How to answer it

Step-by-Step Guide: Gas Laws and Uncertainty

Question (a): Calculating the Volume of Propanone

Step 1: Calculate the Moles of Propanone

Use the formula: n = m / Mr .

Substitute: n = 0.146 / 58 = 2.52 × 10-3 mol .

Step 2: Rearrange the Ideal Gas Equation

Use PV = nRT rearranged for V:

V = (nRT) / P

Convert temperature and pressure:

  • T: 95°C = 368 K
  • P: 103 kPa = 103,000 Pa

Substitute:

V = (2.52 × 10-3 × 8.31 × 368) / 103,000

Evaluate:

Answer: 75 cm3 (accept 74-75 cm3).

Question (b): Decrease in Volume

Step 1: Use Proportionality of Volume and Temperature

Since V ∝ T at constant pressure, use:

V2 = V1 × (T2 / T1)

Substitute values:

  • V1 = 75 cm3
  • T2 = 75°C = 348 K
  • T1 = 95°C = 368 K

Calculation:

V2 = 75 × (348 / 368) = 71 cm3

Step 2: Calculate Decrease in Volume

Decrease = V1 - V2

Answer: 4 cm3

Question (c): Uncertainty in Volume

Step 1: Calculate Percentage Uncertainty

Use the formula:

Percentage uncertainty = (uncertainty / measured value) × 100

Substitute:

(0.001 / 0.146) × 100 = 0.685%

Step 2: Calculate Volume Uncertainty

Use:

Volume uncertainty = (percentage uncertainty / 100) × measured volume

Substitute:

(0.685 / 100) × 75 = 0.5 cm3

Answer: 0.5 cm3 (allow 0.6 cm3 if 89 cm3 is used).

Question (d): Volume of Remaining Gas Mixture

Step 1: Analyse the Reaction

The reaction is:

CH3COCH3(g) + 4 O2(g) → 3 CO2(g) + 3 H2O(l)

Key Points:

  • 1 mole of propanone produces 3 moles of CO2.
  • Water is in liquid form, so its volume is negligible.

Step 2: Calculate CO2 Volume

Given:

  • Initial propanone = 600 cm3
  • CO2 produced = 3 × 600 = 1800 cm3

Step 3: Calculate Remaining Oxygen

Initial oxygen = 2800 cm3

Oxygen used = 4 × 600 = 2400 cm3

Remaining oxygen = 2800 - 2400 = 400 cm3

Step 4: Total Volume

Remaining gases:

  • CO2 = 1800 cm3
  • Oxygen = 400 cm3

Total volume: 1800 + 400 = 2200 cm3

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.