Edexcel A-Level Chemistry AS Paper 1, June 2017: Question 8

11 marks · Medium difficulty · Practical Techniques and Data Analysis

Determine the molar volume of carbon dioxide gas using experimental data from the reaction between calcium carbonate and hydrochloric acid.

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Question

Question text

8 A student wanted to measure the volume of a gas and use the results to find the

volume occupied by one mole of the gas. The following method was used.

Ɣ A sample of calcium carbonate was weighed out in a small plastic container.

Ɣ 20 cm3 of hydrochloric acid of concentration 2.00 mol dm–3 was added to a

conical flask. A small pinch of calcium carbonate was added to the acid.

Ɣ The container was placed in the conical flask and a gas syringe was connected to

the top of the conical flask.

Ɣ The flask was carefully shaken so that the small plastic container fell over, allowing

the acid and calcium carbonate to mix.

The apparatus set up is shown.

The student repeated the experiment five times using different masses of

calcium carbonate on each occasion, with the concentration and volume of the

hydrochloric acid constant.

Experiment number Mass / g Volume of CO / cm3

10.10 23

20.20 44

30.30 67

40.40 96

50.50 115

(a) (i) Write the equation for the reaction between calcium carbonate and

hydrochloric acid. Include state symbols.

(2)

(ii) Calculate the molar mass of calcium carbonate.

(1)

g mol–1

20 (iii) Show that, in each experiment, the hydrochloric acid is in excess.

*P49858A02024* (2)

(b) (i) Plot a graph of volume of carbon dioxide produced against mass of

calcium carbonate on the grid. Include a line of best fit.

(2)

(ii) State how your graph supports the idea that the volume of gas produced 21

depends directly on the mass of calcium carbonate added.*P49858A02124*

(1)

from these data. Give your answer in dm3 to two significant figures.

(2)

(d) Give a reason why the student added a small pinch of calcium carbonate to

the acid before starting the reaction.

(1)

(Total for Question 8 = 11 marks)

Mark scheme

Show the mark scheme

Question

Acceptable Answer Additional Guidance Mark

Number

8 (a) (i) CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) Accept (2)

CaCO (s) + 2H+(aq) →

Ca2+(aq) + H O(l) + CO (g)

Balanced equation (1)

State symbols (1) 2nd mark dependent on first or near

miss.

Reject H2CO3(aq) in equation, but allow

state symbol mark if otherwise correct.

Question

Acceptable Answer Additional Guidance Mark

Number

8 (a) (ii) Example of calculation (1)

Finds molar mass of calcium carbonate Mr of calcium carbonate

= 40.1 + 12 + (16 x 3) =100.1 (g mol-

Allow

= 40 + 12 + (16 x 3) = 100 (g mol-1)

Accept answer with no working

Question

Acceptable Answer Additional Guidance Mark

Number

8 (a)(iii) Example of calculation (2)

calculate moles of calcium carbonate in 0.50 g (1) moles of calcium carbonate =

0.50/100.1 = 0.004995 = 0.0050

(mol)

moles of hydrochloric acid in 20 cm3

moles of hydrochloric acid in 20 cm3 =

AND 20/1000 x 2 = 0.040 (mol)

Show the hydrochloric acid is in excess with

appreciation of 2:1 ratio in equation for reaction (1) 0.04 (moles of hydrochloric acid)

reacts with 0.02 (moles of calcium

carbonate) therefore the acid is in (a

four times) excess.

0.0050 (moles of calcium carbonate)

reacts with 0.010 (moles of

hydrochloric acid) therefore the acid is

in (a four times) excess

Ignore calculations using other masses

of calcium carbonate

Question

Acceptable Answer Additional Guidance Mark

Number

8 (b) (i) 120 (2)

Points plotted accurately Do not

award for

AND reversed 100

axes labelled (1) axes

Points plotted must cover more Volume

than half of graph paper (of CO2)

AND / cm3

Reasonable straight line of best

fit which may extend to the 20

origin (1)

Allow ecf on reasonable line on 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

incorrectly plotted points. Mass (of CaCO3) / g

Question

Acceptable Answer Additional Guidance Mark

Number

8 (b) (ii) Straight line through the origin (therefore volume is directly Allow (1)

proportional to mass) ‘There is a positive correlation.’

Question

Acceptable Answer Additional Guidance Mark

Number

8 (c) Either Example calculation (2)

finds gradient from graph (1) Gradient = volume = 231 (cm3 per gram)

mass

Allow correctly calculated values in the range

= 210 to 250

molar volume given to 2 s.f. with units (1) (Molar Volume = Gradient x Mr)

Molar Volume = 231 x 100.1(or x 100)

= 23 (dm3) (must be 2 s.f)

Answer to 2 s.f. (and units)

Allow TE from any gradient

moles of calcium carbonate (1) Data may be used from any experiment number

eg using data from Experiment 5

Moles of calcium carbonate = 0.50/100.1

= 0.0050

molar Volume (1)

Molar Volume = 115/0.005

= 23 (dm3)

Allow data from a point on the line calculated

using route 2

Question

Acceptable Answer Additional Guidance Mark

Number

8 (d) To saturate the solution with CO2 / to stop the CO2 formed (1)

from dissolving

(Total for Question 8 = 11 marks)

TOTAL FOR PAPER = 80 MARKS

How to answer it

Gas Volume and Moles Study Guide

Edexcel AS Level Chemistry

What this question tests

This multi-part practical chemistry question assesses your ability to write balanced chemical equations with state symbols, calculate molar masses, perform stoichiometric calculations involving limiting reagents and excess, process graphical data (plotting, line of best fit, identifying direct proportionality), calculate molar gas volumes, and understand practical methodology/source of error in gas collection experiments.

Question 8 (a) - Equations, Molar Mass, and Reagent Excess

Part (i): Equation for the reaction

✅ Correct Answer

CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

Alternative ionic: CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) + CO₂(g)

❌ Common Errors

Omitting or misplacing state symbols (remember: the 2nd mark is dependent on the first or near miss).
Writing H₂CO₃(aq) as a product instead of water and carbon dioxide.

Marks: 2 marks (1 for balanced equation, 1 for correct state symbols)

Part (ii): Molar mass of calcium carbonate

📐 Calculation Steps

1. Sum the relative atomic masses ( Ar ) from the periodic table: Ca = 40.1, C = 12.0, O = 16.0.

2. Calculation: 40.1 + 12.0 + (3 × 16.0) = 100.1 g mol⁻¹ (Accept 100 g mol⁻¹).

Marks: 1 mark

Part (iii): Showing hydrochloric acid is in excess

📐 Step-by-Step Calculation

Step 1: Find moles of CaCO₃ in the maximum mass used (0.50 g):

Moles = 0.50 / 100.1 = 0.0050 mol

Step 2: Find moles of HCl added:

Moles = (20 / 1000) × 2.00 = 0.040 mol

Step 3: Compare using the stoichiometric ratio (1 mol CaCO₃ : 2 mol HCl ):

0.0050 mol of CaCO₃ requires 0.0050 × 2 = 0.010 mol of HCl . Since 0.040 mol of HCl is available (which is a 4-fold excess), hydrochloric acid is in excess.

🧠 Exam Technique

Always use the largest mass from the table (0.50 g) when proving a reagent is in excess across the whole experiment series. Clearly state the reacting ratio from your equation!

Marks: 2 marks (1 for calculating moles of both reagents, 1 for comparative conclusion using the 2:1 ratio)

Question 8 (b) - Graphical Analysis

Part (i): Plotting the graph

💡 Key Knowledge

Volume of CO₂ should be plotted on the y-axis against mass of CaCO₃ on the x-axis. Points must be plotted accurately covering more than half the grid, joined by a reasonable straight line of best fit extending toward the origin.

❌ Common Errors

Reversing the axes will immediately lose the plotting accuracy mark. Do not force a non-linear trend; draw a smooth single straight line of best fit.

Marks: 2 marks

Part (ii): Interpreting direct proportionality

✅ Correct Answer

A straight line passing through the origin shows that the volume of gas produced is directly proportional to the mass of calcium carbonate added (or simply state "There is a positive correlation passing through the origin").

Marks: 1 mark

Question 8 (c) - Calculating Molar Volume

📐 Calculation Options

Route 1 (Using Graph Gradient):

1. Calculate gradient: Volume / Mass = e.g., 231 cm³ g⁻¹ (Accept range 210 to 250).

2. Multiply gradient by molar mass ( Mr ) to find molar volume: 231 × 100.1 = 23123 cm³ mol⁻¹ = 23 dm³ mol⁻¹ .

Route 2 (Using a Single Point from Table/Line):

Using Experiment 5 (0.50 g gives 115 cm³):

1. Moles of CaCO₃ = 0.50 / 100.1 = 0.0050 mol

2. Molar Volume = Volume / Moles = 115 / 0.0050 = 23000 cm³ mol⁻¹ = 23 dm³

❌ Common Calculation Traps

Significant Figures: The question explicitly asks for your final answer to two significant figures ( 23 dm³ ). Losing a mark here is entirely avoidable!
Units: Make sure you convert cm³ to dm³ correctly (divide by 1000) if working in cm³ initially.

Marks: 2 marks (1 for finding gradient / moles, 1 for final molar volume with units to 2 s.f.)

Question 8 (d) - Practical Technique

✅ Correct Answer

To saturate the solution with carbon dioxide / to stop the carbon dioxide formed from dissolving into the water in the acid solution before it can enter the gas syringe.

🧠 Examiner Insight

Carbon dioxide is slightly soluble in water. Allowing the small pinch to react beforehand ensures the aqueous solution is saturated, preventing any generated CO₂ gas from dissolving and causing under-reading in the gas syringe.

Marks: 1 mark

Topics

Physical Chemistry · Core Practicals · Topic 5: Formulae, Equations and Amounts of Substance · Core Practical 1: Measuring the molar volume of a gas

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 1, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.