South Axholme School Page 1 AS Paper 1 – Practical Questions MARK SCHEME Amount of Substance M1.(a) Stage 1: appreciation that the acid must be in excess and calculation of amount of solid that permits this Statement that there must be an excess of acid 1 Moles of acid = 50.0 × 0.200 / 1000 = 1.00 × 10 –2 mol 1 2 mol of acid react with 1 mol of calcium hydroxide therefore moles of solid weighed out must be less than half the moles of acid = 0.5 × 1.00 × 10 –2 = 5.00 × 10 –3 mol 1 Mass of solid must be –3 × 74.1 = 1 Stage 2: Experimental method Measure out 50 cm 3 of acid using a pipette and add the weighed amount of solid in a conical flask 1 Titrate against 0.100 (or 0.200) mol dm –3 NaOH added from a burette and record the volume (v) when an added indicator changes colour 1 Stage 3: How to calculate Mr from the experimental data Moles of calcium hydroxide = 5.00 × 10 –3 – (v/2 × conc NaOH) / 1000 = z mol 1 Mr = mass of solid / z 1 Extended response Maximum of 7 marks for answers which do not show a sustained line of reasoning which is coherent, relevant, substantiated and logically structured.
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South Axholme School
Page 1
AS Paper 1 – Practical Questions MARK SCHEME
Amount of Substance
M1.(a) Stage 1: appreciation that the acid must be in excess and calculation of amount of solid that permits this
2 mol of acid react with 1 mol of calcium hydroxide therefore moles of solid weighed out must be less than half the moles of acid = 0.5 × 1.00 × 10–2 = 5.00 × 10–3 mol
1
Mass of solid must be –3 × 74.1 = 1
Stage 2: Experimental method
Measure out 50 cm3 of acid using a pipette and add the weighed amount of solid in a conical flask
1
Titrate against 0.100 (or 0.200) mol dm–3 NaOH added from a burette and record the volume (v) when an added indicator changes colour
1
Stage 3: How to calculate Mr from the experimental data
Moles of calcium hydroxide = 5.00 × 10–3 – (v/2 × conc NaOH) / 1000 = z mol 1
Mr = mass of solid / z 1
Extended response
Maximum of 7 marks for answers which do not show a sustained line of reasoning which is coherent, relevant, substantiated and logically structured.
Correct mass without working scores one mark only.
Allow 1.25 × 10–2 × 10 × 100.1= 12.5 g 1
(iv) (3.27 / 95) × 100
Accept (b(iii) / 95) × 100.
Do not penalise precision. 1
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3.44 g
Do not penalise lack of units.
Using 12.5 g gives 13.2 g
Correct answer without working scores 2 marks. 1
(v) Abundant / readily available
Accept not caustic or alkaline.
Non-corrosive
Accept insoluble so safe to add in excess (owtte). 1
[17]
M12.Include washings or words to that effect / mix contents
Accept „use distilled / deionised water‟.
Allow „weigh directly into flask‟ if washing included. 1
[1]
M13.(a) 2-6 drops / 0.1-0.3 cm3
Accept „a few drops‟ 1
(b) Incorrect volume recorded / space will fill during titration / produces larger titre value
Do not accept „to give an accurate result‟ without further qualification Do not accept references to contamination
1
[2]
M14.(a) Mass of mineral on x-axis;
If axes unlabelled use data to decide if mass of mineral is on the
x-axis. 1
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Sensible continuous scales;
Lose this mark if the plotted points do not cover at least 9 squares by 7. Lose this mark if the graph plot goes off the squared paper. The graph does not have to start at the origin.
1
Plots points correctly ± one square;
Award this mark if the line is close to your line. 1
Draws a best fit straight line
Award this mark if best fit line is consistent with candidate‟s plotted points. Lose this mark if line is kinked or doubled.
1
(b) 1.48 or 1.49 or 1.50 or 1.5 (g);
Accept these answers only Ignore precision of answer. Allow range 1.48 – 1.5
1
(c) 0.0124 (mol);
Accept 0.012, 0.0125. Allow answer without working.
1
(d) (1.49 / 0.0124) = 119.4 – 125.0;
Must divide answer to part (b) by answer to part (c) to score first mark. Allow consequential answer from part (b). Allow answer without working. Ignore precision of answer.
1
(e) Answer to part (e) close to 120.3;
Allow consequential answer from part (d).
Allow correct calculation of x 1
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(f) x must be a whole number; 1
(g) Good / straight line so results good / reliable;
Allow consequential answers from candidate‟s graph Do not allow „so results are accurate‟.
1
Anomaly at 1.34 g;
Allow anomaly clearly indicated on the graph. 1
(h) Ensure reaction / decomposition goes to completion;
Do not allow „to make fair test‟ or „improve reliability‟ Accept to „remove all carbon dioxide and water‟.
1
(i) (i) Percentage errors too high / errors in weighing too high;
Do not allow „to make fair test‟ or „improve reliability‟ Do not allow „errors‟ on its own.
1
(ii) Incomplete decomposition or words to that effect;
Do not allow „to make fair test‟ or „improve reliability‟ Do not allow „takes too long‟ or „wastes chemicals‟ Do not allow „not all of the water removed‟.
1
(j) 39.05 / 18 = 2.170 and 60.95 / 84.3 = 0.723;
Allow Mr of MgCO3.H2O = 138.3 1
MgCO3.3H2O;
54 / 138.3 + 39.05% MgCO3.3H2O without working scores 1 mark.
1
(k) Atom economy for Reaction 1 is (40.3 / 84.3) x 100 = 47.8%
Maximum 1 mark if no working.
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Ignore precision of answers. 1
Atom economy for Reaction 2 is (40.3 / 58.3) x 100 = 69.1% 1
(l) No gas produced in stomach / won’t cause wind;
Do not allow „gas produced‟ on its own. 1
[19]
Bonding Currently no practical questions Energetics
M1.(a) Start a clock when KCl is added to water 1
Record the temperature every subsequent minute for about 5 minutes
Allow record the temperature at regular time intervals untilsome time after all the solid has dissolved for M2
1
Plot a graph of temperature vs time 1
Extrapolate back to time of mixing = 0 and determine the temperature 1
(b) Heat taken in = m × c × ΔT = 50 × 4.18 × 5.4 = 1128.6 J
Do not penalise additional significant figures, but do not allow 0.02
Allow consequential answer from (i). 1
(iii) 0.018(2) ×171.3 = 3.12
Do not penalise precision.
If 0.018 used, answer = 3.08 1
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×10 = 31.2
Do not penalise precision.
Allow this mark if 0.18(2) used directly.
Correct answer without working scores one mark only.
Allow consequential answer on (ii) 1
(f) Barium sulfate / it is insoluble
Do not accept answers based on small amount ingested.
Do not accept barium. 1
[9]
M8.(a) sulfuric acid / H2SO4
1
(b) hydriodic acid / HI OR hydrobromic acid / HBr 1
(c) add dilute ammonia solution
Notes * do not allow „concentrated ammonia‟ or „ammonia’
1
precipitate / ppt disappears / dissolves OR colourless solution forms 1
(d) would react with the acid / no gas evolved in tests 1
[5]
M9.Add (hydrochloric) acid to the mixture;
acid. Allow correct acid eg nitric
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1
to isolate strontium sulphate; Filter
Do not allow „drain‟ or decant‟ 1
[2]
M10.(a) 2NaBr + 2H2SO4 Na2SO4 + Br2 + SO2 + 2H2O
Allow ionic equation
2Br– + 2H2SO4 Br2 + SO42– + SO2 + 2H2O
1
Br– ions are bigger than Cl– ions 1
Therefore Br– ions more easily oxidised / lose an electron more easily (than Cl– ions) 1
(b) This question is marked using levels of response. Refer to the Mark Scheme Instructions for Examiners for guidance on how to mark this question.
Level 3
All stages are covered and the explanation of each stage is generally correct and virtually complete. Stages 1 and 2 are supported by correct equations.
Answer communicates the whole process coherently and shows a logical progression from stage 1 to stage 2 and then stage 3. The steps in stage 3 are in a logical order.
5–6 marks
Level 2
All stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies OR two stages are covered and the explanations are generally correct and virtually complete.
Answer is mainly coherent and shows a progression through the stages. Some steps in each stage may be out of order and incomplete.
3–4 marks
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Level 1
Two stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies, OR only one stage is covered but the explanation is generally correct and virtually complete.
Answer includes some isolated statements, but these are not presented in a logical order or show confused reasoning.
1–2 marks
Level 0
Insufficient correct chemistry to warrant a mark. 0 marks
Indicative chemistry content
Stage 1: formation of precipitates
• Add silver nitrate • to form precipitates of AgCl and AgBr • AgNO3 + NaCl → AgCl + NaNO3
• AgNO3 + NaBr → AgBr + NaNO3
Stage 2: selective dissolving of AgCl
• Add excess of dilute ammonia to the mixture of precipitates • the silver chloride precipitate dissolves • AgCl + 2NH3 → Ag(NH3)2
+ + Cl−
Stage 3: separation and purification of AgBr
• Filter off the remaining silver bromide precipitate • Wash to remove soluble compounds • Dry to remove water
6
(c) Cl2 + 2HO– OCl– + Cl– + H2O 1
OCl– is +1
Cl– is –1
Both required for the mark 1
[11]
M11. (a) (i) M1 iodine OR I2 OR I3–
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Ignore state symbols
Credit M1 for “iodine solution”
M2 Cl2 + 2I – 2Cl – + I2
OR ½ Cl2 + I – Cl – + ½ I2
Penalise multiples in M2 except those shown
M2 accept correct use of I3–
M3 redox or reduction-oxidation or displacement 3
silver chloride(the white precipitate is) M1 (ii)
ignore incorrect formula this markand for must be named M1
M2 Ag+ + Cl – AgCl
For M2 ignore state symbols
Penalise multiples
it dissolves(white) precipitate / M3
colourless solution OR
Ignore references to “clear” alone 3
(b) (i) M1 H2SO4 + 2Cl – 2HCl + SO42–
For M1 ignore state symbols
OR H2SO4 + Cl– HCl + HSO4–
Penalise multiples for equations and apply the list principle
OR H+ + Cl– HCl
M2 hydrogen chloride OR HCl OR hydrochloric acid 2
(ii) M1 and M2 in either order
For M1 and M2, ignore state symbols and credit multiples
M1 2I – I2 + 2e –
OR
8I – 4I2 + 8e –
Do not penalise absence of charge on the electron
Credit electrons shown correctly on the other side of each equation
M2 H2SO4 + 8H+ + 8e – H2S + 4H2O
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OR
SO42– + 10H+ + 8e – H2S + 4H2O
Additional equations should not contradict
(ions) oxidises the iodide/ oxidising agent M3
OR
electron acceptor
M4 sulfur OR S OR S2 OR S8 OR sulphur 4
/ (sodium) hydroxide reacts with / neutralises the –NaOH / OHThe M1 (iii) (lowering its concentration) / acid / HBr +H
OR a correct neutralisation equation for H+ or HBr with NaOH or with hydroxide ion
Ignore reference to NaOH reacting with bromide ions
Ignore reference to NaOH reacting with HBrO alone
M2 Requires a correct statement for M1
(from L to R)equilibrium moves / shiftsThe (position of)
/ acid / HBr that has been removed / lost +replace the Hto •
/ acid / HBr concentration +increase the Hto OR •
/ acid / HBr / product(s) +make more Hto OR •
/ loss of product(s) +oppose the loss of Hto OR •
of product(s) crease in concentrationoppose the deto OR •
equilibrium In M2, answers must refer to the (position of) and is not enough to state simply that it / the shifts / moves
system / the reaction shifts to oppose the change.
M3 The (health) benefit outweighs the risk or wtte
OR
a clear statement that once it has done its job, little of it remains
OR
used in (very) dilute concentrations / small amounts / low doses 3
[15]
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M12. (a) NaBr ONLY
Penalise incorrect case or additional formulae. Ignore names
1
(b) NaF ONLY
Penalise incorrect case or additional formulae. Ignore names
1
from either ONLY one (c) NaF
OR
NaCl
Penalise incorrect case or additional formulae. Ignore names
1
(d) NaI ONLY
Penalise incorrect case or additional formulae. Ignore names
1
[4]
M13.(a) Increase 1
Van der Waal’s forces between molecules 1
Increase with size (or Mr or surface area etc) 1
More energy needed to break (overcome) these forces
(Note max 2 from last three marks if no mention of molecules or ‘molecular’)
1
(b) (i) Brown solution (or yellow or orange)
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1
Cl2 + 2Br → 2C1– + Br2
1
(ii) cream precipitate 1
Br– + Ag+ → AgBr 1
Precipitate dissolves 1
(iii) orange (brown) fumes (gas), White fumes (or misty fumes), choking gas (any 2)