(total 6 marks)

Practical Paper 2 SCT of 29

Q1. This question is about ion testing.

(a) Describe how a student could distinguish between aqueous solutions of potassium nitrate, KNO3, and potassium sulfate, K2SO4, using one simple test-tube reaction.

Reagent ___________________________________________________________

Observation with KNO3(aq) _____________________________________________

___________________________________________________________________

Observation with K2SO4(aq) ____________________________________________

___________________________________________________________________

(3)

(b) Describe how a student could distinguish between aqueous solutions of magnesium chloride, MgCl2, and aluminium chloride, AlCl3, using one simple test-tube reaction.

Reagent ___________________________________________________________

Observation with MgCl2(aq) ____________________________________________

___________________________________________________________________

Observation with AlCl3(aq) _____________________________________________

___________________________________________________________________

(3)

(Total 6 marks)


Q2. An aqueous solution Y is known to contain one type of group 2 metal ion and one type of

negative ion.

Aqueous solutions of sulfuric acid and magnesium nitrate are added to separate samples of solution Y. The observations are shown in the table.

Solution added Observation with solution Y

Sulfuric acid A white precipitate forms

Magnesium nitrate A white precipitate forms

(a) Suggest the identity of the group 2 metal ion present in solution Y.

Write an ionic equation, including state symbols, for the reaction that takes place when sulfuric acid is added to solution Y.

Group 2 metal ion ____________________________________________________

Ionic equation _______________________________________________________

(2)

(b) Suggest the identity of the negative ion present in solution Y.

Write an ionic equation, including state symbols, for the reaction that takes place when magnesium nitrate is added to solution Y.

Negative ion ________________________________________________________

Ionic equation _______________________________________________________

(2)

(Total 4 marks)


Q3. Iodine reacts slowly with propanone in the presence of an acid catalyst according to the

equation

CH3COCH3 + l2 ⟶ CH3COCH2l + Hl

The rate of this reaction can be followed by preparing mixtures in which only the initial concentration of propanone is varied. At suitable time intervals, a small sample of the mixture is removed and titrated with sodium thiosulfate solution. This allows determination of the concentration of iodine remaining at that time.The rate of this reaction can be followed by preparing mixtures in which only the initial concentration of propanone is varied. At suitable time intervals, a small sample of the mixture is removed and titrated with sodium thiosulfate solution. This allows determination of the concentration of iodine remaining at that time.

Five mixtures, A, B, C, D and E, are prepared as shown in Table 1.

Table 1

Mixture A B C D E

Volume of 0.0200 mol dm−3 I2(aq)/cm3 40.0 40.0 40.0 40.0 40.0

Volume of 0.100 mol dm−3 H2SO4(aq)/cm3 25.0 25.0 25.0 25.0 25.0

Volume of 1.00 mol dm−3 CH3COCH3(aq)/cm3

25.0 20.0 15.0 10.0 6.5

Volume of distilled water/cm3 0.0 5.0 10.0 15.0 18.5

(a) Calculate the initial concentration, in mol dm−3, of the propanone in mixture A.

Concentration = _________ mol dm−3

(2)

(b) State and explain why different volumes of water are added to mixtures B, C, D and E.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)


(c) Calculate the volume of 0.0100 mol dm−3 sodium thiosulfate solution required to react with all of the iodine in a 10.0 cm3 sample of mixture E, before the iodine reacts with propanone.

The equation for the reaction in the titration is

2Na2S2O3(aq) + I2(aq) ⟶ Na2S4O6(aq) + 2Nal(aq)

Volume = __________________ cm3

(4)


(d) The results for mixture E are shown in Table 2. V is the volume of 0.0100 mol dm−3 sodium thiosulfate solution needed, at different times, t, to react with the iodine in a 10.0 cm3 sample of E.

Table 2

t/min 5 10 20 30

V/cm3 17.5 17.2 16.6 16.0

Use these data and your answer to part (c) to plot a graph of V (y-axis) against t

(x-axis) for mixture E.

Draw a best-fit straight line through your points and calculate the gradient of this line.

gradient = _____________ cm3 min−1

(5)


(e) The gradients for similar graphs produced by mixtures A, B, C and D are shown in Table 3. Each gradient is a measure of the rate of the reaction between iodine and propanone.

Table 3

Mixture A B C D

Gradient / cm3 min−1 −0.24 −0.20 −0.15 −0.10

Use information from Table 1 and Table 3 to deduce the order with respect to propanone. Explain your answer.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(f) Each sample taken from the reaction mixtures is immediately added to an excess of sodium hydrogencarbonate solution before being titrated with sodium thiosulfate solution.

Suggest the purpose of this addition. Explain your answer.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(Total 17 marks)


Q4. The reaction between propanone and iodine in the presence of hydrochloric acid was studied

at a constant temperature.

CH3COCH3 + I2 → CH3COCH2I + HI

The following rate equation was deduced.

rate = k [CH3COCH3][H+]

(a) Suggest why the order with respect to iodine is zero.

___________________________________________________________________

___________________________________________________________________

(1)

(b) In an experiment the initial concentrations of propanone, iodine and hydrochloric acid were as shown in the table. The initial rate of reaction in this experiment was 8.64 × 10–7 mol dm–3 s–1.

Initial concentration / mol dm–3

CH3COCH3 5.82 × 10–2

I2 1.78 × 10–3

H+ 4.76 × 10–1

Use the data in the table and the rate equation to calculate a value for the rate constant at this temperature. Give units with your answer.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(c) A series of experiments was carried out using concentrations of propanone approximately 100 times the concentrations of iodine and hydrochloric acid.

Suggest the rate equation under these conditions. Explain your answer.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)


Q5. Hydrogen peroxide is a powerful oxidising agent. Acidified hydrogen peroxide reacts with

iodide ions to form iodine according to the following equation.

H2O2(aq) + 2H+(aq) + 2I−(aq) → I2(aq) + 2H2O(l)

The initial rate of this reaction is investigated by measuring the time taken to produce sufficient iodine to give a blue colour with starch solution.

A series of experiments was carried out, in which the concentration of iodide ions was varied, while keeping the concentrations of all of the other reagents the same. In each experiment the time taken (t) for the reaction mixture to turn blue was recorded.

The initial rate of the reaction can be represented as ( ), and the initial concentration of iodide ions can be represented by the volume of potassium iodide solution used.

A graph of log10 ( ) on the y-axis against log10 (volume of KI(aq)) is a straight line. The

gradient of this straight line is equal to the order of the reaction with respect to iodide ions.

The results obtained are given in the table below. The time taken for each mixture to turn blue was recorded on a stopclock graduated in seconds.

Expt. Volume of

KI(aq) / cm3

log10 (volume of KI(aq))

Time / s log10 ( )

1 5 0.70 71 −1.85

2 8 0.90 46 −1.66

3 10 1.00 37 −1.57

4 15 1.18 25 −1.40

5 20 1.30 19 −1.28

6 25 1.40 14 −1.15


(a) Use the results given in the table to plot a graph of log10 ( ) on the y-axis against

log10 (volume of KI(aq)).

Draw a straight line of best fit on the graph, ignoring any anomalous points.

(5)


(b) Determine the gradient of the line you have drawn. Give your answer to two decimal places. Show your working.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(3)

(c) Deduce the order of reaction with respect to iodide ions.

___________________________________________________________________

(1)

(d) A student carried out the experiment using a flask on the laboratory bench. The student recorded the time taken for the reaction mixture to turn blue. State one way this method could be improved, other than by repeating the experiment or by improving the precision of time or volume measurements. Explain why the accuracy of the experiment would be improved.

Improvement ________________________________________________________

___________________________________________________________________

Explanation _________________________________________________________

___________________________________________________________________

(2)

(Total 11 marks)


Q6.

A 0.100 mol dm−3 solution of sodium hydroxide was gradually added to 25.0 cm3 of a solution of a weak acid, HX, in the presence of a suitable indicator.

A graph was plotted of pH against the volume of sodium hydroxide solution, as shown in the figure below.

The first pH reading was taken after 20.0 cm3 of sodium hydroxide solution had been added.

The acid dissociation constant of HX, Ka, = 2.62 × 10−5 mol dm−3

(a) The pH range of an indicator is the range over which it changes colour.

Suggest the pH range of a suitable indicator for this titration.

___________________________________________________________________

(1)

(b) Give the expression for the acid dissociation constant of HX.

Ka =

(1)


(c) Calculate the concentration of HX in the original solution.

Concentration ____________________ mol dm−3

(2)

(d) Calculate the pH of the solution of HX before the addition of any sodium hydroxide.

(If you were unable to calculate a value for the concentration of HX in part (c) you

should use a value of 0.600 mol dm−3 in this calculation. This is not the correct value.)

pH of HX ____________________

(2)

(e) Calculate the pH of the solution when half of the acid has reacted.

pH of solution ____________________

(1)

(f) Plot your answers to part (d) and part (e) on the grid in the figure above.

Use these points to sketch the missing part of the curve between 0 and 20 cm3 of NaOH solution added.

(2)

(Total 9 marks)


Q7. This question is about compounds containing ethanedioate ions.

(a) A white solid is a mixture of sodium ethanedioate (Na2C2O4), ethanedioic acid dihydrate (H2C2O4.2H2O) and an inert solid. A volumetric flask contained 1.90  g of this solid mixture in 250  cm3 of aqueous solution.

Two different titrations were carried out using this solution.

In the first titration 25.0 cm3 of the solution were added to an excess of sulfuric acid in a conical flask. The flask and contents were heated to 60 °C and then titrated with

a 0.0200 mol dm−3 solution of potassium manganate(VII). When 26.50 cm3 of potassium manganate(VII) had been added the solution changed colour.

The equation for this reaction is

2MnO4− + 5C2O4

2− + 16H+ → 2Mn2+ + 8H2O + 10CO2

In the second titration 25.0 cm3 of the solution were titrated with a 0.100 mol dm−3 solution of sodium hydroxide using phenolphthalein as an indicator. The indicator changed colour after the addition of 10.45 cm3 of sodium hydroxide solution.

The equation for this reaction is

H2C2O4 + 2OH− → C2O42− + 2H2O

Calculate the percentage by mass of sodium ethanedioate in the white solid.

Give your answer to the appropriate number of significant figures.

Show your working.

Percentage by mass of sodium ethanedioate ____________________ %

(8)


(b) Ethanedioate ions react with aqueous iron(III) ions in a ligand substitution reaction.

Write an equation for this reaction.

Suggest why the value of the enthalpy change for this reaction is close to zero.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(c) Draw the displayed formula of the iron complex produced in the reaction in part (b)

Indicate the value of the O—Fe—O bond angle.

State the type of isomerism shown by the iron complex.

Bond angle _________________________________________________________

Type of isomerism ____________________________________________________

(3)

(d) Ethanedioate ions are poisonous because they react with iron ions in the body. Ethanedioate ions are present in foods such as broccoli and spinach.

Suggest one reason why people who eat these foods do not suffer from poisoning.

___________________________________________________________________

___________________________________________________________________

(1)

(Total 14 marks)


Q8. Benzoic acid can be prepared from ethyl benzoate.

Ethyl benzoate is first hydrolysed in alkaline conditions as shown:

A student used the following method.

Add 5.0 cm3 of ethyl benzoate (density = 1.05 g cm−3, Mr = 150) to 30.0 cm3 of aqueous

2 mol dm−3 sodium hydroxide in a round-bottomed flask.

Add a few anti-bumping granules and attach a condenser to the flask. Heat the mixture under reflux for half an hour. Allow the mixture to cool to room temperature.

Pour 50.0 cm3 of 2 mol dm−3 hydrochloric acid into the cooled mixture.

Filter off the precipitate of benzoic acid under reduced pressure.

(a) Suggest how the anti-bumping granules prevent bumping during reflux.

___________________________________________________________________

___________________________________________________________________

(1)

(b) Show, by calculation, that an excess of sodium hydroxide is used in this reaction.

(2)

(c) Suggest why an excess of sodium hydroxide is used.

___________________________________________________________________

___________________________________________________________________

(1)

(d) Suggest why an electric heater is used rather than a Bunsen burner in this hydrolysis.

___________________________________________________________________

___________________________________________________________________

(1)


(e) State why reflux is used in this hydrolysis.

___________________________________________________________________

___________________________________________________________________

(1)

(f) Write an equation for the reaction between sodium benzoate and hydrochloric acid.

___________________________________________________________________

(1)

(g) Suggest why sodium benzoate is soluble in cold water but benzoic acid is insoluble in cold water.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(h) After the solid benzoic acid has been filtered off, it can be purified.

Describe the method that the student should use to purify the benzoic acid.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(6)

(i) In a similar experiment, another student used 0.040 mol of ethyl benzoate and obtained 5.12 g of benzoic acid.

Calculate the percentage yield of benzoic acid.

Suggest why the yield is not 100%.

percentage yield ____________________ %

Suggestion _________________________________________________________

___________________________________________________________________

___________________________________________________________________

(3)

(Total 18 marks)


Q9. Test-tube reactions can be used to identify the functional groups in organic molecules.

You are provided with samples of each of the four compounds.

Describe how you could distinguish between all four compounds using the minimum number of tests on each compound.

You should describe what would be observed in each test.

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

_______________________________________________________________________

(Total 6 marks)

Q10. Draw a fully labelled diagram of the apparatus to show how propan-1-ol can be converted into propanoic acid in the laboratory.

(Total 4 marks)


Q11. In a titration experiment, a good technique is essential for an accurate result to be obtained.

(a) Suggest a reason for removing the funnel after it has been used for filling the burette.

___________________________________________________________________

___________________________________________________________________

(1)

(b) Suggest one other source of error in using the burette to carry out a titration.

___________________________________________________________________

___________________________________________________________________

(1)

(c) During the titration, the inside of the conical flask is rinsed with distilled water.

Suggest why rinsing improves the accuracy of the titre.

___________________________________________________________________

___________________________________________________________________

(1)

(d) Explain why adding this extra water does not change the volume of EDTA solution that is required in the titration.

___________________________________________________________________

___________________________________________________________________

(1)

(Total 4 marks)


Mark schemes

Q1. (a) BaCl2 / Ba(OH)2 / Ba(NO3)2 / BaX2 or names

Ignore acidification but CE = 0/3 if H2SO4

If reagent incorrect or blank then CE =0/3

If Ba2+ or wrong formula, lose M1 and mark on 1

colourless solution / no (visible) change (nvc) / no ppt / no (visible) reaction

Ignore nothing happens and no observation 1

white precipitate / white solid 1

(b) NaOH / sodium hydroxide / other Group 1 hydroxides

If reagent incorrect or blank then CE =0/3

If reagent incomplete, lose M1 and mark on 1


(white) ppt which dissolves in excess (NaOH)

If reagent is excess NaOH, allow colourless solution for M3 1

Alternative Method

Name or formula of Group 1 carbonate 1


(white) precipitate and effervescence 1

[6]

Q2.

(a) Ba2+ OR Sr2+

1

Award M1 if barium named in M1 then used Ba2+ in the equation

M2 SO42− (aq) + Ba2+ (aq) ⟶ BaSO4 (s)

1

(b) OH−

Award M1 if hydroxide named in M1 then used OH− in the equation

1

Mg2+ (aq) + 2OH− (aq) ⟶ Mg(OH)2 (s) 1


[4]

Q3. (a) Amount of propanone = (25.0 × 1) / 1000 = 0.025 mol

1

Concentration in mixture A = 0.025 / (90.0/1000) = 0.278 mol dm−3

1

(b) To make volumes constant for all mixtures. 1

So that volume of propanone is proportional to concentration. 1

(c) Amount of iodine in mixture E = (40.0 × 0.02) / 1000 = 8.0 × 10−4 mol 1

Amount of iodine in sample = 8.0 × 10−4 × (10 / 90) = 8.89 × 10−5 mol

As 10 cm3 sample taken from total volume of 90 cm3

1

Amount of thio required = 2 × 8.89 × 10−5 = 1.78 × 10−4 mol 1

Volume of thio = (1.78 × 10−4 / 0.01) × 1000 = 17.8 cm3

1

(d) Scale

Graph must cover at least half the grid and axes must be plotted in correct orientation but ignore labeling.

1

Points

Must be correctly plotted to within ± half a small square. 1

Best-fit straight line

Must be the best-fit line possible if point(s) are plotted incorrectly. Penalise doubled or kinked lines.

1

Gradient = y / x 1

= −0.060

Allow −0.059 to −0.061 Ignore any units given.

1

(e) Gradients / rates are proportional to volumes / concentrations of propanone or B to D show that gradient / rate halves when vol / concentration halves.

1

So first order with regard to propanone. 1


(f) To stop / quench the reaction at that time.

Ignore NaHCO3 reacts with the HI produced. 1

By removing the acid catalyst for the reaction (by neutralisation). 1

[17]

Q4. (a) Iodine is not involved in (or before) the rate determining / slow(est) / limiting step (in

the mechanism)

Ignore, iodine does not appear in the rate equation or iodine concentration does not affect the rate

1

(b)

Mark for answer 1

mol–1 dm+3 s–1

Mark units separately, i.e. only these units but can be in any order

1

(c) Rate = k [H+]

If wrong or missing CE = 0 1

(Large excess of propanone) so [CH3COCH3] is (effectively) constant 1

[5]

Q5.

(a) Log (1 / time) on the y-axis + log (vol) on x-axis

If axes unlabelled use data to decide that log (1 / time) is on

the y-axis 1

Sensible scales

Lose this mark if the plotted points do not cover at least half of the paper

Lose this mark if the graph plot goes off the squared paper

Lose this mark if plots a non-linear / broken scale

Lose this mark if uses an ascending y-axis of negative

numbers 1

Plots points correctly ± one square 1

Line through the points is smooth

Lose this mark if the candidate’s line is doubled


1

Line through the points is best fit – ignores last point

Must recognise that point at 25 cm3 is an anomaly

If wrong graph, mark consequentially on anomaly if correctly plotted.

A kinked graph loses smooth and best fit marks 1

(b) Uses appropriate x and y readings

Allow taken from table or taken or drawn on graph

Must show triangle on graph or such as 1

Correctly calculates gradient 0.95 ± 0.02

Ignore positive or negative sign

Correct answer only with no working scores this mark 1

Answer given to 2 decimal places 1

(c) First order or order is 1

Allow consequential answer from candidate’s results 1

(d) Thermostat the mixture / constant temperature / use a water bath or Colorimeter / uv-visible spectrometer / light sensor to monitor colour change

1

Reaction / rate affected by temperature change or Eliminates human error in timing / more accurate time of colour change

1

[11]

Q6. (a) 7–10.2

any range (i.e. 2 values) within this range 1

(b)

ALLOW H3O+ for H+ and A for X

IGNORE [H+]2/[HX]

must be square brackets

IGNORE state symbols 1

(c) Amount NaOH = (24.0 × 0.100)/1000 = 2.40 × 10−3 mol

(= amount HX)

Conc HX = 2.40 × 10−3/0.025 = 0.0960 mol dm−3


ecf for M1/0.025

(d) (Ka = 2.62 × 10−5 = [H+]2/0.0960)

[H+] = √(2.62 × 10−5 × 0.0960) (= 1.59 × 10−3 mol dm−3)

ecf from part (c) [H+] = √(2.62 × 10−5 × ans to part (c))

From alternative data

[H+] = √(2.62 × 10−5 × 0.600) (= 3.96 × 10−3 mol dm−3)

(pH = –log 1.59 × 10−3 =) 2.80 (must be 2 or more dp) 1

pH = 2.40 (must be 2 or more dp)

M2 dependent on a calculation of [H+] 1

(e) (pH at half-neutralisation = pKa)

= –log 2.62 × 10−5 = 4.58 (must be 2 or more dp)

ALLOW 1dp if already penalised in part (d) 1

(f) Both points plotted correctly and line touches both points

ecf from (d) and (e) within 1 small square 1

Line steeper at start then levels (to show buffering)

Mark independently 1

[9]

Q7.

(a) Moles MnO4− = 5.30 × 10−4

1

Moles in 25cm3 sample / pipette C2O42− ( from acid and salt)

= 5.30 × 10−4 × 5/2 = (1.325 × 10−3) 1

Moles NaOH = ( = 1.045 × 10−3) 1

So moles C2O42− from acid in 25cm3 sample / pipette

= 1.045 × 10−3 ÷ 2 = 5.225 × 10−4

1

Hence moles C2O42− in sodium ethanedioate in 25 cm3

= 1.325 × 10−3 – 5.225 × 10−4 (= 8.025 × 10−4) 1

So moles C2O42− in sodium ethanedioate in original sample

= 8.025 × 10–4 × 10 (= 8.025 × 10−3) 1

Mass Na2C2O4 = 8.025 × 10−3 × 134(.0) = 1.075(35) g So % sodium ethanedioate in original sample

1


× 100 = 56.6 % to 3 sig fig 1

The first CE is penalised by 2 marks; further errors are penalised by one mark each

M2 = M1 × 5/2

M4 = M3 ÷ 2

M5 = M2 – M4 (do not allow if negative and do not allow = M4-M2)

If no subtraction, max = 5 (M1, M2, M3, M4 and M6)

If incorrect subtraction, max = 6 (M1, M2, M3, M4, M6 and M7)

M6 = M5 × 10

(M6 can be scored by multiplying M2 and M4 by 10 before subtraction (giving 1.325 × 10−2 – 5.225 × 10−3 = 8.025 × 10−3 )

M7 = M6 × 134

M8 = (M7/1.90) × 100 Allow 56.5 – 56.8%

(b) [Fe(H2O)6]3+ + 3C2O42− → [Fe(C2O4)3]3− + 6H2O

1

There are 6 Fe–O bonds broken and then made / same number and type of bond being broken and made.

1

(c)

Ignore all charges even if wrong

Ignore absence of square brackets

Candidates do not need to show 3D shape 1

90° or 180° 1

optical 1

(d) The ethanedioic acid is only present in small quantities/low

concentration in these foods. 1

[14]


Q8. (a) allows smaller bubbles to form / prevents the formation of (very) large bubbles

ALLOW provides large surface area for bubbles to form on

IGNORE ‘air’

NOT no bubbles form / prevents bubbles forming 1

(b) (Mass of ester = 1.05 × 5.0 = 5.25g) amount of ester = 5.25 / 150.0 = 0.0350 mol

1

amount of NaOH = 30 × 2 / 1000 = 0.06 mol 1

OR

(Mass of ester = 1.05 × 5.0 = 5.25g) amount of ester = 5.25 / 150.0 = 0.0350 mol

1

Vol of 0.035 mol of NaOH = (0.035/2) × 1000 = 17.5 cm3

(so 30 cm3 used is an excess) 1

OR

amount of NaOH = 30 × 2 / 1000 = 0.06 mol 1

0.06 mol of ester = 9 g = 8.57 cm3

(only 5 cm3 used so NaOH in excess) 1

Mark independently Max 2

(c) To ensure that the ester is completely hydrolysed / to ensure all the ester reacts

ALLOW to ensure the other reagent has completely reacted 1

(d) Many organic compounds / the ester / ethanol are flammable

ALLOW prevent ignition of any flammable vapours formed 1

(e) Reflux allows reactant vapours (of volatile organic compounds) to be returned to the reaction mixture / does not allow any reactant vapour to escape

IGNORE reference to products 1

(f) C6H5COONa + HCl → C6H5COOH + NaCl

Allow ionic equation.

ALLOW molecular formulae (C7H5O2Na and C7H6O2 )

ALLOW skeletal benzene ring 1

(g) Sodium benzoate soluble because it is ionic

IGNORE polar


1

Benzoic acid insoluble because: despite the polarity of the COOH group / ability of COOH to form H-bonds, the benzene ring is non-polar.

ALLOW ‘part of molecule’ or ‘one end’ for COOH 1

(h) Dissolve crude product in hot solvent/water

ALLOW ethanol

If no M1 max = 4 1

of minimum volume

ALLOW reference to saturated soln as alternative to ‘min vol’ 1

Filter (hot to remove insoluble impurities)

IGNORE use of Buchner funnel here 1

Cool to recrystallise

apply list principle for each additional process in an incorrect method but IGNORE additional m.pt determination

1

Filter under reduced pressure / with Buchner/Hirsch apparatus 1

wash (with cold solvent) and dry 1

(i) 5.12 / 122 (= 0.042 mol)

method mark 1

(0.042/0.04) × 100 = 105 %

ecf for M1/0.04

or calculation that 0.04 mol of benzoic = 4.88 g (M1) so

% yield = (5.12/4.88) × 100 = 105% 1

Product not dried / impurities present in product

Only allow M3 if M2>100% 1

[18]

Q9. This question is marked using Levels of Response. Refer to the Mark Scheme Instructions for Examiners for guidance.

Level 3 (5 – 6 marks) All stages are covered and each stage is generally correct and virtually complete. Answer is communicated coherently and shows a logical progression from Stage 1 to Stages 2 and 3 to distinguish all the compounds with results for all remaining compounds stated. Describing subsequent organic test on product (unnecessary) - limits to lower mark in


level

Level 2 (3 – 4 marks) All stages are covered but stage(s) may be incomplete or may contain inaccuracies OR two stages are covered and are generally correct and virtually complete. Answer is communicated mainly coherently and shows a logical progression from Stage 1 to Stages 2 and 3. Describing subsequent organic test on product (unnecessary) - limits to lower mark in level

Level 1 (1 – 2 marks) Two stages are covered but stage(s) may be incomplete or may contain inaccuracies OR only one stage is covered but is generally correct and virtually complete. Answer includes isolated statements but these are not presented in a logical order.

Level 0 (0 marks) Insufficient correct chemistry to gain a mark.

Indicative chemistry content

Stage 1: An initial test to separate into two groups (2 groups of 2 OR 1 group of 3 and 1 group of 1)

Stage 2: A second test to distinguish within a group or to separate into two further groups

Stage 3: A third test leads to a set of results/observations which distinguishes between all 4 compounds

Tests must include reagent and observation which identifies compound(s)

−COOH a) NaHCO3 / Na2CO3 (or correct alternative) b) effervescence /gas turns limewater milky c) K and /or M but not L and/or N

-OH and -CHO d) acidified K2Cr2O7

e) solution turns green f) K and/or L and/or N but not M

-CHO g) Fehlings OR Tollens h) red ppt OR silver mirror i) N only but not K and/or L and/or M

-Br j) Silver nitrate k) cream ppt l) L and/or N but not K and/or M

Isolated tests on individual compounds - max LEVEL 2 Isolated tests not linked to any compound – max LEVEL 1 Penalise observation if deduction wrong, but allow observation if deduction incomplete

Alternative tests

-COOH -COOH -OH only

a) named alcohol & a) named indicator m) named carboxylic


H2SO4

b) sweet smell (of ester)

c) K and /or M but not L and/or N

b) correct colour c) K and /or M but not

L and/or N

acid & H2SO4

n) sweet smell (of ester)

o) K and/or L but not M and /or N

Test Tests for

K L M N

a) NaHCO3 / Mg / Indicator

K M ✓ ✘ ✓ ✘

d) K2Cr2O7 / H+ K L N ✓ ✓ ✘ ✓

g) Fehlings / Tollens

N ✘ ✘ ✘ ✓

j) AgNO3 see Note *

L N ✘ ✓ ✘ ✓

a) named alcohol & H2SO4

K M ✓ ✘ ✓ ✘

m) named carboxylic acid & H2SO4

K L ✓ ✓ ✘ ✘

Note * allow NaOH then HNO3, AgNO3 as one test; but treat NaOH, AgNO3 without acid as incomplete,so can mark on.

[6]

Q10. Acidified (or a suitable acid) potassium dichromate (or suitable oxidising agent)

Penalise missing ‘acidified’ once in paper 1

Heat source 1

Flask with vertical water condenser

Ignore additional distillation condenser 1

No gaps in the apparatus apart from at the top where it must be open

The top of the condenser must not be sealed or covered in any way

1

[4]


Q11. (a) As a droplet from the funnel could enter the burette / affect volume / readings / titre

1

(b) Air bubble in jet or wtte

Do not allow misreading burette or overshooting end point. 1

(c) Ensures all reagents are able to react / mix / come into contact

Accept no reagent is left unreacted on sides of flask

Do not allow any reference to ‘removal’ of the solution unless it is clear that it is added to the flask.

1

(d) The added water does not affect the mols / amount of reagents / reactants / solution Z

Do not allow mols of solution or mols in the flask.

Allow water does not react with the reagents / water is not one of the reactants

Do not allow ‘water is not involved’ 1

[4]

(total 6 marks)

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