this question is about compounds containing ethanedioate...

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 mixturein 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 aconical flask. The flask and contents were heated to 60 °C and then titrated with a0.0200 mol dm−3 solution of potassium manganate(VII). When 26.50 cm3 of potassiummanganate(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 solutionof sodium hydroxide using phenolphthalein as an indicator. The indicator changed colourafter 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)

1

of 35Catalyst Tutors

(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)


A 0.100 mol dm−3 solution of sodium hydroxide was gradually added to 25.0 cm3 of a solution ofa 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 thefigure 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)

2

(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 NaOHsolution added.

(2)

(Total 9 marks)


The diagram shows some compounds made from a halogenoalkane.

(a) Draw the displayed formula of compound J.

(1)

3

(b) Name the mechanism for Reaction 2 and give an essential condition used to ensure thatCH3CH2CH2NH2 is the major product.

Name of mechanism __________________________________________________

Condition ___________________________________________________________

(2)

(c) Calculate the mass, in grams, of CH3CH2CH2NH2 produced from 25.2 g of CH3CH2CH2Brin Reaction 2 assuming a 75.0% yield.


Mass ____________________ g

(3)


(d) When Reaction 2 is carried out under different conditions, a compound with molecularformula C9H21N is produced.

Draw the skeletal formula of the compound.

Identify the functional group in the compound including its classification.

Skeletal formula

Functional group including classification ____________________

(2)

(e) Identify the reagent and conditions used in Reaction 3.

___________________________________________________________________

(1)

(f) Name and outline a mechanism for Reaction 3.


Mechanism

(4)

(Total 13 marks)


This question is about sodium fluoride.

(a) Complete the Born–Haber cycle for sodium fluoride by adding the missing species on thelines.

(2)

4


(b) Use the data in the table and your completed Born–Haber cycle from part (a) to calculatethe enthalpy of lattice formation of sodium fluoride.

ΔHθ / kJ mol−1

Na(s) ⟶ Na(g) +109

Na(g) ⟶ Na+(g) + e− +494

F2(g) ⟶ 2F(g) +158

F(g) + e− ⟶ F−(g) −348

Na(s) + F2(g) ⟶ NaF(s) −569

Enthalpy of lattice formation ____________ kJ mol−1

(2)

(c) Suggest how the enthalpy of lattice formation of NaCl compares with that of NaF

Justify your answer.

How enthalpies of formation compare ____________________________________

___________________________________________________________________

Justification _________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(3)


(d) Calculate the volume, in cm3, of fluorine gas at 298 K and 100 kPa required to produce 1.00g of sodium fluoride by reaction with an excess of sodium.

The gas constant R = 8.31 J −1 mol−1

Volume ___________________ cm3

(4)

(Total 11 marks)

This question is about halogenoalkanes.

(a) Chlorine atoms are formed in the upper atmosphere when ultraviolet radiation causes C–Clbonds in chlorofluorocarbons (CFCs) to break.

Write two equations to show how chlorine atoms catalyse the decomposition of ozone.

1. _________________________________________________________________

___________________________________________________________________

2. _________________________________________________________________

___________________________________________________________________

(2)

5

(b) Chloroethane reacts with potassium hydroxide in the presence of propan-1-ol to formethene.

State the role of potassium hydroxide and the role of propan-1-ol in the reaction.

Role of potassium hydroxide ___________________________________________

Role of propan-1-ol ___________________________________________________

(2)


(c) Name and outline a mechanism for the reaction in part (b) between chloroethane andpotassium hydroxide to produce ethene.


Mechanism

(4)

(d) The structure of polymer A is shown.

Draw the structure of the monomer used to form polymer A.

(1)

(e) Chemical analysis shows that a chlorofluoroalkane, B, contains by mass 51.6% fluorine,32.1% chlorine and no hydrogen.

Chlorine exists as two isotopes, 35Cl and 337Cl, in the ratio 3:1Fluorine only exists as one isotope, 19F.

A mass spectrum of B is obtained using electron impact ionisation. The mass spectrumshows three molecular ion peaks at m/z = 220, 222 and 224.

Determine the formula of each of the three molecular ions of B.

Predict and explain the ratio of the relative abundancies of each of the three molecular ionpeaks at m/z = 220, 222 and 224.

To gain full marks you must show all your working.

(6)

(Total 15 marks)


Chemists design synthetic routes to convert one organic compound into another.

Buta-1,3-diene, C, is converted into compound F as shown in the diagram below.

(a) State the IUPAC name of compound F.

___________________________________________________________________

(1)

6

(b) Deduce the structure of compound D.

For each of the conversions in steps 1 and 2, suggest a reagent for the conversion andname the mechanism.

Suggest the type of reaction occurring in step 3.

Structure of D

Step 1 _____________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Step 2 _____________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

Type of reaction in Step 3 ______________________________________________

___________________________________________________________________

(6)


(c) Compound F can also be made from compound G.

State a reagent (or combination of reagents) that can be used in a test-tube reaction todistinguish between F and G.

Describe what you would observe when the reagent is added to each compound and thetest tube is shaken.

Reagent(s) _________________________________________________________

___________________________________________________________________

Observation with F ___________________________________________________

___________________________________________________________________

___________________________________________________________________

Observation with G ___________________________________________________

___________________________________________________________________

___________________________________________________________________

(3)

(d) Compounds F and G react to form a polymer.

Draw the repeating unit of the polymer.

(2)


(e) In an experiment, 0.930 kg of purified F were obtained from 1.11 dm3 of G(density 1.04 g cm−3).

Calculate the percentage yield.


Percentage yield _____________ %

(4)

(f) One reason for a yield of less than 100% in part (e) is that G reacts to form a number ofother compounds.

The other compounds are all liquids at room temperature.

Name the technique that should be used to separate and collect each of these othercompounds from the reaction mixture.

Include in your answer a description of the apparatus.

Your description of the apparatus can be either a description in words or a labelled sketch.

Name of technique ___________________________________________________

Apparatus

(4)

(Total 20 marks)


EDTA is a useful laboratory chemical and is found in a wide variety of commercial productsincluding detergents. It is very soluble in water and is often used in its ionic form EDTA4− asshown in the diagram below.

(a) EDTA4− can act as a multidentate ligand.

Explain the meanings of the terms multidentate and ligand with reference to the reactionof EDTA4− with [Cu(H2O)6

2+](aq) ions to form a complex ion.

Draw on the diagram above a separate circle around each atom that bonds to the Cu2+ ionin this complex ion.

Multidentate ________________________________________________________

___________________________________________________________________

Ligand _____________________________________________________________

___________________________________________________________________

(3)

7

(b) Copper(II) compounds may be used as fungicides in vineyards. When used in this way,copper(II) ions can enter the water supply and cause problems because they are toxic inhigh concentrations.

The water supply near a vineyard can be tested for copper(II) ions by forming a blueaqueous complex with EDTA4− ions. The concentration of this complex can be determinedusing a colorimeter.

Outline the practical steps that you would follow, using colorimetry, to determine theconcentration of this complex in a sample of water.

(3)


(c) The concentration of copper(II) ions, in the sample of water, determined by colorimetry was7.56 × 10−5 mol dm−3.

This result was checked by titrating a sample of the water with a solution containingEDTA4−(aq) ions.

The EDTA4−(aq) used in the titration had a concentration of 1.00 × 10−3 mol dm−3.

Write an equation for the reaction between [Cu(H2O)6]2+ and EDTA4− ions.

Calculate the volume of the EDTA4− solution needed to react with a 25.0 cm3 sample of thewater.

Justify whether this titration will give an accurate value for the concentration of copper(II)ions. If necessary, suggest a practical step that would improve the accuracy.

(5)

(Total 11 marks)


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

8

(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 withall 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 shownin 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 ofsodium hydrogencarbonate solution before being titrated with sodium thiosulfate solution.

Suggest the purpose of this addition.Explain your answer.

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

___________________________________________________________________

(2)

(Total 17 marks)


Mark schemes

(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) gSo % 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 byone 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 beforesubtraction (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%

1


(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 bondbeing broken and made.

1

(c)

Ignore all charges even if wrong

Ignore absence of square brackets

Candidates do not need to show 3D shape1

90° or 180°1

optical1

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

concentration in these foods.1

[14]

(a) 7–10.2

any range (i.e. 2 values) within this range1

(b)

ALLOW H3O+ for H+ and A for X

IGNORE [H+]2/[HX]

must be square brackets

IGNORE state symbols1

2

(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 square1

Line steeper at start then levels (to show buffering)

Mark independently1

[9]

(a)

Must be displayed1

3

(b) Nucleophilic substitution1

Excess NH3

Ignore aqueous, alcoholic, conc, dil, temp, heat, pressure1


(c) Amount of CH3CH2CH2Br 25.2/122.9 (=0.205) (mol)M1

Amount of CH3CH2CH2NH2 M1 × 0.75 (= 0.154) (mol)M2

Mass CH3CH2CH2NH2 M2 × 59.0 = 9.07g Must be 3sfM3

If either Mr incorrect or used incorrectly then only award 1 mark for75% yield calculation

(ignore rounding to 123 for CH3CH2CH2Br)

OR

Max mass amine = M1 × 59.0 (= 12.1) (g)

Actual mass = M2 × 0.75 = 9.07g Must be 3sf

Allow 9.09 but if 9.08 check for AE

18.9 scores 1 for 75%

(d)

Must be skeletal

Ignore lone pair1

tertiary amine or 3° amine (only award if a tertiary amine shown)1

(e) NaOH/ ethanol or KOH / ethanol (both required)NOT aqueous

Ignore heat, temp, conc., dil,

Accept alcoholic for ethanol1


(f) (Basic) Elimination

Also credit E1 mechanism

1

M1 arrow from lone pair on O of hydroxide to correct H (or to space mid way betweenhydroxide O and H)

M2 arrow from C-H bond to C-C bond following attack by OH− on the correct H

M3 arrow from C-Br bond to Br

If nucleophilic substitution shown then allow M3 only in mechanism

If wrong haloalkane used then Max 2 for mechanism

M3 curly arrow for loss of Br− & structure of carbocation

M1 arrow from lone pair on O of hydroxide to H (or to space midway between hydroxide O and H) (same as E2)

M2 arrow from C-H bond to C-C bond (same as E2)3

[13]


(a)

or

M2 Na(g) + F(g)

M1 Na(s) + F(g)

or

M2 = Na(g) + F(g)

M1 = Na(g) + 0.5F2(g)2

4

(b) LE = −109 − 494 − 158/2 + 348 − 5691

= −903 (kJ mol−1)1

(c) NaCl enthalpy of lattice formation will be less negative OR less exothermic1

Cl− ion bigger (than F−)1

So attraction of halide ion to Na+ is weaker

Allow so ionic bonding is stronger1


(d) Amount of NaF = 1/42 = 2.38 × 10−2 mol1

Amount of F2 = (2.38 × 10−2)/2 = 1.19 × 10−2 mol1

V = nRT/p = (1.19 × 10−2 × 8.31 × 298)/1000001

V = 2.95 × 10−4 m3

= 295 cm3

1

[11]

(a) Cl• + O 3 ⟶ ClO• + O 2

Allow dot in free-radical on either O or Cl.1

ClO• + O 3 ⟶ 2O2 + Cl•1

5

(b) (KOH acts as a) base1

(propan-1-ol acts as a) solvent.

Allow product of reaction between KOH and propan-1-ol /CH3CH2CH2O− acts as base.

1

(c) (Name of mechanism) Elimination1

mechanism: 3 arrows (1 mark each).

3


(d)

1


(e) This question is marked using levels of response. Refer to the Mark Scheme Instructionsfor Examiners for guidance on how to mark this question.

Level 3

All stages are covered & the explanation of each stage is generally correct and virtuallycomplete. Stages 1 and 2 are supported by correct data. Answer communicates the wholeprocess coherently and shows a logical progression from stage 1 to stage 2 and then stage3.

Steps in stage 3 are in logical order and working is shown. If there is no working for ratio orstatement of ratio then full marks cannot be awarded.

If the formulae of the three molecular ions are not correct (2d) then the student can’taccess Level 3 (any incorrect chemistry drops the student to the bottom mark withinthe level they have achieved).

5-6 marks

Level 2

Stage 2 is attempted (2a-2c do not need to be explicitly stated) but the calculation maycontain inaccuracies OR the explanation may be incomplete OR first two stages arecovered and the explanations are generally correct and virtually complete. Answer ismainly coherent and shows a progression through the first two stages. Some steps in eachstage may be incomplete.

If percentage of carbon is missing or incorrect (1a) then student can’t access Level2.

3-4 marks

Level 1

Stage 1 needs to be attempted but may contain inaccuracies.

OR

Stage 3 attempted but may contain inaccuracies / molecular formula not determined.

Answer includes some isolated statements, but these are not presented in a logical orderor show confused reasoning.

1-2 marks

Level 0

Insufficient correct chemistry to warrant a mark.0 marks

Indicative Chemistry content

Stage 1 (determines empirical formula)

1a 16.3% carbon

1b Divide by Ar

1c Divide by smallest (0.904)

1d Convert ratio in simplest integer (x 2)


C Cl F

= 1.358 = 0.904 = 2.716

3 2 6

Stage 2 (determines formulae of three molecular ions)

2a For E.F. Mr (corresponds to the molecule) = 221

2b since Mr = 221 lies within molecular ion range 220–224

2c Thus empirical formula = molecular formula

2d Three correct formulae

2e Correct Mr for each of three molecules

220 222 224

Stage 3 (explains the ratio of 3 molecular ion peaks)

3a Working

2b Correct (simplified) ratio 9:6:1

Note: 9:3:1 will be a common incorrect answer (max 5).

Working:

220 222 224

and

[15]

(a) 2,3-dimethylbutane(-1,4-)dioic acid

Penalise other numbers.

Ignore hyphens, commas, spaces.1

6


(b)

Allow displayed formula

1

Step 1:

HBr1

Electrophilic addition1

Step 2:

KCN

Not HCN, not KCN with acid.1

Nucleophilic substitution1

Step 3:

Hydrolysis1


(c)

Mark

Reagent 1

Observation with F 1

Observation with G 1

K2Cr2O7 & H2SO4 (allowacidified)

Mg Na2CO3 or NaHCO3

F: no visible change F: effervescence F: effervescence

G: orange to green G: no visible change G: no visible change

Named alcohol and conc. sulfuric acidNamed carboxylic acid and conc.

sulfuric acid

F: pleasant smell F: no visible change

G: no visible change G: pleasant smell

(d)

OR

Two ester groups.1

One unit only.Must have trailing bonds.Ignore n and brackets.

1


(e) Mass of G = (1.11 × 103) cm3 × 1.04 g cm−3 = 1154 g

65.1 scores 4 marks.1

Amount of G = 9.78 mols

1

Amount of F (actual) = =

6.37 mol

OR

Expected mass of F = 9.78 × (Mr =)

146 = 1428 g1

% yield = × 100 = 65.1(%)

OR

% yield = × 100 = 65.1(%)

M4 answer must be to 3 significant figures.1


(f) Fractional distillation

A rough labelled sketch illustrating these points scores the marks.1

Apparatus for fractional distillation must clearly work withfractionating column.

1

Fractionating column and thermometer.1

Condenser / water jacket.

Ignore heat source.1

[20]

(a) Multidentate − EDTA can form many / six dative bonds with central cation.1

Ligand − lone pair (on N or O of EDTA) can form dative bond with copper(II) ions.1

6 circles drawn on EDTA4− structure − 2 × N and 4 × −O1

7

(b) Calibrate a colorimeter / produce a calibration curve.1

By testing the colorimeter with solutions of copper-EDTA complex of known concentration.1

Add excess EDTA salt to the sample.1


(c) [Cu(H2O)6]2+ + EDTA4− ⟶⟶⟶⟶ [Cu(EDTA)]2− + 6H2O1

Amount of copper(II) = (25.0 × 7.56 × 10−5) /1000 = 1.89 × 10−6 mol1

Volume of EDTA4− = (1.89 × 10−6 / 0.001) × 1000 = 1.89 cm3

1

This is too small to be accurate.1

Dilute the EDTA4− solution / use larger volume of river water.1

[11]

(a) Amount of propanone = (25.0 × 1) / 1000 = 0.025 mol1

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

1

8

(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 mol1

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 mol1

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 incorrect 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 / x1

= −0.060Allow −0.059 to −0.061Ignore any units given.

1

(e) Gradients / rates are proportional to volumes / concentrations of propanone or B to D showthat 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]


this question is about compounds containing ethanedioate...

Documents