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SL & HL Answers to Paper 3 Section A Experimental work Question 7

7. (a) The student has not included the water of crystallisation in the sum for the molar mass. The correct molar mass should be 126.05 g mol-1. [1] The correct concentration = 12.03 ÷ 126.05 = 0.0954 mol dm−3. [1]

(b) A volumetric flask. (The flask is filled until the bottom of the meniscus lies on the graduated mark.) [1]

(c) It will make the oxalic acid slightly weaker so more of the acid will be required to neutralise the sodium hydroxide. This means that the concentration of the sodium hydroxide will be determined to be slightly higher than the true value. [1]

(d) It will have no effect. It does not alter the amount of sodium hydroxide present so the same amount of oxalic acid solution will be required to neutralise it. [1]

(e) The sodium hydroxide solution will have a much higher conductivity as it is a strong base so is completely ionized in solution. Oxalic acid is a weak acid. As it is only slightly dissociated, the concentration of ions in solution will be much lower, which will make the conductivity much less. [1]

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SL & HL Answers to Paper 3 Section A Experimental work Question 6

6. (a) Amount of magnesium = 0.196 ÷ 24.31 = 8.063 x 10−3 mol Amount of oxygen = (0.299 − 0.196) ÷ 16.00 = 6.438 x 10−3 mol (8.063 x 10−3) ÷ (6.438 x 10−3) = 1.25 so simplest ratio for Mg:O is 5:4. [1]

(b) Add a few drops of dilute hydrochloric acid and test to see whether hydrogen gas is evolved. [1] (Note that both Mg and MgO react with hydrochloric acid but only magnesium gives hydrogen as a product. The ‘pop’ test can be used to identify the hydrogen.)

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

or for any dilute mineral acid: Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) [1]

(c) Repeat the experiment by reheating the product until a constant weight of product is obtained. [1]

(d) Place a known mass of magnesium in a porcelain ‘boat’ inside a glass combustion tube and pass oxygen through the tube. Heat the magnesium strongly and have a filter before the exit tube to stop any of the magnesium oxide escaping. or show using a diagram

[2]

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SL & HL Answers to Paper 3 Section A Experimental work Question 5 5. (a) Place the manganese dioxide powder in a small tube standing vertically inside the flask which can be knocked over to start the reaction. [1]

(b) Initially there will be a colourless liquid. When the MnO2 is added the black solid will be dispersed throughout the liquid and the liquid will effervesce as gas is evolved. The effervescence will decrease as the reaction proceeds and when the reaction is complete the MnO2 will settle as a black precipitate on the bottom of the flask leaving the clear solution above. [2]

(c) From the gradient (shown on the graph below) the rate of reaction after one minute = 36 ÷ 1.0 = 36 cm3 min−1. [1]

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(d) Although the mass increases tenfold the surface area does not and the hydrogen peroxide only reacts on the surface of the heterogeneous catalyst. [1]

(e) 1 mol of gas at STP occupies 22700 cm3 (from data booklet or use PV = nRT) Amount of oxygen evolved at STP = 100 ÷ 22700 = 4.41 x 10−3 mol Two mol of H2O2 are required to produce 1 mol of O2 Amount of H2O2 present initially = 2 x (4.41 x 10−3) = 8.82 x 10−3 mol [1]

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SL & HL Answers to Paper 3 Section A Experimental work Question 4

4. (a) To prevent any of the ethanol from evaporating (particularly when the spirit lamp was hot after the combustion), as this would have affected the mass. [1]

(b) Heat evolved by combusting 0.553 g = 4.18 x 0.250 x 10.4 = 10.868 kJ [1] M(C2H5OH) = (2 x 12.01) + (6 x 1.01) + 16.00 = 46.08 g mol-1 ΔH = − (46.08 ÷ 0.553) x 10.868 = − 906 kJ mol−1 (minus as exothermic) [1]

(c) Experimental error = ((1367 – 906) ÷ 1367) x 100 = 33.7% [1]

(d) Any two from: The water in the beaker needed to be stirred to ensure a uniform temperature throughout.

The heat content of the glass, thermometer and stirrer (if used) needed to be taken into account as these were also being heated up.

The experimental error was obtained by comparing the result obtained with the standard enthalpy of combustion but the reaction was not carried out under standard conditions. [2 max]

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SL&HLAnswerstoPaper3SectionAExperimentalworkQuestion3

3.(a)Asaltbridgeneedstobeaddedtocompletethecircuitbyallowingtheflowofionsbetweenthetwosolutions.[1]

(b)Theelectronswillflowfromthemagnesiumtothecopper.(Magnesiumismorereactivethancoppersowillloseelectronstoformmagnesiumionsandflowthroughtheexternalcircuittoreducethecopper(II)ionstocoppermetal).[1]

Mg(s)+Cu2+(aq)→Mg2+(aq)+Cu(s)[1]

(c)Thesolutionwillloseitsbluecolourandareddishbrownprecipitatewillbeformed.[1]

(d)Itwillbesmalleraszincislessreactivethanmagnesium.[1]

(e)Nochangewilltakeplace(ascopperdoesnotreactwithzincasitisbelowzincintheactivityseries).[1]

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SL&HLAnswerstoPaper3SectionAExperimentalworkQuestion2

2.(a)Na2CO3(s)+2HCl(aq)→2NaCl(aq)+CO2(g)+H2O(l)[1]orCO3

2−(s)+2H+(aq)→CO2(g)+H2O(l)

(b)Thecarbondioxidegasevolvedcauseseffervescencesothecottonwoolpreventsanyoftheliquidescapingfromtheflask.[1]

(c)Theendpointoccurswhentheadditionofonedropofthesodiumhydroxidesolutioncausesafaintpinkcolourtoremain.[1](NotethatthecolourchangesforcommonindicatorsaregiveninSection22ofthedatabooklet.)

(d)AmountofNaOHin12.60cm3of0.100moldm-3NaOH=(12.60x0.100)÷1000=1.26x10−3molSince1molofNaOHreactswith1molofHClAmountofHClin10.0cm3sample=1.26x10−3molAmountofHClin100cm3volumetricflask=theamountremainingafterthereaction=10x1.26x10−3=1.26x10−2mol[1]

(e)Initialamountofhydrochloricacidadded=50.0x1.00÷1000=0.0500molAmountthatreactedwiththewashingsodacrystals=0.0500–0.0126=0.0374molSince2molofHClreactwith1molofwashingsodacrystalsAmountofwashingsodacrystalsplacedinflask=½x0.0374=0.0187mol[1]

(f)0.0187molofwashingsodacrystalshasamassof5.35g1molofwashingsodacrystalshasamassof5.35÷0.0187=286.1gM(Na2CO3.xH2O)=(2x22.99)+12.01+(3x16.0)+x(18.02)=105.99+18.02xNumberofmolesofwaterofcrystallisation=x=(286.1–105.99)÷18.02=10.0[1]

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SL&HLAnswerstoDataresponsequestion7onCompoundXa.i.Fromtheelementalanalysis

Element Amount/mol Simplestratio

Carbon 68.84/12.01=5.73 7

Hydrogen 4.96/1.01=4.91 6

Oxygen 26.2/16.00=1.64 2

theempiricalformulaisC7H6O2[1]ThemolecularionpeakgivesMas122gmol-1sothemolecularformulaisC7H6O2[1]ii.Thehydrogenatomsinthemoleculeexistintwodifferentchemicalenvironments.[1]FiveHatomsareinthesameenvironmentandoneisinadifferentenvironment.[1]iii.C6H5COOH.[1]b.HX(aq)+H2O(l)⇌ X−(aq)+H3O+(aq)orHX(aq)⇌ X−(aq)+H+(aq)[1](Notethattogainthemarktheequationmusthavereversiblearrowsasthevaluefortheequilibriumconstantshowsthatitisaweakacid.)c.RecrystalliseCompoundXfromdistilledwater.[1](Dissolveinboilingwater,filtertoremoveanyinsolubleimpurities,leavetocoolthenfilteroffthecrystalsandwashwithalittlecolddistilledwaterbeforedrying.)d.i.Condensationoresterification.[1]ii.C7H6O2+C3H8O→H2O+C10H12O2sothemolecularformulaisC10H12O2[1]

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SL&HLAnswerstoDataresponsequestion8oncoala.Forgecoal[1]b.iCorrectcalculationofmolaramounts[1];35708kJkg-1[1]ii.Thecalculationinb.iistheoreticalasitassumesstandardconditionsandthatcompletecombustionofallthreeelementsoccurswhichwillnotbethecasewhencoalburnsnormally.[1]c.44.01gofcarbondioxideformedwhen12.01gofcarboncombustcompletelyMassofcoalrequiredtoproduce1.442x1010tonnesofCO2

=1.442x1010x(12.01÷44.01)x(100÷87.0)=4.523x109tonnes.[1]MassofcoalnotreleasingCO2intotheatmosphere=7.695x109–4.523x109=3.172x109tonnesPercentageofcoalnotreleasingCO2intotheatmosphere=((3.172x109)÷(7.695x109))x100=41.2%.[1]d.Carbondioxideconsistsofsimplenon-polarO=C=Omoleculeswhichhaveweakforcesofattractionbetweenthem.[1]Silicondioxideconsistsofagiantcovalenttetrahedralstructureandittakesmuchenergytobreakdownthelattice.[1]e.i.SO2(g)+H2O(l)→H2SO3(aq)orSO3(g)+H2O(l)→H2SO4(aq)[1]ii.Asrainfallsthroughtheairitdissolvescarbondioxidetoformcarbonicacid.CarbonicacidisaweakacidwithapHnolowerthan5.65.Forraintobe‘acidicrain”itmustcontainagreaterconcentrationofhydrogenionsthanarepresentincarbonicacidsothepHmustbebelow5.6.[1]

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SL&HLAnswerstoDataresponsequestion5onvitaminCa.I.

[1]

I2+2e−→ 2I−[1]

ii.Itcontainspolar–OHgroupswhichcanformhydrogenbondswithwatermolecules.[1]b.i.3.04grequired82dropsso1.00gwouldrequire82÷3.04=27drops[1]Interpolationofthegraphgives0.65mgofvitaminC,sotheconcentrationinfreshredpepperis0.65mgg-1[1]ii.27dropsofiodinesolutionreactswith0.65mgofvitaminCM(vitaminC)=176.12gmol−1soonedropsreactswith0.65÷(1000x176.12x27)=1.37x10−7molofvitaminC[1]OnemolofvitaminCreactswithonemolofiodineAmountofiodineinonedrop=1.37x10−7mol,andthereare103x1000dropsinonedm3Concentrationofiodinesolution=103x1000x1.37x10−7=0.014moldm−3.[1]c.Applyingtherulesfordeterminingoxidationstatesgivesanaverageoxidationstateof–1/3foriodineintheI3

−ion.AssumingthisisthecorrectvaluethendisproportionationhasoccurredasIinI2hasbeenreducedfromzeroto–1/3andoxidisedfrom–1intheiodideionto–1/3.[1]TherulesdonottakeintoaccountthefactthatiodineisbondedtoitselfintheI3

−ionandalsoassumefalselythatthecovalentbondsinacomplexionbehaveasiftheyareionic.SomechemistsregardtheI3

−ionasbeingamixtureofiodineandiodide.Ifthisisassumedthennoredoxreactionisoccurring.[1]