Edexcel Jun 2010 Paper 3 Q3 (with explained solutions) |

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Edexcel Jun 2010 Paper 3 Q3

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PMT 3 The purity of a sample of potassium iodate(V) was determined by titration. The steps of the experimental procedure are as follows. 1. 0.100 g of the sample was dissolved in water in a beaker and the solution made up to 100 cm3 in an appropriate flask. 2. A 10.0 cm3 portion of this solution of potassium iodate(V) was transferred to a conical flask. 3. An excess of both potassium iodide solution and sulfuric acid were then added to the conical flask. This produced a solution, T, containing iodine. 4. Solution T was titrated with 0.0200 mol dm3 sodium thiosulfate solution using a suitable indicator. 5. Steps 2, 3 and 4 were repeated twice. (a) (i) Name the piece of apparatus used to remove the 10.0 cm3 portions of potassium iodate(V) solution (step 2). (1)(ii) Name the indicator you would use for the titration and give the colour change you would expect to see (step 4). (2) IndicatorColour change fromto(b) The following results were obtained for the titrations. Titration number Final burette reading / cm3 Initial burette reading / cm3 19.50 5.00 33.20 19.50 46.95 33.20 Titre / cm3 (i) Complete the table by calculating the titres. (1) *N35693A0716* Turn over<br />
PMT (ii) Explain why the correct value for the mean titre is 13.73 cm3. (1)(iii) Calculate the number of moles of sodium thiosulfate in the mean titre. (c) The ionic equation for the reaction between iodine and sodium thiosulfate in the titration is shown below. I2(aq) + 2S2O3 2(aq)2I(aq) + S4O6 2(aq) Calculate the number of moles of iodine in solution T using this equation and your answer to (b)(iii). (d) The ionic equation for the reaction of iodate(V) ions with iodide ions is shown below. IO3 (aq) + 5I(aq) + 6H+(aq)3I2(aq) + 3H2O(l) Using this equation and your answer to (c), calculate the number of moles of iodate(V) ions which reacted to produce solution T. (1) (1) (1) *N35693A0816*<br />
PMT (e) (i) Name the appropriate flask used in step 1. (1)(ii) Describe how you would make up exactly 100 cm3 of potassium iodate(V) solution in this flask, ready for step 2. (3)(iii) Calculate the number of moles of potassium iodate(V) in 100 cm3 of the solution, using your answer to (d). (iv) Calculate the mass of potassium iodate(V) in the sample. [Assume the molar mass of potassium iodate(V) is 214 g mol1] (v) Calculate the percentage purity of the sample. (f) Suggest the most significant hazard in step 3. (1) (1) (1) (1)(Total for Question 3 = 16 marks) *N35693A0916* Turn over<br />

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Qu ties on Number 3 (a)(i Question Number 3 (a)(ii) Question Nu emb r 3 (b)(i Acceptable Answers Re t jec Mark (Glass/graduated/volumetric/bulb)pipette (and pipette filler) Accept any recognisable spelling of pipette eg pipet, pipette etc Burette/ pipette and burette/ measuring cyl inder/ teat p ipette/ i dro pip t Pipate/pipotte pp ng et e Acceptable Answers Rej cte Mark Starch (solution) (1) Accept startch Blue-black/blue/black to colourless (1) Accept purple/blue-black to colourless Second mark depends on first tors eg l orange olph- Other indica Methy /phen thalein Colour blue- black/ black less to blue/ Acceptable Answers Reject Mark 14.5(0), 13.7(0), 13.75 All three needed for the mark 6CH0701 1006<br />
Question Number 3 (b)(i i) Question Number 3 (b)(iii) Question Number 3 (c) Acceptable Answers Reject Mark e s cond and thi The first result is discarded/ ignored/ not included/a range finder OR Only use last two values OR Th value not within 0.2 cm3 Accept: Th OR Actual correct average calculation to give 13.73/13.725 rd are concordant / first is is the average of the second and third runs Acceptable Answers Reject Mark 13.73 x 0.0200 = 2.746 x 10-4 /0.0002746(mol) 1000 Accept 2.7/2.75 x 10-4 Note that 13.725 gives 2.745 x 10-4 Allow TE from different calculated average titre 1 sf Acceptable Answers Reject Mark 1.373 x10-4 /0.0001373(mol) Also 1.35/1.37/1.375 /1.4 x 10-4 Accept answer to (b)(iii) 2 1 sf (unless already penalised anywhere in this question) 6CH0701 1006<br />
Qu ties on Number 3 (d) Question Number 3 (e)(i) Question Number 3 (e)(i i) Question Number 3 (e (i ) ii) Acceptable Answers Re t jec Mark 4.58 x 10-5 /0.0000458(mol) 4.57667/4.577 x 10-5 etc Als Ac cept answer to (c) o 4.583/4.57/4.6(0) x 10-5 1 sf (unless already penalised an this question) h re in yw e Acceptable Answers Re t jec Mark Vo lumetric/graduated (flask) (100 cm3) round bottomed flask/ conical flask/ measuring cylinder Acceptable Answers Reject Mark ns er solution and rinsings/wash These marks are independent of flask used in (i) Tra f Ma Mix g end) ( in / inverting / shaking (th ke p to the mark (1) is must be at the ings (1) 1) Make sure it is all transferred Acceptable Answers Reject Mark -4 /0.000458 (mol) 4.58 x 10 Also 4.57/4.6 x 10-4 Accept Answer to (d) x 10 1 sf (unles already penalised anywhere in this question) 6CH0701 1006<br />
Question Number 3 (e)(i v) Question Number 3 (e)(v) Question Number 3 (f) Acceptable Answers Reject Mark -4 4.58 x 10 x 214 = 0.098 /0.98012 g Also 0.097941/0.0979 etc Accept answer to (e)(iii) x 214 Ign e or SF 1 sf (unless already penalised anywhere in this question) Acceptable Answers Reject Mark f ( nless 1 s u already penalised anywhere in this question) 0.098 x 100 0.10 = 98 /98.0/98.01/98.012 % Also 97.941/97.94/97.9 % Accept answer to (e)(iii) x100 0.10 correct answer with no working scores (1) Ignore SF Acceptable Answers Reject Mark Sulfuric acid is corrosive/irritant/irritable/burns (skin) Sulfuric acid is harmful/ hazardous/ toxic 6CH0701 1006<br />
Qu ties on Number 4 (a)(i Question Number 4 (a)(ii) Acceptable Answers Re t jec Mark t any clear indication that they appreciate / 0.1 80 0.8 (= 0.125) From the equation 1 mol butan-1-ol gives 1 mol of 1-bromobutane / ratio 1:1 OWTTE (1) As 80% yield, 0.125 mol of butan-1-ol gives 0.125 x 0.8 mol = 0.1 mol of 1-brombutane (1) Accep the proportion calculation and the mole ratio Examples: Number of moles of butan-1-ol = 0.1 x 100 OR butane Nu = 80 100 (= 0.1) In both these examples butan-1-ol/ 1-bromobutane as appropriate, must be present to g The numerical expression alone would gain (1) OR As above examples but additionally using molar masses to calculate masses e mb r of moles of 1-bromo x 0.125 ain (2) Acceptable Answers Reject Mark 25 = 1.4 /11.42/11.420/11.419753 cm (1) 74 x 0.125 (1) = 9.25 g) 9. 0.81 ie Accept 11.4 /11.42/11.420/11.419753 cm3 (2) (with no working) ign re sf unless only one 1sf 6CH0701 1006<br />

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