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CIE Nov 2015 v4 Paper 3 Q1
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![In this experiment you will determine the relative atomic mass, Ar, of magnesium by a titration method. PMT FB 1 is 2.00 mol dm3 hydrochloric acid, HCl. FB 3 is 0.120 mol dm3 sodium hydroxide, NaOH. magnesium ribbon bromophenol blue indicator (a) Method Reaction of magnesium with FB 1Pipette 25.0 cm3 of FB 1 into the 250 cm3 beaker. Weigh the strip of magnesium ribbon and record its mass.Coil the strip of magnesium ribbon loosely and then add it to the FB 1 in the beaker. Stir the mixture occasionally and wait until the reaction has finished. mass of magnesium =g Dilution of the excess acidTransfer all the solution from the beaker into the volumetric flask. Make the solution up to the mark using distilled water. Shake the flask to mix the solution before using it for your titrations. Label this solution of hydrochloric acid FB 2. TitrationFill the burette with FB 2. Rinse the pipette out thoroughly. Then pipette 25.0 cm3 of FB 3 into a conicalask. Add several drops of bromophenol blue indicator. Perform a rough titration, by running the solution from the burette into the conicalask until the mixture just becomes yellow. Record your burette readings in the space below.The rough titre iscm3. Carry out as many accurate titrations as you think necessary to obtain consistent results. Make sure any recorded results show the precision of your practical work. Record in a suitable form below all of your burette readings and the volume of FB 2 added in each accurate titration. II III IV VI VII [7]UCLES 2015 9701/34/O/N/15<br />
(b) From your accurate titration results, obtain a suitable value for the volume of FB 2 to be used in your calculations. Show clearly how you have obtained this value. PMT 25.0 cm3 of FB 3 requiredcm3 of FB 2. [1] (c) Calculations Show your working and appropriate signi cantgures in thenal answer to each step of your calculations. (i) Calculate the number of moles of sodium hydroxide present in 25.0 cm3 of solution FB 3. moles of NaOH =mol (ii) Give the equation for the reaction of hydrochloric acid, HCl, with sodium hydroxide, NaOH. State symbols are not required.Deduce the number of moles of hydrochloric acid in the volume of FB 2 you calculated in (b). (iii) Calculate the number of moles of hydrochloric acid in 250 cm3 of FB 2. moles of HCl =mol (iv) Calculate the number of moles of hydrochloric acid in 25.0 cm3 of FB 1. moles of HCl in 250 cm3 of FB 2 =mol moles of HCl in 25.0 cm3 of FB 1 =molUCLES 2015 9701/34/O/N/15 [Turn over<br />
(v) In (a), you reacted 25.0 cm3 of FB 1 with your weighed piece of magnesium. After the reaction, the unreacted hydrochloric acid was used to prepare 250 cm3 of FB 2. Use your answers to (iii) and (iv) to calculate the number of moles of hydrochloric acid that reacted with the magnesium ribbon. PMT moles of HCl reacting with Mg =mol (vi) Complete the equation below, for the reaction of magnesium with hydrochloric acid. State symbols are required. Mg + HClMgCl 2 +Use your answer to (v) to calculate the number of moles of magnesium used. (vii) Use your answer to (vi) to calculate the relative atomic mass, Ar, of magnesium. moles of Mg =mol Ar of Mg =[6] (d) (i) State one observation that proves that the hydrochloric acid in FB 1 was in excess for the reaction with the magnesium ribbon.(ii) A student carried out exactly the same experiment but used 1.00 g of magnesium ribbon. State and explain why the students experiment could not be used to determine the value for the Ar of magnesium. Include a calculation in your answer.[3] [Total: 17]UCLES 2015 9701/34/O/N/15<br />](https://learnah.org/wp-content/uploads/2018/08/November_2015_v4_QP_-_Paper_3_CIE_Chemistry_A-level_1-0.jpg)
![Page 2 Mark Scheme Syllabus Paper Cambridge International AS/A LevelOctober/November 2015 9701 34 Question (a) Indicative material Mark Total I The following readings are recorded (cid:127) mass of Mg used (cid:127) (cid:127) two burette readings and the titre for the rough titration initial and final burette readings for two (or more) accurate titrations II Titre values recorded for accurate titrations and appropriate headings for the accurate titration table and cm3 units. (cid:127) (cid:127) (cid:127) initial / start (burette) reading / volume final / end (burette) reading / volume titre or volume / FB 2 and used / added (not differenceor total) unit: / cm3 or (cm3) or in cm3 (for each heading) (cid:127) If cm3 units are not given in the headings, every entry in the table must have the correct unit. 50(.00) is used as an initial burette reading III All accurate burette readings are to the nearest 0.05 cm3. Do not award this mark if: (cid:127) (cid:127) more than one final burette reading is 50.(00) (cid:127) (cid:127) any burette reading is greater than 50.(00) there is only one accurate titration. IV There are two uncorrected accurate titres within 0.10 cm3 (cid:127) Do not award this mark if, having performed two titres within 0.10 cm3, a further titration is performed which is more than 0.10 cm3 from the closer of the initial two titres, unless a further titration, within 0.10 cm3 of any other, has also been carried out. (cid:127) Do not award the mark if any accurate burette readings (apart from initial 0 cm3) are given to zero dp two (or more) accurate identical titres then two (or more) accurate titres within 0.05 cm3, then Examiner checks and corrects titre subtractions where necessary. Examiner selects the best titres using the hierarchy: (cid:127) (cid:127) two (or more) accurate titres within 0.10 cm3, etc.These best titres are used to calculate the mean titre, expressed to nearest 0.01 cm3. Examiner calculates the difference () between the mean titres obtained by the candidate and the Supervisor. Accuracy marks are awarded as shown.Cambridge International Examinations 2015<br />
Page 3 Mark Scheme Syllabus Paper Cambridge International AS/A LevelOctober/November 2015 9701 34 Question Indicative material Mark Total (b) if0.50 cm3 if 0.50 <1.00 cm3 if 1.00 <1.50 cm3 Award V, VI and VII Award V and VI Award V, only, Spread penalty: if the two best (corrected) titres used by the Examiner were0.50 cm3 apart, cancel one accuracy mark. Candidate must take the average of two (or more) titres that are within a total spread of not more than 0.20 cm3. Working / explanation must be shown or ticks must be put next to the two (or more) accurate readings selected. The mean should normally be quoted to 2 decimal places rounded to nearest 0.01 cm3. Two special cases where the mean may not be to 2 dp: allow mean to 3 dp only for 0.025 or 0.075, e.g. 26.325; allow mean to 1 dp if all accurate burette readings were given to 1 dp and the mean is exactly correct. e.g. 26.0 and 26.2 = 26.1 is correct but 26.0 and 26.1 = 26.1 is incorrect. Note: the candidates mean will sometimes be marked correct even if it was different from the mean calculated by the Examiner for the purpose of assessing accuracy. (c) (i) mol NaOH = 0.12025.0 / 1000 = 0.003(00) (ii) (cid:127) NaOH + HClNaCl +H2O (cid:127) Answer to (ii) must be the same as in (i) (iii) + (iv) Correct expressions required in both (iii) and (iv) (Correct expression = correct figures shown) (iii) : no moles of HCl (in 250 cm3) = (ii)250 / (b) (iv) : no moles of HCl (in 25.0 cm3) = 2.0025.0 / 1000 (= 0.05) (v) Correct expression: Mol HCl used = (iv)(iii) (vi) Equation and correctly calculates answer for number of moles Mg: (cid:127) Mg(s) + 2HCl(aq)MgCl2(aq) + H2(g) (cid:127) No of moles Mg = 0.5(v) (vii) Ar = mass of Mg used / (vi) (d) (i) All solid / magnesium dissolved / disappeared / reacted (owtte) or indicator turned from blue to yellow when FB 2 added (to alkali) (ii) (If 1.0 g Mg is used) Mg would be in excess / acid would be the limiting reagent / all the acid would be used upCambridge International Examinations 2015 [7] [1] [6]<br />
Page 4 Mark Scheme Syllabus Paper Cambridge International AS/A LevelOctober/November 2015 9701 34 Question Indicative material Mark Total Qn 1 Reference to moles of both acid and Mg (or other correct calculation) Calculation to show that Mg would be in excess n(Mg) = 1 / (c)(vii)) n(HCl) needed = 0.082 mol or only 0.05 mol present / 24.3 = 0.041 mol (allow 1 / 24 or 1 Total [17]Cambridge International Examinations 2015<br />](https://learnah.org/wp-content/uploads/2018/08/November_2015_v4_MS_-_Paper_3_CIE_Chemistry_A-level_1-0.jpg)
