Edexcel Jun 2014 (IAL) Paper 5 Q21 (with explained solutions) |

Skip to content

Edexcel Jun 2014 (IAL) Paper 5 Q21

Answers available below

21 Limestone is an important industrial raw material. The chemically important component of limestone is calcium carbonate and the proportion of this compound determines the quality of the material. One method of determining the proportion of calcium carbonate in limestone involves dissolving the limestone in acid and precipitating the calcium ions as calcium ethanedioate. The precipitate is filtered and then re-dissolved in dilute sulfuric acid. The concentration of ethanedioate ions is determined by titration with a solution of potassium manganate(VII) of known concentration. In one such determination, 1.77 g of limestone produced 250 cm3 of calcium ethanedioate solution and 25.0 cm3 samples of this solution were titrated at about 70 C with potassium manganate(VII) solution of concentration 0.0205 mol dm3. The mean titre was 24.55 cm3. (a) (i) State the appearance of the titration mixture at the end-point. (1)(ii) Complete the ionic equation for this titration. 2MnO4+ 5C2O4 2 + H+ (cid:314) Mn2+ + CO2 + H2O *(iii) Calculate the percentage by mass of calcium carbonate in the limestone. Show your working and give your final answer to three significant figures. (1) (5) *P42979A01732* 17 Turn over<br />
(b) (i) Before the calcium ethanedioate is re-dissolved, it is washed in deionized or distilled water. Explain why this step is necessary and the effect its omission would have on the titration. (2)(ii) The table below summarises the tolerances on the measuring apparatus used in this experiment. Complete the table by calculating the percentage error for each piece of apparatus to two significant figures. (2) Apparatus Balance Value 1.77 g Volumetric flask 250 cm3 Pipette Burette 25 cm3 24.55 cm3 Maximum total error on the stated value Percentage error on the stated value 0.01 g 0.12 cm3 0.06 cm3 0.10 cm3 18 *P42979A01832*<br />
(iii) Using the method in (a), it was found that 2.00 g of a different sample of limestone contained 0.015 mol of calcium carbonate. Use the data from (b)(ii), and the fact that the solubility of calcium ethanedioate in water is 0.0067 g dm3, to calculate whether the fact that some calcium ethanedioate remains in solution has a significant effect on the accuracy of this second calcium carbonate determination. Assume that the combined volume of the filtrate and the washings is approximately 500 cm3. (3)(Total for Question 21 = 14 marks) *P42979A01932* 19 Turn over<br />

Show answer

Acceptable Answers Question Number 21(a)(i) (pale) pink OR First permanent pink Ignore Colourless to Acceptable Answers Question Number 21(a)(ii) 2MnO4+ 5C2O4 10CO2 + 8H2O Reject purple Mark Reject Mark 2 + 16H+2Mn2+ + WCH0501 1406<br />
Acceptable Answers Question Number 21(a)(iii) Amount MnO4= 24.55 x .0205 x 103 (1) = 5.03275 x 104 mol (ans *) Amount C2O4 = 5.03275 x 104 x 5/2 (1) 2 in 25 cm3 = ans. * x 5 /2 = 1.2581875 x 103 mol In 250 cm3 = 1.2581875 x 102 mol (ans **) (1) = amount Ca2+ = amount CaCO3 Mass CaCO3 = (ans**) x 100.1 = 1.2581875 x 102 x 100.1 (1) = 1.2594457 g (ans***) % CaCO3 = 100 x (ans***) / 1.77 = 71.15512 = 71.2 ( % ) (1) ALLOW Final answer 71.1 / 71.2 / 71.3 scores 5 marks Final answer must be to 3 SF ( max 4 if not) Until final answer ignore SF except 1 SF (penalise once) TE at each stage unless mass CaCO3 > 1.77 NOTE Use of ethanedioate mass of 88 in step 4 gives final answer of 62.6% ( max 4 ) Use of calcium ethanedioate mass of 128.1 / 128 in step 4 gives final answer of 91.0% ( max 4 ) Reject Mark WCH0501 1406<br />
Question Number Acceptable Answers Reject Mark 21(b)(i) Excess ethanedioate ( ions in the solution) must be removed (1) ALLOW Remove ethanedioic acid Otherwise more KMnO4 will be used (in the titration) / bigger titre (1) MP2 dependent on MP1 Impurities Acid Question Number 21 (b)(ii) Acceptable Answers Apparatu Value Maximum total error on the stated value Percentag e error on the stated value Balance 1.77 g 0.01 g 0.56 (0.56497) Volumetric flask 250 cm3 0.12 cm3 0.048 Pipette 25 cm3 0.06 cm3 Burette 24.55 cm3 0.10 cm3 0.24 0.41 (0.40733) All % calculations correct (2) Any two or three calculations correct (1) 1 mark lost if 2 or more correct answers are not given to 2 SF Reject Mark WCH0501 1406<br />
Acceptable Answers Question Number 21(b)(iii) First mark Reject Mark EITHER Max. mass of CaC2O4 precipitated = 0.015 x 128.1 = 1.9215 g OR 0.0067/2 = 0.00335 g remains in solution (1) Second Mark % error = 100 x 0.00335 / (1.9215 + 0.00335) = 100 x 0.00335 / 1.92485 = 0.174040 = 0.174 % ALLOW % error = 100 x 0.00335 / 1.9215 = 0.174343 = 0.174 % If Mr (CaC2O4) = 128 used = 0.174479 % (1) Third Mark Error comparable to / smaller than apparatus uncertainty / less than the worst / less than the balance / less than the total And so acceptable (1) IGNORE SF but penalise incorrect rounding once NOTE No TE for mark 2 from mark 1 BUT TE for mark 3. Accept reverse argument for large percentage. Total for Question 21 = 14 marks WCH0501 1406<br />
Question Number Acceptable Answers 22(a) A = PCl5 / phosphorus(V) chloride / phosphorus pentachloride / PCl3 / phosphorus(III) chloride / phosphorus trichloride / SOCl2 / thionyl chloride / thionyl dichloride B = benzene / C6H6 or ring structures C = bromine / Br2 (1) (1) (1) Question Number Acceptable Answers 22(b)(i) CH3CH2Br + AlBr3CH3CH2 + + AlBr4(1) ALLOW + C2H5 + sign anywhere on formula of electrophile (AlBr3 is an) electron pair acceptor / lone pair acceptor / Lewis acid / Friedal-Crafts catalyst (1) ALLOW polarizes C-Br bond IGNORE Halogen carrier Mark Mark Reject Bromine water / Bromine and FeBr3 Reject Accepts electrons Just catalyst WCH0501 1406<br />

Question 19 20 21 22 23 24

About us | Need Help? | Feedback

Proudly powered by WordPress | Theme: Sydney by aThemes.