Chemistry 9701/51 October/November 2022: question paper and mark scheme

Question

A student attempts to determine the percentage by mass of magnesium chloride in the solid mixture containing magnesium chloride, MgCl 2, and anhydrous magnesium nitrate, Mg(NO3)2, using the following method. step 1 Accurately weigh about 1.5 g of the solid mixture and record the mass. step 2 Dissolve the solid mixture in distilled water. step 3 Add an excess of silver nitrate solution. step 4 Filter the solid mixture and wash the precipitate collected with distilled water. step 5 Dry the precipitate in an oven. step 6 Weigh the precipitate and record the mass. In this process only the chloride ions from the magnesium chloride form a precipitate with the silver nitrate solution. MgCl 2(aq) + 2AgNO3(aq) → Mg(NO3)2(aq) + 2AgCl (s) One student in the class obtains the following results. mass of solid mixture = 1.52 g mass of AgCl solid after drying = 3.63 g (a) (i) Calculate the amount, in mol, of magnesium chloride present in the sample. amount of magnesium chloride = …… mol [1] (ii) Use your answer to (i) to calculate the percentage by mass of magnesium chloride in the sample. (If you were unable to answer (i) use 0.0102 mol. This is not the correct answer.) percentage by mass = …… [2] (b) (i) Suggest what the student could do in step 2 to ensure the solid dissolves as quickly as possible. . …… . …… . …… [1] (ii) Explain why the precipitate was washed with distilled water before it was dried. . …… . …… . …… [1] (iii) Suggest why the precipitate is dried in an oven and not by direct heating with a Bunsen burner. . …… . …… . …… [1] (c) (i) In step 1, a small beaker was weighed, using a balance accurate to two decimal places, and its mass recorded. The sample was placed in the beaker and the mass of the beaker increased by 1.52 g. Calculate the percentage error in measuring the mass of this sample. Show your working. percentage error = …… [1] (ii) Other than by changing the balance, state how this percentage error could be reduced. . …… [1] (iii) State what could be done in step 5 to ensure that the precipitate was completely dried. . …… . …… [1] (d) Another student in the class did not dry their silver chloride. State how this would affect the value of the percentage by mass of magnesium chloride in the sample. Explain your answer. . …… . …… [1] [Total: 10]

Question

Charles’ law states that for a fixed mass of gas at constant pressure, its volume is proportional to its absolute temperature. Most gases are non-ideal and do not obey this law, but at lower pressures and high temperatures some gases are close to ideal behaviour. One gas that behaves like this is oxygen. Oxygen can be prepared by decomposing hydrogen peroxide with the catalyst manganese(IV) oxide, MnO2. The equation for the decomposition of hydrogen peroxide is shown. 2H2O2(aq) → 2H2O(l) + O2(g) Safety hazard: hydrogen peroxide is corrosive to skin and can cause serious eye damage. gas syringe thermometer beaker oxygen water heat Fig. 2.1 Once the apparatus is assembled the volume of oxygen in the gas syringe is 2 cm3. There are 80 cm3 of oxygen remaining in the flask. The total volume of oxygen is 82 cm3. Charles’ law is investigated by the following method. step 1 Once assembled allow the apparatus to reach room temperature. step 2 Record this temperature and the total volume of oxygen reading on the syringe. step 3 Gently heat the apparatus until the temperature reaches 30 °C and record the total volume of oxygen. step 4 Repeat at intervals of 5 °C until the temperature reaches 70 °C. Question 2 continues on the next page. The student carried out the experiment and obtained the following results: Table 2.1 temperature absolute total volume of / °C temperature / K oxygen gas / cm3 24 297 82 30 303 84 35 308 88 40 313 88 45 318 89 50 323 91 55 328 93 60 333 95 65 338 97 70 343 98 (a) Other than the wearing of safety goggles, give a safety precaution that the student must take during the preparation of oxygen. . …… . …… [1] (b) (i) Complete the following diagram to show how the student can obtain oxygen by gas collection over water for use in the experiment shown in Fig. 2.1. H2O2(aq) MnO2 catalyst [2] (ii) Suggest how the student could ensure they collect pure oxygen gas in the conical flask. . …… . …… [1] (c) (i) Plot a graph on the grid to show the relationship between volume of oxygen and absolute temperature. Use a cross (×) to plot each data point. Draw a line of best fit. 100 98 96 94 92 total volume of oxygen gas / cm3 90 88 86 84 82 80 290 300 310 320 330 340 350 absolute temperature / K [2] (ii) Determine the gradient of your line of best fit. State the coordinates of both points you used in your calculation. These must be selected from your line of best fit. Give the gradient to three significant figures. coordinates 1 …… coordinates 2 …… gradient = …… cm3 K–1 [2] (d) (i) On the graph, circle the point which you believe to be the most anomalous. [1] (ii) Suggest a possible explanation for this anomaly. . …… . …… [1] (e) (i) Identify the independent variable. . …… [1] (ii) Suggest how the experiment could be made to be more reliable. . …… [1] (f) The ideal gas equation is shown. pV = nRT p = the pressure of the gas in Pa; V = the volume of gas in m3; n = the number of moles of gas; R = the universal gas constant 8.31 J mol–1 K–1 and T = absolute temperature in K Using this equation, describe how the gradient of the graph you have plotted would be affected by using a smaller volume of oxygen at the start of the experiment. Explain your answer. . …… . …… . …… . …… . …… [1] [Total: 13]

Question

Potassium bromate(V) reacts with potassium bromide and sulfuric acid to form potassium sulfate, bromine and water according to the following equation. KBrO3(aq) + 5KBr(aq) + 3H2SO4(aq) → 3K2SO4(aq) + 3Br2(aq) + 3H2O(l) A student is investigating how the rate of this reaction is affected by changing the concentration of the reactants in turn. This is done by keeping the total volume of mixture constant and adding different, small volumes of each reagent. The reaction produces bromine which is orange in colour. The student times the reaction and then determines the rate as 1 . time The rate equation for the reaction is of the form: rate = k [KBrO3]x[KBr]y[H2SO4]z k is the rate constant for the reaction and x, y and z are the respective orders of the reaction for each reagent. The student carried out the experiment and obtained the following data. Table 3.1 [KBrO3] [KBr] [H2SO4] rate of reaction mixture / mol dm–3 / mol dm–3 / mol dm–3 / s–1 A 0.025 0.125 0.075 0.059 B 0.050 0.125 0.075 0.117 C 0.025 0.250 0.075 0.118 D 0.025 0.125 0.150 0.235 E 0.050 0.250 0.150 0.941 (a) (i) Suggest how the student might time the reaction and judge the end point of the reaction for each mixture. . …… . …… . …… [1] (ii) By comparing the data for the mixtures deduce the values of x, y and z. [2] (b) The student carried out each reaction using a boiling tube (capacity 50 cm3) and varied the concentration by adding different volumes of each reagent. For example, in mixture A, 5.0 cm3 of KBrO3(aq) is required. Name a suitable piece of apparatus which could be used to measure this volume. . …… [1] (c) Suggest why the reagents are heated to the same temperature before mixing. . …… [1] (d) The solution of sulfuric acid used in each mixture was of concentration 0.150 mol dm–3. This acid was prepared from a solution of concentration 1 mol dm–3. Briefly describe how to make the more dilute solution, stating the capacity of any apparatus used. . …… . …… . …… . …… [2] [Total: 7] Important values, constants and standards molar gas constant R = 8.31 J K–1 mol–1 Faraday constant F = 9.65 × 104 C mol–1 Avogadro constant L = 6.022 × 1023 mol–1 electronic charge e = –1.60 × 10–19 C molar volume of gas Vm = 22.4 dm3 mol–1 at s.t.p. (101 kPa and 273 K) Vm = 24.0 dm3 mol–1 at room conditions ionic product of water Kw = 1.00 × 10–14 mol2 dm–6 (at 298 K (25 °C)) specific heat capacity of water c = 4.18 kJ kg–1 K–1 (4.18 J g–1 K–1) The Periodic Table of Elements Group 1 2 13 14 15 16 17 18 1 2 H He hydrogen helium Key 1.0 4.0 3 4 atomic number 5 6 7 8 9 10 Li Be atomic symbol B C N O F Ne lithium beryllium name boron carbon nitrogen oxygen fluorine neon 6.9 9.0 relative atomic mass 10.8 12.0 14.0 16.0 19.0 20.2 11 12 13 14 15 16 17 18 Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon 23.0 24.3 3 4 5 6 7 8 9 10 11 12 27.0 28.1 31.0 32.1 35.5 39.9 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 at www.cambridgeinternational.org after the live examination series. K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton 39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 12 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon 85.5 87.6 88.9 91.2 92.9 95.9 – 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3 55 56 57–71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 lanthanoids Cs Ba Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn caesium barium hafnium tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth polonium astatine radon 132.9 137.3 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 – – – 87 88 89–103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 actinoids Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Nh Fl Mc Lv Ts Og francium radium rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium roentgenium copernicium nihonium flerovium moscovium livermorium tennessine oganesson – – – – – – – – – – – – – – – – – 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 lanthanoids La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium 138.9 140.1 140.9 144.4 – 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.1 175.0 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 actinoids Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium – 232.0 231.0 238.0 – – – – – – – – – – – Assessment International Education Copyright Acknowledgements Booklet. 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