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CIE IGCSE Chemistry 0620 Paper 6 Experimental Skills Practice Paper (6 Questions, 50 Marks, 60 Minutes)
IGCSE Chemistry 0620 Paper 6 Skills
CIE IGCSE Chemistry 0620 Paper 6 experimental skills paper: 6 questions worth 50 marks in 60 minutes on titration and indicators, gas collection and molar volume.
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Paper details Level: IGCSE | Subject: Chemistry 0620 | Time: 60 minutes | Marks: 50 | Questions: 6
Section A Titration, gases and water of crystallisation (25 marks)
Questions 1-3
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | (a) 25.0 cm3 of sulfuric acid of unknown concentration is titrated with 0.100 mol/dm3 sodium hydroxide using phenolphthalein; the mean titre of three runs is 23.60 cm3. The reaction is H2SO4 + 2NaOH -> Na2SO4 + 2H2O. Calculate the concentration of the acid in mol/dm3, showing each step and the unit. (5 marks) (b) Explain why the burette must be rinsed with the standard solution, why the conical flask is rinsed with distilled water only, what the white tile is for, and why the titration is repeated and averaged. (4 marks) |
Answer: (a) n(NaOH) = 0.100 x 23.60 / 1000 = 0.002360 mol. From the 1:2 ratio, n(H2SO4) = 0.002360 / 2 = 0.001180 mol, so c(H2SO4) = 0.001180 / 0.0250 = 0.04720 mol/dm3. (b) Water left in the burette would dilute the standard solution and make the titre too large, so it is rinsed with that solution. Water in the flask only dilutes the acid without changing its number of moles, so distilled water is correct; rinsing with acid would make the result too large. The white tile makes the first permanent pink easier to see. Repeating and averaging reduces random error. Marking notes: 1:2 mole ratio, unit mol/dm3, end point colourless to pink lasting 30 s. |
| 2 | (a) Magnesium ribbon reacts with dilute hydrochloric acid: Mg + 2HCl -> MgCl2 + H2. At room temperature and pressure a gas syringe collects 72.0 cm3 of hydrogen (molar volume 24.0 dm3/mol). Calculate the number of moles of hydrogen, the mass of magnesium that has reacted in grams, and the average rate in cm3/s when 40 cm3 is collected in the first 30 seconds. (5 marks) (b) Explain why the ribbon is cleaned with sandpaper before the experiment, why the syringe reading is taken after one minute, and how the initial rate is found from a volume-time graph. (3 marks) |
Answer: (a) n(H2) = 72.0 / 24000 = 0.00300 mol. The equation gives a 1:1 ratio, so n(Mg) = 0.00300 mol and m(Mg) = 0.00300 x 24.3 = 0.0729 g. Rate = 40 / 30 = 1.33 cm3/s. (b) The ribbon carries a layer of magnesium oxide that slows the reaction and lowers the gas volume, so it is sanded to expose fresh metal. The reaction is exothermic and warms the gas, so the reading is taken after one minute when the gas is back at room temperature; otherwise the volume is too large. The initial rate is the gradient of the tangent drawn at the start of the volume-time curve, in cm3/s. Marking notes: Ar(Mg) = 24.3, 1:1 ratio, unit on rate. |
| 3 | (a) 5.00 g of blue hydrated copper(II) sulfate crystals, CuSO4.xH2O, is heated in a crucible to constant mass, leaving 3.20 g of white solid (Mr: CuSO4 = 159.6, H2O = 18.0). Calculate the mass of water lost, the moles of water, the moles of anhydrous salt, and hence x as an integer. (5 marks) (b) Explain why the solid is heated to constant mass, why it is cooled in a desiccator before weighing, and whether x would be too large or too small if heating were incomplete. (3 marks) |
Answer: (a) Mass of water = 5.00 – 3.20 = 1.80 g. n(H2O) = 1.80 / 18.0 = 0.1000 mol. n(CuSO4) = 3.20 / 159.6 = 0.02005 mol. x = 0.1000 / 0.02005 = 4.99, so x = 5 and the formula is CuSO4.5H2O. (b) Heating to constant mass, with two weighings within 0.01 g, shows that all the water has been driven off. A hot crucible would absorb moisture from the air, so it is cooled in a desiccator. If heating were incomplete, the residue would be too heavy, the moles of anhydrous salt too large and x too small. Marking notes: subtract masses first, divide each by its Mr, round x to a whole number. |
Section B Electrolysis, rates and chromatography (25 marks)
Questions 4-6
| # | Question | Answer and marking notes |
|---|---|---|
| 4 | (a) A current of 2.0 A is passed for 15 minutes through copper(II) sulfate solution using copper electrodes at both sides; copper is deposited on the cathode (1 mole of electrons = 96500 C, Ar: Cu = 63.5). Calculate the charge in coulombs, the moles of electrons, and the moles and mass of copper deposited. (5 marks) (b) Explain why the mass of the anode decreases and why the blue colour of the solution stays almost the same after long electrolysis. (3 marks) |
Answer: (a) Q = I x t = 2.0 x 900 = 1800 C. n(e-) = 1800 / 96500 = 0.018653 mol. Since Cu2+ + 2e- -> Cu, n(Cu) = 0.018653 / 2 = 0.009326 mol and m(Cu) = 0.009326 x 63.5 = 0.592 g. (b) Copper atoms at the anode lose electrons and pass into solution as Cu2+ (Cu -> Cu2+ + 2e-), so the anode loses mass. For every Cu2+ deposited at the cathode one Cu2+ is released at the anode, so the concentration of copper ions in the solution is almost unchanged and the blue colour stays the same. Marking notes: Q = It, 96500 C per mole of electrons, two electrons per copper ion. |
| 5 | (a) Equal masses of the same magnesium ribbon react with 1.0 mol/dm3 and 0.5 mol/dm3 hydrochloric acid. Collecting 50 cm3 of hydrogen takes 20 s and 80 s respectively. Calculate the average rate in each case and the ratio of the rates. (5 marks) (b) Use collision theory to explain how concentration affects the rate, why the mass and surface area of the magnesium must be kept the same, and how you would show that temperature rather than concentration is responsible. (4 marks) |
Answer: (a) Rate 1 = 50 / 20 = 2.5 cm3/s and rate 2 = 50 / 80 = 0.625 cm3/s; the ratio is 2.5 / 0.625 = 4, so doubling the concentration makes the reaction about four times faster. (b) A higher concentration means more hydrogen ions per unit volume, so collisions with the magnesium surface are more frequent and the rate rises. Mass and surface area set the amount of exposed metal, so they must be fixed to keep the comparison fair and change only one variable. Keep concentration and magnesium the same and raise the temperature from 20 C to 40 C; a clear increase in rate then shows that temperature matters. Marking notes: rate in cm3/s, collision frequency, control of variables. |
| 6 | (a) Three food colourings are separated by paper chromatography. The solvent front is 8.4 cm from the start line and the centres of the three spots are 3.5, 5.9 and 7.1 cm from it. Calculate the three Rf values to 3 decimal places and decide which reference dyes the sample contains (reference B has Rf = 0.42, reference C has Rf = 0.70). (5 marks) (b) Explain why the start line is drawn in pencil rather than ballpoint pen, why the solvent level must stay below the start line, and why the paper is dried before the spots are measured. (3 marks) |
Answer: (a) Rf = distance moved by spot / distance moved by solvent front: 3.5 / 8.4 = 0.417; 5.9 / 8.4 = 0.702; 7.1 / 8.4 = 0.845. The first two values, 0.417 and 0.702, match reference B (0.42) and reference C (0.70), so the sample contains B and C; the third spot has no matching reference. (b) Graphite in pencil does not dissolve in the solvent, whereas ballpoint ink dissolves and travels up the paper, spoiling the result. If the solvent covered the start line the dyes would simply dissolve into the solvent and never separate. While the paper is wet the dyes are still moving, so positions are only stable and measurable after drying. Marking notes: Rf has no unit and is below 1, compare with references, pencil start line. |
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