Copy of PLMGSS 2024 4E Chem P3 Ans

1 Answer to Sec 4 Prelim Exams 2024 Qn Answer Mark 1(a)(i) Final burette reading /cm3 18.70 28.90 28.90 Initial burette reading /cm3 0.00 10.00 10.00 Volume of P used /cm3 18.70 18.90 18.90 Reading chosen ✓ ✓ A-Accuracy (average titre within 0.20 cm3 of teacher’s average tire value) [2] (average titre within 0.10 cm3) [1] C- Concordance (two accurate titre values are within 0.20cm3) [1] T- [2] ✓ The table headings and units has to be correct ✓ All burette readings for all accurate titres in titration table are recorded to nearest 0.05 cm3 [5] 1(a)(ii) Average volume of P = (18.90 + 18.90) / 2 = 18.90 cm3 [1] 1(b)(i) Number of moles of sodium hydroxide = 0.200 x 18.90/1000 = 0.00378 mol Number of moles of HA = 0.00378 mol [1] 1(b)(ii) In 25 cm3, there are 0.00378 moles of HA. In 1000 cm3, there are (0.00378/25) x 1000 = 0.1512 mol Concentration in cleaning product = 0.1512 / 100/1000 = 1.51 mol.dm3 Working – [1], answer – [1] 1(c)(i) 1) Measure 25.0 cm3 of aqueous sodium carbonate using a measuring cylinder and pour the solution into a conical flask. 2) Weigh 1.0 g of solid HA using an electronic balance and place the solid into a small vial. Carefully lower the solid HA into the conical flask, ensuring the chemicals do not mix. 3) Set up the apparatus as shown above. 4) Pull the string to allow the solid HA to react completely with the aqueous sodium carbonate. 5) Assume all the solid HA has reacted with the aqueous sodium carbonate (in excess). 6) Record the total volume of carbon dioxide gas collected in the measuring cylinder when reaction is completed (no more effervescence or more further change in the volume of CO2) in the conical flask. 7) Calculate the number of moles of CO2 produced using the formula: volume (in cm3) / 24 000 cm3 = x mol 8) Using the mole ratio HA:Na2CO3 = 2:1 Number of moles of HA = 2x mol 9) Using the formula, number of moles of HA = mass of HA / Mr of HA 2x = 1/Mr of HA Mr of HA = 1/2x Apparatus (measuring cylinder, electronic balance) – [1] Measurement: excess aqueous sodium carbonate – [1] Measurement: Total volume of carbon dioxide collected – [1] Calculation for number of moles of HA – [1] Calculation for Mr– [1]

2 1(c)(ii) Since CO2 is soluble in water, the total volume of CO 2 collected would be lower than actual, which will cause the number of moles of CO2 as well as number of moles of HA to be lower than expected. As a result, the Mr of HA will be larger than actual. [1] [1] [1] 1(c)(iii) Change: Collect the volume of CO2 produced by using a gas syringe instead of displacement of water. Explanation: Volume of CO2 will be the true volume as water is absent in the collection of CO2. [1] 2(a)(i) metal carbonate X: carbonate of metal R Y: carbonate of metal S Colour of carbonate before heating white white Colour change during heating/cooling Yellow when hot White when cold white Mass of test-tube and contents before heating /g 16.31 16.24 Mass of test-tube and contents after heating for 1 min /g 16.14 16.24 Mass of test-tube and contents after heating for 2 min /g 16.09 16.24 Mass of test-tube and contents after heating for 3 min /g 16.07 16.24 Total mass loss after heating /g 0.24 0.00 [2] [2] [1] 2(a)(ii) More reactive metal: S Less reactive metal: R Carbonate Y does not decompose at all , showing that it is more thermally stable. Hence, metal S is more reactive. Carbonate X decomposes over the 3 min of heating , showing that it is less thermally stable. Hence, metal R is less reactive. [1] [1] 2(b)(i) Add aqueous ammonia and sodium hydroxide to the solutions. Observe the colour of ppt, and and their solubility in excess reagents [1] [1] 2(b)(ii) Solution with metal ions of R (prepared using carbonate X) Solution with metal ions of S (prepared using carbonate Y) Test White ppt, soluble in excess aqueous ammonia, giving colourless solution White ppt, soluble in excess sodium hydroxide, giving colourless solution White ppt, insoluble in excess sodium hydroxide no ppt with aqueous ammonia Metal R: zinc, Metal S: calcium [2] [1]

3 3(a) experiment volume of aqueous iron(II) sulfate /cm3 volume of distilled water /cm3 initial temperature /OC highest temperature /OC change in temperature /OC 1 10.0 0.0 30.0 46.5 16.5 2 8.0 2.0 29.5 44.5 15.0 3 6.0 4.0 29.0 41.0 12.0 4 4.0 6.0 29.5 37.0 7.5 5 2.0 8.0 29.5 33.0 3.5 [1] 3(b) Axis – [1] Scale – [1] Points plotted correctly – [1] Best fit line – [1] Volume of iron(II) sulfate / cm3

4 3(c)(i) The larger the volume of iron(II) sulfate used, the greater the change in temperature. The larger the volume of iron(II) sulfate used, the higher the concentration of the iron(II) sulfate solution. This leads to a faster reaction and a large temperature change. [1] [1] 3(c)(ii) 1000 cm3 – 0.500 mol ? cm3 – 0.0035 mol Volume of iron(II) sulfate solution with the same number of moles as 0.0035 / 0.500 x 1000 = 7 cm3 Expected temperature is 12.4 OC. [1] [1] 3(c)(iii) 10 cm3 of 0.500 mol/dm 3 iron(II) sulfate leads to a change in temperature of 16.5 oC. 10 cm3 of 0.1 mol/dm 3 iron(II) sulfate will lead to a change in temperature of 16.5 x 2 = 33.0 oC. [1]