RVHS Prelim P2 Student ANS

River Valley High School 9647/02/PRELIM/15 2015 Preliminary Examination II RIVER VALLEY HIGH SCHOOL YEAR 6 PRELIMINARY EXAMINATION (II) CANDIDATE NAME Suggested Solutions CLASS 6 CENTRE NUMBER S INDEX NUMBER H2 CHEMISTRY 9647/02 Paper 2 Structured Questions 14 September 2015 2 hours Candidates answer on the Question Paper. Additional Materials: Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, class, centre number and index number on all the work you hand in. Write in dark blue or black pen on both sides of paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all questions in the space provided. A Data Booklet is provided. Do NOT write anything on it. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Paper 2 Question Number 1 2 3 4 5 6 Total Marks 12 14 14 8 9 15 72 Paper 1 40 Paper 3 80 Total 192 This document consists of 20 printed pages.

2 River Valley High School 9647/02/PRELIM/15 2015 Preliminary Examination II 1 (a) (i) Pb2+(aq) + 2Cl (aq) PbCl2(s) [1] (ii) Q: silver(I) ion Explanation: Ag+ (aq) forms a white, insoluble/sparingly soluble salt, AgCl, with Cl(aq). AgCl (s)  Ag+(aq) +Cl (aq) ----(1) [Ag+] decreases as Ag+ reacts with NH3(aq) forming a colourless complex, [Ag(NH3)2]+(aq). Equilibrium position of (1) shifts to the right and thus AgCl dissolves [2] (ii) Chemical Test: To 1 cm 3 of sample in a test -tube, NaOH(aq), dropwise follow by excess. Expected observation: White ppt formed is soluble in excess NaOH(aq) to give a colourless solution. Reject: Aq NH3 because it is one of the reagents used in a(i) [2] (b) 1. Place the 10 cm 3 of aqueous sample provided in a clean 100 cm3 beaker. 2. Using a 10 cm 3 measuring cylinder , add 10 cm 3 of 0.10 mol dm -3 potassium chloride solution . Stir well to ensure complete precipitation. 3. Using a 10 cm 3 measuring cylinder, add 10 cm3 (accept 2 cm3, i.e. 20 times of nCl - used, to 10 cm 3 ) of 1.0 mol dm -3 aqueous ammonia. Stir well to ensure any AgC l precipitated will dissolve. 4. Filter the mixture into a clean 100 cm3 conical flask, and collect the insoluble white precipitate of PbCl2 as residue. 5. To the filtrate in the conical flask, add a dropper full of aqueous potassium chloride to check if there is Pb 2+ remaining in the solution. If white ppt is formed, repeat steps 2 to 4. 6. With the ppt still on the filtration setup, use deionized water to wash the ppt and discard the washings. 7. Weigh a clean crucible/boiling tube/small beaker. Transfer quantitatively all the ppt into this crucible/boiling tube/small beaker. 8. Dry the ppt by using infrared lamp/oven/press dry with filter paper and leave in a dessicator for some time / heat the ppt in a [5]

3 River Valley High School 9647/02/PRELIM/15 2015 Preliminary Examination II boiling tube with known mass for 10 minutes (any one method of drying ). 9. Cool and weigh the crucible/boiling tube/small beaker and its content. 10. Repeat step 8 and 9 until the mass of crucible/boiling tube/small beaker and its content is less than 0.01g in different. (c) (i) 3- lead(II) 3 2 3 dmmol r wastewatein ion lead(II) of ionconcentrat mol sample cm 10 in n r wastewatedm 5 in ions lead(II) of amount edprecipitat PbCl of amount sample cm 10 in ions lead(II) of amount 5 4 43 1034.7 51067.3 1067.3)2(5.35207 102.0         [2] [Total: 12] 2 (a) 90 [2] (b) Both BMIM +PF6  and NaC l have giant ionic lattice structure in solid state. Both anion ( PF6 ) and cation size of BMIM+PF6  are significantly larger than that of Na + and Cl, and they have the same charge . Since charge density of Na + and C l is larger, m ore energy is required to overcome the stronger electrostatic forces of attraction between Na + and Cl compared to those between BMIM+ and PF6 . [2] (c) (i) In a solution of the weak base at equilibruim: B + H2O ⇌ BH+ + OH 𝐾𝑏 = [𝐵𝐻+][𝑂𝐻−] [𝐵] 𝑝𝐾𝑏 = −𝑙𝑔𝐾𝑏 [2] 

4 River Valley High School 9647/02/PRELIM/15 2015 Preliminary Examination II (ii) nbase = 25 1000 × 0.1 = 0.00250 mol nacid = 20 1000 × 0.125 = 0.00250 mol Hence, an acidic salt solution is formed. [salt] = 0.0025 45 × 1000 = 0.0556 mol dm3 Ka of conjugate acid = 10−(14−6.95) = 8.91 × 10−8 8.91 × 10−8 =√ [𝐻+] 0.0556 2 [H+] = √(8.91 × 10−8)(0.0556) 2 = 7.06 × 10−5 mol dm3 pH = −𝑙𝑔(7.06 × 10−5) = 4.15 [3] (d) Metal 1: Al Metal 2: Zn Metal 3: Fe [2] (e) (i) [Fe(H2O)6]2+ [1] (ii) Fe(OH)2(s) [1] (iii) Fe(OH)2 + 6CN ⇌ [Fe(CN)6]4 + 2OH (or similar reaction) [1] [Total: 14] 3 (a) (i) N2O4(g)  2NO2(g) [1] (ii) Since ln [N2O4] [N2O4]0 = kt, rate constant, k =  gradient At 60 s, ln [N2O4] [N2O4]0 = 1.51 k =  gradient =  1.51 60 = 2.52 × 102 s1 *must use given data points from graph [1]

5 River Valley High School 9647/02/PRELIM/15 2015 Preliminary Examination II (iii) t½ = ln2 k = ln2 2.52 × 102 = 28 s [1] (iv) Physical method: The rate of decomposition of N 2O4 can be experimentally determined by measuring the amount of light transmitted through the reaction vessel / absorbed by reaction mixture at regular time intervals , where the colour intensity is proportional to the concentration of NO2. The rate of decomposition can be determined by plotting graph of absorbance/turbidity against time and finding the gradient of the tangent of graph. OR The rate of decomposition of N 2O4 can be experimentally determined by measuring the change in total volume of reaction mixture at regular time intervals (of 20s), where the increase in volume is proportional to the volume of N2O4 reacted. The rate of decomposition can be determined by plotting graph of change in total volume against time and finding the gradient of the tangent of graph. *do not accept measuring volume of NO2 produced or N2O4 left OR initial rate method: The rate of decomposition of N 2O4 can be experimentally determined by measuring the change in total pressure /volume at regular time intervals (of 20 s ) using varying initial concentrations of N2O4. The rate of decomposition can be determined by comparing changes in initial [N 2O4] with proportional changes in total pressure/volume. *time for experiment should be short , not beyond 20 s, if mentioned [2] (b) (i) Time / s [N2O4] / mol dm3 0 1.00 20 0.60 40 0.35 60 0.22 80 0.22 100 0.22 [3]

6 River Valley High School 9647/02/PRELIM/15 2015 Preliminary Examination II (working not required) From (a)(ii), k = 2.52 × 102 s1 Using ln [NxOy] 1.00 = kt, at 20s, [N2O4] = 0.60 mol dm3 OR At 40s where [N2O4] = 0.35, k = 2.62 × 102 s1 Using ln [NxOy] 1.00 = kt, at 20s, [N2O4] = 0.59 mol dm3 OR From (a)(iii), t½ = 28 s [N2O4] = 1.00  0.5 20 28 = 0.61 mol dm3 (ii) [N2O4] at equilibrium = 0.22 mol dm3 Degree of dissociation = 10.22 1 ×100% = 78.0 % [1] (iii) [NO2] at equilibrium = (10.22) × 2 = 1.56 mol dm3 OR N2O4 ⇌ 2NO2 Initial conc / mol dm3 1.00 0 Change / mol dm3 0.78 +1.56 Eqm conc / mol dm3 0.22 1.56 Kc = [NO2]2 [N2O4] = 1.562 0.22 = 11.1 mol dm3 [3] [N2O4] / mol dm3 Time / s 1.00 0.50 60 28 0.22

7 River Valley High School 9647/02/PRELIM/15 2015 Preliminary Examination II (c) (i) [1] (ii) Phosphorus is able to expand its octet configuration as it has energetically accessible vacant 3d subshells, whereas nitrogen does not. [1] [Total: 14] 4 (a) SiO2 has a giant covalent structure