RI Prelim P1 (Ans)

1 © Raffles Institution 2020 9729/01/S/20 2020 RI H2 Chemistry Prelim Paper 1 – Suggested Solutions Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer C C D A A B D A B C B A B C C MCQ worked solutions Q1 (Ans: C) 28.0 m 25.0 131 27.0 132 20.0 134 131.3128.0 25.0 27.0 20.0         m = 129 Q2 (Ans: C) 5 X− + XOn−  halogen-containing product (either X2 or XO−) Step 1: Balance X; 5 X− + XOn−  3 X2 OR 5 X− + XOn−  6 XO− Step 2: Check if amt of electrons gained = amt of electrons lost Option A 5 X− + XO−  3 X2 amt of electrons gained by 1 mol of XO− to form ½ mol X2 = 1 mol amt of electrons lost by 5 mol of X− to form 5 2 mol of X2 = 5 mol not balanced Option B 5 X− + XO2−  6 XO− amt of electrons gained by 1 mol of XO2− to form 1 mol of XO− = 2 mol amt of electrons lost by 5 mol of X− to form 5 mol of XO−= 10 mol not balanced Option C 5 X− + XO3−  3 X2 amt of electrons gained by 1 mol of XO3− to form ½ mol X2 = 5 mol amt of electrons lost by 5 mol of X− to form 5 2 mol of X2 = 5 mol balanced Option D 5 X− + XO4−  6 XO− amt of electrons gained by 1 mol of XO4− to form 1 mol of XO− = 6 mol amt of electrons lost by 5 mol of X− to form 5 mol of XO− = 10 mol not balanced Question 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer C D D B D A B A D B D C B A D

2 © Raffles Institution 2020 9729/01/S/20 Q3 (Ans: D) ntotal = nAr + nXe PtotalVtotal/RT = PArVAr/RT + PXeVXe/RT At constant T, PtotalVtotal = PArVAr + PXeVXe P  (5 + 10) = (150  5) + (350  10) P = 283 kPa At constant V and n, pressure  Temperature 600 T 283 298 T = 632 K Q4 (Ans: A) U92 235  Rb37 96 + Cs55 135 + 4 XA Z 235 = 96 + 135 + 4Z  Z = 1 92 = 37 + 55 + 4A  A = 0 X is n0 1  neutron Q5 (Ans: A) Horizontal axis: P4O10 (pH 3), SiO2 and Al2O3 (pH 7 as they do not dissolve in water), MgO (pH 9) Vertical axis: P4 (simple molecular) < Mg (giant metallic) < Al (giant metallic) < Si (giant covalent) Q6 (Ans: B) The graph shows increased solubility in acidic and alkaline solutions  compound is amphoteric, i.e. Al(OH)3 With acid: Al(OH)3(s) + 3HCl(aq)  AlCl3(aq) + 3H2O(l) With base: Al(OH)3(s) + NaOH(aq)  Na+[Al(OH)4]–(aq)

3 © Raffles Institution 2020 9729/01/S/20 Q7 (Ans: D) Graph shows a sharp decrease in second IE from W to X  second electron removed from W is from an inner electronic shell  W is in Group 1 Statement 1 Not correct. V is in Group 18. However, if V is helium then U is hydrogen, but hydrogen does not have a second IE. As such, U to Z must be fluorine to silicon (with atomic numbers less than 20). Label Group No. Identity U 17 F V 18 Ne W 1 Na X 2 Mg Y 3 Al Z 4 Si Statement 2 Correct. Group 1 metals (e.g