RI 2021 Y5 Timed Practice Solutions

© Raffles Institution 2021 9729/J/21 2021 Y5 H2 Chemistry July Common Test – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer D B A C A D D B C D B A A B C MCQ worked solutions 1 Amount of CO2 formed = 39.6 / 44 = 0.9 mol Amount of H2O formed = 16.2 / 18 = 0.9 mol Since X and Y have the same empirical formula, the mole ratio of C : H in the products is the same as the mole ratio of C : H in X and Y.  mole ratio of C : H = 0.9 : 2(0.9) = 1 : 2  empirical formula of X and Y is CH2. Let the molecular formula of X be CnH2n. Then the molecular formulae of Y will be C2nH4n. By considering the total amount of CO2 formed from the combustion of 0.1 mol of CnH2n and 0.1 mol of C2nH4n, 0.1 × (n + 2n) = 0.9 n = 3  Y has molecular formula C6H12. 2 Let the other species be X. Oxidation number of Mn in MnO42− = +6 Oxidation number of Mn in MnO4− = +7 oxidation: MnO42− ⎯→ MnO4− + e− 1 mol of MnO42− loses 1 mol of electron to form 1 mol of MnO4−. Since this is a disproportionation reaction, MnO42− undergoes oxidation to form MnO4− and reduction to form X.  Oxidation number of Mn in X should be lower than +6.  Option D is incorrect as oxidation number of Mn in Mn2O7 = +7 X Oxidation number of Mn No. of electrons gained by 1 mol of MnO42− to form 1 mol of X Since number of electrons gained = number of electrons lost, ratio of X : MnO4− A Mn(OH)2 +2 4 mol 1 : 4 B MnO2 +4 2 mol 1 : 2 C Mn2O3 +3 3 mol 1 : 3 Only option B gives the correct ratio between the two manganese-containing products.

© Raffles Institution 2021 9729/J/21 3 HCl(aq) + NaOH(aq) ⎯→ NaCl(aq) + H2O(l) Let the volume of each solution used be V dm3. Amount of HCl = V mol Amount of NaOH = 1.5V mol Since the mole ratio of HCl : NaOH = 1 : 1, HCl is limiting. In the resulting solution: Amount of NaOH unreacted = 0.5V mol Amount of NaCl formed = V mol Final total volume = 2V dm3 [NaOH] = 0.5V / 2V = 0.25 mol dm−3 [NaCl] = V / 2V = 0.50 mol dm−3 4 Species Electronic Configuration Electrons-in-boxes diagram (for highest energy subshell) Number of unpaired electrons N− 1s22s22p4 2 F2+ 1s22s22p3 3 Cr2+ 1s22s22p63s23p63d4 4 Co2+ 1s22s22p63s23p63d7 3 5 Option A is correct (as the statement is incorrect). A neutron has no charge while an electron is negatively charged. Hence, they are not oppositely charged. Option B incorrect (as the statement is correct). A neutron is found in the nucleus at the centre of the atom while an electron is found around the nucleus. Hence, they are found in different parts of the atom. Option C is incorrect (as the statement is correct). A neutron has a relative mass of 1 while an electron has a relative mass of approximately 1/1840. Hence, the neutron has a greater mass than the electron. Option D is incorrect (as the statement is correct). The number of electrons in an atom must equal to the number of pr