RVHS H1 Chemistry P1 Soln 2024

River Valley High School 2024 JC 2 H1 Chemistry 8873 Prelim Exam Paper 1 Worked Solutions 1 A The decomposition involves: n → p+ + e− where the newly-formed proton stays in the nucleus. As the proton number of the atom increases by 1 following the decomposition, the element will no longer be the same. Options B and D are eliminated. Option A involves an increase in proton number by 1 going from K to Ca, hence it is the correct answer. 2 C Option A: By definition, one mole of a substance contains exactly 6.02  1023 (or Avogadro number) elementary entities. Option A is incorrect because of the phrase “same number of atoms as there are in 12.000 g of carbon-12”. Option B: By definition, relative isotopic mass (Ar) is the mass of one mole of atoms of an isotope (of a certain element) relative to 1 12 the mass of one mole of 12C atoms. Option B is incorrect because in the formula given, the numerator used is “average mass of all isotopes of lithium”. Option C: By definition, relative atomic mass (Ar) is the average mass of one mole of atoms of an element relative to 1 12 the mass of one mole of 12C atoms. Note that some textbooks define relative atomic mass as “the average mass of one atom of an element relative to 1 12 the mass of one 12C atom”. Hence, option C is correct. Option D: By definition, relative molecular (Mr) mass is the average mass of one mole of molecules relative to 1 12 the mass of one mole of 12C atoms. Option D is incorrect because in the formula given, the numerator used is “average mass of one atom of E”.

3 A Statement 1: Correct According to Aufbau’s build up principle, in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy level before occupying higher-energy levels. Statement 2: Incorrect The order of filling the orbitals is 1s, 2s, 2p, 3s, 3p, 4s, 3d. Statement 3: Incorrect The shell with principal quantum number 2 only has 2s and 2p orbitals. Hence, it can hold a maximum of 8 electrons. 4 D A: Due to the overlap of unhybridised 2p orbitals, mobile delocalised electrons are found in the lattice structure. B: Each carbon atom forms 3 sigma bonds with 3 other carbon atoms. C: Instantaneous dipole-induced dipole interactions exist between each graphite plane. D: Conduction of electricity occurs due to the overlap of unhybridised 2p orbitals. The overlapping occurs perpendicular to the axis of the unhybridised 2p orbitals. 5 A A: There is an unpaired electron on N atom in NO2 molecule. B: The Cu+ ion has an electronic configuration of 1s2 2s2 2p6 3s2 3p6 3d10 4s0. C: The Li+ ion has an electronic configuration of 1s2 2s0. D: After heterolytic fission, Cl+ and Cl ions are formed, which do not have a single unpaired electron. 6 B When indicator P turns yellow, it means pH of solution is higher than 5. When indicator Q turns yellow, it means pH of solution is less than 5.7. Therefore, pH range of the solution is between 5 to 5.7.

7 A A: Recall that H + does not contain any electrons so the total number of electrons in both CO32 and HCO3 are the same (32 electrons). B: The blood buffer system consists of H2CO3 and HCO3. C: While CO32 can only exhibit basic property (as the conjugate base of HCO3), HCO3can both exhibit acidic (as the conjugate acid of CO32) and basic property (as the conjugate base of H2CO3). D: They are a conjugate acid -base pair as they differ by 1 proton (H +). A random combination of a weak acid and a weak base does not constitute a conjugate acid-base pair. 8 D The volume for equivalence point is 10 cm 3. Thus, the molar ratio of acid and base used is 1: 1 which implies that the acid is monobasic. Therefore options A and B are not possible. Since the equivalence point is above 7, this will mean that the titration is between a strong base and a weak monobasic acid. Therefore the answer is D. 9 B Option A is correct: Anionic size decreases in the order of P3− > S2− > Cl−. Option B is incorrect: The element with the highest melting point is silicon, which has giant covalent structure. Option C is correct: Aluminium has giant metallic lattice structure. Due to the highest number of delocalised valence electrons (per atom), aluminium has the highest electrical conductivity. Option D is correct: Sulfur exists as S8 molecules. 10 A Step I: MgO is basic in nature and does not react with NaOH. Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2Na[Al(OH)4](aq) SiO2 only reacts with concentrated alkali. Step II: Filtrate contains Na[Al(OH)4]. Residue contains MgO and SiO2. Step III: MgO(s) + 2H+(aq) → Mg2+(aq) + H2O(l) SiO2 is acidic in nature and does not react with HCl. Step IV: Filtrate contains Mg2+(aq). Residue contains SiO2.

11 D Option A: Correct for both Group 1 and Group 17. Down both Group 1 and Group 17, the valence electron to be removed is further away from the nucleus. Hence, the ionisation energies decrease down the groups. Option B: Correct for both Group 1and Group 17. Group 1 elements act as reducing agents and get oxidised themselves. Oxidation involves the loss of electrons. Down Group 1, it is easier to lose the valence electron as it is further from the nucleus. Hence, reducing power increases down Group 1. Group 17 elements act as oxidising agents and get reduced themselves. Reduction involves the gain of electrons. Down Group 17, it is harder to gain electrons because the electrons gained are placed further away from the nucleus and experience weaker nuclear attraction. Hence, the oxidising power of Group 17 elements decreases down the group. Option C: Correct for both Group 1 and Group 17. In general, Group 1 elements are good reducing agents while Group 17 elements are good oxidising agents. Option D: Incorrect because a cross the period, atomic radius decreases . Thus a Group 1 element should have larger atomic radius than a Group 17 element in the same Period.

12 C Amount of CO2 = 48.0 24 000 = 0.00200 mol Amount of sodium percarbonate = 10.00.100 1000 = 0.00100 mol (Na2CO3)xy(H2O2) : CO2 x : 1 0.00200 : 0.00100 2 : 1 Therefore, x = 2 Amount of H2O2 = 0.00100y mol Amount of KMnO4 = 24.00.0500 1000 = 0.00120 mol Given the ratio is 2 : 5 0.00120 2 0.00100y 5= y = 3 Therefore, y3 x2= 13 B Amount of ethanol produced = (100  1000)/46.0 = 2173 mol Amount of carbon dioxide required = 2  2173 = 4347 mol Volume of carbon dioxide required = 4347  22.7 = 98676.9 = 98700 dm3

14 C Option A: Incorrect Amount of potassium chloride = 74.6 / 74.6 = 1 mol Amount of ions = 1  2 = 2 mol No. of ions = 2  6.02  1023 = 1.20  1024 Option B: Incorrect Amount of CO2 = 44/44 = 1 mol Amount of atoms = 1  3 = 3 mol No. of atoms = 3  6.02  1023 = 1.81  1024 Option C: Correct Amount of nitric acid, HNO3 = 0.5 mol Amount of ions = 0.5  2 = 1 mol No. of ions = 1  6.02  1023 = 6.02  1023 Option D: Incorrect Amount of delocalised electrons = 3 mol No. of delocalised electrons = 3  6.02  1023 = 1.81  1024 15 B H2S(g) + ½O2(g) → H2O(g) + S(s) Hr = −243.0 – (−20.5) = −222.5 kJ mol−1

16 A Option A: Correct. Bond dissociation energy: HF(g) → H(g) + F(g) +565 kJ mol−1 The reverse of the above reaction results in 565 kJ of energy released. Option B: Incorrect because change of state from liquid to gas is endothermic. Heat is absorbed and not released. Option C: Incorrect because the equation is showing the endothermic process of bond dissociation of H F. The equation is shown under option A. Option D: Incorrect. The equation is showing the enthalpy of formation of HF. Using the bond energies found in Data Booklet, H = (0.5)(+436) + (0.5)(+158) − (+562) = −265 kJ mol−1 17 B Statement 1: Correct Introducing a catalyst will lower the Ea and hence increase the rate constant for both the forward and backward reactions. Statement 2: Correct Increa