2022 CJC H2 CHEM Prelim P1 QP w Ans FINAL COPY

1 9729/01 CJC JC2 Preliminary Examination 2022 CANDIDATE NAME CLASS 2T CHEMISTRY 9729/01 Paper 1 Multiple Choice 15 September 2022 1 hour Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, class and NRIC/FIN number on the Answer Sheet in the spaces provided. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 31 printed pages. Catholic Junior College JC 2 Preliminary Examinations Higher 2 WORKED SOLUTIONS

2 9729/01 CJC JC2 Preliminary Examination 2022 + – R S T 1 Which statement about 27 g of Al is always correct? A It contains the same number of atoms as 1 12 g of 12C. B It contains the same number of atoms as 24 dm3 of krypton gas at room temperature and pressure. C It contains the same number of hydrogen ions as 1 dm3 of 1 mol dm-3 aqueous sulfuric acid. D It contains the same number of atoms as 28 g of nitrogen gas. 2 The following are flight paths of charged particles as they pass through an electric field at the same speed. Which of the following correctly identifies X, Y and Z? X Y Z A 14N‾ 16O2+ 28Si2+ B 14N‾ 14C+ 28Si4+ C 15O+ 14C+ 14N+ D 15O‾ 14C+ 28Si+ X Y Z Topic: Mole Concept No of Al atoms in 27 g = 27 27.0 x 6.02 x 1023 = 6.02 x 1023 A No of C atoms in 1 12.0 g of 12C = 1 12 12.0 x 6.02 x 1023 = 4.18 x 1021 B No of Kr atoms in 24 dm3 of Kr = 24 24.0 x 6.02 x 1023 = 6.02 x 1023 N2  2 C H2SO4  2H+ No of H+ ions in 1 dm3 of aq H2SO4 = 1 x 1 x 2 x 6.02 x 1023 = 1.20 x 1024 D N2  2N No of N atoms in 28 g of N2 = 28 28.0 x 2 x 6.02 x 1023 = 1.20 x 1024 Answer: B

3 9729/01 CJC JC2 Preliminary Examination 2022 [Turn over 3 Use of the Data Booklet is relevant to this question. Species containing one or more unpaired electrons can be attracted by an external magnetic field and are said to be paramagnetic. Which of the following species is paramagnetic? 1 Cr3+ 2 Cu+ 3 Ni2+ A 3 B 1 and 2 C 1 and 3 D 1, 2 and 3 Concept: Angle of deflection particles 14N‾ 14C+ 28Si4+ m/z 1/14 1/14 4/28 = 1/7 14N‾ is negatively charged, so it is attracted to anode while 14C+ and 28Si4+ are attracted to cathode. Since angle of deflection is charge/mass ratio, 14N‾ and 14C+ have roughly the same angle of deflection (paths X and Y), while 28Si4+ has almost double the angle of deflection (path Z). Answer: B Concept: Electronic Configuration Cr3+: [Ar]3d3 Cu+: [Ar]3d10 (Does not have any unpaired electrons) Ni2+: [Ar]3d8 Answer: C

4 9729/01 CJC JC2 Preliminary Examination 2022 4 In which of the following sequences are the species quoted in order of decreasing boiling points? A RbCl, KCl B HF, HCl C K, Ca D CO, CO2 Concept: Boiling points Option A: Both of them have giant ionic lattice structure with strong electrostatic forces of attractions between cations and anions. The strength of ionic bonding α qq rr +− +−  + Rb+ has larger cationic size than K+, hence RbCl has weaker ionic bonding than KCl. Option B: HF has a higher boiling point than HCl as it has stronger intermolecular hydrogen bonding while HCl has weaker permanent dipole-permanent dipole between its molecules. Option C: Both K and Ca have giant metallic lattice structure with strong electrostatic forces of attractions between cations and sea of delocalised electrons. The strength of metallic bonding α the number of valence electrons involved in the delocalisation. K: 4s1 Ca: 4s2 Ca has higher boiling point than K because it has more valence electrons than K involved in the delocalisation. Option D: Both of them have simple molecular structure. CO 2 has larger electron cloud size compared to CO. CO2 has higher boiling point as it has stronger intermolecular instantaneous dipole-induced dipole attractions between its molecules compared to the weaker intermolecular permanent dipole-permanent dipole between CO molecules. Answer: B

5 9729/01 CJC JC2 Preliminary Examination 2022 [Turn over 5 People drink beverages containing caffeine to relieve or prevent drowsiness and to improve cognitive performance. Caffeine Which statement about caffeine is true? A Caffeine molecule has a planar structure. B The  bond in C=C is formed by sideways overlap of 2p orbitals. C The nitrogen atom in caffeine, N1 is basic. D There are 16 sigma bonds in a caffeine molecule. Concept: Chemical Bonding Option A is wrong. There are sp3 carbon atoms which have tetrahedral geometry. Option B is correct. The pi bond is formed by sideways overlap of paral lel unhybridised p orbitals. Option C is wrong. The lone pair of electrons in the unhybridised p orbital of N1 has overlapped sideways with the adjacent  electron cloud of C=O bond resulting in the delocalisation of lone pair of electrons. Hence, lone pair of electrons is unavailable for protonation. Option D is wrong. There are 25  bonds in a caffeine molecul e. There is one  bond and one  bond in a double bond. Answer: D

6 9729/01 CJC JC2 Preliminary Examination 2022 6 Which of the following substances conduct electricity due to delocalised electrons? 1 Graphite 2 Solid magnesium 3 Molten sodium chloride A 1 only B 1 and 2 only C 2 and 3 only D 1, 2 and 3 7 Which of the following statements is incorrect? A When methane gas is subjected to low pressure, it liquefies. B Tyre pressure readings are higher on a hot day. C The boiling point of water is lower than 100ºC at a higher altitude. D The density of an ideal gas at constant pressure is inversely proportional to the temperature. Concept: Gaseous State A: Gas should be subjected to high pressure for it to be liquified. B: P ∝ T, when tyre is hot, T increases hence pressure is higher. C: At a higher altitude than sea level, the atmospheric pressure is lower than atmospheric pressure at sea level, thus boiling point will be lowered as P ∝ T. D:  = PM RT density is inversely proportional to temperature. Answer: A Concept: Physical properties of compounds Answer: B Molten sodium chloride conducts electricity due to the presence of free mobile ions.

7 9729/01 CJC JC2 Preliminary Examination 2022 [Turn over 8 For the oxides of Period 3 elements (Na to P), which property decreases from Na 2O to P4O10? A melting point B covalent character C pH when mixed with water D solubility in aqueous alkali Oxide formula Na2O MgO Al2O3 SiO2 P4O6 and P4O10 Structure Giant ionic Giant covalent Simple covalent (molecular) Melting Point / oC 1280 2900 2040 1610 24 580 Acid/base behaviour Basic: Na2O + 2H+ → 2Na+ + H2O Basic: MgO + 2H+ → Mg2+ + H2O Amphoteric: Al2O3 + 6H+ → 2Al3+ + 3H2O Al2O3 + 2OH- + 3H2O → 2Al(OH)4- Acidic: forms SiO32- with base SiO2 + 2NaOH → Na2SiO3 + H2O SiO2 + CaO → CaSiO3 Acidic P4O6 + 8NaOH → 4Na2HPO3 + 2H2O P4O10 + 12NaOH → 4Na3PO4 + 6H2O Rxn w/H2O Eqn Na2O + H2O → 2NaOH MgO + H2O ⇌ Mg(OH)2 (slightly alkaline as only mi