2024 YIJC JC2 H1 CHEM PE P1 (worked solutions)

©YIJC [Turn over YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 1 CANDIDATE NAME SUGGESTED ANSWERS CG H1 GROUP 1 / 2 / 3 CHEMISTRY Paper 1 Additional Materials: Multiple Choice Answer Sheet Data Booklet 8873 /01 13 September 2024 1 hour READ THESE INSTRUCTIONS FIRST This document consists of 19 printed pages. Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name and class in the spaces at the top of this page and on the Answer Sheet in the spaces provided unless this has been done for you. 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.

©YIJC [Turn over 2 1 Which of the species in their gaseous state shown below will be deflected by the smallest angle in an electric field? A 1H2 B 4He2+ C 9Be2+ D 11B3+ Answer: C angle of deflection ∝ charge mass of the particle A 1H2 Not deflected since it is not charged B 4He2+ 2 4 = 0.5 C 9Be2+ 2 9 = 0.222 D 11B3+ 3 11 = 0.273 2 What is the electronic configuration of vanadium, 23V? A 1s2 2s2 2p6 3s2 3p6 3d4 4s1 B 1s2 2s2 2p6 3s2 3p6 3d5 C 1s2 2s2 2p6 3s2 3p6 4s2 4p3 D 1s2 2s2 2p6 3s2 3p6 3d3 4s2 Answer: D V atom has 23 protons and 23 electrons. 1s2 2s2 2p6 3s2 3p6 3d3 4s2 3 Use of the Data Booklet is relevant to this question. How many atoms of gold, 79Au, are there in a pure gold coin with a mass of 3.94g? A 1.20 × 1021 B 2.00 × 1021 C 1.20 × 1022 D 2.00 × 1022

©YIJC [Turn over 3 Answer: C Amount of Au = 3.94  197 = 2 × 10−2 mol No. of atoms = (2 × 10−2) × (6.02 × 1023) = 1.20 × 1022 4 10cm3 of gaseous hydrocarbon is burnt completely in 70cm 3 of oxygen. When cooled to room temperature, the gaseous volume was 50cm 3. A further decrease of 40cm 3 in the gaseous volume was observed when the gaseous mixture is passed through aqueous potassium hydroxide. What is the molecular formula of the hydrocarbon? A C2H6 B C3H8 C C4H8 D C5H12 Answer: C Reduction of gaseous volume due to reaction with KOH(aq) → CO2(g) is an acidic gas and reacts with KOH(aq) via acid-base reaction → volume of CO2(g) = 40 cm3, Volume of oxygen unreacted = 50 − 40 = 10 cm3 Volume of O2 reacted = 70 – 10 = 60 cm3 CxHy : O2 CO2 : H2O stoichiometric ratio from balanced equation 1 mol (x + y 4) mol x mol volume ratio 10 cm3 60 cm3 40 cm3 1 cm3 6 cm3 4 cm3 mole ratio 1 mol 6 mol 4 mol x = 4, (x + y 4) = 6, y = 8 The molecular formula of the hydrocarbon is C4H8 5 A student added 12.5cm 3 of 0.0500moldm –3 sodium hydroxide to 25.0cm 3 of 0.100moldm –3 hydrochloric acid. What is the concentration of hydrochloric acid remaining in the reaction mixture? A 0.0333moldm–3 B 0.0500moldm–3 C 0.066moldm–3 D 0.0750moldm–3

©YIJC [Turn over 4 Answer: B NaOH + HCl → NaCl + H2O Amount of NaOH = 12.5 1000 × 0.0500= 6.25 × 10–4 mol (limiting reagent) Amount of HCl = 25.0 1000 × 0.100 = 2.50 × 10–3 mol Number of moles of HCl remaining = (2.50 × 10–3) – (6.25 × 10–4) =1.875 × 10–3 mol Concentration of HCl remaining = 1.875 x 10-3 37.5 1000 = 0.0500 mol dm–3 6 Substance T • is soluble in polar solvent. • does not conduct electricity in solid but conducts electricity when molten. • is a solid at room temperature. What could be the structure of T? A giant molecular B giant ionic C giant metallic D simple molecular Answer: B Ionic compounds are soluble in polar solvent due to the formation of ion-dipole interactions. They are able to conduct electricity in the molten state as the ions are mobile and acts as charge carrier. Ionic compounds generally have high melting point and hence are usually solid at room temperature.

©YIJC [Turn over 5 7 Hydrazine, N2H4, has some properties that are similar to those of ammonia, NH3. Which statement explain why hydrazine molecules are more soluble in water than ammonia molecules? A Hydrazine is more polar than ammonia. B The covalent bonding in hydrazine is stronger than that in water. C There are stronger permanent dipole interactions between hydrazine and water than between ammonia and water. D There are more hydrogen bonds between hydrazine and water than between ammonia and water. Answer: D * Bonds between hydrazine/ammonia with water: hydrogen bond (not covalent) * Covalent bond refers to the bonding between N-H. No. of H-bonds form in ammonia = 1 (only one lone pair in electronegative atom N) No. of H-bonds form in hydrazine = 2 (two lone pairs in electronegative atom N) Therefore hydrazine able to form more hydrogen bonds with water molecules.

©YIJC [Turn over 6 8 Which row correctly describes the change in the properties for Group 1 elements down the group? reducing power ionic radius first ionisation energy A increases increases increases B decreases decreases increases C increases increases decreases D increases decreases decreases Answer: C Reducing power As effective nuclear charge decreases down the group, the attraction of nucleus for electrons decreases and hence ease of losing electrons increases. Thus, the reducing power increases as well. Ionic radius Down the group, each element has 1 more principal quantum shell and hence bigger ionic radius. Ionisation energy As effective nuclear charge decreases down the group, the attraction of nucleus for electrons decreases and hence ease of losing electrons increases. Thus, the energy needed to remove electrons (ionisation energy) decreases. 9 The proton number of the element X is less than 20. The aqueous chloride of X is acidic and gives a white precipitate when neutralised with NaOH(aq). The precipitate dissolves on reacting either with NaOH(aq) or with H2SO4(aq). In which Group of the Periodic Table is X likely to be found? A 2 B 13 C 14 D 15 Answer: B Since the aqueous chloride of X is acidic, X can be from Group 13, 14 or 15 of the Periodic Table. The white precipitate is a hydroxide of the element. Since the hydroxide is amphoteric, X is likely to be in Group 13.

©YIJC [Turn over 7 10 Pure germanium is an important element in the electronic industry and is used in transistors. To manufacture pure germanium, the metal is separated from other metals by the formation of germanium tetrachloride, GeCl4, followed by fractional distillation. GeCl4 is a liquid at room temperature and has similar properties to SiCl4. Which statement about germanium tetrachloride is correct? A The bond angle in germanium tetrachloride is 120o. B Germanium tetrachloride is an ionic compound. C Germanium tetrachloride will conduct electricity. D Germanium tetrachloride is hydrolysed by water. Answer: D A Incorrect. GeCl4 is tetrahedral in shape and the bond angle in germanium tetrachloride is 109.5o. B Incorrect. Germanium tetrachloride is a covalent compound. C Incorrect. All valence electrons in Ge is used for bonding and there are no mobile charge carrier. D Correct. Germanium tetrachloride is