NJC 2022 H2 Chemistry Prelim P2 Ans

NJC/H2 Chem Preliminary Examination/02/2022 1 [Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 2 Structured Questions Candidates answer on Question Paper. Additional Materials: Data Booklet 9729/02 29 August 2022 2 hours READ THE INSTRUCTIONS FIRST Write your subject class, registration number and name on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use paper clips, highlighters, glue or correction fluid. Answers all questions. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 /15 2 /21 3 /15 4 /10 5 /14 Paper 2 Total /75 Marks Weightings Paper 1 /30 15% Paper 2 /75 30% Overall Percentage Paper 3 /80 35% Paper 4 /55 20% Grade This document consists of 23 printed pages and 1 blank page.

NJC/H2 Chem Preliminary Examination/02/2022 2 Answer all the questions in the spaces provided. 1 (a) Chemists have recently established that four molecules of water are required for the dissociation of a single molecule of HCl (reported in Science, 2009). Given that 1.00 dm 3 of water contains 55.6 mol of H 2O, calculate the number of molecules of hydrogen chloride, HC l, that should therefore dissociate in 1.00 dm 3 of water. [1] No. of molecules of HCl = ¼ × 55.6 × 6.02 × 1023 = 8.3678 × 1024 = 8.37 × 1024 [1] (b) Commercial concentrated hydrochloric acid, HCl, fumes strongly on exposure to moist air and thus is known as ‘fuming hydrochloric acid’. 1.00 cm 3 of fuming hydrochloric acid was transferred with a graduated pipette to a 100 cm3 volumetric flask. The volume was made up to 100 cm 3 with deionised water and the solution was labelled F. 10.0 cm 3 of solution F was neutralised by 24.75 cm 3 of 0.0500 mol dm−3 of aqueous sodium hydroxide. Calculate the concentration of HCl in the fuming hydrochloric acid in mol dm−3. [2] Amount of NaOH = 0.02475 × 0.0500 = 0.0012375 mol [1] Amount of HCl in volumetric flask = 10 × 0.0012375 = 0.012375 mol [HCl] = 0.012375 / 0.00100 = 12.4 mol dm−3 [1] Must scale up, no ECF (c) (i) The halide ions, X – (where X = Cl, Br, I), show clear trends in their physical and chemical properties. State and explain the relative thermal stabilities of the hydrogen halides, HX. ………………………………………………………………………………………...…... …………………………………………………………………………………..………... [2] Thermal stability decreases from HCl to HI. [1] H-X Bond energy decrease[ ½ ] OR Less effective overlapping between the orbitals of H atom and Halogen atom as size of halogen atom increases [ ½ ] hence less energy to break the H-X decreases. [ ½ ]

NJC/H2 Chem Preliminary Examination/02/2022 3 [Turn over The bar chart below shows the boiling points of HC l, HBr and H I. The boiling point of HF is not shown. (ii) Explain why HI has a higher boiling point than HCl and HBr. ……………………………………………………...……………………………………... …………………………………………………………………………………………….. [1] HI, HCl and HBr has instantaneous dipole-induced dipole attraction between molecules. HI has the largest electron cloud size/most number of electrons [ ½ ], hence largest energy is required to overcome the greater instantaneous dipole-induced dipole attraction than in that HCl and HBr. [½] (iii) Complete the bar chart above to show the boiling point of HF. Explain your answer. ……………………………………………………...……………………………………... …………………………………………………………………………………………….. [2] HF has the highest boiling point shown on graph [1] More energy is required to overcome the stronger hydrogen bonding between HF molecules[1] than the instantaneous dipole-induced dipole attraction (and permanent dipoles) in HCl, HBr, HI

NJC/H2 Chem Preliminary Examination/02/2022 4 (d) Indium and aluminium are elements in Group 13 of the Periodic Table. Indium has very similar chemical properties to aluminium. ● Indium reacts vigorously with hydrochloric acid to form a colourless gas and a soluble salt. ● Indium oxide, In2O3, is amphoteric. In2O3 is a white solid which dissolves in excess aqueous NaOH. ● Gaseous indium bromide, In2Br6, contains coordinate bonds. (i) Identify the formula of the salt formed when indium reacts with hydrochloric acid. ………………………………………… [1] InCl3 [1] (ii) Construct an equation for the reaction of In2O3 with excess aqueous NaOH. ………………………………………………………………………..…………………… [1] In2O3 + 2NaOH + 3H2O → 2NaIn(OH)4 In2O3 + 2 OH− + 3H2O → 2 In(OH)4− [1] (iii) Draw a diagram that clearly shows the types of bonds in a molecule of In2Br6 (g). [1] [1] Must show dative bond from Br to In or label as dative (e) In the periodic table, e lement Y is above element Z in the same group. Elements Y and Z are subjected to a series of reactions shown in the flowchart below. Y 0.020 mol excess Cl2 AgCl (s) 11.54 g excess AgNO3(aq) YCln (l) 0.020 mol water two immiscible layers of water and YCln Z 0.020 mol excess Cl2 ZCln (l) 0.020 mol water filter filtrate white residue +

NJC/H2 Chem Preliminary Examination/02/2022 5 [Turn over (i) Calculate the value of n in ZCln. [1] Amount of AgCl = 11.54 ÷ 143.4 = 0.0805 mol ratio Z:Cl is 1:4 n = 4 [1] (ii) Suggest the identities of elements Y and Z and explain why YCln and ZCln react differently with water. ………………………………………………………………………..…………………… ………………………………………………………………………..…………………… ………………………………………………………………………..…………………… [2] Y: Carbon [ ½ ] and Z: Silicon [ ½ ] Si in SiCl4 has empty energetically accessible d orbitals / vacant orbitals in the valence shell [ ½ ] to accept the lone pair from water and hence can undergo hydrolysis while C in CCl4 does not have. [ ½ ] (iii) Write a balanced equation for the reaction of ZCln with water. ………………………………………………………………………..…………………… [1] ZCl4 + 4 H2O → Z(OH)4 + 4 HCl or SiCl4 + 4 H2O → Si(OH)4 + 4 HCl or SiCl4 + 4 H2O → SiO2•2H2O + 4 HCl SiCl4 + 2 H2O → SiO2 + 4HCl [1] [Total:15] 2 (a) Oil spills are hazardous as they affect the marine ecosystem, and the marine life-forms existence gets threatened. It becomes increasingly important to employ various clean up methods for tackling the menace that oil spills pose to the marine ecosystem. (i) Oil consists mainly of hydrocarbons. Sheets of sorbents , usually made of materials like polypropene, can be used to absorb oil. The structure of polypropene is given below. Structure of polypropone Sorbent sheets

NJC/H2 Chem Preliminary Examination/02/2022 6 Suggest how sorbents are able to absorb and effectively remove oil from water. ……………………………………………………………………………………... ……………………………………………………………………………………... ……………………………………………………………………………………... ……………………………………………………………………………………... [2] The sorbents have large Mr with large electron cloud size/ non-polar [½] and can form strong [ ½ ] instantaneous dipole – induced dipole interactions [½ ] between the oil and material of the sorbent and hence allowing oil to be removed. The sorbents do not form favourable interactions / weaker idid [½] with water and hence do not absorb water. Must state “strong” or to the effect (ii) Another method for removing the oil spill that is floating on the sea surface is to ignite and burn it off. One of the major components of oil is decane, C10H22. Such burning of oil will