CJC Prelim P1 SOL

1 CHEMISTRY 9729/01 Paper 1 Multiple Choice 29 August 2017 Additional Materials: Multiple Choice Answer Sheet Data Booklet READ THESE INSTRUCTIONS FIRST Write your name, HT group and NRIC/FIN number on the Answer Sheet in the spaces provided. Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. 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 25 printed pages and 1 blank page. Catholic Junior College JC 2 Preliminary Examinations Higher 2 MARK SCHEME

2 9729/01/CJC JC2 Preliminary Examination 2017 For each question there are four possible answers, A, B, C and D. Choose the one you consider to be correct. 1 When an unknown organic compound is burned completely in excess oxygen, 90 cm3 of gaseous products is collected. When cooled to room temperature, the gaseous volume decreased to 50 cm3. A further decrease of 40 cm3 in the gaseous volume was observed when the gaseous mixture is passed through aqueous potassium hydroxide. What is the possible identity of the organic compound? 1 CH2CH2 2 CH3CO2H 3 CH3CH2CH3 4 CH2CHCH2OH A 1 and 3 only B 1 and 4 only C 1, 2 and 4 only D 2 and 4 only Answer: C First reduction of gaseous vol due to cooling  hot H2O(g) has condensed to form H2O(l) at r.t.p.  thus, vol of H2O(g) = 90 – 50 = 40 cm3 Second reduction of gaseous vol due to reaction with KOH(aq)  CO2(g) is an acidic gas and reacts with KOH(aq) via acid-base reaction  vol of CO2(g) = 40 cm3 Thus, CO2(g) : H2O(g) Thus, C : H = 40 : 40 = 1 : 2 = 1 : 1 Thus the organic compound must have a ratio C : H = 1: 2. 1 CH2CH2  C2H4  correct 2 CH3CO2H  C2H4O2  correct 3 CH3CH2CH3  C3H8  incorrect 4 CH2CHCH2OH  C3H6O  correct Concept: MCS: rxn stoichiometry, molar volume of gas at r.t.p, Start with volume of CO2 given and use the equation to work out the mole ratio.

3 9729/01/CJC JC2 Preliminary Examination 2017 [Turn over 2 Use of the Data Booklet is relevant to this question. A vanadium salt of unknown oxidation state was dissolved in water to form a solution of 0.500 mol dm –3. It was found that 20.4 cm 3 of this solution will react with 1 .00 g of zinc powder to form vanadium(II) solution. What is the possible identity of the vanadium salt used? A V2+ B V3+ C VO2+ D VO2+ Answer: D Let the vanadium salt of unknown oxidation state be Vx+ Amt of zinc used = 1  65.4 = 0.0153 mol Amt of Vx+ reacted = 20.4 1000 × 0.500 = 0.0102 mol Ratio of Zn : Vx+ = 0.0153 : 0.0102 = 3 : 2 Given that Zn  Zn2+ + 2e–, 3 mol of Zn will donate 6 mol of e– and 2 mol of Vx+ will accept 6 mol of e–. Therefore, 1 mol of Vx+ will accept 3 mol of e–. Since Vx+ is reduced to V2+, its original oxidation state is +5. species V2+ V3+ VO2+ VO2+ Oxidation state +2 +3 +4 +5 Thus, only VO2+ is the possible identities of the unknown salt solution. Concept: Redox: determination of oxidation state

4 9729/01/CJC JC2 Preliminary Examination 2017 3 The table below gives some data about four ions. ions number of neutrons number of nucleons Q– 16 33 R+ 19 39 S2– 17 33 T2+ 18 35 Which of the following pairs consists of ions that are isoelectronic? A Q– and S2– C S2– and T2+ B R+ and S2– D Q– and T2+ 4 Which bond angle is present in a molecule of alanine, H2NCH(CH3)CO2H, but is not present in its zwitterion? A 90 B 107 C 109 D 120 Answer: B Isoelectronic species have the same number of electrons. particle number of neutrons number of nucleons number of protons number of electrons Q– 16 33 33 – 16 = 17 17 + 1 = 18 R+ 19 39 39 – 19 = 20 20 – 1 = 19 S2– 17 33 33 – 17 = 16 16 + 2 = 18 T2+ 18 35 35 – 18 = 17 17 – 2 = 15 ANS: A Concept: Atomic Structure: determination of no. of protons and electrons from given species, isoelectronic species Concept: Chemical Bonding: Structure of zwitterion of amino acid Predict bond angles based on number of bond and lone pairs, and shape around central atom.

5 9729/01/CJC JC2 Preliminary Examination 2017 [Turn over Zwitterion of alanine is +H3NCH(CH3)COO- In the zwitterion, –NH2 group (trigonal pyramidal about N, bond angle = 107°) becomes –NH3+ (tetrahedral about N, bond angle = 109°) and –CO2H (trigonal planar about sp 2 hybridised c arbon, 120 o) group becomes CO 2- (trigonal planar about sp 2 hybridised carbon, 120o). Option A: bond angle of 90° is not present in both alanine and its zwitterion. 5 What are the types of chemical bonds present in solid phenylammonium chloride, C6H5NH3Cl? 1 dative covalent bonds 2 ionic bonds 3 hydrogen bonds A 2 only B 1 and 2 only C 2 and 3 only D 1, 2 and 3 Answer: B Concept: Chemical Bonding: Identify type of chemical bonds within an ionic compound, consisting of polyatomic ions. Identify structure to be giant ionic, with covalent bonds within the polyatomic cation, C6H5NH3+

6 9729/01/CJC JC2 Preliminary Examination 2017 C6H5NH3Cl is an ionic compound, consisting of polyatomic cations, C 6H5NH3+ and Cl-. Hence ionic bonds exist between C 6H5NH3+ and Cl - ions while covalent bond and dative covalent bonds exist between the C, N and H atoms within the C6H5NH3+ cation. Hydrogen bonds are not present as no lone electron pair is present on N atom in C6H5NH3+. 6 Which one of the following shows the standard enthalpy change of formation of carbon monoxide? A C(s) + ½O2(g) → CO(g) B C(s) + CO2(g) → 2CO(g) C C(g) + ½O2(g) → CO(g) D C(g) + CO2(g) → 2CO(g) Answer: A The standard enthalpy change of formation, Hfo, of a substance (usually a compound) is defined as the enthalpy change when one mole of the substance is formed (Hence options B and D are incorrect as 2 moles of CO are formed) from its elements under standard conditions of 298 K and 1 bar. (Elements must be in most stable physical form.) The most stable physical form of carbon is graphite (solid). Hence option C is incorrect and the answer is option A. Concept: Chemical Energetics, Definition of enthalpy change of formation.

7 9729/01/CJC JC2 Preliminary Examination 2017 [Turn over 7 Nitrogen dioxide, NO2, has an unpaired electron and dimerises to form N2O4. 2NO2(g) → N2O4(g) Which of the following statements about the spontaneity of the reaction is true? A The reaction is only spontaneous at low temperatures. B The reaction is only spontaneous at high temperatures. C The reaction is spontaneous at all temperatures. D The reaction is non-spontaneous at all temperatures. ∆H of the reaction is negative since the reaction involves bond formation which is exothermic. ∆S of the reaction is negative as there is a decrease in the number of moles of gaseous particles. ∆G = ∆H - T∆S Since ∆H is negative and -T∆S is positive, a higher temperature would cause ∆G to become more positive and less spontaneous. Hence the reacti on is only spontaneous at low temperatures. 8 Use of the Data Booklet is relevant to this question. Gas canisters used in camping stoves contain partially liquefied hydrocarbon. A canister was connected to a gas syringe and the valve opened slightly to allow some gas into the syringe. 0.200 g of the gas occupied a volume of 96.0 cm 3 at a temperature of 30.0 C