SRJC H2 Chem 2012 Prelim P1 Soln

Section A Answer all questions 1 To determine the percentage of nitrogen present in a snack, 1.0 g of the snack was boiled with concentrated sulphuric acid to convert all the nitrogen into ammonium sulphate. The ammonium salt obtained was then boiled with excess aqueous sodium hydroxide to liberate the ammonia, which was passed into 25.0 cm 3 of 0.20 mol dm –3 hydrochloric acid. The unreacted hydrochloric acid required 20.0 cm 3 of 0.10 mol dm –3 aqueous sodium hydroxide for complete neutralisation. What is the percentage by mass of nitrogen in the snack? A 2.8% B 4.2% C 7.2% D 8.4% Answer: B Amt of NH3 =( 25 1000x 0.20) – ( 20 1000x 0.10) = 0.003 mol 2NH3≡ 1(NH4)2SO4 ≡ 2N Mass of nitrogen in snack = 0.003 x 14.0 = 0.042 g % by mass = 0.042 1.0 x 100 = 4.2% 2 In an experiment, 25 .0 cm3 of 0.2 0 mol dm –3 solution of K 2AO4 reacted exactly with 25.0 cm 3 of 0.10 mol dm –3 a queous sodium sulfate(IV). The half- equation for the oxidation of the sulfate(IV) ion is shown below. SO32– (aq) + H2O (l) ® SO42– (aq) + 2H+ (aq) + 2e– Calculate the final oxidation state of A. A +2 B +3 C +4 D +5 Answer: D 2K2AO4 ≡ SO32– No. of e gained by A in K2AO4 = No. of e lost by S in SO32– = 2e 2A6+ + 2e à 2An+ 12 – 2 = 2n n = +5

3 Two elements D and E have the following properties. · D and E form ionic compounds Na2D and Na2E respectively. · Element E forms EF6 molecules whereas D is not able to do so. Which pair of electronic configurations for D and E is correct? D E A [He] 2s2 2p2 [Ne] 3s2 3p4 B [He] 2s2 2p2 [Ne] 3s2 3p2 C [He] 2s2 2p4 [Ne] 3s2 3p2 D [He] 2s2 2p4 [Ne] 3s2 3p4 Answer: D 4 The diagram below shows liquid trichloromethane and liquid benzene flowing from burettes 1 and 2 respectively. What would happen to the flow of the liquids trichloromethane and benzene when a negatively-charged rod is brought near to each of them? Liquid trichloromethane Liquid benzene A Deflected towards the rod Deflected towards the rod B Undeflected Deflected towards the rod C Deflected towards the rod Undeflected D Undeflected Undeflected Answer: C Liquid trichloromethane is polar. The partial positive charge can be attracted by the negatively charged rod. Benzene is non-polar and does not have partial charges. Burette 1 Liquid trichloromethane Burette 2 Liquid Benzene

5 The value of pV is plotted against p for two gases, G and H, where p is the pressure and V is the volume of the gas. Which of the following could be the identities of the gases? Gas G Gas H A 0.5 mol of H2 at 25 ºC 0.5 mol of H2 at 50 ºC B 0.5 mol of H2 at 25 ºC 1 mol of SO2 at 25 ºC C 0.5 mol of SO2 at 25 ºC 0.5 mol of SO2 at 50 ºC D 0.5 mol of SO2 at 25 ºC 1 mol of H2 at 25 ºC Answer: D Amount of gas H should be twice the amount of gas G. According to the shape of the curves, gas G should be a less ideal gas than gas H. 6 During an inspection, a small spacecraft of capacity 20 m3 was connected to another of capacity 50 m 3. Before connection, the pressure inside the smaller craft was 40 atm and that inside the larger one was 150 atm. Given that all measurements were made at the same temperature, What is the pressure of the system after the connection? A 78 atm B 95 atm C 119 atm D 190 atm Answer: C Amt of gas in small spacecraft = RT RT RT pV 800) 20)(40 ( == Amt of gas in large spacecraft = RTRT RT pV 7500) 50)(150 ( == total amt of moles of gas = RTRT 83007500 800=+ Pressure in the combined arrangement = ( ) ( ) atm RTRT V nRT 11950 20 8300 =+= 2x pV p Gas H Gas G 0 x

7 In which of the following pairs of compounds will compound II have a higher boiling point than compound I? I II A Br Cl B CH3CH2CH2CH2CH3 C(CH3)4 C CH3CH2CH2COOH CH3CH2CH2OH D C C CH2Cl H CH2ClCH3 C C CH2Cl H CH2Cl CH3 Answer: D A: I has higher Mr than II and hence has more extensive intermolecular VDW forces of attraction and a higher boiling point. B: I is linear and II is branched and hence has more extensive intermolecular VDW forces of attraction and a higher boiling point. C: I has more extensive intermolecular hydrogen bonding than II and hence a higher boiling point. D: I is non-polar and has temporary dipole-ind uced dipole interactions while II is polar and has permanent dipole-dipole interactions. Hence II has a higher boiling point.

8 The conversion of compound X into Z was exothermic and proceeded by two steps , where Y was the intermediate. The steps involved were: Step 1: X ® Y Step 2: Y ® Z It was found that Step 1 is the rate-deter mining step. Which diagram represents the energy level diagram for the reaction? Answer: A

9 Pure nitrosyl chloride, NOC l gas, was heated at 320°C in a 2.0 dm 3 vessel. At equilibrium, 30% of the NOC l gas had dissociated according to the equation below and the total pressure was P atm. 2NOCl (g) 2NO (g) + Cl2 (g) What is value of the equilibrium constant, Kp? A 17.9 𝑝𝑝 B 41.7 𝑝𝑝 C 0.0120p D 0.0130p Answer: C 2NOCl (g) 2NO (g) + Cl2 (g) Initial partial pressure/atm x 0 0 Change in partial pressure / atm -0.3x +0.3x +0.15x Equilibrium partial pressure / atm 0.7x 0.3x 0.15x 0.6087p 0.2609p 0.1304p 0.7x + 0.3x + 0.15x = p x = 0.8696p Kp = (0.2609𝑝𝑝)(0.1304𝑝𝑝)2 (0.6087𝑝𝑝)2 = 0.01197 = 0.0120p

10 The pH change when 0.100 mol dm -3 CH3COOH is added drop -wise to 10.0 cm 3 of 0.100 mol dm-3 NaOH (aq) is shown below. At which point on the graph does pH = p Ka, where Ka is the acid dissociation constant of the weak acid? Answer: D Regions C and D are where there is an excess of weak acid CH 3COOH as well as the salt CH3COO-Na+ that is formed. Hence, buffer region. At Region D, Amt of CH3COO-Na+ formed = =10 0.100 0.0011000 x mol Amt of excess CH3COOH added = - =(20 10) 0.100 0.0011000 x mol Amt of CH3COO-Na+ and CH3COOH is the same, hence this is the point where there is maximum buffering capacity and pH = pKa. A B C D

11 In an experiment, 70 cm 3 of water at 25ºC was brought to boiling point by burning butane in excess oxygen. Given that the standard enthalpy change of combustion of butane is -2877 kJ mol -1, calculate the volume of butane needed if this process is only 85% efficient. Assume that the specific heat capacity of water is 4.2 J g -1 K-1 and that 1 mole of gas occupies 24 dm3 under the given conditions. A 0.0721 dm3 B 0.156 dm3 C 0.184 dm3 D 0.216 dm3 Answer: D Since Q’ = mc∆T = 70 × 4.2 × (100.0 - 25.0) = 22050 J Apparent amount of heat absorbed by water, Q’ = 85 100 Q (Actual amount of heat evolved by burning butane) Actual amount of heat evolved, Q = 100 85 x 22050 = 25941 J ∆Hcө(CH3CH2CH2CH3) = - 25941 n = -2877 x 103 Amount of butane = 9.017 x 10-3 mol Volume of butane = 9.017 x 10-3 x 24 = 0.216 dm3 12 Which of the following changes does not alter the E q value measured for a C l2/Cl- half- cell that is under standard conditions? A Adding water into the half-cell. B Placing the half-cell in an ice bath. C Adding copper(II) ions into the half-cell. D Introducing an inert gas into the half- cell at a pressure of 1 atm through a separate inlet from the Cl2 gas inlet. Answer: D A: Dilution causes the concentration of Cl- ions to be lower than 1 mol dm-3. B: The ice bath lowers the temperature of the half-cell to less than 298 K. C: Cu2+ ions will form a complex with C l- ions and lower the concentration of C l- ions to less than 1 mol dm-3. D: Introducing an inert gas through a separate inlet does not affect the pressure of C l2 gas hence doing so does not affect the Eq value of the half-cell.

13 Which statement concerning the chlorine-c