2021 YIJC Prelim P1 Solution

©YIJC [Turn over YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CG INDEX NO CHEMISTRY Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet Data Booklet 9729/01 16 September 2021 1 hour MCQ Answer Key: 1 2 3 4 5 6 7 8 9 10 D B B A C D C B C C 11 12 13 14 15 16 17 18 19 20 D A A C D A B B C D 21 22 23 24 25 26 27 28 29 30 A A A D B D C D A B

2 ©YIJC [Turn over 1 A 1.00 g alkane D was burnt in an excess of oxygen, and the gases that were produced were first passed through a U-tube containing phosphorus pentoxide and another U-tube containing NaOH(s) as shown in the diagram. The phosphorus pentoxide U -tube increased in mass by 1.55 g, and the NaOH(aq) bottle increased in mass by 3.03 g. All volumes were measured at room temperature and pressure. What is the molecular formula of the alkane D? A CH4 B C2H5 C C2H6 D C4H10 Answer: D Amount of H atoms = 2  0.0861 = 0.1722 mol Amount of C atoms = 0.0689 mol Ratio of H : C = 0.1722 0.0689 = 2.4992 : 1 = 5 : 2 Since the molecular formula of an alkane is CnH2n+2, the molecular formula of alkane D must be C4H10. 2 Use of the Data Booklet is relevant to this question. What do the ions 36S2 and 37Cl have in common? A Both ions have more electrons than neutrons. B Both ions have the same electronic configuration. C Both ions contains the same number of nucleons. D 36S2 has a smaller angle of deflection than 37Cl in an electric field. Answer: B Option A: 36S2 has 16 protons, 20 neutrons and 18 electrons, 37Cl has 17 protons, 20 neutrons and 18 electrons. Option B: Both ions are 1s2 2s2 2p6 3s2 3p6 Option C: 36S2 has 16 protons + 20 neutrons = 36 nucleons 37Cl has 17 protons + 20 neutrons = 37 nucleons Option D: angle of deflection = charge/mass, 36S2= 0.0556 > 37Cl , 0.0270 Oxygen gas used oxygen gas Phosphorous pentoxide NaOH(s) pellets Alkane D

3 ©YIJC [Turn over 3 Boyle’s law states that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure. Which of the following statements shows application of Boyle’s law? 1 Human lungs, inhalation and exhalation. 2 Spraying paint from a can. 3 Working of hot air balloon. A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 1 only Answer: B Option 1: correct. While inhaling, the lungs are filled with air; therefore, they expand. The volume increases hence, the pressure level goes down. Similarly, when the lungs are evacuated of air, they shrink; therefore, the volume reduces and the pressure increases. Option 2: correct. When the top of the can is pressed, the volume inside the can gets reduced and the paint is thrown out with great pressure. Since the pressure has an inverse relationship with the volume, Boyle’s law can be observed in action. Option 3: Incorrect. On ignition of the fuel, the air inside the envelope heats up. This hot air expands as per Charles's law. As the temperature of the air increases, the volume of the air also increases and consequently, the density decreases. This makes the envelope lighter than the atmospheric air surrounding it. The buoyant force pushes the lighter envelope up in the air, and it flies.

4 ©YIJC [Turn over 4 Carmine is a red colorant extracted from the bodies of dead female insects, used in food colouring and lipsticks. The proposed structure of carmine is as shown. The Al+ ion is situated in the centre of a planar arrangement of numbered oxygen atoms. Which of the following descriptions of the bonds between Al+ and the numbered O atoms is most likely to be correct? O atoms numbered 1 O atoms numbered 2 A co-ordinate co-ordinate B co-ordinate ionic C ionic co-ordinate D ionic ionic Answer: A Oxygen (in period 2) has 6 valence electrons and it cannot expand its octet. Oxygen 1 has lone pair available to form co-ordinate bond with Al3+, hence oxygen has full octet. Oxygen 2 comes from O(phenoxide), forms co-ordinate bond with Al3+(high charge density) , hence overall charge on Al is 1+. +3 -2(-1)= +1

5 ©YIJC [Turn over 5 The table shows the boiling point of some halogenoalkanes. compound boiling point/ °C CH3CH2Cl 12.3 CH3CH2Br 34.8 CH3CH2I 70.0 Which of the following correctly explains the difference in the boiling point? 1 the electronegativity difference between the halogen and carbon increases from C−Cl to C−I 2 the strength of permanent dipole -permanent dipole attraction increases from C−Cl to C−I 3 the strength of instantaneous dipole-induced dipole attraction increases from CH3CH2Cl to CH3CH2I 4 the bond energy of C−X bond decreases from C−Cl to C−I A 1 and 2 only B 2 and 4 only C 3 only D 4 only Answer: C N.B. H-bonding > pd-pd> id-id only if size of electron cloud of molecules are similar. 1 the electronegativity difference between the halogen and carbon should decrease from C−Cl to C−I. Statement does not explain for the trend of increasing boiling point from CH 3CH2Cl to CH3CH2I. 2 the strength of permanent dipole-permanent dipole attraction decreases from C−Cl to C−I. The statement of option 2 is incorrect and does not explain for the trend of increasing boiling point from CH3CH2Cl to CH3CH2I. 3 the strength of instantaneous dipole -induced dipole attraction increases from CH 3CH2Cl to CH3CH2I. Statement is correct as the total number of electrons increases from CH 3CH2Cl to CH3CH2I and due to the increase in id -id attraction, the boiling point increases from CH 3CH2Cl to CH3CH2I. 4 the bond energy of C-X bond decreases from C−Cl to C− I. Statement is correct but boiling does not break the C−X bond, so this does not explain for the trend of increasing boiling point from CH3CH2Cl to CH3CH2 I. 6 The radioactive decay of element X is a first-order reaction. It take 16 days for element X to decay to 25% of its initial value. What fraction of element X would remain after 800 days? A 1 210 B 1 250 C 1 280 D 1 2100 Answer: D Half-life = 8 days (10.50.25, 16 days 2 half-lives) 80 days = 100 x 8 days = 100 half-lives Fraction of isotope remaining = ( 1 2) 100 = 1 2100

6 ©YIJC [Turn over 7 The Boltzmann distribution of kinetic energies for the following equilibrium N2O4(g) ⇌ 2NO2(g) H = +57 kJ mol1 is shown graphically below as temperature is increased. Which statements can be drawn from the graph? 1 At all energies, the number of molecules of N2O4 of a given value increases. 2 The maximum of the curve lowers and shifts to the right. 3 The reaction is first order with respect to [N2O4]. 4 The number of molecules with energies equal or greater than Ea increases. A 1, 2 and 3 only B 1 and 2 only C 2 and 4 only D 3 and 4 only Answer: C 1 Incorrect as at the lower energy regions the number of molecules decreases when the temperature increases. 2 Correct as seen for the graph, at higher temperature, T2, the maximum of the curve lowers and shifts to the right. 3 Incorrect as you cannot deduce order of reaction from Boltzmann curve. 4 Correct. The number of molecules with energies above Ea increases as represented by area under the curve for T2 is greater than T1. number of molecules energy Ea = activation energy T1 T2 T2 > T1

7 ©YIJC [Turn over 8 Methanol can be synthesised from hydrogen and carbon monoxide using a suitable catalyst at 480 K and a pressure of 3 x 106 Pa. 2H2(g) + CO(g)  CH3OH(g) H = –90.6 kJ mol–1 The reaction mixture reached equilibrium under the above conditions. The graph below