RVHS 2023 J2 H2 CM Prelim P1 - P3 Suggested Solutions

RIVER VALLEY HIGH SCHOOL 2023 JC 2 H2 Chemistry 9729 Preliminary Examination (Paper 1 - 3) Suggested Solutions Paper 1 Worked solutions 1 D Statement A: Incorrect. This statement would explain why the 3rd IE of aluminium is higher than that of magnesium. Statement B: Incorrect. Ionisation energy is dependent on nuclear charge and shielding effect, not octet configuration. Statement C: While it is a factual statement, it does not explain the significant difference in 3rd IE between the two species. Statement D: Al2+: 1s2 2s2 2p6 3s1 vs Mg2+: 1s2 2s2 2p6 Electron to be removed from Mg2+ is from n = 2 principal quantum shell and is closer and more strongly attracted to the nucleus due to a significantly smaller shielding effect, while that to be removed from Al2+ is from n = 3 principal quantum shell. 2 C charge mass ratio of proton (1H+) = 1 1 = 1 charge mass ratio of particle to be deflected = 1 × 8.012.0 = 2 3 charge mass ratios of: 12C3 : 3 12 = 1 4 (incorrect) 10B2 : 2 10 = 1 5 (incorrect) 6Li4: 4 6 = 2 3 (correct) 3He : 1 3 (incorrect) 3 D Tritium isotope: 1p, 1e, 2n Helium isotope: 2p, 2e, 1n Option 1 is incorrect. Option 2 is incorrect. Option 3 is incorrect (charged sub-atomic particles refer to protons and electrons). 4 B

A B C D 5 B A is correct. Ice cannot conduct electricity. B is incorrect. Water has a simple covalent structure. When it is in the form of ice, hydrogen bonds are formed between the molecules of water. However, it is not a giant covalent structure. C is correct. Each O atom in water molecule has 4 bond pairs. So bond angle is 109.5° D is correct. Density of ice is lower due to its open structure. 6 B = Let initial volume be 3V2 => final volume is V2 = (Note: T needs to be converted to K) p2 = 34.1 atm 7 C When temperature increases from 25 oC to 62 oC, Kw increases, signifiying that there is more dissociation of water (to form H+ and OH−). Since H2O  H+  OH−, n(H+) = n(OH−)  value of pH = value of pOH At 62 oC, pKw = −lg(1.00  1013) = 13  pH = 6.5 and pOH = 6.5 (i.e., [H+] = [OH−] = 3.162  107 mol dm3) 8 A S < 0 since amount of gaseous molecules decrease from 1.5 to 1. H = 396  (297) =  99 kJ mol1 < 0 9 B Statement A: Incorrect. Energy level of products is higher than that of reactants. Statement B: Correct. Hsoln = LE + Hhyd Both LE and Hhyd are negative values, so LE > 0, Hhyd <0. For Hsoln to be > 0, magnitude of LE has to be > magnitude of Hhyd. Statement C: Reaction is endothermic, heat is taken in. Statement D: Incorrect definition of Hsoln 10 A Statement 1: Correct. H = 12/18  44 = 29.3 kJ mol1 Statement 2: Incorrect. G = 0 during phase change. Statement 3: Correct. G = HTS, S = (29.3)/(373) = 78.6 J K1 11 B CH4 + 2O2  CO2 + 2H2O HCHO + O2  CO2 + H2O COV 2 = 10 cm3 = Vorg total OV 2 left = 18 – 10 = 8 cm3 OV 2 used = 20 – 8 = 12 cm3 Let CHV 4 = x, HCHOV = 10 – x Since CH4  2O2, HCHO  O2

2x + (10 – x) = 12 x = 2 Ratio is 4:1 12 C No of mole of Chlorine= (90/1000) / 24 = 0.00375 mol Mass of NaClO = 0.00375  (23.0+ 35.5+ 16.0) = 0.27948 g Volume of solution = 0.2794 / 0.005 = 55.87 cm3 13 D rate = k [O2] [NO2]2 rate = k [O2] [NO2]2 = k (0.1) (0.1)2 = 0.1 k =100 rate = 100 [O2] [NO2]2 14 B Half life = ln 2/ k = 1386s (0.5)n = 0.4 n = 1.32 time = 1386  1.32 = 1829s = 30.5 min 15 C At constant temperature, when pressure increase, % of product decreases. Eqm position shift to the left to reduce the pressure by reducing the no. of moles of particles. Hence left side of the eqm should have less no. of moles of gaseous particles. At constant pressure, as T increase, % product decrease. This implies that forward reaction is endothermic reaction. 16 A s = 3.60  10–5 mol dm–3 Ksp = 4s3 = 4(3.60  10–5)3 = 1.866  10–13 mol3 dm–9 pOH = 2.0 [OH] = 102 mol dm–3 Let solubility of salt in buffer solution of pH 12.0 be a mol dm–3 (102 + 2a)2(a) = 1.866  10–13 Since a is small, 102 + 2a  102 (102)2(a) = 1.866  10–13 a = 1.87  10–9 17 B CaCO3 CaF2 Ksp 8.7  10−9 4.0  10−11 Solubilit y (8.7  10−9)1/2 = 9.32  10−10 (4.0  10−11/4)1/3 = 4.00  10−11 Option 1: Correct. Option 2: Correct. Adding NaF will increase [F]. IP of CaF2 increases. Option 3: Incorrect. Ksp is affected only by T. Option 4: Incorrect. Let concentration of Ca2+ in solution be y mol dm-3 IP of CaCO3 = (y)(1) = y IP of CaF2 = (y)(1)2 = y Since CaF2 has a lower Ksp value, CaF2 will precipitate out 1st.

18 C H2PO4 – + HBO3 2–  HPO4 2– + H2BO3 – Acid Base Acid Base Since forward reaction is favoured, the acid and base on the LHS are stronger. 19 D A: Incorrect. Kw, is the product of Ka of citric acid and Kb of citrate anion. B: Incorrect. The pH of a buffer solution remains relatively unchanged when small amount of water is added to it OR the pH of a buffer solution reaches pH 7 when large amount of water is added to it. C: Incorrect. The pH of a buffer solution remains relatively unchanged when small amount of acid or base is added to it. D: Correct. 20 A Red: 2H2O + 2e–  H2 + 2OH– Ox: 2Cl–  Cl2 + 2e– Amount of NaCl = (58.5 × 1000) / 58.5 = 1000 mol = Amount of Cl– Amount of Cl2 = 1000 / 2 = 500 mol Mass of Cl2 = 500 × 71 = 35.5 kg Amount of H2 = 1000 / 2 = 500 mol Mass of H2 = 500 × 2 = 1 kg Amount of NaOH = 1000 mol Mass of NaOH = 1000 × 40 = 40 kg 21 D Ox: M  Mn+ + ne– Red: ClO– + H2O + 2e–  Cl– + 2OH– E = 0.81 V Under standard conditions, lg [1 mol dm3 Mn+] = 0 Let E(Mn+/M) be x 1.2 = 0.81  x x = 0.39 V 22 C This question is about probable products from a single propagation step. Options A and C: products attained through a single propagation step Options B and D: products required multisteps Bonds broken Bonds formed H (bonds broken – bonds formed) A CH CH 0 C CCl CC 10 23 C Refer to Kinetic notes Pg 35

24 D 2NaOH + H2SO4  H2O + Na2SO4 NaOHn reacted/ mol NaOHn remains/ mol H SOn 24 required/ mol Remarks W 0.01  2 = 0.02 0.05 – 0.02 = 0.03 0.015 Rxn: Hydrolysis of amides product: (COO)2 X 0.01 0.05 – 0.01 = 0.04 0.02 Rxn: Hydrolysis of nitrile product: CH3COO Y 0 0.05 0.025 Aryl halides do not undergo hydrolysis 25 C The anion generated from CH3CO2CH3 in the presence of a strong base is CH2CO2CH3 which is . Starting materials A Starting materials include an aldehyde, but incorrect nucleophile. B Starting materials include an aldehyde, but incorrect nucleophile. C Starting materials include an aldehyde (R is CH2CH3) and correct nucleophile. D Starting materials include the correct nucleophile, but a ketone.

26 B Statement 1: Incorrect. 1 mole of rosmarinic acid reacts with 7 moles of Br2. Statement 2: Correct. Statement 3: Incorrect. No orange ppt due to absence of carbonyl C=O. 27 A  T undergoes reduction with NaBH4, only gains 2 H atoms, T contains only 1 C=O, either ketone or aldehyde.  T undergoes oxidation with alkaline iodine, o T contains COCH3 and CH(OH)CH3. o T gives Cx1Hy3Oz+2 2 , loss of Cl T undergoes alkaline hydrolysis (using warm alkali) T contains acyl chloride, COCl instead of alkyl chloride, RCl (alkyl chloride requires strong heating. COCl reacts to give COOH, deprotonates to give COO.  T does not react with dry SOCl2, T contains COCH3 instead of CH(OH)CH3. o With alkaline iodine, COCH3 gives an organic product COO. 28 A Statement B: Incorrect product. Dilute HNO3 should give mono-substituted product. Statement C: Incorrect reaction. The reaction is known as elimination. Statement D: Incorrect conditions. With limited CH3NH2, the product will be poly-substituted. Other conditions required: ethanolic medium, heat. 29 D A: Incorrect. L is a bidentat