SRJC H2 CHEM P1 ANS

1 SERANGOON JUNIOR COLLEGE General Certificate of Education Advanced Level Higher 2 CHEMISTRY 9746/01 Prelim Examination Vetting Copy MCQ Solutions Mole concept (Dilution calculation, back titration) 1 3.920 g of an oxide of formula MO was completely dissolved in 30.0 cm 3 of 2.00 mol dm -3 sulphuric acid. The resulting solution was made up to 100 cm 3. 25.0 cm 3 of this solution was neutralised by 27.5 cm 3 of 0.100 mol dm-3 sodium hydroxide. What is the relative atomic mass of M? A 47.9 B 54.9 C 55.9 D 101.0 Solution: C Amount of NaOH = 327.5 0.1 2.75 101000   mol Amount of unreacted H2SO4 in 25.0 cm3 of solution = 1.375 x 10─3 mol. Amount of unreacted H2SO4 in 100 cm3 of solution = 33100 1.375 10 5.5 1025.0    mol. Initial amount of H2SO4 = 30.0 2.0 0.061000 mol Therefore, the amount of H2SO4 that reacts with MO = 0.06 – 5.5x10-3= 0.0545 mol Given that 1 mol of H 2SO4 reacts with 1 mol of MO ( M has an oxidation state of +2), therefore, no of moles of MO in 3.920g of MO = 0.0545 mol Given no. of moles of MO = mass of MO/ molar mass of MO 0.0545 = 3.920/(molar mass of M + 16.0) Molar mass of M = (3.920/0.0545) – 16.0 = 55.9 g/mol

2 REDOX (Use of half equation to deduce mole of reactants) 2 A commercial production of iodine involves the reduction of a solut ion of iodate(V) ions, IO3 , with the theoretical quantity of hydrogen sulphite ions, HSO 3  in acidic condition. The hydrogen sulphite ions are oxidised to sulphate ions, SO 4 2 while the iodate( V) ions are reduced to iodine, I2. Given H2O(l) + HSO3 ─(aq) → SO4 2─(aq) + 3H+(aq) + 2e─ How many moles of hydrogen sulphite ions are needed to r educe one mole of iodate(V) ions? A 0.4 B 1 C 2.5 D 5 Solution: C Reduction: IO3െ is reduced to I2. Reduction half-equation: 2IO3െ(aq) + 12H+(aq) + 10eെ → I2(aq) + 6H2O(l) Oxidation: HSO3 ─ is oxidised to SO4 2─ Oxidation half-equation: H2O(l) + HSO3 ─(aq) → SO4 2─(aq) + 3H+(aq) + 2e─ ( x 5) Overall equation: 2IO3െ(aq) + 5HSO3 ─(aq) → I2(aq) + H2O(l) + 5SO4 2─(aq) + 3H+(aq) Atomic Structure (deduce no of protons, electrons and neutrons in atoms or ions) 3 One of the isotopes of carbon is carbon-14. Carbon-14 is radioactive and is used in carbon dating by archaeologists. Which one of the following species is isotonic and isoelectronic with respect to an atom of carbon-14? A 13C B 14N+ C 16O2+

3 D 17F+ Solution: C Number of protons in Carbon-14= 6p Number of neutrons in Carbon-14 = 14 – 6 = 8n Number of electrons = 6e Gas La w (Interpretation of graph) A 13C 6p 7e 7n B 14N+ 7p 6e 7n C 16O2+ 8p 6e 8n D 17F+ 9p 8e 8n 4 Which of the following shows a graph of  P against T for an ideal gas? (P = pressure;  = density; T = temperature in C). Solution: B pV = nRT  pV = m Mr RT  p =( m V ) RT Mr  P  = RT Mr Since in Ideal gas equation temperature is expressed as Kelvin  P A T  P T B  P  P C T T D It is a linear graph.  P At 0C = 273K, hence p/ is not zero. T B

4  P = (2 7 3RT Mr ) hence when T is at 0oC, there will be a value for  P Chemical Bonding (Identification of Bond angle) 5 Polyurethane is used in coatings, insulators and adhesiv es. Polyurethane What are the values of the bond angles marked x and y in polyurethane? x y A 90 90 B 120 120 C 107 120 D 109.5 90 C O N H CC H H H H N H C O OCC H H H H Solution: C Around N: 3 bond pair and 1 lone pair  shape is trigonal pyramidal and bond angle is 107 Around C: 3 bond pair and 0 lone pair  shape is trigonal planar and bond angle is 120 O y x n

5 Chemical Equilibria (Calculatio n of Kp) 6 Zinc oxide reacts with hydrogen according to the following equation. ZnO (s) + H2 (g) Zn (s) + H2O (g) At temperature T and a total pr essure of 10 atm, it is fo und that initial amounts of 1 mole each of zinc oxide and hydrogen gas produce 0.01 mole each of zinc and steam at equilibrium. What is the approximate value of K p at temperature T? A 10-4 B 10-2 C 10 D 100 Solution: B ZnO(s) + H2 (g) Zn(s) + H2O(g) Initial no. of mol 1 1 0 0 Change in mol -0.01 -0.01 +0.01 +0.01 Equilibrium no. of mol 0.99 0.99 0.01 0.01 [Omit ZnO and Zn in Kp expression as they are solid.] Kp = 2 2 HO H P P = 0.01/1 10 0.99 /1 10   = 10-2

6 Reaction Kinetics 7 Hydrogen peroxide reacts with acidified iodide ions liberating iodine according to the equation below: H 2O2 (aq) + 2H+ (aq) + 2I- (aq)  I2 (aq) + 2H2O (l) The kinetics of this reaction were investigated and it was found to have the following rate equation: rate = k [H 2O2] [I-] Two series of experiments were conducted giving rise to Graph A and Graph B. Graph A Graph B rate of reaction/ mol dm -3 s-1 y x time Which of the following shows the co rrect labeling of the x-axis for Graph A and y-axis for Graph B? x-axis for Graph A y-axis for Graph B A [H2O2] [I -] / mol2 dm-6 [H +] / mol dm-3 B [I -] [H+] / mol2 dm-6 [ I 2] / mol dm-3 C [H2O2] / mol dm-3 [I -] / mol dm-3 D [H+] / mol dm-3 [H 2O2] / mol dm-3 Solution: A rate = k [H2O2] [I-]

7 Graph of rate vs. [H2O2] [I-] is a linear graph which passes through the origin. Rate of reaction is independent of [H+]. Thus graph of [H+] vs time is a downwards sloping straight line graph of constant gradient. Ionic Equilibrium (Buffer Solution component) 8 An enzyme was found to operate at maxi mum efficiency in an aqueous solution buffered at pH 5. Whi ch of the following would give the necessary buffer solution when dissolved in 10 dm3 of water? A 1 mol of NaOH and 1 mol of CH3COOH B 1 mol of CH3COOH and 1 mol of CH3COO–Na+ C 1 mol of HCl and 1 mol of CH3COO–Na+ D 1 mol of CH3COO–NH4 + Solution: B Components of acidic buffer: weak acid and salt of conjugate base Ionic Equilibrium (Ksp) 9 When the concentration of Ca2+ ions in water is greater than 10-5 mol dm-3, an insoluble scum would be formed with soap. To prevent scum forming with s oap, what minimum mass of sodium carbonate should be added to 1 dm3 of water? [Solubility product of CaCO3 = 5.0 x 10-9 mol2 dm-6] A 0.037 g B 0.042 g C 0.050 g D 0.053 g Solution: D The scum formed is CaCO3 (s) CaCO3 (s) Ca2+ (aq) + CO3 2- (aq) Ksp of CaCO3 = [Ca2+][CO3 2-] 5.0 x 10-9 = (10-5) [CO3 2-] (there are presence of 10 -5 mol dm-3 of Ca2+ present in water)

8 [CO3 2-] = 5.0 x 10-4 mol dm-3 Na2CO3 (aq)  2 Na+ (aq) + CO3 2- (aq) [Na2CO3] = [CO3 2-] = 5.0 x 10 -4 mol dm-3 Minimum mass of Na2CO3 added to 1 dm3 of water = 5.0 x 10-4 x (2 x 23.0 + 12.0 + 3 x 16.0) = 0.053 g Thermoch emistry (Calculation involving enthalpy of change of neutralisation) 10 The enthalpy change of neutralisat ion of dilute sulphuric acid by dilute sodium hydroxide is the heat liberated when A 1 mole of H2SO4 reacts with ½ mole of NaOH B 1 mole of H2SO4 reacts with 1 mole of NaOH C 1 mole