2019 JPJC Prelim H2 Chem P4 ANS

 Jurong Pioneer Junior College 9729/04/J2 PRELIMINARY EXAM/2019 [Turn Over JURONG PIONEER JUNIOR COLLEGE 2019 JC2 H2 CHEMISTRY (9729) Preliminary Examination Paper 4 (Suggested Answers) 1 Determination of the identity of the halogen, X, in CH2XCO2H (a) (ii) Titration results Titration number 1 2 Final burette reading / cm3 24.90 34.95 Initial burette reading / cm3 0.00 10.00 Volume of FA3 used / cm3 24.90 24.95   (iii) average volume of FA 3 used , VFA3 = 24.90 24.95 2  = 24.93 cm3 (b) (i) n(H2SO4) in 10.0 cm3 of FA 2 = 2.00  10.0 1000 = 0.0200 mol = n(H2SO4) in 250 cm3 FA 3 [H2SO4] in FA 3 = 0.020  250 1000 = 0.0800 mol dm3 or Using c1V1 = c2V2, [H2SO4] in FA 3 = 2.00 10.0 250  = 0.0800 mol dm3 (ii) n(H2SO4) reacted in titration = 24.93 0.08001000  = 0.001996 mol Since 1H2SO4  2 NaOH, n(NaOH) in 25.0 cm3 of FA 1 = 0.001992  2 = 0.00399 mol (iii) n(NaOH) in 250 cm3 of FA 1 = 0.00399 250 25.0 = 0.0399 mol = n(NaOH) left unreacted after reaction with W n(NaOH) added to W = 0.40 = 0.100 mol n(NaOH) reacted with W = 0.100  0.0399 = 0.0601 mol

 Jurong Pioneer Junior College 9729/04/J2 PRELIMINARY EXAM/2019 (iv) From equation 1 and 2, since 1W  2NaOH, n(W) in 4 g = ½  0.0601 = 0.03004 mol Mr of W = 4  0.03004 = 133.2  133 (no units) Ar of X = 133.2 – 59 = 74.2 (no units) X is bromine/ Br (c) (i) max. % error in volume of FA3 used = 2(±0.05) 10024.95  =  0.401 % (ii) Error : Mass measurement of W was not precise as the mass was given to nearest g. Modification : Use a more precise weighing balance that can measure to 3.d.p. or Error : Substitution of halogeno group may be incomplete. Modification : Heat W with NaOH for a longer period of time. or Error : Loss of product through heating. Modification : Heat the reaction mixture under reflux 2 An experiment to investigate the behaviour of acids and bases in aqueous solutions (a) Determination of the enthalpy change of reaction between FA 4 and FA 5 Total volume of FA 5 added/ cm3 Maximum temperature/ C 0.00 29.0 2.00 30.1 4.00 30.8 6.00 31.5 8.00 32.0 10.00 32.5 12.00 32.8 14.00 32.6 16.00 32.3 18.00 32.0 20.00 31.7 22.00 31.5 24.00 31.2 26.00 31.0 28.00 30.8 30.00 30.6

 Jurong Pioneer Junior College 9729/04/J2 PRELIMINARY EXAM/2019 [Turn Over (b) (i) (ii) & (iii) correctly reads Tmax to ½ small square + correctly calculates Tmax + correctly reads Vequivalence to ½ small square. (iv) n(NaHCO3) used = 25.0 ×1.001000 = 0.0250 mol = n(NaOH) reacted [NaOH] in FA 5 = 0.0250  12.4 1000 = 2.02 mol dm3 (v) From the graph, Vequivalence = 12.4 cm3. heat evolved, q = (25.0 + 12.4)(4.18)(3.85)= 603 J H3 = (603  103)  0.0250 = 24.1 kJ mol1 (c) The reaction between FA 4 and FA 2 (d) (40.0 28.6) (15.0 28.9) (40.0 15.0) averageT     = 28.7 C