JJC_H2_CHEM_P2_Ans(1)

[Turn over JURONG JUNIOR COLLEGE 2015 JC 2 PRELIMINARY EXAMINATION Higher 2 Paper 2 (solutions) 1 Preparation of 5 diluted solutions from 0.0500 mol dm 3 Ca2+ (aq) Dilution and measurement: 1. Using a burette, run Vcm 3 of Ca 2+ stock solution into a 25.0 cm 3 graduated flask. 2. Make up to the graduated mark with deionised water. 3. Stopper and shake the flask to obtain a homogeneous solution. 4. Repeat steps 1-3 using the volumes specified in the table above in the preparation of solutions 2 to 5. 5. Place each of the prepared solutions into the atomic absorption spectrophotometer and record the absorbance value for each solution. 6. Plot absorbance value for each solution against concentration of Ca 2+ Solution [Ca2 (aq)] / mol dm3 Volume of Ca2+ stock solution, V / cm3 1 0.0250 12.50 2 0.0200 10.00 3 0.0150 7.50 4 0.0100 5.00 5 0.00500 2.50

2 © Jurong Junior College 9647/02/ J2 PRELIMINARY EXAMINATION/2015 1 Preparation of saturated solution of Ca(OH) 2(aq) 1. Using a measuring cylinder, place 100 cm 3 of deionised water into a 250 cm3 beaker. 2. Using a spatula, add a few tips of solid Ca(OH) 2 into the flask. Stir to dissolve all the solids using a glass rod. Keep adding more solids, with stirring after each addition, until some solids are left undissolved. 3. To ensure that the solution is saturated, stir the solution for a while and leave the conical flask containing the solution to stand in a water bath at 25 oC for some time. There must be some solids left undissolved. 4. To remove undissolved solids, filter the saturated solution into a clean, dry conical flask using a dry filter funnel and a piece of dry filter paper. 5. Place the saturated Ca(OH) 2 solution into the atomic absorption spectrophotometer and record the absorbance value for the solution. Using the calibration line, read the corresponding concentration of Ca2+, for the absorbance value for the saturated Ca(OH)2 solution. (b) Solubility product of Ca(OH)2 = 4 x3 mol3 dm9

3 © Jurong Junior College 9647/02/ J2 PRELIMINARY EXAMINATION/2015 [Turn over 2 (a) (b) Nucleophile (c) (d) Nucleophilic addition (e) (f) (i) Since half life are constant (t½  76 min), the reaction is first order wrt [HCN]. (ii) When [CH2=NH] is halved, rate is halved. The reaction is first order wrt [CH2=NH]. (iii) Rate = k[HCN][CH2=NH] (iv) Rate = k’[HCN] where k’=k[CH2=NH] k’ = ln2 /t1/2 = ln2 / 76 = 0.00912 min–1 k = 0.00912 / 0.2 = 0.0456 mol-1 dm3 min-1 Or Initial rate = gradient = 0.01/ 124 = 0.0000807 mol dm–3 min–1 0.0000807 = k(0.01)(0.2) k = 0.0403 mol-1 dm3 min-1 (v) 78 min

4 © Jurong Junior College 9647/02/ J2 PRELIMINARY EXAMINATION/2015 3 (a)   22 2 c 32 [Pb ][Cr ]K [Cr ] units: mol dm3 (b) (i) [Pb2+(aq)] = ½ (2.96  104) = 1.48  104