SAJC_H2_CHEM_P1_Ans Prelim

1 2016 SAJC H2 CHEM PRELIM PAPER 1 (Worked solutions) 1 2 3 4 5 6 7 8 9 10 C B A D B D C C C C 11 12 13 14 15 16 17 18 19 20 A D B A A B A B D B 21 22 23 24 25 26 27 28 29 30 B B B B B C B A A C 31 32 33 34 35 36 37 38 39 40 D D D C D D B B A D 1. Ratio of methane: ethene = 40 : 60 = 2 : 3 Let the total volume of mixture be y cm3 i.e. fraction of methane in mixture = 5 2 ; fraction of ethene in mixture = 5 3 CH4 + 2O2  CO2 + 2H2O C2H4 + 3O2  2CO2 + 2H2O Volume/cm3 5 2 y 5 2 y 5 3 y 2 x ( 5 3 y) Hence, total volume of CO2 = 5 2 y + 2 ( 5 3 y) = 5 8y cm3 2. From 2MnO4- (aq) + 5C2O42- (aq) + 16H+ (aq)  2Mn2+ (aq) + 10CO2 (g) + 8H2O (l), Reacting mole ratio of MnO4- : C2O42- = 2 : 5 --- (*) 1 mol of NaHC2O4.H2C2O4 gives 2 mol of C2O42-. 1.0 x 10-3 mol of NaHC2O4.H2C2O4 gives 2.0 x 10-3 mol of C2O42-. From (*), no. of moles of MnO4- reacted with 2.0 x 10-3 moles of C2O42- = 5 2 x 2.0 x 10-3 = 0.008 mol Hence, volume of KMnO4 solution required = 2.0 008.0 = 0.400 dm3 = 40 cm3

2 3. pV = nRT pV = rM m RT Option A: Shape of graph is correct. rM m R is the gradient. Since X has a higher Mr, its gradient is less steep. (Correct) Option B: Shape of graph is correct. From above, since X has a higher Mr, its gradient should be less steep. Hence graph is wrong. Option C: Shape of graph is wrong. The graphs should be a horizontal line. Since T is constant, pV is constant for each compound. Option D: Shape of graph is wrong. The graphs should be a vertical line. Since T is constant, pV is constant for each compound. 4. Element A belongs to Group V. This is because the first big jump in ionisation energy occurs between the 5th and 6th I.E., indicating that the 6th electron is removed from the inner quantum shell. Valence configuration: ns2np3 Element B belongs to Group IV. This is because the first big jump in ionisation energy occurs between the 4th and 5th I.E., indicating that the 5th electron is removed from the inner quantum shell. Valence configuration: ns2np2 Element C may belongs to Group VIII. This is because there is no noticeable big jump between the ionisation energies. Element D belongs to Group II. This is because the first big jump in ionisation energy occurs between the 2 nd and 3rd I.E., indicating that the 3 th electron is removed from the inner quantum shell. Hence its valence electronic configuration is ns2 and has no p electrons in the valence shell. 5. Option A is correct as the 2 hydrogen bonds and 2 O-H covalent bonds would form a bond angle of 109.5. Option B is wrong as two lone pairs from each oxygen atom are involved in hydrogen bonding. Option C is correct as intermolecular forces of attraction are wea ker than strong covalent bonds. Option D is correct and this is the explanation why ice floats on water.

3 6. Option A is wrong as 1st ionisation energy is endothermic. Option B is wrong as lat