SAJC Prelim P1 ANS

JC2 Prelims 2009 H2 Chemistry (Paper 1) Worked Solutions 1. Ans: A MO + H2SO4  MSO4 + H2O H2SO4 + 2NaOH  Na2SO4 + 2H2O No. of moles of H2SO4 = 11000 100  = 0.1 mol No. of moles of excess H2SO4 = ½ × No. of moles of NaOH = ½ 11000 45 21 . = 0.0107 mol No of moles of H 2SO4 reacted with MO = no. of moles of MO = 0.1- 0.0107 =0.0892 mol M r of MO = 0892 0 0 5 . . = 56.0 Ar of M = 40.0 2. Ans: C Let the volume of CH4 be x. Then the volume of C2H6 = 70- x Volume of carbon dioxide gas formed in total = x + 2(70-x) = 140 – x Change in volume of residual gases after passing through NaOH = volume of carbon dioxide formed = 130 -35 = 95 cm 3 Hecne, 140- x = 95 x = 45 cm 3 Volume of methane = 45 cm3 ; volume of ethane = 25 cm3

3. Ans: C A: Si: [Ne] 3s2 3p2 2 unpaired electrons in the 3p orbital B: S [Ne] 3s2 3p4 3 unpaired electrons in the 3p orbital C: Fe 2+ : [Ar]3d6 4 unpaired electrons in the 3d orbital D: Cr 3+: [Ar]3d3 3 unpaired electrons in the 3d orbital 4. Ans: B Element X is in Group III while element Y is in Group VI. Hence the formula formed between X and Y is X 2Y3. 5. Ans: C NO2, SO2 are simple covalent molecules with weak van der Waals forces, while water, also a simple molecule, has hydrogen bonding. SiO2 is a giant covalent molecule with strong covalent bonds between Si and O atoms, hence it requires the largest amount of energy to overcome the strong covalent bonds. Hence, it has the highest energy. 6. Ans: A In ice, each water molecule is hydrogen-bonded to four other water molecules, in a tetrahedral arrangement, hence having a bond angle of 109.5°. 7. Ans: B no. of moles of gas in small spacecraft = RT RT RT pV 800) 20 )( 40 ( no. of moles of gas in large spacecraft = RTRTRT pV 7500) 50 )( 150 ( total no. of moles of gas = RTRT 83007500 800 Pressure in the combined arrangement =    atm RTRT V nRT 11950 20 8300  8. Ans: D pV=nRT  p nRTV  when the mass is fixed, n is constant. when pressure is constant, V is proportional to T in Kelvin. Since the x-axis shown is T in °C, the straight line should cut the x-axis at -273 9. Ans: C Q=mcT = (25+25)(4.2)(20) no. of mol of H 2O formed = 0.04 numerical value of H = 04 . 0 20 2 . 4 50 04 . 0  Q

10. Ans: C H O O H  2H (g) + 2O(g) Hatm = energy required to break 2 O-H bonds and 1 O-O bond = + (2)(460) + 150 = +1070 kJ mol-1 11. Ans: A Students need to balance their own half equation as follows: MnO4 - + e  MnO4 2- H2O + HCOO-  CO3 2- + 3H+ + 2e 3 4 4 4 4 505 . 0 1