2024 Prelims JPJC H2 Chem P1 (Ans)

Jurong Pioneer Junior College 2024 H2 Chemistry Paper 1 Worked Solutions 1 D The definition for atomic mass is the ratio of the average mass of one atom of an element to one-twelfth the mass of one atom of 12C. In option D, the mass of one mole of atoms of an element has already considered all the isotopes and their relative abundances. 2 C A Al2+ ions 1s22s22p63s1; F+ ions. 1s22s22p4 Incorrect as Al3+ has 1 outer shell electron vs F+ 6 outer shell electron. 2nd IE of F is greater than the 3rd IE of Al as 3s electron is Al2+ is higher in energy and further away from the nucleus (hence lower nuclear attraction) then the 2p electron in F+. B Incorrect as the 3rd IE involves the removal of 3rd electron from Al is from 3rd principal quantum shell as compared to the removal of 3 rd electrons from the 2 nd PQM for the remaining 4 species. ✓C Na3+ 1s22s22p4+ vs Ne2+ 1s22s22p4+ Same no of electrons but as Na 3+ has more protons than Ne, hence higher nuclear charge, thus nuclear attraction of outermost electrons in Na 3+ is higher, thus higher 4th IE for Na compared to 3rd IE for Ne . D Incorrect as the removal of electrons are removed from different PQM. 3 Due to the no of bond pairs and lone pairs of electrons in each molecule – BCl3 : 3 Bond Pairs ; PH3: 3 Bond Pairs & 1 Lone pair The shape of BCl3 and PH3 are trigonal planar and trigonal pyramidal respectively 4 D (1, 2 and 3) ✓1: Only id –id attraction can be formed between hydrocarbon molecules and H 2O molecules which is weaker than the hydrogen bonds between H2O molecules. Hence, the energy released upon forming the less favourable interaction is not sufficient to compensate the energy required to break the stronger hydrogen bonds. Hence, hydrocarbon molecules are not solvated by water and hence the two layers a re immiscible. ✓2: See (1). ✓3: The stronger hydrogen bonds between H 2O molecules pulls the molecules closer to each other and hence, the volume of water is smaller. Given similar mass, the density of water is higher and hence, it will be below the hydrocarbon layer. P H H H

5 B Under constant n and T, the ideal gas equation is simplified to pV = k (where k is a constant and k = nRT). A Rearranging pV = k such that y = 1 p and x = V ( i.e. 1 p = k’V) , a graph of y = mx is obtained (i.e. straight line passing through origin). ✓B Since pV is a constant and x = pV, a graph of x = c is obtained (i.e. vertical line) C Rearranging pV = nRT such that y = p and x =  (i.e. p = r RT M  ), a graph of y = mx is obtained (i.e. straight line passing through origin). D Rearranging pV = nRT such that y = pV T and x = p (i.e. pV T = nR = constant), a graph of y = c is obtained (i.e. horizontal line) 6 C Mol ratio of O2 = 150 500 =0.3 Hence amt of O2 = 0.3 x 1.2 = 0.36 mol Amt of N2 = 5.76 24 =0.24 mol Hence amt of Ar = 1.2 – 0.36 – 0.24 =0.6 mol Thus pAr = 0.6 1.2 × 500 =250 kPa 7 A 1 only When