JPJC Prelim 2025 Paper 1 Worked Solutions_H2 Chem

© Jurong Pioneer Junior College [Turn Over ANSWER SCHEME 1 C ( ) 5 -3 mpV= RTMr 18.010 ×5.7×10 = (8.31)(273)Mr Mr = 71.6 Let a be the %mass of 35Cl2 ( 𝒂 𝟏𝟎𝟎 × 𝟕𝟎) + [𝟏 − 𝒂 𝟏𝟎𝟎 × 𝟕𝟒] = 𝟕𝟏. 𝟔 a = 60 2 B From the first 2 properties, E must be a main group metal. Hence, option C and D are incorrect. Since it was obtained by removing electrons from the same orbital, E must be either a metal in Group 1 or 2. 3 A Since its shape is trigonal pyramidal, that must mean that around O there are 3 bp and 1lp. The 3 bp are formed between one electron from F and 1 from O and the lp belongs to O. With only 5 valence electrons around O than the usual 6, the entire ion is short of 1 electron and hence n = 1 4 C CH4 + 2O2 → CO2 + 2H2O 10 cm3 CH4  10 cm3 CO2 C2H6 + 3.5O2 → 2CO2 + 3H2O 20 cm3 C2H6  40 cm3 CO2 5 B Amount of MnO4- = 37.5 0.021000 = 7.5 ×10-4 MnO4- + 8H+ + 5e ⇌ Mn2+ + 4H2O Amount of MnO4- : amount of electrons: amount of G ion : 7.5  10−4 : 3.75  10−3 :1.25  10−3 3: 1 Given that ion of G is oxidised to GO3-, O.N. +5, Original O.N of ion of G is +5 –3 = +2 50 cm3 CO2

2 © Jurong Pioneer Junior College 9729/01/J2 PRELIM EXAM/2025 6 B Assuming ideal gas behaviour (i.e. pV = nRT), 1) value of pV of L = nLRTL = 3x value of pV of J = nJRTJ = x 2) J deviates from Ideal gas behaviour more than L, indicating more significant/stronger intermolecular forces of attraction between J molecules than L (i.e. 𝐧𝐋𝐓𝐋 𝐧𝐉𝐓𝐉 = 𝟑) 1: LL JJ nT nT = ( ) ( ) + + 0.5 273 50 0.5 273 25 = 1.08 ✓2: J=NH3 (stronger hydrogen bonds between NH3 molecules) L = CH4 (weaker id-id attractions between CH4 molecules) LL JJ nT nT = ( ) ( ) + + 1.5 273 25 0.5 273 25 = 3 3: J = H2 (weaker id-id attractions between CH4 molecules) L=SO2 (stronger pd-pd bonds between NH3 molecules) LL JJ nT nT = ( ) ( ) + + 0.75 273 25 0.25 273 25 = 3 7 A K: 1s2 2s2 2p6 3s2 3p6 4s1 Cl: 1s2 2s2 2p6 3s2 3p5 ✓1: K has one more quantum shell, valence electrons are further away from nucleus. Increase in shielding effect by inner shell electrons outweighs the increase in nuclear charge hence, decreased attraction between nucleus and valence electrons / decreased n uclear attraction. Thus, the atomic radius of potassium is greater than atomic radius of chlorine atom. 2: Similarly, the 4s electron is further away from the nucleus, decreased nuclear attraction between nucleus and valence electrons, thus, less energy is required to remove an outermost/valence electron. 3 K+: 1s2 2s2 2p6 3s2 3p6 Cl-: 1s2 2s2 2p6 3s2 3p6 Both are isoelectronic, with potassium having a bigger nuclear charge. Hence, there is great nuclear attraction between nucleus and valence electrons. Thus the ionic radius of potassium is smaller than that of chlorine.

3 © Jurong Pioneer Junior College 9729/01/J2 PRELIM EXAM/2025 [Turn Over 8 D reaction 1 Br2(aq) + 2NaOH(aq) → NaBr(aq)