2024 Prelims EJC H2 Chem P1 (Ans)

2024 JC2 Preliminary Examination H2 Chemistry 9729 Paper 1 Worked Solution 1 mass of Fe in 0.32 g of Fe2O3 2 55.8 0.32 g 0.2238 g2 55.8 3 16.0 =  = +  0.2238% by mass of Fe in ore 0.6 37.3% = =  B 2 2Y 25.0 0.400 0.0100 mol1000n + =  = 4MnO 40.00 0.10 0.00400 mol1000n − =  = 5Y2+ ≡ 2MnO – 4 MnO – 4 is an oxidising agent, per mole, it takes in 5 moles of electrons: MnO – 4 + 8H+ + 5e–  Mn2+ + 4H2O 5Y2+ ≡ 10e–, each mole of Y 2+ gives out 2 moles of electrons.  Final oxidation of Y is +4.  A 3 nuclide charge, ratiomass, q m 24Mg+ 1 0.041724 = 48Ti3+ 3 0.062548 = 59Co2+ 2 0.033959 = 101Ru4+ 4 0.0396101= Since 48Ti3+ has the largest q m ratio amongst the four nuclides, it will experience the greatest deflection.  B 4 A: bent B: trigonal planar C: square planar D✓: octahedral  D 5 Using 1 1 2 2 1 2 21 1 2 2 1 pV p V V T ppT T V T=  =   ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 3 227 273 K100 cm40 kPa 100 400 cm 227 273 K 8 kPa 227 273 K400 cm20 kPa 100 400 cm 127 273 K 20 kPa p p +=   ++ = +=   ++ = K L final pressure of mixture 8 kPa 20 kPa 28 kPa pp=+ =+ = KL  A 6 1✓: Lattice energy is the energy released when 1 mole of ionic compound , KCl(s), is formed from its constituent gaseous ions, K +(g) and C l–(g). The lattice energy of potassium chloride is ∆H3. 2: The enthalpy change of solution is the energy change when 1 mole of substance in its standard state , KCl(s), is completely dissolved in a solvent to give a solution of infinite dilution, K +(aq) and C l–(aq). The enthalpy change of solution of potassium chloride is ∆H4. 3✓: Enthalpy change of formation is the energy change when 1 mole of a substance in its standard state , KCl(s), is formed from its constituent elements in their standard states , K(s) and Cl2(g). The enthalpy change of formation of solid potassium chloride is (∆H1 + ∆H2 + ∆H3).  B 7 1✓: (G1 + G2) is G for dissolving of AgCl(s) in NH 3(aq); (G3 + G4) is G for dissolving of AgBr(s) in NH3(aq). As solid AgC l is more soluble than solid AgBr in NH 3(aq), the former must be more exergonic, i.e. G is more negative. Thus, (G1 + G2) < (G3 + G4). 2✓: Eqm 2 and 4 are identical, with C l– and Br– being spectator ions: Ag+(aq) + 2NH3(aq)  [Ag(NH3)2]+(aq) Hence, G2 = G4. 3: From 2, G2 = G4. 4✓: Since solid AgCl is more soluble than solid AgBr in water, G1 must be more exergonic, i.e. more negative, than G3. Thus G1 < G3.  A 8 A: Forward reaction of reaction 1 leads to the formation of Ag atoms, causing the photochromic glass to darken. Reaction 2 and 3 together, is the backward reaction of reaction 1, catalysed by Cu + and Cu 2+. This removes Ag atoms, causing the glass to becomes transparent again. Forward reaction of reaction 1 must be much faster than the backward reaction (reaction 2 + reaction 3) when intensity of UV light is high to ensure the gl