GESS 4Exp Pure Chem Prelim 2024 Answers

1 GESS Sec 4 Chemistry (6092) Preliminary Exams 2024 Suggested Solutions PAPER 1 Question Number Key Question Number Key 1 A 21 B 2 B 22 A 3 B 23 D 4 C 24 D 5 B 25 B 6 B 26 A 7 C 27 B 8 D 28 A 9 A 29 A 10 D 30 B 11 A 31 D 12 A 32 C 13 C 33 D 14 A 34 B 15 A 35 B 16 C 36 C 17 D 37 D 18 A 38 D 19 A 39 B 20 C 40 D

2 Question Answer Explanation 1 A Option A is correct as the burette can measure variable volumes of a liquid in a titration. Option B is incorrect because the 0.24 dm3 is equivalent to 240 cm3 which exceeds the capacity of the gas syringe. Option D is incorrect as a pipette can only measure a fixed volume of a liquid depending on its size, in this case 25.0 cm3. 2 B Q has a lower Rf value than R as it travelled a shorter distance from the start line. 3 B The mixture is heated using a water bath, hence any substance with a boiling point above 100 ºC (i.e. C & D) cannot be distilled. Substance A is a gas at room temperature. 4 C The transition from (s) → (l) → (g) is endothermic as energy is absorbed to overcome forces of attraction between the particles. 5 B A mixture of solid and liquid exists at the melting point/freezing point, which occurs at 50 ºC. 6 B Since the ion has a charge of 3+, it means that Z would have lost 3 electrons, and is therefore in Group 13. Since the ion has 2 electron shells, Z must have had three electron shells in total and is hence in Period 3. 7 C Sea water is a mixture; sodium chloride is a compound and chlorine is an element. 8 D In graphite, each carbon atom is only bonded to three other carbon atoms, leaving one electron per carbon atom not involved in bonding. The electrons which are not involved in bonding are delocalised, acting as mobile charged carriers. Metals such as aluminium have a sea of delocalised electrons which act as mobile charged carriers. Ionic compounds conduct electricity due to the presence of mobile ions. 9 A The electronic structure of oxygen is 2,6. It needs 2 more electrons to have a fully-filled valence shell. Each oxygen atom forms 2 covalent bonds. The electronic structure of chlorine is 2,8,7. It needs 1 more electron to have a fully-filled valence shell. Each chlorine atom forms 1 covalent bond. This results in a structure with the formula Cl2O. 10 D Both silicon and carbon are in Group 14, hence SiO2 and diamond both have a giant covalent structure where the atoms around Si and C are in a tetrahedral arrangement, i.e. each Si and C atom forms 4 covalent bonds. For giant covalent structures, there are no intermolecular forces of attraction as all the atoms are joined via covalent bonds. Melting requires breaking these covalent bonds. 11 A nO2 = 500/24000 = 0.020833 mol 1 mol contains 6.02 × 1023 molecules. Hence, number of molecules = 0.020833 × 6.02 × 1023 = 1.254 × 1022

3 12 A The sum of the relative molecular masses of Mg atoms in a molecule of chlorophyll-a is !.#$%&&×893 = 24.02. Since the Ar of a Mg atom is 24, there is only one Mg atom. 13 C concentration of saline in g/dm3 = &.$!""!""" = 9 g/dm3 concentration of saline in mol/dm3 = $!'(').) = 0.1538 mol/dm3 14 A nH2 = *! = 4 mol nO2 = *'! = 0.25 mol (limiting reactant) nH2O formed = 0.5 mol mass of H2O = 0.5 × 18 = 9 g 15 A Quicklime (CaO) or slaked lime (Ca(OH)2) are both used by farmers to neutralise the excess acids in the soil, raising the pH. 16 C The pH at which the indicator changes colour must fall within the region of drastic pH change in the titration graph. This is where the ‘step’ occurs, in this case it spans the range of around pH 5 to 11. 17 D Copper is an unreactive metal, hence it does not react with acids. To prepare copper(II) sulfate, we add excess copper(II) oxide, copper(II) carbonate or copper(II) hydroxide to ensure all the acid is reacted. We then filter the mixture to remove the excess solid, collect the filtrate and carry out crystallisation on the filtrate. 18 A Ammonia turns moist red litmus paper blue, while chlorine bleaches both blue and red litmus papers. 19 A Y must be an acid since a gas is produced when reacted with a carbonate. Y also contains nitrate ions as ammonia gas is given off when the solution is warmed with aluminium and sodium hydroxide. 20 C In the chapter of Acids and Bases, we learnt that alkalis dissociate in water to give OH– ions: NaOH(aq) → Na+(aq) + OH–(aq) NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq) We also know that most metal hydroxides are insoluble in water. Hence, the green precipitate is Fe(OH)2. The green precipitate slowly turns reddish-brown in air (oxidation). 4 Fe(OH)2(s) + O2(g) + 2 H2O(l) → 4 Fe(OH)3(s) 21 B CO acts as a reducing agent as it causes Fe2O3 to be reduced to Fe. This is seen in the oxidation state of iron decreasing from +3 in Fe2O3 to 0 in Fe. 22 A positive (+) electrode X negative (–) electrode Y Ions attracted Br– Pb2+ Half equation 2Br–(l) → Br2(g) + 2e– Pb2+(l) + 2e– → Pb(l) Nature of reaction Oxidation Reduction Anode or cathode? Anode Cathode

4 Observation Effervescence of reddish-brown gas Grey/silvery globules of molten lead metal float to the surface Overall reaction PbBr2(l) → Pb(l) + Br2(g) 23 D In electroplating, the object to be electroplated is always placed at the cathode (negative electrode), so that reduction of cations at the cathode will result in a solid deposit/coating around the object. The anode (positive electrode) is the metal used for electroplating. Oxidation at the anode replenishes the electrolyte with the cations. anode: Ag(s) → Ag+(aq) + e– cathode: Ag+(aq) + e– → Ag(s) 24 D In a simple cell, the larger the difference in reactivity of the two metal electrodes, the greater the voltage. Hence the voltage of cell 1 < cell 2. In both cells, magnesium is the more reactive metal and hence undergoes oxidation: Mg(s) → Mg2+(aq) + 2 e–. Mg2+ ions enter the electrolyte, which does not result in a colour change. Electrons flow from the more reactive metal (magnesium) to the other electrode. 25 B Trends going down Group 1: • Melting and boiling points decrease • Density increases • Reactivity increases 26 A Statement 1 is true. Melting and boiling points increases down the group. Since X is a liquid, it must be Br2. Y, being a solid, must be below X in the Periodic Table. Statement 2 is false. All Group 17 elements are diatomic, meaning they are diatomic molecules which consist of two atoms covalently bonded. Statement 3 is false. Reactivity decreases down the group. Y is less reactive than X. Hence Y cannot displace X from its solution. 27 B The two properties are characteristic of transition metals. Option B is the only one where all examples are transition metals. 28 A This is a question on sacrificial protection. In the presence of an acid, the more reactive metal is preferentially oxidised: M(s) + 2 H+(aq) → M2+(aq) + H2(g) Given that all beakers contain zinc, the beaker where Zn is least likely to be oxidised is where there is a more reactive metal present, i.e. Mg. 29 A The more reactive the metal, the more thermally stable the compound and the harder it is to decompose the carbonate. Z is the most stable carbonate (Na2CO3) while Y is the least stable carbonate (CuCO3). 30 B Energy absorbed during bond breaking = 4BE(H–H) + BE(N–N) = 4(390) + 160 = 1720 kJ/mol Energy released during bond forming = BE(NºN) + 2BE(H–H) = 945 + 2(436) = 1817 kJ/mol

5 Enthalpy change = +1720 – 1817 = –97 kJ/mol 31 D nO2 = 4 mol mole ratio of O2 : e– = 1:4 ne– = 16 mol mole ratio of e– : H2 = 2:1 nH2 = 8 mol Mass of H2 = 8 × 2 = 16 g 32 C Option A, B and D are incorrect as a lower temperature or different particle size will not change the yield. HCl is the limiting reactant. Hence to produce curve Q, a lower number of moles of HCl is used. 33 D P represents the enthalpy change for the backward reaction. Q represents the enthalpy change for the forward reaction. R represents the activation energy for the forward reaction. 34 B The temperature