2025 TKGS Chemistry Prelim Paper 1 MS

2025 S4 Preliminary Examination Suggested Solutions Tanjong Katong Girls’ School 1 Answer Key 1 2 3 4 5 6 7 8 9 10 C B C D C A C D C D 11 12 13 14 15 16 17 18 19 20 D B D B B D B B A A 21 22 23 24 25 26 27 28 29 30 A D A B A C C D B C 31 32 33 34 35 36 37 38 39 40 A C A A D C B A B D 1 C Why C: To make a dry acidic gas like SO₂ from solid + acid, you need a delivery tube → drying agent that doesn’t react with SO ₂ (conc. H ₂SO₄ or anhydrous CaCl₂) → downward delivery (SO₂ is denser than air). Why not A/B/D: A: Basic CaO will react with acidic SO2 gas. B: Using the filter funnel will allow the SO 2 gas to escape. D: Upward delivery method is used for gases less dense than air. 2 B Why B: Iodine is a substance that sublimes (changes directly from solid to purple vapour without melting). Sodium chloride (NaCl) does not sublime; it has a very high melting point and remains solid. By heating the mixture gently, iodine vapour can be driven off, collected and cooled back into solid iodine crystals. Sodium chloride stays behind as a solid. Why not A/C/D: A: Both iodine and sodium chloride are solids, not dissolved in a solution, so crystallisation is not suitable. C: Works only to separate insoluble solid from liquid, not two solids. D: Used to separate liquids with different boiling points, not solid mixtures like this. 3 C Step 1: Interpreting statement 1 Z shows four distinct spots on the chromatogram. That means Z contains at least four different components. It cannot be “no more than four” because there could be more that did not separate under the given solvent conditions. So, statement 1 must be “Z contains at least four alcohols”. Step 2: Interpreting statement 2 If spots from X, Y and Z line up at the same Rf values, that suggests those mixtures may contain a common alcohol. Therefore, it is possible they share the same substance. So, statement 2 = “X, Y and Z could contain the same alcohol”. 4 D Statement 1: Molecules gain kinetic energy and temperature increases. Wrong. During melting, the temperature stays constant at 0 ºC. The particles don’t gain kinetic energy; instead, energy goes into overcoming intermolecular forces of attraction. Statement 2: Energy is gained to overcome the intermolecular forces of attraction. Correct. Heat energy supplied during melting is used to overcome intermolecular forces of attraction. Statement 3: Molecules gain sufficient energy to move from fixed positions. Correct. Solid ice particles are in fixed positions. When melting, they break free and can slide past each other. 5 C The lighter the gas, the faster it diffuses. Cl₂: 35.5 × 2 = 71 CO₂: 12 + (16 × 2) = 44 CH₄: 12 + (1 × 4) = 16 NO₂: 14 + (16 × 2) = 46 The smallest Mr = 16 (CH4). Hence, CH4 diffuses the fastest.

Tanjong Katong Girls’ School 2 6 A  A Proton charge = +1 Electron charge = –1 Equal magnitude, opposite signs.  B Neutron has 0 charge, not +1.  C Neutron = 1 (relative mass). Electron = 1 1840 so it’s much lighter, not heavier.  D Proton = 1 and Electron = 1 1840 7 C  A Isotopes of an element have the same number of electrons → same electron ic configuration → same chemical properties. They may differ in physical properties (e.g. density, boiling point), but not in chemistry reactions.  B Isotopes differ in number of neutrons. They have the same chemical properties but physical properties are different.  C mass number = protons + neutrons Different isotopes differ in number of neutron s hence, they have different mass numbers. Different isotopes have the same number of electrons hence, they have the same electronic configuration, same chemical properties.  D If the proton numbers are different, they are different elements instead not isotopes. 8 D  1 W is 2+ not 2–.  2 Should be W3Y2  3 WZ2 is MgCl2 9 C  1 Giant covalent (e.g. diamond, SiO₂) → high mp but do not conduct electricity at all (except graphite).  2 Example: CO₂, I₂. Weak intermolecular forces → low melting point. No free ions/free delocalised electrons → does not conduct electricity  3 Metals conduct due to delocalised electrons in both states. Different metals have different melting points. 10 D Use trial and error method 11 D Why right: Molecular formula + Ar values gives Mr directly. Why others wrong: Empirical formula or % by mass alone isn’t enough to get molecular mass. 12 B % Cu in CuFeS₂ = 64 64+56+2(32) × 100 = 34.782% If ore has 1% Cu, % = (1 ÷ 34.782) x 100% = 2.88% 13 D no. mol of B = 0.78 11 = 0.070909 no. mol of C = 0.22 12 = 0.018333 Ratio of B:C ≈ 4:1 B4C 14 B no. mol of Q molecules = 1 𝑥 no. of Q molecules = 1 𝑥 × 𝐿 = 𝐿 𝑥 no. of Q atoms = 2 x 𝐿 𝑥 = 2𝐿 𝑥 15 B To neutralise acidic soil, add a base (alkali). Reasoning: • Calcium hydroxide is a strong base used in agriculture (“liming”). • Neutralisation increases pH value towards neutral. 16 D Acidic oxides (e.g. CO2) react with alkalis Amphoteric oxides (e.g. ZnO) react with both acids and alkalis. Basic oxides (e.g. CuO) do not react with alkali. 17 B Insoluble base/carbonate + acid → filter off excess solid → evaporate filtrate → crystallise salt. • CuSO4 can be made by reacting an insoluble copper(II) oxide/carbonate with dilute sulfuric acid. • Ammonium chloride is prepared by titration method. • Barium sulfate is insoluble; it forms via precipitation, not crystallisation. • Potassium chloride is prepared by titration method.

Tanjong Katong Girls’ School 3 18 B Higher pressure → higher yield of NH3 ; also increases reaction rate. Higher temperature increases reaction rate but reduces yield. Catalyst increases reaction rate only; no change to yield. 19 A Reaction of metal ions with NH₃(aq) vs excess NH₃(aq): Zn²⁺: white Zn(OH)₂ ppt with few drops; dissolves in excess NH₃ to give colourless solution. Al³⁺: white Al(OH) ₃ ppt with few drops; does not dissolve in excess NH₃ (unlike in excess NaOH) After dissolving alloy in HCl you have Zn² ⁺(aq) and Al³⁺(aq). Adding excess NH₃: Zn ppt dissolves, Al(OH)₃ remains→ white ppt observed. 20 A Fe²⁺ + OH⁻ → green Fe(OH)₂(s). Fe³⁺ would give brown Fe(OH)₃ ppt. NH₄⁺ + warm NaOH → NH₃ gas (alkaline, turns red litmus blue). Nitrate test (Al foil in warm alkali): forms NH₃→ turns red litmus blue. Green ppt ⇒ Fe²⁺, not Fe³⁺. Heating with NaOH produced no gas ⇒ no NH₄⁺. After adding Al foil and warming, a gas formed that turned red litmus blue ⇒ NO₃⁻ present. 21 A Electrorefining of copper, electrolyte: CuSO₄(aq). Anode (impure copper): Cu(s) → Cu²⁺(aq) + 2e⁻ (anode loses mass). Cathode (pure copper): Cu²⁺(aq) + 2e⁻ → Cu(s) (cathode gains mass). Electrolyte colour: remains blue (Cu²⁺ continually replenished at anode as it is removed at cathode).  B / D Solution does not become colourless; Cu² ⁺ stays present.  C Solution does not get increasingly blue overall; [Cu²⁺] is maintained. 22 D Cathode (reduction of O₂): O₂ + 2H₂O + 4e⁻ → 4OH⁻ ⇒ [OH⁻] increases → pH value increases. Anode (oxidation of H₂): H₂ + 2OH⁻ → 2H₂O + 2e⁻ ⇒ OH⁻ consumed → pH value decreases. 23 A Concept to recall: At inert anodes in dilute solutions, OH⁻ tends to be oxidised to O₂ unless the halide is concentrated and lower in the discharge series. Reasoning: Dilute NaCl: OH⁻ oxidised → O₂ (Cl₂ is favoured only when concentrated). Concentrated NaBr: Br⁻ oxidised → Br₂. Dilute Na₂SO₄: sulfate is hard to oxidise; OH⁻ oxidised → O₂. 24 B Concept to recall: Group number equals number of valence electrons in the atom. Reasoning: 1: 2,3 ⇒ Group 13. 2: 2,8,3 ⇒ Group 13. 3: (2,8,8)³ ⁻ corresponds to P³⁻ (came from 2,8,5; Group 15 as an atom). Thus not same group as 1 & 2. 25 A