2026 Chung Cheng Main Prelim 6092 P1 MS

Chung Cheng High School (Main) 2026 Preliminary Examination Sec 4 Chemistry (6092) Paper 1 1 How to use these solutions • For each question, read the Approach line first, then close the solutions and try the question again before reading on. • When you have the answer, read why each wrong option is wrong. • Most MCQ errors come from three habits: reading the question too quickly and missing "not", "incorrect" or a detail; skipping a step in a calculation; and choosing an option that is true but does not answer the question. Answer key Q 1 2 3 4 5 6 7 8 9 10 Ans A D C A B A C C C A Q 11 12 13 14 15 16 17 18 19 20 Ans B B D C A D D B B C Q 21 22 23 24 25 26 27 28 29 30 Ans B C A C D C D B A B Q 31 32 33 34 35 36 37 38 39 40 Ans D B A C B B C B B D Q1. Answer: A Topic: Salt preparation (insoluble base + acid) Approach: Write the steps first: add excess solid CuO to the acid, filter off the excess, then crystallise the filtrate. Then look for the option that has apparatus for every step. • A is correct. Measuring cylinder and beaker to react the acid with excess CuO; filter funnel and paper to remove the unreacted CuO; evaporating dish to crystallise CuCl2 from the filtrate. • B: no evaporating dish, so the salt cannot be recovered from the solution. • C and D: no filtration step, so the excess solid CuO stays in the product. A burette and pipette are for titration, which is only used when both reactants are soluble. Q2. Answer: D Topic: Separation techniques Approach: Take each separation in turn and ask what physical property differs. Rule out options as soon as one column is wrong. • D is correct. 1: water is boiled off and condensed from the dissolved KCl, so distillation. 2: iodine sublimes and iron does not, so sublimation. 3: the water is evaporated to leave solid KCl, so evaporation. 4: ethanol and water are miscible liquids with different boiling points, so fractional distillation. • A: only one liquid is being collected in 1, so fractional distillation is not needed. Crystallisation cannot separate iodine from iron. • B: filtration cannot separate a dissolved solid (KCl) from water in 3. • C: filtration cannot recover water from a solution in 1. Q3. Answer: C Topic: Chromatography Approach: Distance travelled tells you solubility in the solvent. Then test each statement against what RF and the start line mean. • C is correct. X travels further up the paper than Y, so X is more soluble in the solvent. • A: the solvent must be below the start line. If it were above Z, the spots would dissolve straight into the solvent in the beaker. • B: RF = distance moved by substance ÷ distance moved by solvent. It is a ratio, so it stays the same when the paper is longer. • D: a locating agent is needed only for colourless substances, not always. Q4. Answer: A Topic: Diffusion and relative molecular mass Approach: At the same temperature and pressure, gases diffuse at the same rate only if they have the same Mr. Work out every Mr before looking at the pairs. gas CO N2O C2H4 C3H8 N2 Mr 28 44 28 44 28 • Pair 1 is a match: N2 and C2H4 are both 28. • Pair 2 is a match: CO and N2 are both 28. • Pair 3 is a match: N2O and C3H8 are both 44. • Pair 4 is not a match: C2H4 is 28 but C3H8 is 44, so C2H4 diffuses faster (lower Mr, faster diffusion). • A is correct (1, 2 and 3). Q5. Answer: B Topic: Atomic structure of ions Approach: Tabulate protons, neutrons and electrons for each ion. Electrons = protons − charge. ion protons neutrons electrons Fe2+ 26 56 − 26 = 30 24 Co3+ 27 59 − 27 = 32 24 Mn2+ 25 56 − 25 = 31 23 • B is correct. Only Fe2+ and Co3+ have 24 electrons. • A: all three have different numbers of neutrons. • C: all three are different elements, so they have different numbers of protons. • D: Fe2+ and Mn2+ both have a nucleon number of 56, so they have the same relative mass. Q6. Answer: A Topic: Covalent bonding (lone pairs) Approach: Draw each molecule and look for lone pairs. An atom with a lone pair has outer electrons that are not used in bonding. • A is correct. C forms four bonds and uses all four outer electrons. Each H forms one bond and uses its one electron. • B: F in HF has three lone pairs. • C: N in NH3 has one lone pair. • D: O in CH3OH has two lone pairs. Q7. Answer: C Topic: Ionic lattice structure Approach: In an ionic lattice, oppositely charged ions sit next to each other. Look along every edge of each cube for two ions of the same colour side by side. • C is correct. Black and white ions alternate along every edge, so each Ca2+ ion is next to S2− ions only. • A, B and D: each has at least one edge where two ions of the same charge are next to each other. Like charges repel, so this arrangement is not stable. Q8. Answer: C Topic: Giant covalent and simple molecular structures Approach: Check each statement for the one detail that is wrong: a number, or a particle type. • C is correct. Poly(ethene) is made of molecules. Melting overcomes the intermolecular forces between the chains; the covalent bonds are not broken. • A: each Si atom is bonded to four oxygen atoms. Silicon dioxide is covalent, so there are no oxide ions. • B: each carbon atom in diamond is bonded to four other carbon atoms. • D: graphite conducts because of delocalised electrons between the layers, not ions.

Chung Cheng High School (Main) 2026 Preliminary Examination Sec 4 Chemistry (6092) Paper 1 2 Q9. Answer: C Topic: Writing formulae from ions Approach: Get the charge on each ion first. NaSCN tells you thiocyanate is SCN−; copper(II) is Cu2+. Balance the charges. • C is correct. One Cu2+ needs two SCN− ions: Cu(SCN)2. • A and D: a 1 : 1 ratio would need Cu+, which is copper(I). • B: Cu2SCN would need a thiocyanate ion with a 2− charge. Q10. Answer: A Topic: Moles and covalent bonding Approach: Balance the equation to find moles of water, then count shared electrons per water molecule. Note "electrons", not "bonds". • C3H6 + 4.5O2 → 3CO2 + 3H2O, so 0.5 mol propene gives 3 × 0.5 = 1.5 mol water. • Each water molecule has two O-H single bonds, and each bond shares two electrons, so 4 shared electrons per molecule. • A is correct. 1.5 × 4 = 6 mol of shared electrons = 6 × 6 × 1023 = 3.6 × 1024. • B: 2 mol, which uses 0.5 mol of water (the propene) instead of 1.5 mol. • C: 3 mol, which counts the bonds and not the electrons. • D: 1 mol, which makes both errors. Q11. Answer: B Topic: Moles and extraction of metals Approach: Find moles of Fe3O4, then remember each formula unit contains three Fe atoms. • Mr(Fe3O4) = (3 × 56) + (4 × 16) = 232, so 116 g is 0.5 mol. • B is correct. Moles of Fe = 3 × 0.5 = 1.5 mol; mass = 1.5 × 56 = 84 g. • A: 1 mol of Fe, which fits neither the mole ratio nor the mass given. • C: 2 mol of Fe, a wrong mole ratio. • D: 3 mol of Fe, which forgets that only 0.5 mol of Fe3O4 is present. Q12. Answer: B Topic: Titration and formula of a salt Approach: Find the mole ratio of hydroxide to acid. The ratio tells you how many OH− the metal hydroxide has, and so the charge on M. • Moles of hydroxide = 10 ÷ 1000 × 2.0 = 0.0200 mol. Moles of H2SO4 = 50 ÷ 1000 × 0.20 = 0.0100 mol. • Ratio 2 : 1, so 2MOH + H2SO4 → M2SO4 + 2H2O. M forms M+ ions. • B is correct. Two M+ balance one SO42−: M2SO4. • A: MSO4 would need M2+, which reacts 1 : 1 with H2SO4. • C: M2(SO4)3 would need M3+. • D: M(SO4)2 would need M4+. Q13. Answer: D Topic: Percentage purity Approach: Build the calculation in four steps: moles of water, moles of NH4NO3 (use the 1 : 2 ratio), mass of NH4NO3, then divide by the mass of the impure sample. • Moles of H2O = 28 ÷ 18. Moles of NH4NO3 = 28 ÷ 18 × ½ = 28 ÷ 36. • Mass of NH4NO3 = 28 ÷ 36 × 80, where 80 is Mr(NH4NO3). • D is correct. Percentage purity = (28 ÷ 36) × (80 ÷ 90) × 100. • A: misses the 1 : 2 mole ratio (uses 28 ÷ 18). • B and C: 90 ÷ 80 is upside down. Divide th