Year 12 WJEC Chemistry: Unit Test Mock Paper Analysis | Year 12 WJEC 化学：单元测试模拟卷解析

📚 Year 12 WJEC Chemistry: Unit Test Mock Paper Analysis | Year 12 WJEC 化学：单元测试模拟卷解析

Welcome to this in-depth analysis of a mock Unit test for Year 12 WJEC Chemistry. This mock paper covers essential AS topics, including atomic structure, mole calculations, bonding, energetics, kinetics, equilibria, redox, organic chemistry, periodicity, and intermolecular forces. By working through these questions and model answers, you will strengthen your understanding of key concepts, improve exam technique, and gain confidence for the real assessment.

欢迎深入解析这份 Year 12 WJEC 化学单元模拟试卷。本模拟卷涵盖 AS 核心主题，包括原子结构、摩尔计算、化学键、能量学、动力学、平衡、氧化还原、有机化学、周期性和分子间力。通过逐题讲解和答案分析，你将巩固关键概念，提升解题技巧，并为真实考试树立信心。

1. Q1: Atomic Structure, Isotopes & Mass Spectrometry | 原子结构、同位素与质谱

In the mock paper, you are asked to calculate the relative atomic mass of boron given isotopic abundances. Boron-10 (¹⁰B) has an abundance of 19.9%, boron-11 (¹¹B) 80.1%. Using the formula Ar = Σ(abundance × isotopic mass)/100, we obtain Ar = (19.9 × 10 + 80.1 × 11) / 100 = (199 + 881.1)/100 = 1080.1/100 ≈ 10.80.

模拟卷中，要求根据同位素丰度计算硼的相对原子质量。硼-10 (¹⁰B) 丰度 19.9%，硼-11 (¹¹B) 80.1%。用公式 Ar = Σ(丰度 × 同位素质量)/100，得到 Ar = (19.9×10 + 80.1×11)/100 = (199+881.1)/100 ≈ 10.80。

The mass spectrum of boron shows two peaks at m/z 10 and 11. Peak 11 is approximately four times taller, consistent with the 80% abundance of ¹¹B. In the spectrometer, the ions are deflected according to their m/z ratio; lighter ions (¹⁰B⁺) are deflected more, but the detector records intensity proportional to the number of ions.

硼的质谱图显示 m/z 10 和 11 两个峰。峰11 高度约为峰10 的四倍，与 ¹¹B 的 80% 丰度一致。在质谱仪中，离子根据其质荷比偏转；较轻离子 (¹⁰B⁺) 偏转更大，但检测器记录的强度与离子数量成正比。

2. Q2: Mole Calculations & Titration | 摩尔计算与滴定

The titration reaction is Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. Moles of HCl used = c × V = 0.100 mol dm⁻³ × 23.50/1000 dm³ = 2.35 × 10⁻³ mol. From the 1:2 stoichiometry, moles of Na₂CO₃ = moles of HCl / 2 = 1.175 × 10⁻³ mol.

滴定反应为 Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂。所用 HCl 的摩尔数 = c×V = 0.100 mol dm⁻³ × 23.50/1000 dm³ = 2.35×10⁻³ mol。根据 1:2 化学计量比，Na₂CO₃ 摩尔数 = HCl 摩尔数/2 = 1.175×10⁻³ mol。

Concentration of Na₂CO₃ = moles / volume (in dm³) = 1.175 × 10⁻³ mol / 0.0250 dm³ = 0.0470 mol dm⁻³. Always convert cm³ to dm³ by dividing by 1000 and give final answer to 3 significant figures reflecting the data.Published by TutorHao | Year 12 Chemistry Revision Series | aleveler.com

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