Typical Example Questions in IB & AQA Science: Detailed Solutions | IB 与 AQA 科学典型例题详解

📚 Typical Example Questions in IB & AQA Science: Detailed Solutions | IB 与 AQA 科学典型例题详解

In both IB Diploma and AQA A-level science courses, students must master a variety of quantitative problem-solving skills. This article presents a selection of typical worked examples from Chemistry and Physics, showing step-by-step solutions that mirror the style of IB and AQA exam questions. By studying these examples, you will reinforce your understanding of core concepts and improve your ability to tackle similar problems under exam conditions.

在 IB 文凭和 AQA A-level 科学课程中，学生必须掌握各种定量问题解决技能。本文精选了化学和物理中若干典型例题，按 IB 与 AQA 考试风格分步详解。通过学习这些例题，你能巩固核心概念的理解，并提高在考试环境下解答类同问题的能力。

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1. Mole Calculations | 摩尔计算

Example: Calculate the mass of water produced when 4.0 g of hydrogen gas reacts completely with excess oxygen. (Molar masses: H = 1.01 g mol⁻¹, O = 16.00 g mol⁻¹)

例题：计算 4.0 g 氢气与过量氧气完全反应时生成的水的质量。(摩尔质量：H = 1.01 g mol⁻¹, O = 16.00 g mol⁻¹)

Step 1: Write the balanced chemical equation: 2H₂ + O₂ → 2H₂O.

步骤1：写出配平的化学方程式：2H₂ + O₂ → 2H₂O。

Step 2: Calculate moles of H₂: n(H₂) = mass / M = 4.0 g / 2.02 g mol⁻¹ = 1.98 mol.

步骤2：计算 H₂ 的摩尔数：n(H₂) = 质量 / 摩尔质量 = 4.0 g / 2.02 g mol⁻¹ = 1.98 mol。

Step 3: Use the mole ratio from the equation. 2 mol H₂ produce 2 mol H₂O, so n(H₂O) = n(H₂) = 1.98 mol.

步骤3：利用方程式中的摩尔比。2 mol H₂ 生成 2 mol H₂O，因此 n(H₂O) = n(H₂) = 1.98 mol。

Step 4: Convert moles of H₂O to mass: mass = n × M = 1.98 mol × 18.02 g mol⁻¹ = 35.7 g (3 s.f.).

步骤4：将 H₂O 摩尔数转化为质量：质量 = n × M = 1.98 mol × 18.02 g mol⁻¹ = 35.7 g（三位有效数字）。

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2. Empirical and Molecular Formulae | 经验式与分子式

Example: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Determine its empirical formula. If its molar mass is 180 g mol⁻¹, find the molecular formula.

例题：某化合物含碳 40.0%，氢 6.7%，氧 53.3%（质量百分数）。求其经验式。若其摩尔质量为 180 g mol⁻¹，求分子式。

Step 1: Assume a 100 g sample, so masses are C: 40.0 g, H: 6.7 g, O: 53.3 g.

步骤1：假设样本为 100 g，则质量分别为 C：40.0 g，H：6.7 g，O：53.3 g。

Step 2: Convert masses to moles: n(C) = 40.0 g / 12.01 g mol⁻¹ = 3.33 mol; n(H) = 6.7 g / 1.01 g mol⁻¹ = 6.63 mol; n(O) = 53.3 g / 16.00 g mol⁻¹ = 3.33 mol.

步骤2：转化为摩尔数：n(C) = 40.0 g / 12.01 g mol⁻¹ = 3.33 mol；n(H) = 6.7 g / 1.01 g mol⁻¹ = 6.63 mol；n(O) = 53.3 g / 16.00 g mol⁻¹ = 3.33 mol。

Step 3: Divide by the smallest number of moles (3.33) to get the ratio: C : H : O = 1 : 1.99 : 1 ≈ 1 : 2 : 1. Empirical formula = CH₂O.

步骤3：除以最小摩尔数 (3.33) 得到比例：C : H : O = 1 : 1.99 : 1 ≈ 1 : 2 : 1。经验式 = CH₂O。

Step 4: Calculate the empirical formula mass: 12.01 + 2×1.01 + 16.00 = 30.03 g mol⁻¹. The molar mass is 180 g mol⁻¹, so the multiplier n = 180 / 30.03 ≈ 6. Molecular formula = (CH₂O)₆ = C₆H₁₂O₆.

步骤4：计算经验式质量：12.01 + 2×1.01 + 16.00 = 30.03 g mol⁻¹。摩尔质量为 180 g mol⁻¹，倍数 n = 180 / 30.03 ≈ 6。分子式 = (CH₂O)₆ = C₆H₁₂O₆。

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3. Stoichiometry and Limiting Reactants | 化学计量与限量反应物

Example: 19.6 g of H₂SO₄ react with 10.0 g of NaOH. Which reactant is in excess and what mass of Na₂SO₄ is produced? (Molar masses: H₂SO₄ = 98.1 g mol⁻¹, NaOH = 40.0 g mol⁻¹, Na₂SO₄ = 142.1 g mol⁻¹)

例题：19.6 g H₂SO₄ 与 10.0 g NaOH 反应。哪种反应物过量？生成 Na₂SO₄ 的质量是多少？(摩尔质量：H₂SO₄ = 98.1 g mol⁻¹, NaOH = 40.0 g mol⁻¹, Na₂SO₄ = 142.1 g mol⁻¹)

Step 1: Balanced equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.

步骤1：平衡方程式：H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O。

Step 2: Moles of H₂SO₄ = 19.6 g / 98.1 g mol⁻¹ = 0.200 mol. Moles of NaOH = 10.0 g / 40.0 g mol⁻¹ = 0.250 mol.

步骤2：H₂SO₄ 的摩尔数 = 19.6 g / 98.1 g mol⁻¹ = 0.200 mol。NaOH 的摩尔数 = 10.0 g / 40.0 g mol⁻¹ = 0.250 mol。

Step 3: From the equation, 1 mol H₂SO₄ reacts with 2 mol NaOH. 0.200 mol H₂SO₄ would require 0.400 mol NaOH, but only 0.250 mol NaOH is available – NaOH is the limiting reactant. H₂SO₄ is in excess.

步骤3：根据方程式，1 mol H₂SO₄ 与 2 mol NaOH 反应。0.200 mol H₂SO₄ 需要 0.400 mol NaOH，但仅有 0.250 mol NaOH，所以 NaOH 是限量反应物，H₂SO₄ 过量。

Step 4: Moles of Na₂SO₄ produced = moles of NaOH / 2 = 0.250 / 2 = 0.125 mol. Mass of Na₂SO₄ = 0.125 mol × 142.1 g mol⁻¹ = 17.8 g.

步骤4：生成的 Na₂SO₄ 摩尔数 = NaOH 摩尔数 / 2 = 0.250 / 2 = 0.125 mol。Na₂SO₄ 质量 = 0.125 mol × 142.1 g mol⁻¹ = 17.8 g。

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4. Acid-Base Titrations | 酸碱滴定

Example: 25.0 cm³ of 0.100 mol dm⁻³ HCl is neutralised by 23.6 cm³ of NaOH solution. Calculate the concentration of the NaOH solution.

例题：25.0 cm³ 浓度为 0.100 mol dm⁻³ 的 HCl 被 23.6 cm³ NaOH 溶液中和。求 NaOH 溶液的浓度。

Step 1: Write the neutralisation equation: HCl + NaOH → NaCl + H₂O. Mole ratio is 1:1.

步骤1：写出中和方程式：HCl + NaOH → NaCl + H₂O。摩尔比为 1:1。

Step 2: Calculate moles of HCl used: n(HCl) = c × V = 0.100 mol dm⁻³ × (25.0 / 1000) dm³ = 0.002

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