Answers to the numerical questions in this book | IGCSE 化学数值题解答

📚 Answers to the numerical questions in this book | IGCSE 化学数值题解答

This article provides complete worked solutions for the numerical problems featured in our IGCSE Chemistry Numerical Practice Workbook. Every section targets a key quantitative topic, walking you through the logic step by step and presenting the final answers in a clear, exam-ready format.

本文提供了我们IGCSE化学数值练习册中所有数值题的完整解答。每个小节聚焦一个关键的定量主题，逐步梳理计算逻辑，并以清晰、符合考试要求的格式给出最终答案。

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1. Relative Atomic Mass and Molar Mass | 相对原子质量与摩尔质量

The relative atomic mass (Aᵣ) of an element is the average mass of its atoms compared to 1/12 of the mass of a carbon‑12 atom. To find the molar mass (Mᵣ) of a compound, simply add the Aᵣ values of all atoms in the formula.

元素的相对原子质量 (Aᵣ) 是其原子平均质量与碳‑12 原子质量的 1/12 之比。要计算化合物的摩尔质量 (Mᵣ)，只需将化学式中所有原子的 Aᵣ 值相加即可。

Example from the book: Calculate the molar mass of aluminium sulfate, Al₂(SO₄)₃.

本书例题：计算硫酸铝 Al₂(SO₄)₃ 的摩尔质量。

Solution: Aᵣ(Al) = 27, S = 32, O = 16. Mᵣ = 2×27 + 3×(32 + 4×16) = 54 + 3×(32 + 64) = 54 + 3×96 = 54 + 288 = 342 g/mol.

解答：Aᵣ(Al) = 27, S = 32, O = 16. Mᵣ = 2×27 + 3×(32 + 4×16) = 54 + 3×(32+64) = 54 + 3×96 = 54 + 288 = 342 g/mol。

Therefore, one mole of Al₂(SO₄)₃ has a mass of 342 g. This value is used throughout the unit on moles.

因此，1摩尔 Al₂(SO₄)₃ 的质量为 342 g。该数值贯穿整个摩尔单元使用。

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2. Mole Calculations from Mass | 从质量计算摩尔数

The number of moles n is found by dividing the mass m of a substance by its molar mass M.

物质的摩尔数 n 可通过质量 m 除以其摩尔质量 M 得到。

n = m / M

Question 2.4: How many moles are present in 22 g of carbon dioxide, CO₂? (Mᵣ of CO₂ = 44)

问题 2.4：22 g 二氧化碳 (CO₂) 中含有多少摩尔？(CO₂ 的 Mᵣ = 44)

Step 1: Identify the mass, m = 22 g. Step 2: Mᵣ = 44 g/mol. Step 3: n = 22 / 44 = 0.50 mol.

步骤 1：确定质量 m = 22 g。步骤 2：Mᵣ = 44 g/mol。步骤 3：n = 22 / 44 = 0.50 mol。

Answer: 0.50 mol of CO₂ molecules.

答案：0.50 mol CO₂ 分子。

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

An empirical formula shows the simplest whole‑number ratio of atoms in a compound. The molecular formula is a multiple of the empirical formula.

经验式表示化合物中原子的最简整数比。分子式是经验式的整数倍。

Question 3.7: A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its Mᵣ is 60. Find the empirical and molecular formulae.

问题 3.7：某化合物含有 40.0% 碳、6.7% 氢和 53.3% 氧 (质量分数)。其 Mᵣ = 60。求经验式和分子式。

Element / 元素	% mass	÷ Aᵣ / 除以Aᵣ	Ratio / 比	Simplest ratio / 最简比
C	40.0	40.0 / 12 = 3.33	3.33 / 3.33 = 1	1
H	6.7	6.7 / 1 = 6.7	6.7 / 3.33 ≈ 2	2
O	53.3	53.3 / 16 = 3.33	3.33 / 3.33 = 1	1

Empirical formula = CH₂O. Its empirical formula mass = 12 + 2×1 + 16 = 30. The given Mᵣ = 60, so the multiplier is 60 / 30 = 2. Molecular formula = C₂H₄O₂.

经验式 = CH₂O。经验式质量 = 12 + 2×1 + 16 = 30。已知 Mᵣ = 60，所以倍数 = 60 / 30 = 2。分子式 = C₂H₄O₂。

Final answers: empirical formula CH₂O; molecular formula C₂H₄O₂ (which could be ethanoic acid or methyl methanoate).

最终答案：经验式 CH₂O；分子式 C₂H₄O₂（可能是乙酸或甲酸甲酯）。

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4. Reacting Masses and Limiting Reactants | 反应质量与限量试剂

In these problems you use the balanced equation to relate the moles of reactants to the moles of products. If masses of more than one reactant are given, the limiting reactant determines how much product is formed.

在这类问题中，你需要通过配平的化学方程式，将反应物的摩尔数与产物的摩尔数联系起来。如果给出多种反应物的质量，限量试剂将决定生成产物的量。

Question 4.2: 2.4 g of magnesium ribbon is added to an excess of dilute hydrochloric acid. What mass of magnesium chloride is produced? Equation: Mg + 2HCl → MgCl₂ + H₂

问题 4.2：将 2.4 g 镁条加入过量的稀盐酸中。生成的氯化镁质量是多少？方程式：Mg + 2HCl → MgCl₂ + H₂

Aᵣ: Mg = 24, Cl = 35.5. Mᵣ of MgCl₂ = 24 + 2×35.5 = 95 g/mol. Moles of Mg = 2.4 / 24 = 0.10 mol. From the equation, 1 mol Mg gives 1 mol MgCl₂, so moles of MgCl₂ = 0.10 mol. Mass of MgCl₂ = 0.10 × 95 = 9.5 g.

相对原子质量: Mg = 24, Cl = 35.5。MgCl₂ 的 Mᵣ = 24 + 2×35.5 = 95 g/mol。Mg 的摩尔数 = 2.4 / 24 = 0.10 mol。由方程式，1 mol Mg 生成 1 mol MgCl₂，因此 MgCl₂ 的摩尔数 = 0.10 mol。MgCl₂ 的质量 = 0.10 × 95 = 9.5 g。

Answer: 9.5 g of magnesium chloride.

答案：9.5 g 氯化镁。

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5. Gas Volumes at RTP | 室温常压下的气体体积

At room temperature and pressure (RTP, 25 °C and 1 atm), one mole of any gas occupies 24 dm³ (or 24 000 cm³). Use this to interconvert moles and gas volume.

在室温常压 (RTP, 25 °C, 1 atm) 下，1摩尔任何气体的体积为 24 dm³ (或 24 000 cm³)。利用这一关系进行摩尔数与气体体积的换算。

Question 5.5: When 10.0 g of calcium carbonate is heated strongly, it decomposes: CaCO₃ → CaO + CO₂. Calculate the volume of carbon dioxide produced at RTP.

问题 5.5：当 10.0 g 碳酸钙强热分解时: CaCO₃ → CaO + CO₂。计算在 RTP 下生成的二氧化碳体积。

Mᵣ of CaCO₃ = 40 + 12 + 3×16 = 100 g/mol. Moles of CaCO₃ = 10.0 / 100 = 0.100 mol. From the equation, 1 mol CaCO₃ produces 1 mol CO₂, so n(CO₂) = 0.100 mol. Volume of CO₂ = 0.100 × 24 = 2.4 dm³.

CaCO₃ 的 Mᵣ = 40 + 12 + 3×16 = 100 g/mol。CaCO₃ 的摩尔数 = 10.0 / 100 = 0.100 mol。由方程式，1 mol CaCO₃ 生成 1 mol CO₂，因此 n(CO₂) = 0.100 mol。CO₂ 的体积 = 0.100 × 24 = 2.4 dm³。

Answer: 2.4 dm³ (or 2400 cm³) of carbon dioxide.

答案：2.4 dm³ (或 2400 cm³) 二氧化碳。

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6. Concentration of Solutions | 溶液浓度

Concentration c (mol/dm³) is the number of moles of solute dissolved in 1 dm³ of solution. It can be calculated using c = n / V, where V is the volume in dm³.

浓度 c (mol/dm³) 是溶解在 1 dm³ 溶液中的溶质摩尔数。可通过 c = n / V 计算，其中 V 为体积，单位 dm³。

Question 6.8: 5.85 g of sodium chloride is dissolved in water to make 250 cm³ of solution. Calculate the concentration in mol/dm³. (Na = 23, Cl = 35.5)

问题 6.8：将 5.85 g 氯化钠溶于水，配成 250 cm³ 溶液。计算浓度 (mol/dm³)。(Na = 23, Cl = 35.5)

Mᵣ of NaCl = 23 + 35.5 = 58.5 g/mol. Moles of NaCl = 5.85 / 58.5 = 0.100 mol. Volume in dm³ = 250 / 1000 = 0.250 dm³. Concentration c = 0.100 / 0.250 = 0.40 mol/dm³.

NaCl 的 Mᵣ = 23 + 35.5 = 58.5 g/mol。NaCl 的摩尔数 = 5.85 / 58.5 = 0.100 mol。体积 (dm³) = 250 / 1000 = 0.250 dm³。浓度 c = 0.100 / 0.250 = 0.40 mol/dm³。

Answer: 0.40 mol/dm³. You can also quote the concentration in g/dm³ as 23.4 g/dm³.

答案：0.40 mol/dm³。也可表示为质量浓度 23.4 g/dm³。

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7. Titration Calculations | 滴定计算

Titration results let you calculate the unknown concentration of an acid or alkali using the concept that at the endpoint, moles of H⁺ equal moles of OH⁻.

滴定结果可用于计算酸或碱的未知浓度，其原理是在终点时 H⁺ 的摩尔数等于 OH⁻ 的摩尔数。

Question 7.3: 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide solution is neutralised by 21.5 cm³ of hydrochloric acid. Calculate the concentration of the acid. NaOH + HCl → NaCl + H₂O

问题 7.3：25.0 cm³ 的 0.100 mol/dm³ 氢氧化钠溶液被 21.5 cm³ 盐酸中和。计算该盐酸的浓度。NaOH + HCl → NaCl + H₂O

Moles of NaOH = c × V = 0.100 × (25.0 / 1000) = 0.00250 mol. The 1:1 ratio means moles of HCl = 0.00250 mol. Volume of HCl (dm³) = 21.5 / 1000 = 0.0215 dm³. Concentration of HCl = 0.00250 / 0.0215 = 0.1163 mol/dm³ (≈ 0.116 mol/dm³ to 3 s.f.).

NaOH 的摩尔数 = c × V = 0.100 × (25.0 / 1000) = 0.00250 mol。1:1 的反应比意味着 HCl 的摩尔数也为 0.00250 mol。HCl 的体积(dm³) = 21.5 / 1000 = 0.0215 dm³。HCl 浓度 = 0.00250 / 0.0215 = 0.1163 mol/dm³ (约为 0.116 mol/dm³，保留三位有效数字)。

Answer: 0.116 mol/dm³ (or 0.1163 mol/dm³ before rounding).

答案：0.116 mol/dm³ (或未舍入时为 0.1163 mol/dm³)。

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8. Energy Changes – Enthalpy | 能量变化 – 焓变

Enthalpy change ΔH is often measured by carrying out a reaction in solution and measuring the temperature change. The heat absorbed or released is Q = m c ΔT, where m is the total mass of solution, c is the specific heat capacity (usually 4.2 J/g°C), and ΔT is the temperature change. ΔH = –Q / n.

焓变 ΔH 通常通过溶液中的反应和温度变化来测量。吸收或放出的热量 Q = m c ΔT，其中 m 为溶液总质量，c 为比热容 (通常取 4.2 J/g°C)，ΔT 为温度变化。ΔH = –Q / n。

Question 8.4: In an experiment, 50 cm³ of 1.0 mol/dm³ HCl is mixed with 50 cm³ of 1.0 mol/dm³ NaOH. The temperature rises by 6.5 °C. Calculate the enthalpy of neutralisation. Assume the density of the solution is 1 g/cm³ and c = 4.2 J/g°C.

问题 8.4：实验中，将 50 cm³ 1.0 mol/dm³ HCl 与 50 cm³ 1.0 mol/dm³ NaOH 混合。温度上升了 6.5 °C。计算中和焓。假设溶液密度为 1 g/cm³，c = 4.2 J/g°C。

Total volume = 100 cm³ → mass m = 100 g. Q = 100 × 4.2 × 6.5 = 2730 J. Moles of HCl = 1.0 × (50 / 1000) = 0.050 mol (and same for NaOH). The reaction is 1:1, so moles of water formed = 0.050 mol. ΔH = –2730 / 0.050 = –54 600 J/mol = –54.6 kJ/mol (exothermic).

总体积 = 100 cm³ → 质量 m = 100 g。Q = 100 × 4.2 × 6.5 = 2730 J。HCl 的摩尔数 = 1.0 × (50 / 1000) = 0.050 mol (NaOH 同)。反应比为 1:1，因此生成水的摩尔数为 0.050 mol。ΔH = –2730 / 0.050 = –54 600 J/mol = –54.6 kJ/mol (放热)。

Answer: ΔH = –54.6 kJ/mol.

答案：ΔH = –54.6 kJ/mol。

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9. Rate of Reaction – Numerical Data | 反应速率 – 数值数据

The rate of a reaction can be calculated from the change in concentration, mass or volume of a product per unit time. Graphs are often used to find the initial rate.

反应速率可通过单位时间内反应物浓度、质量或产物体积的变化来计算。通常利用曲线图求初始速率。

Question 9.2: In the reaction between marble chips and acid, 2.0 g of chips produced 480 cm³ of CO₂ in 5 minutes. Calculate the average rate in cm³/s and in mol/s. (Molar volume at RTP = 24 000 cm³/mol)

问题 9.2：大理石碎片与酸的反应中，2.0 g 碎片在 5 分钟内产生了 480 cm³ CO₂。计算平均速率，分别以 cm³/s 和 mol/s 表示。(RTP 下摩尔体积 = 24 000 cm³/mol)

Time in seconds = 5 × 60 = 300 s. Average rate (volume) = 480 / 300 = 1.6 cm³/s. Moles of CO₂ = 480 / 24 000 = 0.020 mol. Average rate (moles) = 0.020 / 300 = 6.67 × 10⁻⁵ mol/s.

时间(s) = 5 × 60 = 300 s。平均速率 (体积) = 480 / 300 = 1.6 cm³/s。CO₂ 的摩尔数 = 480 / 24 000 = 0.020 mol。平均速率 (摩尔) = 0.020 / 300 = 6.67 × 10⁻⁵ mol/s。

Answer: 1.6 cm³/s and 6.67 × 10⁻⁵ mol/s.

答案：1.6 cm³/s 和 6.67 × 10⁻⁵ mol/s。

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10. Electrolysis – Quantitative Aspects | 电解 – 定量方面

In electrolysis, the quantity of product formed depends on the electric charge passed. Charge Q (coulombs) = current I (A) × time t (s). The amount of substance deposited or liberated is given by: n = Q / (z F), where z is the number of electrons transferred per ion and F = 96 500 C/mol.

在电解中，生成物的量取决于通过的电量。电量 Q (库仑) = 电流 I (A) × 时间 t (s)。析出或放出的物质的量由下式给出：n = Q / (z F)，其中 z 为每个离子转移的电子数，F = 96 500 C/mol。

Question 10.6: A current of 0.50 A is passed through copper(II) sulfate solution for 30 minutes. Calculate the mass of copper deposited on the cathode. (Cu²⁺ + 2e⁻ → Cu; Aᵣ = 63.5)

问题 10.6：将 0.50 A 的电流通过硫酸铜(II) 溶液 30 分钟。计算在阴极上析出的铜的质量。(Cu²⁺ + 2e⁻ → Cu; Aᵣ = 63.5)

Time in seconds = 30 × 60 = 1800 s. Q = I × t = 0.50 × 1800 = 900 C. For copper, z = 2. Moles of electrons = 900 / 96 500 = 0.00933 mol. Moles of Cu deposited = moles of electrons / z = 0.00933 / 2 = 0.00466 mol. Mass of Cu = 0.00466 × 63.5 = 0.296 g ≈ 0.30 g (2 s.f.).

时间(s) = 30 × 60 = 1800 s。Q = I × t = 0.50 × 1800 = 900 C。对于铜，z = 2。电子的摩尔数 = 900 / 96 500 = 0.00933 mol。析出铜的摩尔数 = 电子的摩尔数 / z = 0.00933 / 2 = 0.00466 mol。Cu 的质量 = 0.00466 × 63.5 = 0.296 g ≈ 0.30 g (两位有效数字)。

Answer: approximately 0.30 g of copper.

答案：约 0.30 g 铜。

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11. Redox Titrations | 氧化还原滴定

Redox titrations often involve potassium manganate(VII) or iodine. The volume of oxidising agent is used to find the concentration of a

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