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Structure 1 · SolutionsStructure 1 · 解析

The Particulate Nature of Matter: Solutions物质的粒子性：解析

Companion to the Structure 1 Practice SetStructure 1 练习题的解析配套

EASY MEDIUM HARD Paper 1 Paper 2 Paper 3 HL HL

Topics Structure 1.1 to 1.5考点 Structure 1.1 至 1.5HL

PART I  ·  PAPER 1第一部分  ·  第一卷Multiple Choice: Worked Answers选择题：详细解析

Multiple Choice选择题

Q1MEDIUMPaper 11.1 Kinetic Theory

Ice at −10°C heated at constant pressure to steam at 110°C: kinetic energy and separation?恒压下冰从 −10°C 加热到 110°C 蒸汽：动能与间距如何变化？

Answer:答案： (B)
Kinetic energy tracks temperature. It rises while the ice warms (−10°C → 0°C) and while the steam warms (100°C → 110°C), but during melting and boiling the temperature is fixed, so all added energy goes into overcoming intermolecular forces rather than raising $E_k$. Particle separation changes little on melting (solid → liquid particles are still close together) but increases enormously on boiling (liquid → gas particles become widely spaced), so the largest jump in separation is during boiling.动能与温度同步变化：冰升温（−10°C → 0°C）与蒸汽升温（100°C → 110°C）阶段动能增大；但熔化与沸腾期间温度不变，吸收的能量全部用于克服分子间作用力，而非提高 $E_k$。粒子间距在熔化时变化很小（固 → 液粒子仍彼此靠近），但在沸腾时急剧增大（液 → 气粒子间距变得很大），因此间距增大最多的阶段是沸腾。
Insight洞见 Heating-curve questions are really testing one idea: temperature (and hence $E_k$) is flat exactly where the particle arrangement is changing. The mark-losing trap is assuming a smooth, continuously rising temperature graph. Sketch or picture the plateaus at the melting and boiling points before answering.加热曲线题本质只考一个概念：温度（进而 $E_k$）恰好在粒子排列发生变化时保持不变。常见丢分陷阱是默认温度图是平滑连续上升的。作答前先在脑中画出熔点与沸点处的平台。
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Q2MEDIUMPaper 11.2 Isotope Notation

Composition of $_{35}^{80}\mathrm{Br}^-$?$_{35}^{80}\mathrm{Br}^-$ 的组成？

Answer:答案： (B)
Protons = $Z$ = 35 (the bottom-left number). Neutrons = $A - Z$ = $80 - 35 = 45$. The $-1$ charge means the ion has one more electron than a neutral atom: $35 + 1 = 36$ electrons. Trap (C) confuses the mass number with the neutron count directly; trap (A) subtracts an electron instead of adding one.质子数 = $Z$ = 35（左下角数字）。中子数 = $A - Z$ = $80 - 35 = 45$。$-1$ 电荷表示该离子比中性原子多一个电子：$35 + 1 = 36$ 个电子。陷阱 (C) 把质量数直接当作中子数；陷阱 (A) 是减去而不是加上一个电子。
Insight洞见 The nuclear-symbol triangle never changes: protons $= Z$, neutrons $= A - Z$, electrons $= Z \mp$ (charge). For anions, electrons $> Z$; for cations, electrons $< Z$. Write the three numbers next to the symbol before reading the answer options: doing so removes the sign-error trap entirely.核符号三角关系永远不变：质子数 $= Z$，中子数 $= A - Z$，电子数 $= Z \mp$（电荷）。阴离子电子数 $> Z$；阳离子电子数 $< Z$。作答前先在核符号旁写出这三个数字，就能彻底避开符号陷阱。
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Q3MEDIUMPaper 11.3 Electron Configuration

Which species is [Ar] 3d$^{10}$ 4s$^1$?哪个物种是 [Ar] 3d$^{10}$ 4s$^1$？

Answer:答案： (A)
[Ar] 3d$^{10}$ 4s$^1$ has $18 + 10 + 1 = 29$ electrons, matching neutral Cu. Cu is the required exception where a 4s electron is promoted to complete a stable, fully filled 3d$^{10}$ sublevel (expected [Ar] 3d$^9$ 4s$^2$ is not what forms). Cu$^+$ (28 electrons) loses the 4s electron entirely to give [Ar] 3d$^{10}$. Zn (30 electrons, no exception) is [Ar] 3d$^{10}$ 4s$^2$. Cr (24 electrons) is [Ar] 3d$^5$ 4s$^1$, the other required exception.[Ar] 3d$^{10}$ 4s$^1$ 共有 $18 + 10 + 1 = 29$ 个电子，与中性 Cu 相符。Cu 是必考例外：一个 4s 电子被提升，以形成稳定、全充满的 3d$^{10}$ 亚层（并非预期的 [Ar] 3d$^9$ 4s$^2$）。Cu$^+$（28 个电子）完全失去 4s 电子，得到 [Ar] 3d$^{10}$。Zn（30 个电子，无例外）为 [Ar] 3d$^{10}$ 4s$^2$。Cr（24 个电子）为 [Ar] 3d$^5$ 4s$^1$，是另一个必考例外。
Insight洞见 Only two neutral-atom exceptions exist on the IB syllabus (Cr and Cu), both driven by the extra stability of a half-filled or fully filled d-sublevel. The trap is applying this "promotion" logic to the corresponding ions: once Cu loses an electron to become Cu$^+$, the 4s orbital is simply empty ([Ar] 3d$^{10}$), not "3d$^9$ 4s$^1$", because ions never re-promote electrons back into 4s.IB 大纲中只有两个中性原子例外（Cr 与 Cu），均源于半充满或全充满 d 亚层带来的额外稳定性。陷阱在于把这种"电子提升"的逻辑套用到对应的离子上：Cu 失去一个电子变成 Cu$^+$ 后，4s 轨道直接空出（[Ar] 3d$^{10}$），而不是"3d$^9$ 4s$^1$"，因为离子中电子绝不会被重新提升回 4s。
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Q4MEDIUMPaper 11.4 Mole Calculations

Total atoms in 0.100 mol $\mathrm{Al_2(SO_4)_3}$?0.100 mol $\mathrm{Al_2(SO_4)_3}$ 中的总原子数？

Answer:答案： (C)
One formula unit of $\mathrm{Al_2(SO_4)_3}$ contains $2 + 3 + (4 \times 3) = 2 + 3 + 12 = 17$ atoms. Amount of atoms $= 0.100 \times 17 = 1.70~\mathrm{mol}$.一个 $\mathrm{Al_2(SO_4)_3}$ 式量单位含 $2 + 3 + (4 \times 3) = 2 + 3 + 12 = 17$ 个原子。原子的物质的量 $= 0.100 \times 17 = 1.70~\mathrm{mol}$。
$$N = n \times N_A = 1.70 \times 6.02 \times 10^{23} = 1.02 \times 10^{24}~\text{atoms}$$
Trap (A) uses only the number of formula units ($0.100 \times N_A$), forgetting to multiply by 17.陷阱 (A) 只用了式量单位的数目（$0.100 \times N_A$），忘记乘以 17。
Insight洞见 "Number of particles" questions hide a units trap: read carefully whether the question asks for formula units, molecules, ions, or atoms, since each requires a different multiplier applied to $n \times N_A$. For an ionic formula like $\mathrm{Al_2(SO_4)_3}$, count every atom in every ion (including the 4 oxygens inside each sulfate) before multiplying."粒子数"类题目暗藏单位陷阱：仔细分辨题目问的是式量单位、分子、离子还是原子，因为每种都需要在 $n \times N_A$ 上乘以不同的倍数。对于 $\mathrm{Al_2(SO_4)_3}$ 这类离子化合物，务必先数清每个离子内的每个原子（包括每个硫酸根内的 4 个氧），再进行乘法。
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Q5HARDPaper 11.5 Real Gas Deviation

Gas compressed to high P, low T; measured V larger than ideal predicts. Best explanation?气体在低温下压缩至高压；实测体积大于理想值。最佳解释？

Answer:答案： (B)
The ideal gas model assumes molecules occupy negligible volume. At very high pressure the molecules are forced close together, and their own finite volume becomes a significant fraction of the container volume: there is less truly "empty" space left to compress than the ideal equation assumes, so the real volume measured is larger than $V = nRT/P$ predicts. (A) describes the low-pressure, low-temperature deviation (attractive forces reduce measured pressure, making real $V$ smaller than ideal at moderate conditions), the opposite regime to this question. (C) is not a real-gas assumption in the IB syllabus: real gas collisions are still treated as effectively elastic.理想气体模型假设分子自身体积可忽略。在极高压下分子被迫彼此靠近，其自身有限的体积占容器体积的比例变得不可忽略，真正"空"出来可供压缩的空间比理想方程假设的要少，因此实测体积大于 $V = nRT/P$ 的预测值。(A) 描述的是低压低温下的偏差（吸引力使实测压强偏低，从而在中等条件下使实测 $V$ 小于理想值），与本题所处的高压区间相反。(C) 并非 IB 大纲中真实气体的假设内容：真实气体碰撞仍被视为近似弹性碰撞。
Insight洞见 Real-gas deviation questions test which of the two broken assumptions is dominant: at high pressure, molecular volume dominates (real $V$ > ideal $V$); at low temperature (and moderate pressure), intermolecular attraction dominates (real $V$ < ideal $V$, and pressure reads lower than ideal). Identify the condition in the stem first, then pick the matching assumption: don't default to "attractive forces" for every deviation question.真实气体偏差题考查的是两条被破坏的假设中哪一条占主导：在高压下，分子体积占主导（实测 $V$ > 理想 $V$）；在低温（中等压强）下，分子间吸引力占主导（实测 $V$ < 理想 $V$，压强读数也低于理想值）。先判断题干给出的条件，再选择对应的假设：不要每道偏差题都默认答案是"吸引力"。
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Q6HARDPaper 1HL1.3 Successive IE

IEs 0.90, 1.76, 14.85, 21.0, 25.0, 32.0 MJ mol$^{-1}$. Which group?电离能 0.90、1.76、14.85、21.0、25.0、32.0 MJ mol$^{-1}$。属于哪一族？

Answer:答案： (B)
IE$_1 \to$ IE$_2$: $1.76/0.90 \approx 2.0$, a normal-sized increase. IE$_2 \to$ IE$_3$: $14.85/1.76 \approx 8.4$, an enormous jump. A huge jump between IE$_n$ and IE$_{n+1}$ signals that the $n$th electron was the last one in an outer (valence) shell, and the $(n+1)$th electron must be pulled from a full inner shell held far more tightly. Here the jump occurs after the second electron, so X has 2 valence electrons → Group 2.IE$_1 \to$ IE$_2$：$1.76/0.90 \approx 2.0$，属正常幅度的增大。IE$_2 \to$ IE$_3$：$14.85/1.76 \approx 8.4$，跃升幅度巨大。IE$_n$ 与 IE$_{n+1}$ 之间的巨大跃升表明第 $n$ 个电子是外层（价）壳层的最后一个电子，而第 $(n+1)$ 个电子必须从结合得更牢固的内层满壳层中移除。此处跃升发生在第二个电子之后，说明 X 有 2 个价电子 → 第 2 族。
Insight洞见 Don't scan for the single biggest ratio between adjacent IEs by eye: compute each ratio (or at least compare magnitudes) systematically, because a "large" jump on an unfamiliar scale can be misjudged. The position of the jump (after electron $n$) directly gives the group number for main-group elements: jump after IE$_1$ → Group 1, after IE$_2$ → Group 2, and so on.不要仅凭肉眼寻找相邻电离能之间比值最大的那一处：应系统地计算每个比值（或至少比较数量级），因为在不熟悉的量级下"大"跃升容易被误判。跃升出现的位置（第 $n$ 个电子之后）直接给出主族元素的族数：跃升出现在 IE$_1$ 之后 → 第 1 族，出现在 IE$_2$ 之后 → 第 2 族，依此类推。
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PART II  ·  PAPER 2第二部分  ·  第二卷Structured Response: Worked Solutions结构化解答题：详细解析

Structured Response结构化解答题

SR 1MEDIUMPaper 2HLGallium Mass Spectrum

Mass spectrum of gallium: $m/z = 69$ (60.11%), $m/z = 71$ (39.89%).镓的质谱：$m/z = 69$（60.11%），$m/z = 71$（39.89%）。

(a) Isotopes are atoms of the same element (identical number of protons / atomic number) that have different numbers of neutrons (and hence different mass numbers).同位素是质子数（原子序数）相同、但中子数不同（因而质量数不同）的同一元素的原子。
(b) Weighted average using the abundances as percentages:以丰度百分比为权重求加权平均：
$$A_r = \dfrac{(69 \times 60.11) + (71 \times 39.89)}{100} = \dfrac{4147.59 + 2831.19}{100}$$
$$A_r = \dfrac{6978.78}{100} = 69.79~(\text{to 2 d.p.})$$
(c) Chemical properties are determined by the number and arrangement of electrons (which in turn is set by the number of protons, $Z$). Since $^{69}$Ga and $^{71}$Ga have the same number of protons and electrons, they have identical electron configurations and hence identical chemical behavior. Physical properties such as density depend on the mass of the atoms; because the isotopes differ in neutron number (and hence mass), a sample of $^{71}$Ga is slightly denser than the same number of $^{69}$Ga atoms.化学性质由电子的数目与排布决定（而电子数又由质子数 $Z$ 决定）。由于 $^{69}$Ga 与 $^{71}$Ga 质子数与电子数相同，二者电子构型相同，化学行为也相同。密度等物理性质取决于原子的质量；由于两同位素中子数（因而质量）不同，相同原子数的 $^{71}$Ga 样品密度略高于 $^{69}$Ga。
(d) Ga has $Z = 31$, so a neutral Ga atom has 31 protons and 31 electrons. For the $^{69}$Ga isotope, neutrons $= 69 - 31 = 38$. The 3+ charge on $^{69}\mathrm{Ga^{3+}}$ means 3 electrons have been removed: electrons $= 31 - 3 = 28$.Ga 的 $Z = 31$，故中性 Ga 原子有 31 个质子与 31 个电子。对 $^{69}$Ga 同位素，中子数 $= 69 - 31 = 38$。$^{69}\mathrm{Ga^{3+}}$ 的 3+ 电荷表示失去了 3 个电子：电子数 $= 31 - 3 = 28$。
Protons = 31, neutrons = 38, electrons = 28.质子数 = 31，中子数 = 38，电子数 = 28。
Insight洞见 Mass-spectrum $A_r$ questions and ion-composition questions are graded almost entirely on method: always show the weighted-average setup explicitly (numerator with both terms, divide by 100) rather than jumping to a decimal, and always state protons/neutrons/electrons as three separate labelled numbers. A bare final number without labels often loses the method mark even if correct.质谱 $A_r$ 计算题与离子组成题几乎完全按方法给分：务必明确写出加权平均的计算式（分子含两项，除以 100），而不是直接跳到小数结果；并且务必将质子数/中子数/电子数分别标注为三个独立数字。只写一个无标注的最终数字，即便结果正确，往往也会丢失方法分。
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SR 2MEDIUMPaper 2HLVanadium Electron Config + IE

Vanadium, V (Z = 23): configuration, spectra, and ionization energy.钒 V（Z = 23）：构型、光谱与电离能。

(a) Full configuration (23 electrons):完整构型（23 个电子）：
1s$^2$ 2s$^2$ 2p$^6$ 3s$^2$ 3p$^6$ 4s$^2$ 3d$^3$
(b) Transition metals lose 4s electrons before 3d electrons when forming cations (in ions, 3d lies lower in energy than 4s). V$^{3+}$ has $23 - 3 = 20$ electrons: both 4s electrons are lost first, then one 3d electron.过渡金属形成阳离子时先失去 4s 电子，再失去 3d 电子（在离子中 3d 能量低于 4s）。V$^{3+}$ 有 $23 - 3 = 20$ 个电子：先失去两个 4s 电子，再失去一个 3d 电子。
V$^{3+}$: [Ar] 3d$^2$V$^{3+}$：[Ar] 3d$^2$
(c) Any two of: (i) the spectrum consists of sharp, separated lines rather than a smooth continuous band, showing that only certain specific photon energies (and hence only certain energy gaps) are emitted; (ii) each line corresponds to one specific transition between two discrete energy levels, so the finite, countable set of lines implies a finite, countable set of levels; (iii) the lines converge (get closer together) at higher energy, consistent with energy levels becoming more closely spaced as $n$ increases, a feature a continuous-energy model could not explain.以下任写两点：(i) 光谱由分立的清晰谱线组成，而非连续光带，说明只发射特定的光子能量（即特定的能级差）；(ii) 每条谱线对应两个离散能级间的一次特定跃迁，谱线数目有限可数，意味着能级数目也有限可数；(iii) 谱线在高能端逐渐汇聚（间距变小），与能级随 $n$ 增大而愈发靠近的特征一致，这是连续能量模型无法解释的现象。
(d) $IE = hf \times N_A$ (energy per photon $\times$ Avogadro constant, converting per-atom to per-mole):$IE = hf \times N_A$（单光子能量 $\times$ 阿伏伽德罗常数，将单原子能量换算为每摩尔）：
$$IE = (6.63 \times 10^{-34})(5.47 \times 10^{15})(6.02 \times 10^{23})$$
$$IE = (3.627 \times 10^{-18})(6.02 \times 10^{23}) = 2.18 \times 10^{6}~\mathrm{J\,mol^{-1}} = 2180~\mathrm{kJ\,mol^{-1}}$$
(e) After each electron is removed, the remaining electrons are attracted by the same nuclear charge but shared among fewer electrons, so the effective nuclear charge experienced by each remaining electron increases. Each subsequent electron is held more tightly and takes more energy to remove.每移除一个电子后，剩余电子仍受相同核电荷吸引，但共享该电荷的电子数减少，因此每个剩余电子感受到的有效核电荷增大。后续每个电子结合得更紧，需要更多能量才能移除。
Insight洞见 $IE = hf$ gives the energy to ionize one atom; multiplying by $N_A$ converts to a molar quantity in J mol$^{-1}$, which then needs $\div 1000$ to reach the conventional kJ mol$^{-1}$ reporting unit. Forgetting either the $\times N_A$ step or the final unit conversion is the single most common mark loss on this HL calculation. Write out both conversions explicitly rather than doing them mentally.$IE = hf$ 给出的是电离单个原子所需的能量；乘以 $N_A$ 才换算为摩尔量（单位 J mol$^{-1}$），随后还需 $\div 1000$ 才能得到常规使用的 kJ mol$^{-1}$ 单位。忘记乘以 $N_A$ 或忘记最后的单位换算，是这道 HL 计算题中最常见的丢分点。务必把两步换算都明确写出，而不是心算跳过。
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SR 3HARDPaper 2Formulas + Concentration + Avogadro

C: 54.5%, H: 9.1%, O: 36.4% by mass; $M = 88.0~\mathrm{g\,mol^{-1}}$.按质量计 C: 54.5%、H: 9.1%、O: 36.4%；$M = 88.0~\mathrm{g\,mol^{-1}}$。

(a) Assume 100 g; convert each mass to moles, then divide by the smallest:设总质量为 100 g；将各质量换算为摩尔数，再除以最小者：
$$n_C = \dfrac{54.5}{12.01} = 4.538 \quad n_H = \dfrac{9.1}{1.01} = 9.010 \quad n_O = \dfrac{36.4}{16.00} = 2.275$$
$$C: \dfrac{4.538}{2.275} \approx 2 \quad H: \dfrac{9.010}{2.275} \approx 4 \quad O: \dfrac{2.275}{2.275} = 1$$
Empirical formula: $\mathrm{C_2H_4O}$.实验式：$\mathrm{C_2H_4O}$。
(b) Empirical formula mass $= (2 \times 12.01) + (4 \times 1.01) + 16.00 = 44.06~\mathrm{g\,mol^{-1}}$.实验式式量 $= (2 \times 12.01) + (4 \times 1.01) + 16.00 = 44.06~\mathrm{g\,mol^{-1}}$。
$$\text{Multiplier} = \dfrac{88.0}{44.06} \approx 2.0$$
Molecular formula: $\mathrm{C_4H_8O_2}$.分子式：$\mathrm{C_4H_8O_2}$。
(c) $n = m/M = 2.20 / 88.0 = 0.0250~\mathrm{mol}$. Volume $= 250~\mathrm{cm^3} = 0.250~\mathrm{dm^3}$.$n = m/M = 2.20 / 88.0 = 0.0250~\mathrm{mol}$。体积 $= 250~\mathrm{cm^3} = 0.250~\mathrm{dm^3}$。
$$C = \dfrac{n}{V} = \dfrac{0.0250}{0.250} = 0.100~\mathrm{mol\,dm^{-3}}$$
(d) Using $C_1V_1 = C_2V_2$:利用 $C_1V_1 = C_2V_2$：
$$C_2 = \dfrac{C_1 V_1}{V_2} = \dfrac{0.100 \times 25.0}{100.0} = 0.0250~\mathrm{mol\,dm^{-3}}$$
(e) Balancing the combustion of $\mathrm{C_4H_8O_2}$: 4 carbons need 4 $\mathrm{CO_2}$; 8 hydrogens need 4 $\mathrm{H_2O}$. Right-hand oxygens $= (4 \times 2) + (4 \times 1) = 12$; the compound itself supplies 2, so $10$ more O atoms (5 $\mathrm{O_2}$) are needed:配平 $\mathrm{C_4H_8O_2}$ 的燃烧：4 个碳需要 4 个 $\mathrm{CO_2}$；8 个氢需要 4 个 $\mathrm{H_2O}$。右侧氧原子数 $= (4 \times 2) + (4 \times 1) = 12$；化合物自身提供 2 个，故还需 10 个 O 原子（即 5 个 $\mathrm{O_2}$）：
$$\mathrm{C_4H_8O_2(g) + 5O_2(g) \to 4CO_2(g) + 4H_2O(g)}$$
By Avogadro's law, equal volumes of gas at the same temperature and pressure contain equal numbers of moles, so the volume ratio equals the mole ratio directly: compound vapor : $\mathrm{CO_2}$ = 1 : 4.由阿伏伽德罗定律，同温同压下等体积气体含有相同的摩尔数，故体积比直接等于摩尔比：化合物蒸气 : $\mathrm{CO_2}$ = 1 : 4。
Insight洞见 A four-part mole "chain" question like this is graded step-by-step, and each part's answer feeds the next. An early rounding choice (here, rounding the C:H:O ratio to 2:4:1 rather than carrying 1.99:3.96:1.00) is expected and correct, but only if you show the un-rounded numbers first. The trap in part (e) is balancing the equation from the empirical formula $\mathrm{CH_2O}$ instead of the true molecular formula $\mathrm{C_4H_8O_2}$: always carry the molecular formula forward once you have it.像本题这样的四步"摩尔链"题按步骤逐一给分，且前一小题的答案会用于下一小题。把 C:H:O 比值从 1.99:3.96:1.00 四舍五入为 2:4:1 是预期且正确的做法，但前提是先展示未取整的数值。(e) 部分的陷阱是用实验式 $\mathrm{CH_2O}$ 而非真实分子式 $\mathrm{C_4H_8O_2}$ 来配平方程：一旦求得分子式，后续必须沿用它。
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SR 4HARDPaper 2Ideal Gas + Real Gas Deviation

0.176 g of liquid X vaporized: 62.2 cm³ at 373 K, 100 kPa.0.176 g 液体 X 汽化：373 K、100 kPa 下体积为 62.2 cm³。

(a) Convert to SI units: $P = 100~\mathrm{kPa} = 100{,}000~\mathrm{Pa}$, $V = 62.2~\mathrm{cm^3} = 6.22 \times 10^{-5}~\mathrm{m^3}$, $T = 373~\mathrm{K}$.换算为 SI 单位：$P = 100~\mathrm{kPa} = 100{,}000~\mathrm{Pa}$，$V = 62.2~\mathrm{cm^3} = 6.22 \times 10^{-5}~\mathrm{m^3}$，$T = 373~\mathrm{K}$。
$$n = \dfrac{PV}{RT} = \dfrac{(100{,}000)(6.22 \times 10^{-5})}{(8.31)(373)} = 2.01 \times 10^{-3}~\mathrm{mol}$$
(b) Molar mass from $M = m/n$:由 $M = m/n$ 求摩尔质量：
$$M = \dfrac{0.176}{2.01 \times 10^{-3}} = 87.7~\mathrm{g\,mol^{-1}}$$
(c) Pressure is constant, so the combined gas law reduces to Charles's law:压强不变，气体联合定律简化为查理定律：
$$\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \;\Rightarrow\; V_2 = V_1 \times \dfrac{T_2}{T_1} = 62.2 \times \dfrac{298}{373} = 49.7~\mathrm{cm^3}$$
(d) At the lower temperature (298 K), the particles have less average kinetic energy, so the hydrogen-bonding intermolecular forces between X molecules become significant relative to that kinetic energy: the ideal-gas assumption of "no intermolecular forces" breaks down. In fact, X may partially (or fully) condense back to liquid before reaching 298 K if this temperature is close to or below its boiling point, meaning the calculated gas volume in (c) would never actually be observed.在较低温度（298 K）下，粒子的平均动能减小，X 分子间的氢键作用相对该动能变得显著，理想气体"无分子间作用力"的假设失效。事实上，若 298 K 接近或低于 X 的沸点，X 可能在降温过程中部分（甚至全部）重新凝结为液体，这意味着 (c) 中计算出的气体体积实际上根本不会被观测到。
(e) Description of the sketch: the ideal-gas line is a horizontal reference at the ideal molar volume for all pressures. The real curve for X starts close to the ideal line at low pressure, dips below it at low-to-moderate pressure (attractive hydrogen-bonding forces pull molecules together, giving a smaller real volume than ideal), reaches a minimum, then rises steeply above the ideal line at very high pressure (the molecules' own finite volume dominates, leaving less compressible space than ideal). The two breaking assumptions are: (i) "no intermolecular forces", violated at low-to-moderate pressure, causing the dip below ideal; (ii) "negligible molecular volume", violated at high pressure, causing the rise above ideal.图像描述：理想气体线是所有压强下摩尔体积的水平参考线。X 的真实曲线在低压时接近理想线，在中低压时低于理想线（吸引性氢键使分子相互靠拢，真实体积小于理想值），到达最小值后，在极高压时又急剧高于理想线（分子自身有限体积占主导，可压缩空间比理想情形少）。被破坏的两条假设为：(i) "无分子间作用力"，在中低压时失效，导致曲线低于理想线；(ii) "分子体积可忽略"，在高压时失效，导致曲线高于理想线。
Insight洞见 Ideal-gas "determine $M$" problems (parts a-b) are a two-step chain: $n$ from $PV = nRT$, then $M = m/n$. The most common arithmetic slip is forgetting to convert cm$^3$ to m$^3$ ($\times 10^{-6}$) or kPa to Pa ($\times 1000$) before substituting: the equation only balances in SI units. In part (e), resist the urge to describe real gases as uniformly "bigger than ideal" or "smaller than ideal": the direction of deviation flips with pressure, and a fully correct sketch must show both the dip and the rise.理想气体"求 $M$"类题目（a-b 部分）是两步链条：先由 $PV = nRT$ 求 $n$，再由 $M = m/n$ 求摩尔质量。最常见的运算失误是代入前忘记将 cm$^3$ 换算为 m$^3$（$\times 10^{-6}$）或将 kPa 换算为 Pa（$\times 1000$），该方程只有在 SI 单位下才成立。在 (e) 部分，切勿笼统地把真实气体描述为"始终大于理想值"或"始终小于理想值"：偏差方向会随压强反转，完整正确的图像必须同时体现先降后升的形态。
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PART III  ·  PAPER 3 HL第三部分  ·  第三卷 HLData-Based: Worked Solutions数据题：详细解析

Data-Based Questions (HL)数据题（HL）

P3-1HARDPaper 3 HLHLRubidium Mass Spectrum + Reactivity

Rb mass spectrum: $m/z = 85$ (72.17%), $m/z = 87$ (27.83%). $\mathrm{2Rb(s) + 2H_2O(l) \to 2RbOH(aq) + H_2(g)}$.Rb 质谱：$m/z = 85$（72.17%），$m/z = 87$（27.83%）。$\mathrm{2Rb(s) + 2H_2O(l) \to 2RbOH(aq) + H_2(g)}$。

(a) Weighted average:加权平均：
$$A_r = \dfrac{(85 \times 72.17) + (87 \times 27.83)}{100} = \dfrac{6134.45 + 2421.21}{100} = \dfrac{8555.66}{100} = 85.56~(\text{to 2 d.p.})$$
(b) Using $A_r = 85.56~\mathrm{g\,mol^{-1}}$ as the molar mass:以 $A_r = 85.56~\mathrm{g\,mol^{-1}}$ 作为摩尔质量：
$$n(\mathrm{Rb}) = \dfrac{m}{M} = \dfrac{5.00}{85.56} = 0.0584~\mathrm{mol}$$
$$N(\mathrm{Rb}) = n \times N_A = 0.0584 \times 6.02 \times 10^{23} = 3.52 \times 10^{22}~\text{atoms}$$
(c) The equation shows $2\mathrm{Rb} : 1\mathrm{H_2}$, so $n(\mathrm{H_2}) = \tfrac{1}{2}\,n(\mathrm{Rb}) = \tfrac{1}{2}(0.0584) = 0.0292~\mathrm{mol}$. Apply $PV = nRT$ with $P = 100{,}000~\mathrm{Pa}$, $T = 298~\mathrm{K}$:方程给出 $2\mathrm{Rb} : 1\mathrm{H_2}$，故 $n(\mathrm{H_2}) = \tfrac{1}{2}\,n(\mathrm{Rb}) = \tfrac{1}{2}(0.0584) = 0.0292~\mathrm{mol}$。代入 $PV = nRT$，取 $P = 100{,}000~\mathrm{Pa}$，$T = 298~\mathrm{K}$：
$$V = \dfrac{nRT}{P} = \dfrac{(0.0292)(8.31)(298)}{100{,}000} = 7.24 \times 10^{-4}~\mathrm{m^3}$$
$$V = 724~\mathrm{cm^3}$$
(d) Full configuration of Rb (37 electrons):Rb 的完整构型（37 个电子）：
1s$^2$ 2s$^2$ 2p$^6$ 3s$^2$ 3p$^6$ 4s$^2$ 3d$^{10}$ 4p$^6$ 5s$^1$
Rb$^+$ (36 electrons) is formed by removing the single 5s electron:Rb$^+$（36 个电子）由移除唯一的 5s 电子形成：
1s$^2$ 2s$^2$ 2p$^6$ 3s$^2$ 3p$^6$ 3d$^{10}$ 4s$^2$ 4p$^6$  (same as Kr与 Kr 相同)
The single 5s electron is in a shell of its own, outside the complete, noble-gas-like $n=4$ set of sublevels. Removing it gives a full and highly stable octet/noble-gas configuration at low energy cost; removing a second electron would mean breaking into the full 4p sublevel, which requires vastly more energy, so Rb reliably forms only the 1+ ion.这个唯一的 5s 电子独占一个壳层，处于完整、类惰性气体的 $n=4$ 亚层组之外。移除它只需较低能量即可获得完整而高度稳定的八隅体/惰性气体构型；若要移除第二个电子，则需打破全充满的 4p 亚层，所需能量急剧增大，因此 Rb 稳定地只形成 1+ 离子。
(e) IE$_1$ removes the lone 5s electron: it is the outermost electron, well shielded by the full $n = 1$ to $n = 4$ core, so it is removed relatively easily. IE$_2$ must remove an electron from the full 4p sublevel, a "core" shell much closer to the nucleus and far less shielded, requiring dramatically more energy. The huge IE$_1$/IE$_2$ jump therefore confirms that rubidium has exactly one electron in its outermost (5th) shell, separated by a large energy gap from a stable 8-electron ($n=4$) core, consistent with its position in Group 1.IE$_1$ 移除的是唯一的 5s 电子：它是最外层电子，被完整的 $n = 1$ 到 $n = 4$ 内核很好地屏蔽，因此相对容易移除。IE$_2$ 则必须从全充满的 4p 亚层（一个更靠近原子核、屏蔽程度低得多的"内核"壳层）移除电子，所需能量急剧增大。IE$_1$/IE$_2$ 之间的巨大跃升由此证实：铷在其最外层（第 5 层）恰有一个电子，且与稳定的 8 电子（$n=4$）内核之间存在很大的能量间隔，这与铷位于第 1 族的事实一致。
Insight洞见 This question chains four Structure 1 ideas that IB Paper 3 loves to combine: mass spectrum → $A_r$ → moles → stoichiometric gas volume → electron configuration → ionization energy. The single biggest trap across all of it is a stoichiometry slip in part (c): the equation is $2\mathrm{Rb} : 1\mathrm{H_2}$, not 1:1, so halving (not copying) the moles of Rb is essential before applying $PV = nRT$. Always re-read the coefficients from the given equation rather than assuming a 1:1 ratio.本题串联了 IB Paper 3 最喜欢组合考查的四个 Structure 1 概念：质谱 → $A_r$ → 摩尔数 → 化学计量气体体积 → 电子构型 → 电离能。整题最大的陷阱出现在 (c) 部分的化学计量：方程给出的是 $2\mathrm{Rb} : 1\mathrm{H_2}$，而非 1:1，因此在代入 $PV = nRT$ 之前，必须将 Rb 的摩尔数减半（而非直接照搬）。务必重新核对所给方程中的系数，而不是默认为 1:1。
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P3-2HARDPaper 3 HLHLMolar Mass Determination + Error Analysis

Vapor density method for liquid Y at 372 K, 101 kPa; three trials of volume and mass.在 372 K、101 kPa 下用蒸气密度法测定液体 Y；三次实验记录体积与质量。

(a) Mean volume $= (57.8 + 58.0 + 58.4)/3 = 58.1~\mathrm{cm^3}$. Mean mass $= (0.0865 + 0.0870 + 0.0872)/3 = 0.0869~\mathrm{g}$.平均体积 $= (57.8 + 58.0 + 58.4)/3 = 58.1~\mathrm{cm^3}$。平均质量 $= (0.0865 + 0.0870 + 0.0872)/3 = 0.0869~\mathrm{g}$。
(b) Convert to SI: $V = 58.1~\mathrm{cm^3} = 5.81 \times 10^{-5}~\mathrm{m^3}$, $P = 101{,}000~\mathrm{Pa}$, $T = 372~\mathrm{K}$.换算为 SI：$V = 58.1~\mathrm{cm^3} = 5.81 \times 10^{-5}~\mathrm{m^3}$，$P = 101{,}000~\mathrm{Pa}$，$T = 372~\mathrm{K}$。
$$n = \dfrac{PV}{RT} = \dfrac{(101{,}000)(5.81 \times 10^{-5})}{(8.31)(372)} = 1.90 \times 10^{-3}~\mathrm{mol}$$
$$M = \dfrac{m}{n} = \dfrac{0.0869}{1.90 \times 10^{-3}} = 45.8~\mathrm{g\,mol^{-1}}$$
(c) Percentage error against the accepted value:相对公认值的百分误差：
$$\%~\text{error} = \dfrac{|46.07 - 45.8|}{46.07} \times 100\% = 0.57\%$$
(d) Two systematic errors that push $M$ too high, since $M = m/n$ and $n = PV/RT$ (so anything that makes the calculated $n$ too small makes $M$ too large):由于 $M = m/n$、$n = PV/RT$（因此任何使计算出的 $n$ 偏小的因素都会使 $M$ 偏大），以下两个系统误差会使 $M$ 系统性偏高：
Incomplete vaporization: if the recorded volume is read before every drop of liquid Y has vaporized, the gas-phase moles corresponding to that volume are less than the true moles implied by the full injected mass. Since the mass used in the calculation is the mass of all the liquid injected (not just the vaporized fraction), $n$ calculated from $V$ is too small relative to the mass used, so $M = m/n$ comes out too high.汽化不完全：若在液体 Y 尚未完全汽化前就读取体积，该体积对应的气相摩尔数会小于注入总质量所对应的真实摩尔数。由于计算中使用的质量是全部注入液体的质量（而非仅已汽化的部分），由 $V$ 算出的 $n$ 相对所用质量偏小，导致 $M = m/n$ 偏高。
Heat loss to the syringe barrel: if the gas in the syringe is actually slightly cooler than the recorded bath temperature $T$ (e.g. heat lost through the barrel walls), using the higher recorded $T$ in $n = PV/RT$ divides by a value larger than the true temperature, giving a calculated $n$ that is too small, and hence $M$ too high.热量散失到注射器筒壁：若注射器中气体的实际温度略低于记录的水浴温度 $T$（如热量经筒壁散失），在 $n = PV/RT$ 中使用偏高的记录温度会导致除以一个大于真实温度的值，从而使算出的 $n$ 偏小，$M$ 因而偏高。
(e) Use a balance with a finer resolution (e.g. reading to $\pm 0.0001~\mathrm{g}$ instead of $\pm 0.001~\mathrm{g}$) and/or repeat the mass measurement more times and average. The mass difference measured here ($\approx 0.087~\mathrm{g}$) is small, so the balance's reading uncertainty is a large fraction of the measured value; a finer balance directly shrinks the random scatter seen across the three trials.使用精度更高的天平（如读数精确到 $\pm 0.0001~\mathrm{g}$ 而非 $\pm 0.001~\mathrm{g}$），和/或增加质量测量的重复次数并取平均。此处测得的质量差（约 $0.087~\mathrm{g}$）很小，天平的读数不确定度占测量值的比例较大；使用精度更高的天平可直接减小三次实验之间的随机波动。
(f) The ideal gas equation only applies to a sample that is entirely in the gas phase. If the temperature is too close to (or below) the boiling point of Y, some of the sample may remain as liquid inside the syringe. That liquid still contributes to the measured mass but occupies negligible volume and contributes essentially zero moles of gas, so $n$ calculated from $V$ would badly underestimate the true amount corresponding to the mass used: a large, uncontrolled systematic error, not a small one. Heating to just above the boiling point ensures complete vaporization, so every part of the measured mass is genuinely present as gas obeying (approximately) $PV = nRT$, leaving only the smaller, well-understood real-gas deviation as a source of error.理想气体方程只适用于完全处于气相的样品。若温度过于接近（或低于）Y 的沸点，样品中一部分可能仍以液态残留在注射器内。这部分液体仍计入测得的质量，却几乎不占体积、对气体摩尔数的贡献接近零，因此由 $V$ 算出的 $n$ 会大幅低估与所用质量对应的真实物质的量：这是一个巨大且难以控制的系统误差，而非小误差。加热至刚高于沸点可确保完全汽化，使测得质量的每一部分都确实以气体形式存在、（近似）遵循 $PV = nRT$，从而只剩下较小、可预期的真实气体偏差作为误差来源。
Insight洞见 Systematic-error questions are graded on direction, not just naming a plausible error: always trace the error back through $M = m/n$ and $n = PV/RT$ to state explicitly whether $M$ comes out too high or too low, and why. A vague "the balance might be inaccurate" earns little; "if $V$ is under-read, $n$ is under-calculated, so $M = m/n$ is over-calculated" earns the mark. This experiment (the classical vapor-density / Dumas-type method) is a recurring IB Paper 3 context precisely because it links the ideal gas law to experimental error analysis in one coherent story.系统误差类题目的评分点在于方向，而不只是说出一个可信的误差来源：务必沿着 $M = m/n$ 与 $n = PV/RT$ 追溯误差，明确说明 $M$ 会偏高还是偏低，并说明原因。含糊地说"天平可能不准确"得分很少；而"若 $V$ 读数偏低，则 $n$ 被低估，因而 $M = m/n$ 被高估"才能得分。这类实验（经典的蒸气密度法/杜马法）之所以在 IB Paper 3 中反复出现，正是因为它把理想气体定律与实验误差分析串成了一条连贯的逻辑链。
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