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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。
Why the other options attract.其他选项的诱因。 (A) and (C) are one mistake pointing in opposite directions: each takes a single leg of the heating curve and applies it to the whole of it. (A) generalises the warming legs, where the temperature really is climbing, across the melting and boiling plateaus; (C) generalises the plateaus across the warming legs. Neither answer tracks temperature, which is what mean kinetic energy follows, so neither notices that the curve does both things in turn. (D) has the energy story right and the spacing story wrong: melting loosens a rigid lattice into a liquid whose particles are still in contact, whereas boiling has to pull them apart completely, so the large jump in separation sits at the boiling plateau and not the melting one.(A) 与 (C) 其实是同一个错误的正反两面：各自把加热曲线的某一段套用到全过程。(A) 把升温段（温度确实在上升）的规律推广到熔化与沸腾平台；(C) 则把平台的规律推广到升温段。两者都没有以温度为线索——平均动能跟随的正是温度——因此都没看出曲线是两种行为交替出现的。(D) 能量部分答对了，间距部分答错了：熔化只是把刚性晶格松动为粒子仍相互接触的液体，而沸腾必须把粒子彻底分开，因此间距的大幅跃升位于沸腾平台，而非熔化平台。
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。
Why the other options attract.其他选项的诱因。 (A) has the nucleus exactly right and then moves the electron the wrong way: a $1-$ charge is one electron gained, not one lost, so the count is 36 and not 34. (C) keeps the charge reasoning but never performs the subtraction — 80 is the mass number, protons and neutrons together, so the neutrons are $80 - 35 = 45$. (D) reads the symbol upside down, taking the upper number as protons and the lower as neutrons, then repeats (A)'s sign error on top of that. Every number in (D) is wrong at once, which is exactly why it looks internally consistent.(A) 的原子核部分完全正确，只是电子方向弄反了：$1-$ 电荷表示得到一个电子而不是失去一个，因此电子数是 36 而不是 34。(C) 的电荷推理是对的，却始终没做那步减法——80 是质量数，即质子与中子之和，所以中子数为 $80 - 35 = 45$。(D) 把核符号上下读反了，把上方数字当作质子数、下方数字当作中子数，并在此基础上又重复了 (A) 的符号错误。(D) 中每一个数字都同时错了，这恰恰是它看起来自洽的原因。
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。
Why the other options attract.其他选项的诱因。 All three are recognised by pattern rather than counted. (B) Cu$^+$ carries the right d-sublevel but 28 electrons, not 29: forming the cation removes the 4s electron outright, and an ion does not promote it back. (C) Zn is reached by remembering zinc as the 3d$^{10}$ element and stopping there; zinc has 30 electrons and keeps 4s$^2$. (D) Cr matches the other half of the exception pattern — the lone 4s$^1$ — without checking the d exponent, which is 3d$^5$. Counting the electrons in the configuration as given, $18 + 10 + 1 = 29$, settles all three at once.这三个选项都是靠“认模式”而不是靠数电子得到的。(B) Cu$^+$ 的 d 亚层确实对，但它有 28 个电子而不是 29 个：形成阳离子时 4s 电子被直接移走，离子不会把它再“提升”回来。(C) Zn 来自“锌就是那个 3d$^{10}$ 元素”这一印象并就此打住；锌有 30 个电子，且保留 4s$^2$。(D) Cr 匹配的是例外模式的另一半——孤零零的 4s$^1$——却没有核对 d 的指数，铬是 3d$^5$。把题给构型的电子数直接数一遍，$18 + 10 + 1 = 29$，三个选项一次性都能排除。
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。
Why the other options attract.其他选项的诱因。 (A) stops one step early: $0.100 \times N_A$ counts formula units, and the question asks for atoms, so the 17 atoms inside each formula unit are never applied. (B) does apply the 17 and then mishandles the notation: $1.70 \times 6.02 \times 10^{23}$ is $10.2 \times 10^{23}$, and normalising $10.2$ to $1.02$ has to raise the exponent to $10^{24}$. Writing $1.02 \times 10^{23}$ keeps the tidied mantissa and drops the power of ten that paid for it.(A) 早了一步就停下：$0.100 \times N_A$ 数的是式量单位，而题目问的是原子数，因此每个式量单位内的 17 个原子始终没有被乘进去。(B) 确实乘了 17，却在科学记数法上出了错：$1.70 \times 6.02 \times 10^{23}$ 等于 $10.2 \times 10^{23}$，而把 $10.2$ 规范为 $1.02$ 必须同时把指数提高到 $10^{24}$。写成 $1.02 \times 10^{23}$ 等于保留了整理后的尾数，却丢掉了为此付出的那个十的幂。
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。
Why the other options attract.其他选项的诱因。 (A) picks the right menu of real-gas causes and the wrong item from it. Intermolecular attraction is what makes a real gas occupy less volume than predicted; what is reported here is a larger volume under high pressure, which is the molecules' own finite size. Reading “low temperature” in the stem and reaching for attraction skips the step of checking which deviation was actually measured. (D) answers a different question from the one asked: the stem holds the pressure fixed and compares volumes, while (D) makes a claim about pressure at fixed volume — and reverses the direction of the deviation on the way.(A) 选对了真实气体偏差的“菜单”，却点错了菜。分子间吸引力使真实气体占据的体积小于预测值；而本题报告的是高压下体积大于预测值，其成因是分子自身的有限体积。看到题干里的“低温”便伸手去拿吸引力，跳过了“实际测到的是哪一种偏差”这一步核对。(D) 回答的并不是题目所问：题干固定压强、比较体积，而 (D) 谈的是固定体积下的压强——并且在转换过程中把偏差方向也弄反了。
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。
Why the other options attract.其他选项的诱因。 All three miss where the discontinuity is and count something else instead. (A) treats the ordinary rise from the first to the second ionization energy as the break; every element shows that rise, because each electron is pulled from an increasingly positive ion. (C) counts the large third value itself as a third valence electron, when what the jump reports is how many electrons came off before it. (D) reads the later high values as further valence removals, but once the shell is broken every subsequent value is large, so they carry no extra information. The ratios are what separate the two kinds of increase: $1.76/0.90 \approx 2$ is routine, $14.85/1.76 \approx 8$ is not.这三个选项都找错了突变的位置，转而去数了别的东西。(A) 把第一到第二电离能之间的正常增幅当成了突变；任何元素都会出现这一增幅，因为每一个电子都是从正电荷更高的离子中移出的。(C) 把第三个数值本身算作第三个价电子，而突变真正指示的是在它之前已经移走了多少个电子。(D) 把后面几个高数值读作更多的价电子移除，但壳层一旦被突破，其后每一个数值都很大，因而不再携带额外信息。区分这两类增幅要看比值：$1.76/0.90 \approx 2$ 是常规的，$14.85/1.76 \approx 8$ 不是。
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。
Where this goes wrong.错在哪一步。 Part (b) is the one that loses the mark silently. A script that writes $(69 \times 60.11) + (71 \times 39.89) = 6978.78$ and reports that number as $A_r$ has done every step correctly except the last one: the two terms were weighted by percentages, which sum to 100 and not 1, so the sum still needs dividing by 100. $6978.78$ looks plausible as a relative atomic mass — it is not obviously the wrong order of magnitude — which is exactly why the missing division survives a quick sanity check. Part (c) invites a different slip: reasoning that because the two isotopes have different masses, they must have at least slightly different chemical behaviour. Chemical behaviour is set by electron configuration, which is identical for both isotopes of the same element; it is physical properties, sensitive to mass, that differ.第 (b) 问失分往往悄无声息。写出 $(69 \times 60.11) + (71 \times 39.89) = 6978.78$ 并把这个数直接当作 $A_r$ 报出，前面每一步都做对了，只漏了最后一步：两项是用百分比加权的，而百分比之和是 100 而不是 1，因此这个和还需要除以 100。$6978.78$ 看起来完全像一个合理的相对原子质量——数量级并不明显离谱——这正是这处遗漏能躲过粗略检查的原因。第 (c) 问引出的是另一种误区：因为两种同位素质量不同，便推断它们的化学性质至少会有细微差异。化学性质由电子构型决定，同一元素的两种同位素电子构型完全相同；会因质量而不同的是物理性质。
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。
Where this goes wrong.错在哪一步。 Part (b) is the classic reversed-order trap. Vanadium's ground-state configuration is written 4s$^2$ 3d$^3$ by the building-up order, and a script that removes electrons in that same written order takes all three 3d electrons first and one 4s electron second, landing on V$^{3+}$: [Ar] 4s$^1$ instead of [Ar] 3d$^2$. The rule that decides ion formation is which orbital sits lower in energy for an ion, not which subshell was written last when the atom was built up — and for a transition-metal cation that is 3d, so both 4s electrons leave before any 3d electron does. Part (d) has its own trap: computing $IE = hf$ gives the energy to remove one electron from one atom, in joules, and a script that stops there and reports $3.63 \times 10^{-18}$ (or converts only to kJ without the $\times N_A$) has answered a per-atom question when the syllabus quantity is molar.第 (b) 问是典型的顺序颠倒陷阱。钒的基态构型按能级填充顺序写作 4s$^2$ 3d$^3$，若按这个书写顺序移除电子，会先移走全部三个 3d 电子，再移走一个 4s 电子，得到 V$^{3+}$：[Ar] 4s$^1$，而非 [Ar] 3d$^2$。决定离子如何形成的是离子中哪个轨道能量更低，而不是原子构建时最后写下的是哪个亚层——对过渡金属阳离子而言那就是 3d，因此两个 4s 电子都会先于任何 3d 电子失去。第 (d) 问有它自己的陷阱：计算 $IE = hf$ 得到的是移除一个原子中一个电子所需的能量，单位是焦耳，若就此止步、报出 $3.63 \times 10^{-18}$（或只换算成 kJ 而漏掉 $\times N_A$），回答的就是单原子问题，而大纲要求的是摩尔量。
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。
Where this goes wrong.错在哪一步。 The step most likely to go missing is the multiplier in (b). Part (a)'s mole ratio $2:4:1$ reduces cleanly to the empirical formula $\mathrm{C_2H_4O}$, and a script that stops there and reports $\mathrm{C_2H_4O}$ as the molecular formula has treated the empirical-formula mass ($44.06~\mathrm{g\,mol^{-1}}$) as though it already equalled the given molar mass ($88.0~\mathrm{g\,mol^{-1}}$), skipping the check for whether they divide evenly. A second, quieter version of the same slip computes the multiplier correctly as $2.0$ and then forgets to apply it to all three subscripts, reporting $\mathrm{C_4H_4O_2}$ or $\mathrm{C_2H_8O_2}$ instead of scaling every atom count by the same factor.最容易漏掉的一步是 (b) 中的倍乘因子。第 (a) 问的摩尔比 $2:4:1$ 化简后直接给出实验式 $\mathrm{C_2H_4O}$，若就此止步、把 $\mathrm{C_2H_4O}$ 当作分子式报出，等于把实验式式量（$44.06~\mathrm{g\,mol^{-1}}$）当成了题给的摩尔质量（$88.0~\mathrm{g\,mol^{-1}}$）本身，跳过了检验二者是否能整除这一步。另一种更隐蔽的同类失误是正确算出倍乘因子为 $2.0$，却忘了把它应用到全部三个下标，写成 $\mathrm{C_4H_4O_2}$ 或 $\mathrm{C_2H_8O_2}$，而不是把每个原子数都乘以同一个因子。
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。
Where this goes wrong.错在哪一步。 Part (c) hides a ratio inversion. The gas is cooling from 373 K to 298 K at constant pressure, so Charles's law demands the volume shrink; a script that sets up $V_2 = V_1 \times T_1/T_2$ instead of $V_1 \times T_2/T_1$ inverts the temperature ratio and reports a volume larger than 62.2 cm$^3$, the opposite of what cooling a gas at fixed pressure does. Because the ratio $373/298$ and $298/373$ are both unremarkable-looking numbers near 1, the error does not announce itself the way a wrong order of magnitude would. Part (d) has a different trap: naming “intermolecular forces exist” without connecting it to why this gas law calculation stops being trustworthy — the physics is only diagnostic if it is tied to the specific ideal-gas assumption ('no intermolecular forces') that a cooling, hydrogen-bonding real gas violates.第 (c) 问隐藏着一个比值颠倒的陷阱。气体在恒压下从 373 K 降温到 298 K，按查理定律体积应随之缩小；若把式子写成 $V_2 = V_1 \times T_1/T_2$ 而非 $V_1 \times T_2/T_1$，就把温度比颠倒了，得到的体积会大于 62.2 cm³，恰好与恒压降温应有的效果相反。由于 $373/298$ 与 $298/373$ 两个比值看起来都平平无奇、接近 1，这个错误不会像数量级出错那样一眼暴露。第 (d) 问的陷阱不同：只写“存在分子间作用力”而不说明这如何导致本题的气体定律计算失效——只有把物理机制与这个正在降温、存在氢键的真实气体所违反的具体理想气体假设（“无分子间作用力”）联系起来，这条理由才真正成立。
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。
Where this goes wrong.错在哪一步。 Part (e) is graded on which shell break the answer names. IE$_1$ removes the lone 5s electron; a script that explains the size of IE$_1$ by saying only “it is the outermost electron” and stops there has not distinguished Rb from any other metal, since every IE$_1$ removes the outermost electron. What makes Rb's specifically low is that this outermost electron sits in its own shell ($n = 5$), shielded by a complete, noble-gas-like core all the way through $n = 4$: the answer needs the shielding, not just the label “outermost.” IE$_2$ then requires breaking into that full $n = 4$ shell itself, which is a much larger jump than the ordinary successive-electron increase, and an answer that treats IE$_2$ as “just a bit more, because removing an electron from a positive ion is always harder” has missed that this jump is categorically different in size, not merely larger.第 (e) 问的得分点在于说明的是哪一次壳层突破。IE$_1$ 移除的是唯一的 5s 电子；若只写“它是最外层电子”便就此打住，并未把 Rb 与任何其他金属区分开来，因为任何元素的 IE$_1$ 移除的都是最外层电子。真正使 Rb 的 IE$_1$ 特别低的原因是：这个最外层电子独占一个壳层（$n = 5$），并且被一个完整、类惰性气体的 $n = 4$ 内核完全屏蔽——答案需要写出屏蔽这一点，而不只是贴上“最外层”的标签。IE$_2$ 则需要打破这个完整的 $n = 4$ 壳层本身，其跃升幅度远大于普通的逐级电子移除；若把 IE$_2$的增大解释成“只是稍微多一点，因为从正离子上移除电子总是更难”，就没能抓住这次跃升在幅度上是本质不同的，而不仅仅是更大一些。
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。
Where this goes wrong.错在哪一步。 Part (d) asks for systematic errors and the most common wrong answer names a random one instead — typically “the volume readings varied between trials.” That variation is exactly what the three trials and the averaging in (a) are there to smooth out; it pushes individual readings above and below the mean, not the calculated $M$ consistently in one direction, so it cannot explain a result that is consistently 0.57% too high. A systematic error has to survive averaging and push every trial the same way: incomplete vaporization and heat loss to the barrel both make the calculated $n$ too small on every single trial, which is what a script needs to identify and connect back to $M = m/n$ to earn the mark, not a source of scatter that averaging already accounts for.第 (d) 问要求的是系统误差，而最常见的错误答案给出的却是一个随机误差——通常是“各次实验记录的体积有波动”。这种波动正是三次实验并在 (a) 中取平均所要消除的东西；它使单次读数在均值上下波动，而不是使算出的 $M$ 持续偏向同一方向，因此无法解释一个持续偏高 0.57%的结果。系统误差必须能经受住取平均而不消失，并使每一次实验都偏向同一方向：汽化不完全与热量散失到筒壁都会使每一次实验算出的 $n$ 偏小，这才是需要指出、并与 $M = m/n$ 联系起来才能得分的原因，而不是一个取平均就已经处理掉的离散来源。
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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