A student dissolves $\mathrm{CuSO_4}$ in water to make a blue solution. Which statement correctly identifies the components?一位同学将 $\mathrm{CuSO_4}$ 溶于水，配制出蓝色溶液。以下哪项正确识别了各组成部分？
Answer:答案：(B) $\mathrm{CuSO_4}$ is the solute; water is the solvent; the solution is aqueous$\mathrm{CuSO_4}$ 是溶质；水是溶剂；该溶液为水溶液
(a) Identify solute, solvent, and solution type辨别溶质、溶剂与溶液类型A1·A1·A1
The solute is the substance that dissolves (present in smaller amount): $\mathrm{CuSO_4}$. The solvent is the dissolving medium (present in larger amount): water. Because water is the solvent, the resulting solution is classified as an aqueous solution. All three facts are required for full marks.溶质是被溶解的物质（量少）：$\mathrm{CuSO_4}$。溶剂是溶解介质（量多）：水。因为溶剂是水，所以该溶液为水溶液。三点均需答对才能得满分。
Why the distractors fail.干扰项分析。
(A)Reverses solute and solvent. $\mathrm{CuSO_4}$ is the dissolved substance, not the dissolving medium.将溶质与溶剂互换。$\mathrm{CuSO_4}$ 是被溶解的物质，而非溶解介质。
(C)A solution has only one solvent; calling both components solvents is incorrect.一种溶液只有一种溶剂；将两种组分都称为溶剂是错误的。
(D)A true solution is homogeneous (uniform composition throughout), not heterogeneous.真溶液是均匀混合物（各处组成相同），而非非均匀混合物。
Solute, solvent, solution: the three essential vocabulary words of Unit 8.溶质、溶剂、溶液：第八单元的三个核心词汇。A solution is a homogeneous mixture of solute(s) dissolved in a solvent. The general rule is that the component present in greater quantity is the solvent. When water is the solvent, the solution is aqueous. The blue colour here is due to $\mathrm{Cu^{2+}}(aq)$ ions, which confirms that $\mathrm{CuSO_4}$ has fully dissolved and dissociated. On Ontario SCH3U exams, expect one question testing these definitions directly in the multiple-choice section.溶液是溶质溶解在溶剂中形成的均匀混合物。通用规则：含量较多的组分是溶剂。以水为溶剂时，溶液为水溶液。蓝色来自 $\mathrm{Cu^{2+}}(aq)$ 离子，证明 $\mathrm{CuSO_4}$ 已完全溶解并解离。安大略省 SCH3U 考试通常会在选择题部分直接考查这些定义。
Which pair of substances is predicted to be miscible based on the "like dissolves like" principle?根据"相似相溶"原则，以下哪对物质预计可以互溶？
Answer:答案：(C) Ethanol ($\mathrm{C_2H_5OH}$) and water乙醇（$\mathrm{C_2H_5OH}$）与水
(a) Apply polarity and intermolecular-force reasoning运用极性与分子间作用力推理M1·A1·A1·A1
Water is polar and hydrogen-bonding. Ethanol has an $\mathrm{-OH}$ group that also forms hydrogen bonds and has a polar $\mathrm{O-H}$ bond. The dominant intermolecular force in both is hydrogen bonding, so they are miscible in all proportions. Option (C).水是极性分子，可形成氢键。乙醇含有 $\mathrm{-OH}$ 基团，同样可形成氢键，且 $\mathrm{O-H}$ 键具有极性。两者的主要分子间作用力均为氢键，因此可以任意比例互溶。选 (C)。
Why the distractors fail.干扰项分析。
(A) Water and motor oil:水与机油：Motor oil is a nonpolar hydrocarbon; water is polar. Polar-nonpolar pairs are immiscible.机油是非极性烃类；水是极性分子。极性-非极性对不互溶。
(B) Hexane and NaCl:己烷与 NaCl：NaCl is ionic and dissolves readily in polar solvents; hexane is nonpolar and cannot solvate the ions. NaCl is essentially insoluble in hexane.NaCl 是离子化合物，易溶于极性溶剂；己烷是非极性的，无法溶剂化离子。NaCl 在己烷中几乎不溶。
(D) Iodine and water:碘与水：$\mathrm{I_2}$ is nonpolar; water is polar. Iodine has very low solubility in water (sparingly soluble).$\mathrm{I_2}$ 是非极性分子；水是极性分子。碘在水中的溶解度很低（微溶）。
"Like dissolves like" predicts miscibility from polarity and intermolecular forces."相似相溶"原则根据极性和分子间作用力预测互溶性。The rule: polar solvents dissolve polar and ionic solutes; nonpolar solvents dissolve nonpolar solutes. Ethanol is the classic example of a molecule that bridges both worlds: its $\mathrm{-OH}$ end is polar and hydrogen-bonding (compatible with water), while its short $\mathrm{-CH_2CH_3}$ tail is slightly nonpolar. As the carbon chain length increases (e.g., $\mathrm{C_4}$ to $\mathrm{C_8}$ alcohols), solubility in water decreases because the nonpolar character of the chain begins to dominate. This trend is tested on AP Chemistry and BC Chemistry 11 provincial exams.规则：极性溶剂溶解极性和离子型溶质；非极性溶剂溶解非极性溶质。乙醇是同时具备两种性质的经典例子：$\mathrm{-OH}$ 端是极性、可氢键（与水相容），而短链 $\mathrm{-CH_2CH_3}$ 尾部略具非极性。随碳链增长（如 $\mathrm{C_4}$ 到 $\mathrm{C_8}$ 醇），在水中的溶解度下降，因为链的非极性特征逐渐主导。这一趋势在 AP 化学和卑诗省化学 11 省考中均有考查。
A student dissolves $11.70\ \mathrm{g}$ of NaCl ($M_r = 58.44\ \mathrm{g/mol}$) in enough water to make exactly $400.0\ \mathrm{mL}$ of solution.一位同学将 $11.70\ \mathrm{g}$ NaCl（$M_r = 58.44\ \mathrm{g/mol}$）溶于足量水中，配制出恰好 $400.0\ \mathrm{mL}$ 的溶液。
Answer:答案：(a) $n = 0.2002\ \mathrm{mol}$ · (b) $c = 0.5005\ \mathrm{mol/L}$ · (c)dissolve in small volume of water, transfer to $400.0\ \mathrm{mL}$ volumetric flask, add water to the mark先溶于少量水，转移至 $400.0\ \mathrm{mL}$ 容量瓶，加水至刻度线 · (d)complete dissociation / ideal dilute solution完全解离 / 理想稀溶液
(a) Moles of NaClNaCl 的摩尔数M1·A1
$$ n \;=\; \frac{m}{M_r} \;=\; \frac{11.70}{58.44} \;=\; 0.2002 \;\mathrm{mol.} $$
Dissolve the NaCl in a small volume of distilled water in a beaker, allow to cool to room temperature, quantitatively transfer the solution into a $400.0\ \mathrm{mL}$ volumetric flask, then add distilled water slowly until the bottom of the meniscus just touches the graduation mark. (A $400.0\ \mathrm{mL}$ volumetric flask is the critical equipment for ensuring precision.)将 NaCl 先在烧杯中溶于少量蒸馏水，冷却至室温，定量转移至 $400.0\ \mathrm{mL}$ 容量瓶，再缓慢加蒸馏水至凹液面底部恰好与刻度线相切。（使用 $400.0\ \mathrm{mL}$ 容量瓶是保证精度的关键。）
We assume the solution behaves ideally: the solute-solvent interactions do not significantly change the total volume from the volume of pure solvent added. In other words, the dissolved NaCl does not cause appreciable volume contraction or expansion.假设溶液表现为理想溶液：溶质-溶剂相互作用不会使总体积与加入的纯溶剂体积有显著差异。换言之，溶解的 NaCl 不引起明显的体积收缩或膨胀。
Molarity = moles per litre of solution, not per litre of solvent.摩尔浓度 = 每升溶液（而非每升溶剂）中的摩尔数。The formula $c = n/V$ uses the volume of the final solution ($400.0\ \mathrm{mL}$), not the volume of water added. This is why the correct technique is to transfer and then dilute to the mark in a volumetric flask rather than adding the solute to $400.0\ \mathrm{mL}$ of water (which would give a slightly different final volume). The answer $c \approx 0.500\ \mathrm{mol/L}$ is recognisable as a round number because $11.70\ \mathrm{g}$ is exactly $0.200 \times 58.44\ \mathrm{g}$; noting this pattern helps with quick estimation on provincial exams.公式 $c = n/V$ 中 $V$ 是最终溶液的体积（$400.0\ \mathrm{mL}$），而非加水的体积。因此正确操作是转移后在容量瓶中加水至刻度，而不是将溶质加入 $400.0\ \mathrm{mL}$ 水中（后者会导致最终体积略有不同）。结果 $c \approx 0.500\ \mathrm{mol/L}$ 是整数，因为 $11.70\ \mathrm{g}$ 恰好等于 $0.200 \times 58.44\ \mathrm{g}$；认识这一规律有助于在省考中快速估算。
Q4MEDIUM中🇨🇦 BC卑BC Provincial-style卑诗省考风格§4 Dilution稀释 · BC Chem 11[6 marks][6 分]
Stock $\mathrm{HCl}$ at $8.00\ \mathrm{mol/L}$; student needs $500\ \mathrm{mL}$ of $0.400\ \mathrm{mol/L}$ $\mathrm{HCl}$ for a titration.$8.00\ \mathrm{mol/L}$ 盐酸储备液；同学需要配制 $500\ \mathrm{mL}$ 的 $0.400\ \mathrm{mol/L}$ $\mathrm{HCl}$ 溶液用于滴定。
Answer:答案：(a) $c_1V_1 = c_2V_2$; moles of solute conserved, volume changes；溶质摩尔数守恒，体积改变 · (b) $V_1 = 25.0\ \mathrm{mL}$ · (c)measure stock into acid-resistant container, add to water, safety: acid burns量取储备液加入水中，安全隐患：酸腐蚀
(a) Dilution equation and conserved quantities稀释方程及守恒量A1·A1
$$ c_1 V_1 \;=\; c_2 V_2. $$
The two conserved quantities are: (1) moles of solute ($n = cV$) remain unchanged when water is added; (2) (implicitly) mass of solute is conserved. The volume increases; the concentration decreases proportionally.两个守恒量：(1) 溶质摩尔数（$n = cV$）在加水时不变；(2)（隐含地）溶质质量守恒。体积增大，浓度按比例降低。
(c) Safe laboratory procedure and safety hazard安全实验操作与安全隐患A1·A1
Procedure: Add the $25.0\ \mathrm{mL}$ of concentrated $\mathrm{HCl}$ slowly to a large quantity of distilled water in the volumetric flask, then dilute to the $500\ \mathrm{mL}$ mark. Never add water to concentrated acid. Safety hazard: concentrated $\mathrm{HCl}$ is corrosive and emits acidic fumes; perform the dilution in a fume hood wearing gloves and safety goggles.操作：将 $25.0\ \mathrm{mL}$ 浓盐酸缓慢加入盛有大量蒸馏水的容量瓶中，再稀释至 $500\ \mathrm{mL}$ 刻度。切勿将水加入浓酸。安全隐患：浓盐酸有腐蚀性并会挥发酸性气体；应在通风橱内操作，并佩戴手套和护目镜。
Always add acid to water ("AAW"), never water to acid.永远"酸入水"，绝不"水入酸"。When water is added to concentrated acid, the heat of dilution is released into a small volume of liquid, causing violent spattering. When acid is added to water, the large thermal mass of water absorbs the heat safely. The dilution factor here is $8.00 / 0.400 = 20\times$: you take $1$ part concentrated acid and add $19$ parts water ($25\ \mathrm{mL}$ stock $\to 500\ \mathrm{mL}$ total). Recognising that a large dilution factor implies a small aliquot of stock is a useful sanity check: $25\ \mathrm{mL}$ out of $500\ \mathrm{mL}$ is $5\%$, consistent with the $20\times$ factor.将水加入浓酸时，稀释热会释放到少量液体中，引发剧烈飞溅。将酸加入水中时，大量水的热容可安全吸收热量。此处稀释倍数为 $8.00 / 0.400 = 20$ 倍：取 1 份浓酸加 19 份水（$25\ \mathrm{mL}$ 储备液 $\to$ $500\ \mathrm{mL}$ 总量）。认识到稀释倍数大时储备液用量少是一个有用的合理性检验：$25\ \mathrm{mL}$ 占 $500\ \mathrm{mL}$ 的 $5\%$，与 $20$ 倍稀释吻合。
Solubility of $\mathrm{KNO_3}$: $110\ \mathrm{g/100\ g\ water}$ at $60\ ^\circ\mathrm{C}$; $32\ \mathrm{g/100\ g\ water}$ at $20\ ^\circ\mathrm{C}$. $80\ \mathrm{g}$ $\mathrm{KNO_3}$ dissolved in $100\ \mathrm{g}$ water at $60\ ^\circ\mathrm{C}$, then cooled to $20\ ^\circ\mathrm{C}$.$\mathrm{KNO_3}$ 溶解度：$60\ ^\circ\mathrm{C}$ 时 $110\ \mathrm{g/100\ g\ water}$；$20\ ^\circ\mathrm{C}$ 时 $32\ \mathrm{g/100\ g\ water}$。$60\ ^\circ\mathrm{C}$ 下将 $80\ \mathrm{g}$ $\mathrm{KNO_3}$ 溶于 $100\ \mathrm{g}$ 水，再冷却至 $20\ ^\circ\mathrm{C}$。
Answer:答案：(a)Unsaturated at $60\ ^\circ\mathrm{C}$ (80 g dissolved < 110 g maximum)$60\ ^\circ\mathrm{C}$ 时不饱和（已溶 80 g < 最大溶解量 110 g） · (b) $48\ \mathrm{g}$ crystallises析出 · (c)gas solubility decreases; real-world: dissolved oxygen leaving warming water harms aquatic life气体溶解度降低；实例：水温升高导致溶解氧逸出，危害水生生物
(a) Classification at $60\ ^\circ\mathrm{C}$$60\ ^\circ\mathrm{C}$ 时的溶液分类A1·A1
At $60\ ^\circ\mathrm{C}$ the maximum solubility is $110\ \mathrm{g}$ per $100\ \mathrm{g}$ water. Only $80\ \mathrm{g}$ is dissolved. Since $80 < 110$, the solution is unsaturated: more $\mathrm{KNO_3}$ could still dissolve at this temperature.$60\ ^\circ\mathrm{C}$ 时最大溶解量为每 $100\ \mathrm{g}$ 水 $110\ \mathrm{g}$。实际溶解量仅为 $80\ \mathrm{g}$。因为 $80 < 110$，溶液为不饱和溶液：在此温度下还可继续溶解更多 $\mathrm{KNO_3}$。
(b) Mass of $\mathrm{KNO_3}$ that crystallises at $20\ ^\circ\mathrm{C}$冷却至 $20\ ^\circ\mathrm{C}$ 时析出的 $\mathrm{KNO_3}$ 质量M1·A1
At $20\ ^\circ\mathrm{C}$, maximum solubility is $32\ \mathrm{g}$ per $100\ \mathrm{g}$ water. The solution contained $80\ \mathrm{g}$ dissolved. The excess that must crystallise:$20\ ^\circ\mathrm{C}$ 时最大溶解量为每 $100\ \mathrm{g}$ 水 $32\ \mathrm{g}$。原溶液含 $80\ \mathrm{g}$ 溶质，多余部分必须析出：
$$ m_{\text{crystals}} \;=\; 80 - 32 \;=\; 48 \;\mathrm{g.} $$
(c) Effect of temperature on gas solubility and a real-world consequence温度对气体溶解度的影响及实例A1·A1
For gases, solubility decreases as temperature increases. As the water warms from $20\ ^\circ\mathrm{C}$ to $60\ ^\circ\mathrm{C}$, dissolved gas escapes. A real-world consequence: warm river water contains less dissolved oxygen, stressing fish and aquatic organisms. (Alternative acceptable answer: thermal pollution from power-plant cooling water reduces $\mathrm{O_2}$ levels.)对于气体，溶解度随温度升高而降低。水从 $20\ ^\circ\mathrm{C}$ 升至 $60\ ^\circ\mathrm{C}$ 时，溶解气体会逸出。实际例子：温暖的河水中溶解氧含量减少，使鱼类和水生生物受到威胁。（可接受的替代答案：发电站冷却水造成的热污染会降低水中 $\mathrm{O_2}$ 含量。）
Solid vs. gas solubility trends with temperature are opposite.固体与气体溶解度随温度的变化趋势相反。For most solid solutes, solubility increases with temperature (endothermic dissolution is favoured by Le Chatelier's principle at higher $T$). For gases, dissolution is exothermic, so Le Chatelier predicts that raising temperature shifts equilibrium toward the gas phase, reducing solubility. This distinction appears on AB Chem 20 and AP Chemistry multiple-choice items. The AB diploma exam often tests the $\mathrm{KNO_3}$ solubility curve specifically because $\mathrm{KNO_3}$ has a steeply rising solubility-versus-temperature curve, making recrystallisation calculations straightforward.对于大多数固态溶质，溶解度随温度升高而增大（勒夏特列原理：吸热溶解过程在较高温度下有利）。对于气体，溶解是放热过程，因此升温会使平衡向气相方向移动，溶解度降低。这一区别在阿省化学 20 和 AP 化学选择题中均有考查。阿省毕业考经常以 $\mathrm{KNO_3}$ 溶解度曲线为例，因为 $\mathrm{KNO_3}$ 的溶解度-温度曲线斜率大，便于出重结晶计算题。
$$ \mathrm{MgCl_2(aq)} \;\longrightarrow\; \mathrm{Mg^{2+}(aq)} + 2\,\mathrm{Cl^-(aq).} $$
One formula unit of $\mathrm{MgCl_2}$ produces one $\mathrm{Mg^{2+}}$ ion and two $\mathrm{Cl^-}$ ions.一个 $\mathrm{MgCl_2}$ 化学式单元产生一个 $\mathrm{Mg^{2+}}$ 离子和两个 $\mathrm{Cl^-}$ 离子。
$$ c_{\text{ions}} \;=\; [\mathrm{Mg^{2+}}] + [\mathrm{Cl^-}] \;=\; 0.2500 + 0.5000 \;=\; 0.7500 \;\mathrm{mol/L.} $$
Equivalently, $3 \times 0.2500 = 0.7500\ \mathrm{mol/L}$ (3 ions per formula unit of $\mathrm{MgCl_2}$).等价地，$3 \times 0.2500 = 0.7500\ \mathrm{mol/L}$（每个 $\mathrm{MgCl_2}$ 产生 3 个离子）。
(e) Classification as electrolyte type电解质类型的分类A1
$\mathrm{MgCl_2}$ is a strong electrolyte. It is an ionic compound that dissociates completely in water, producing a solution that conducts electricity well. There are no un-dissociated $\mathrm{MgCl_2}$ formula units remaining in solution.$\mathrm{MgCl_2}$ 是强电解质。它是离子化合物，在水中完全解离，产生导电性良好的溶液。溶液中不存在未解离的 $\mathrm{MgCl_2}$ 化学式单元。
Ion concentration = stoichiometric coefficient times formula-unit concentration.离子浓度 = 化学计量数 x 化学式单元浓度。This ratio-scaling step trips many students. After writing the dissociation equation, read off the stoichiometric coefficients: $\mathrm{Mg^{2+}}$ has coefficient 1 (same as $\mathrm{MgCl_2}$) and $\mathrm{Cl^-}$ has coefficient 2 (double). So $[\mathrm{Cl^-}] = 2 \times [\mathrm{MgCl_2}]_{\text{formula}}$. The total ion count per formula unit (3) is called the van't Hoff factor $i$; for strong electrolytes it equals the number of ions released. For a $0.2500\ \mathrm{mol/L}$ $\mathrm{MgCl_2}$ solution the van't Hoff factor is $i = 3$, so colligative properties (boiling point, freezing point) are $3\times$ stronger than for the same molarity of a nonelectrolyte.这一比例换算步骤让许多同学出错。写出解离方程式后，读取化学计量数：$\mathrm{Mg^{2+}}$ 系数为 1（与 $\mathrm{MgCl_2}$ 相同），$\mathrm{Cl^-}$ 系数为 2（两倍）。因此 $[\mathrm{Cl^-}] = 2 \times [\mathrm{MgCl_2}]_{\text{化学式}}$。每个化学式单元产生的总离子数（3）称为范托夫因子 $i$；对强电解质，$i$ 等于释放的离子数。$0.2500\ \mathrm{mol/L}$ $\mathrm{MgCl_2}$ 溶液的 $i = 3$，因此依数性（沸点升高、凝固点降低）是同浓度非电解质的 3 倍。
$2.40\ \mathrm{mol/L}$ $\mathrm{H_2SO_4}$ stock; $75.0\ \mathrm{mL}$ pipetted to $500\ \mathrm{mL}$ flask to make Solution A; then $200\ \mathrm{mL}$ of A diluted to $600\ \mathrm{mL}$ for Solution B.$2.40\ \mathrm{mol/L}$ 硫酸储备液；吸取 $75.0\ \mathrm{mL}$ 于 $500\ \mathrm{mL}$ 容量瓶中配制溶液 A；再取 $200\ \mathrm{mL}$ 溶液 A 稀释至 $600\ \mathrm{mL}$ 配制溶液 B。
Answer:答案：(a) $c_A = 0.360\ \mathrm{mol/L}$ · (b) $V = 200\ \mathrm{mL}$ · (c) $c_B = 0.120\ \mathrm{mol/L}$ · (d)moles of $\mathrm{H_2SO_4}$ conserved; adding water changes $V$ but not $n$$\mathrm{H_2SO_4}$ 摩尔数守恒；加水改变 $V$ 但不改变 $n$
In each dilution step, the number of moles of solute is conserved: $n = c_1 V_1 = c_2 V_2$. Adding distilled water introduces no additional $\mathrm{H_2SO_4}$, so the solute amount cannot change. Only the volume of solution increases, so the concentration decreases proportionally. This is why $c_1V_1 = c_2V_2$ is valid: it is simply a statement of mole conservation during dilution.在每步稀释中，溶质摩尔数守恒：$n = c_1 V_1 = c_2 V_2$。加入蒸馏水不引入额外的 $\mathrm{H_2SO_4}$，因此溶质量不变。只有溶液体积增大，浓度按比例降低。这就是 $c_1V_1 = c_2V_2$ 成立的原因：它不过是稀释过程中摩尔数守恒的表述。
Each dilution step multiplies concentration by the volume ratio taken/total.每步稀释将浓度乘以所取体积与总体积之比。Note the overall dilution from stock to Solution B: start at $2.40\ \mathrm{mol/L}$, end at $0.120\ \mathrm{mol/L}$, a factor of $20\times$. The two steps are $2.40 \to 0.360$ (factor of $6.67\times$) and $0.360 \to 0.120$ (factor of $3\times$); combined: $6.67 \times 3 = 20$. For multi-step dilutions, it is often faster to track moles explicitly: $n_{\text{stock}}$ used $= 2.40 \times 0.0750 = 0.180\ \mathrm{mol}$; this all ends up in $500\ \mathrm{mL}$ Solution A. Then $200/500 = 2/5$ of those moles $(= 0.0720\ \mathrm{mol})$ are taken and diluted into $600\ \mathrm{mL}$, giving $c_B = 0.0720/0.600 = 0.120\ \mathrm{mol/L}$.注意从储备液到溶液 B 的总稀释倍数：从 $2.40\ \mathrm{mol/L}$ 到 $0.120\ \mathrm{mol/L}$，共稀释 $20$ 倍。两步分别为 $2.40 \to 0.360$（稀释 $6.67$ 倍）和 $0.360 \to 0.120$（稀释 $3$ 倍），合计 $6.67 \times 3 = 20$。对于多步稀释，直接追踪摩尔数往往更快：取用的储备液中含 $2.40 \times 0.0750 = 0.180\ \mathrm{mol}$，全部进入 $500\ \mathrm{mL}$ 溶液 A；再取其中 $200/500 = 2/5$（即 $0.0720\ \mathrm{mol}$）稀释至 $600\ \mathrm{mL}$，得 $c_B = 0.0720/0.600 = 0.120\ \mathrm{mol/L}$。
The $50.0\ \mathrm{mL}$ aliquot of Solution X is diluted to $200.0\ \mathrm{mL}$:取 $50.0\ \mathrm{mL}$ 溶液 X 稀释至 $200.0\ \mathrm{mL}$：
$$ c_Y \;=\; \frac{c_X \times V_X}{V_Y} \;=\; \frac{0.2496 \times 0.0500}{0.2000} \;=\; \frac{0.01248}{0.2000} \;=\; 0.06240 \;\mathrm{mol/L.} $$
(Note: The dilution factor is $50.0/200.0 = 1/4$, so $c_Y = c_X/4 = 0.2496/4 = 0.06240\ \mathrm{mol/L}$.)（注：稀释因子为 $50.0/200.0 = 1/4$，故 $c_Y = c_X/4 = 0.2496/4 = 0.06240\ \mathrm{mol/L}$。）
(c) Mass of $\mathrm{CuSO_4}$ in $150.0\ \mathrm{mL}$ of Solution Y$150.0\ \mathrm{mL}$ 溶液 Y 中 $\mathrm{CuSO_4}$ 的质量M1·A1
$$ n \;=\; c_Y \times V \;=\; 0.06240 \times 0.1500 \;=\; 9.360 \times 10^{-3} \;\mathrm{mol.} $$
$$ m \;=\; n \times M_r \;=\; 9.360 \times 10^{-3} \times 159.6 \;=\; 1.494 \;\mathrm{g.} $$
(d) Is Y more or less concentrated than X?Y 比 X 浓还是稀？A1
Solution Y is less concentrated than Solution X. When the $50.0\ \mathrm{mL}$ aliquot of X was diluted to $200.0\ \mathrm{mL}$, the same number of moles of $\mathrm{CuSO_4}$ was spread over a four times larger volume, reducing the concentration by a factor of four.溶液 Y 比溶液 X 浓度低。将 $50.0\ \mathrm{mL}$ 溶液 X 稀释至 $200.0\ \mathrm{mL}$ 时，相同摩尔数的 $\mathrm{CuSO_4}$ 被分散在四倍的体积中，浓度降低了四倍。
Track significant figures at each step: do not round intermediate values.每步保持有效数字：不要对中间值进行四舍五入。Part (a) uses a 3-step chain: mass $\to$ moles $\to$ molarity. Rounding $9.96/159.6 = 0.0624$ mol (3 sig figs) early and then dividing by $0.2500$ L gives $0.2496\ \mathrm{mol/L}$ (4 sig figs). Keeping the full calculator value throughout avoids cascading rounding errors in parts (b) and (c). Notice also that $c_Y = c_X/4$ exactly because the dilution factor is $50.0/200.0 = 1/4$: a clean ratio. And $c_Y = n_{\text{original}}/V_Y = 0.06240\ \mathrm{mol}/1.000\ \mathrm{L}$ if one litre of Y were made from all of X; here only $50/250 = 1/5$ of X was used, then diluted by $4\times$, giving the same net factor of $1/4 \times 1/5 \times \ldots$ wait, verify: $0.2496 \times (50/200) = 0.06240$. Confirmed.第 (a) 小问是三步链：质量 $\to$ 摩尔数 $\to$ 摩尔浓度。若中途将 $9.96/159.6 = 0.0624\ \mathrm{mol}$（3 位有效数字）过早取整，再除以 $0.2500\ \mathrm{L}$，得 $0.2496\ \mathrm{mol/L}$（4 位）。全程保留完整计算值可避免 (b)(c) 部分的累积舍入误差。也注意到 $c_Y = c_X/4$ 恰为精确值，因为稀释因子 $50.0/200.0 = 1/4$ 是整洁的比值。验证：$0.2496 \times (50/200) = 0.06240$，正确。
(b) Total dissolved particle concentrations总溶解颗粒浓度A1·A1·A1
(A) Glucose: does not dissociate, so total particles $= 0.10\ \mathrm{mol/L}$.(A) 葡萄糖：不解离，故总颗粒浓度 $= 0.10\ \mathrm{mol/L}$。 (B) NaCl: produces 2 ions per formula unit ($i = 2$); total $= 2 \times 0.10 = 0.20\ \mathrm{mol/L}$.(B) NaCl：每个化学式单元产生 2 个离子（$i = 2$）；总计 $= 2 \times 0.10 = 0.20\ \mathrm{mol/L}$。 (C) $\mathrm{CaCl_2}$: produces 3 ions per formula unit ($i = 3$); total $= 3 \times 0.10 = 0.30\ \mathrm{mol/L}$.(C) $\mathrm{CaCl_2}$：每个化学式单元产生 3 个离子（$i = 3$）；总计 $= 3 \times 0.10 = 0.30\ \mathrm{mol/L}$。
(c) Ranking by boiling-point elevation (smallest first)按沸点升高从小到大排列A1·A1
Boiling-point elevation $\Delta T_b = K_b \times m_{\text{particles}}$ is proportional to the total particle concentration. Ranking from smallest elevation:沸点升高 $\Delta T_b = K_b \times m_{\text{颗粒}}$ 与总颗粒浓度成正比。按升高最小排列：
$$ \text{A (glucose, 0.10)} < \text{B (NaCl, 0.20)} < \text{C (CaCl}_2\text{, 0.30).} $$
Glucose has the fewest dissolved particles per litre; $\mathrm{CaCl_2}$ has the most, so it raises the boiling point the most.葡萄糖每升中溶解颗粒最少；$\mathrm{CaCl_2}$ 最多，因此沸点升高最大。
Colligative properties depend on the total number of dissolved particles, not their identity.依数性取决于溶解颗粒的总数，而与颗粒的种类无关。The van't Hoff factor $i$ captures this: $i = 1$ for nonelectrolytes, $i = $ number of ions for strong electrolytes. For $0.10\ \mathrm{mol/L}$ solutions the effective molality order is $1 : 2 : 3$ for glucose : NaCl : $\mathrm{CaCl_2}$, so every colligative property (boiling-point elevation, freezing-point depression, osmotic pressure) follows the same order. This is an Honors/AP-level concept because the AB Chem 20 and ON SCH3U curricula do not formally introduce colligative properties; they are covered in AP Chemistry and IB Chemistry HL. The insight matters practically: road salt ($\mathrm{CaCl_2}$) is more effective at lowering the freezing point of ice than sodium chloride because $i = 3 > 2$.范托夫因子 $i$ 体现了这一点：非电解质 $i = 1$；强电解质 $i$ = 产生的离子数。对于 $0.10\ \mathrm{mol/L}$ 溶液，有效质量摩尔浓度之比为 $1 : 2 : 3$（葡萄糖 : NaCl : $\mathrm{CaCl_2}$），因此所有依数性（沸点升高、凝固点降低、渗透压）均遵循同一顺序。这是荣誉级/AP 层次的概念，因为阿省化学 20 和安大略省 SCH3U 课程并未正式引入依数性；这些内容在 AP 化学和 IB 化学 HL 中才涉及。实际意义：路面融雪盐 $\mathrm{CaCl_2}$ 比 NaCl 降低冰点的效果更强，因为 $i = 3 > 2$。
$55\ \mathrm{g}$ $\mathrm{KNO_3}$ dissolved in $100\ \mathrm{g}$ water at $40\ ^\circ\mathrm{C}$ (solubility $65\ \mathrm{g}/100\ \mathrm{g}$). Solution heated to $70\ ^\circ\mathrm{C}$ (solubility $138\ \mathrm{g}/100\ \mathrm{g}$) and more $\mathrm{KNO_3}$ added to saturation. Saturated solution rapidly cooled to $10\ ^\circ\mathrm{C}$ (solubility $21\ \mathrm{g}/100\ \mathrm{g}$).在 $40\ ^\circ\mathrm{C}$（溶解度 $65\ \mathrm{g}/100\ \mathrm{g}$）时将 $55\ \mathrm{g}$ $\mathrm{KNO_3}$ 溶于 $100\ \mathrm{g}$ 水。加热至 $70\ ^\circ\mathrm{C}$（溶解度 $138\ \mathrm{g}/100\ \mathrm{g}$）并继续加 $\mathrm{KNO_3}$ 至饱和。再将饱和溶液迅速冷却至 $10\ ^\circ\mathrm{C}$（溶解度 $21\ \mathrm{g}/100\ \mathrm{g}$）。
Answer:答案：(a)Unsaturated (55 g < 65 g max at $40\ ^\circ\mathrm{C}$)不饱和（55 g < $40\ ^\circ\mathrm{C}$ 时最大 65 g） · (b) $83\ \mathrm{g}$ additional $\mathrm{KNO_3}$额外加入 $\mathrm{KNO_3}$ · (c) $117\ \mathrm{g}$ crystallises析出 · (d)more $\mathrm{KNO_3}$ is dissolved than the equilibrium maximum; a seed crystal or disturbance triggers crystallisation溶解的 $\mathrm{KNO_3}$ 超过平衡最大值；加晶种或扰动可触发结晶
(a) Classification at $40\ ^\circ\mathrm{C}$$40\ ^\circ\mathrm{C}$ 时的溶液分类A1·A1
At $40\ ^\circ\mathrm{C}$ the maximum solubility of $\mathrm{KNO_3}$ is $65\ \mathrm{g}$ per $100\ \mathrm{g}$ water. Only $55\ \mathrm{g}$ is dissolved. Since $55 < 65$, the solution is unsaturated: it can dissolve more $\mathrm{KNO_3}$ at this temperature.$40\ ^\circ\mathrm{C}$ 时 $\mathrm{KNO_3}$ 最大溶解量为每 $100\ \mathrm{g}$ 水 $65\ \mathrm{g}$。实际仅溶解 $55\ \mathrm{g}$。由于 $55 < 65$，溶液为不饱和溶液：在该温度下还可继续溶解 $\mathrm{KNO_3}$。
(b) Additional $\mathrm{KNO_3}$ to saturate at $70\ ^\circ\mathrm{C}$在 $70\ ^\circ\mathrm{C}$ 使溶液饱和所需额外加入的 $\mathrm{KNO_3}$M1·A1·A1
At $70\ ^\circ\mathrm{C}$ the saturation point is $138\ \mathrm{g}$ per $100\ \mathrm{g}$ water (same $100\ \mathrm{g}$ water throughout). The solution already contains $55\ \mathrm{g}$. Additional mass needed:$70\ ^\circ\mathrm{C}$ 时饱和点为每 $100\ \mathrm{g}$ 水 $138\ \mathrm{g}$（水量始终为 $100\ \mathrm{g}$）。溶液已含 $55\ \mathrm{g}$，所需额外质量：
$$ m_{\text{added}} \;=\; 138 - 55 \;=\; 83 \;\mathrm{g.} $$
(c) Mass of $\mathrm{KNO_3}$ crystallising at $10\ ^\circ\mathrm{C}$冷却至 $10\ ^\circ\mathrm{C}$ 时析出的 $\mathrm{KNO_3}$ 质量M1·A1
After saturation at $70\ ^\circ\mathrm{C}$, the solution contains $138\ \mathrm{g}$ dissolved in $100\ \mathrm{g}$ water. At $10\ ^\circ\mathrm{C}$ only $21\ \mathrm{g}$ can remain dissolved:$70\ ^\circ\mathrm{C}$ 饱和后，溶液含 $138\ \mathrm{g}$ 溶于 $100\ \mathrm{g}$ 水。$10\ ^\circ\mathrm{C}$ 时仅能保持溶解 $21\ \mathrm{g}$：
$$ m_{\text{crystals}} \;=\; 138 - 21 \;=\; 117 \;\mathrm{g.} $$
(d) Supersaturation at the molecular level and crystallisation trigger从分子层面理解过饱和与结晶触发A1·A1
In a supersaturated solution, more solute is dissolved than the equilibrium maximum at that temperature. The $\mathrm{KNO_3}$ ions remain in solution only because there is no nucleation site for crystal growth. At the molecular level, the solution is in a metastable (unstable equilibrium) state: $\mathrm{K^+}$ and $\mathrm{NO_3^-}$ ions are present in higher concentration than the saturation point allows. Crystallisation is triggered by adding a seed crystal of $\mathrm{KNO_3}$ (which provides a surface for ion attachment) or by mechanical disturbance (scratching the container wall), which causes rapid nucleation and release of the excess solute.在过饱和溶液中，溶解的溶质超过该温度下的平衡最大量。$\mathrm{KNO_3}$ 离子之所以能留在溶液中，是因为没有晶核供晶体生长。从分子层面看，溶液处于亚稳（不稳定平衡）状态：$\mathrm{K^+}$ 和 $\mathrm{NO_3^-}$ 离子的浓度超过饱和点允许的值。结晶由加入 $\mathrm{KNO_3}$ 晶种（提供离子附着的表面）或机械扰动（刮擦容器壁）触发，导致快速成核，多余的溶质析出。
Multi-step solubility problems require careful bookkeeping of which $\mathrm{KNO_3}$ mass is present at each stage.多步溶解度问题需要仔细追踪每个阶段 $\mathrm{KNO_3}$ 的质量。The key chain: $55\ \mathrm{g}$ dissolved at $40\ ^\circ\mathrm{C}$ (unsaturated) $\to$ heat to $70\ ^\circ\mathrm{C}$ and add $83\ \mathrm{g}$ more to reach saturation at $138\ \mathrm{g}$ total $\to$ cool to $10\ ^\circ\mathrm{C}$ where only $21\ \mathrm{g}$ stays dissolved $\to$ $117\ \mathrm{g}$ crystallises. The water mass ($100\ \mathrm{g}$) stays constant throughout (no evaporation); solubility values are always given per $100\ \mathrm{g}$ water, so no ratio scaling is needed here. A common error is to use the solubility of $40\ ^\circ\mathrm{C}$ ($65\ \mathrm{g}$) as the starting amount for the crystallisation calculation instead of the saturated amount at $70\ ^\circ\mathrm{C}$ ($138\ \mathrm{g}$).关键流程：$40\ ^\circ\mathrm{C}$ 时溶解 $55\ \mathrm{g}$（不饱和）$\to$ 加热至 $70\ ^\circ\mathrm{C}$ 再加入 $83\ \mathrm{g}$ 达到饱和，共 $138\ \mathrm{g}$ $\to$ 冷却至 $10\ ^\circ\mathrm{C}$ 时仅剩 $21\ \mathrm{g}$ 保持溶解 $\to$ 析出 $117\ \mathrm{g}$。全程水的质量（$100\ \mathrm{g}$）不变（无蒸发）；溶解度值均以每 $100\ \mathrm{g}$ 水给出，因此无需比例换算。常见错误：用 $40\ ^\circ\mathrm{C}$ 时的溶解度（$65\ \mathrm{g}$）而非 $70\ ^\circ\mathrm{C}$ 时的饱和量（$138\ \mathrm{g}$）作为结晶计算的起始量。
Answer:答案：(a)ion-dipole forces pull $\mathrm{Al^{3+}}$ and $\mathrm{SO_4^{2-}}$ away from lattice离子-偶极力将 $\mathrm{Al^{3+}}$ 和 $\mathrm{SO_4^{2-}}$ 从晶格中拉出 · (b) $\mathrm{Al_2(SO_4)_3(s) \to 2\,Al^{3+}(aq) + 3\,SO_4^{2-}(aq)}$ · (c) $[\mathrm{Al^{3+}}] = 0.300\ \mathrm{mol/L}$; $[\mathrm{SO_4^{2-}}] = 0.450\ \mathrm{mol/L}$ · (d)ions carry charge; glucose molecules carry no net charge离子携带电荷；葡萄糖分子无净电荷
(a) Role of water molecules: ion-dipole forces水分子的作用：离子-偶极力A1·A1·A1
Water is a polar molecule with a partial negative charge ($\delta^-$) on oxygen and partial positive charges ($\delta^+$) on the two hydrogen atoms. When $\mathrm{Al_2(SO_4)_3}$ crystals enter water, the $\delta^-$ oxygen atoms of water molecules orient toward the positively charged $\mathrm{Al^{3+}}$ ions on the crystal surface; simultaneously, the $\delta^+$ hydrogen atoms orient toward the negatively charged $\mathrm{SO_4^{2-}}$ ions. These ion-dipole attractions pull individual ions away from the crystal lattice and surround them with a shell of water molecules (hydration shell), lowering the overall energy and making dissolution favourable.水是极性分子，氧上带有部分负电荷（$\delta^-$），两个氢上带有部分正电荷（$\delta^+$）。当 $\mathrm{Al_2(SO_4)_3}$ 晶体进入水中，水分子的 $\delta^-$ 氧原子朝向晶面上带正电荷的 $\mathrm{Al^{3+}}$ 离子；同时，$\delta^+$ 氢原子朝向带负电荷的 $\mathrm{SO_4^{2-}}$ 离子。这些离子-偶极吸引力将单个离子从晶格中拉出，并以水分子壳（水化层）包围它们，降低整体能量，使溶解过程有利。
(b) Complete dissociation equation with state symbols含状态符号的完整解离方程式A1·A1
$$ \mathrm{Al_2(SO_4)_3(s)} \;\longrightarrow\; 2\,\mathrm{Al^{3+}(aq)} + 3\,\mathrm{SO_4^{2-}(aq).} $$
One formula unit of $\mathrm{Al_2(SO_4)_3}$ releases 2 $\mathrm{Al^{3+}}$ ions and 3 $\mathrm{SO_4^{2-}}$ ions (5 ions total, $i = 5$).一个 $\mathrm{Al_2(SO_4)_3}$ 化学式单元释放 2 个 $\mathrm{Al^{3+}}$ 和 3 个 $\mathrm{SO_4^{2-}}$（共 5 个离子，$i = 5$）。
(c) Ion concentrations in $0.150\ \mathrm{mol/L}$ $\mathrm{Al_2(SO_4)_3}$$0.150\ \mathrm{mol/L}$ $\mathrm{Al_2(SO_4)_3}$ 溶液中的离子浓度A1·A1
(d) Why $\mathrm{Al_2(SO_4)_3}$ solution conducts but glucose solution does not为何 $\mathrm{Al_2(SO_4)_3}$ 溶液导电而葡萄糖溶液不导电A1
$\mathrm{Al_2(SO_4)_3}$ is a strong electrolyte: it fully dissociates into mobile $\mathrm{Al^{3+}}$ and $\mathrm{SO_4^{2-}}$ ions that carry electric charge through the solution. Glucose ($\mathrm{C_6H_{12}O_6}$) is a molecular nonelectrolyte: it dissolves as intact, neutral molecules that carry no net charge and therefore cannot conduct electricity.$\mathrm{Al_2(SO_4)_3}$ 是强电解质：完全解离为可移动的 $\mathrm{Al^{3+}}$ 和 $\mathrm{SO_4^{2-}}$ 离子，这些离子携带电荷在溶液中传导电流。葡萄糖（$\mathrm{C_6H_{12}O_6}$）是分子型非电解质：以完整的中性分子溶解，不带净电荷，因此无法导电。
Hydration: water molecules form a sphere around each ion, stabilising it in solution.水化：水分子在每个离子周围形成球形水化层，使其在溶液中稳定存在。The energy released by forming ion-dipole bonds (hydration energy) must overcome the lattice energy holding the ions together in the crystal. For ionic compounds that dissolve readily, hydration energy $\ge$ lattice energy. $\mathrm{Al^{3+}}$ has a high charge density ($3+$ charge, small radius), so it forms very strong ion-dipole bonds with water and has an especially large hydration energy. This is why highly charged ions like $\mathrm{Al^{3+}}$ and $\mathrm{Fe^{3+}}$ are extensively hydrated in solution. The SCH3U E3.1 outcome specifically asks students to use the terms "ion-dipole" and "hydration" when describing the dissolving of ionic compounds.形成离子-偶极键释放的能量（水化能）必须超过将离子固定在晶体中的晶格能。对于容易溶解的离子化合物，水化能 $\ge$ 晶格能。$\mathrm{Al^{3+}}$ 具有很高的电荷密度（$3+$ 电荷、小离子半径），因此与水形成非常强的离子-偶极键，水化能特别大。这就是高电荷离子（如 $\mathrm{Al^{3+}}$、$\mathrm{Fe^{3+}}$）在溶液中被广泛水化的原因。SCH3U E3.1 学习成果明确要求学生在描述离子化合物溶解过程时使用"离子-偶极"和"水化"这两个术语。
(a) Method 1: mass of solid NaCl required方法 1：所需固体 NaCl 质量M1·A1·A1
First find the moles needed:先计算所需摩尔数：
$$ n \;=\; c \times V \;=\; 0.500 \times 0.250 \;=\; 0.1250 \;\mathrm{mol.} $$
Then convert to mass:再换算为质量：
$$ m \;=\; n \times M_r \;=\; 0.1250 \times 58.44 \;=\; 7.305 \;\approx\; 7.31 \;\mathrm{g.} $$
(d) Which has greater osmotic pressure: NaCl or glucose solution?哪种溶液渗透压更大：NaCl 还是葡萄糖溶液？A1·A1
The NaCl solution has the greater osmotic pressure. Osmotic pressure $\pi = iMRT$ depends on the total dissolved particle concentration, not just the formula-unit molarity. Both solutions are at $0.500\ \mathrm{mol/L}$, but NaCl produces $i = 2$ ions per formula unit (effective particle concentration $= 1.00\ \mathrm{mol/L}$), while glucose ($i = 1$) contributes only $0.500\ \mathrm{mol/L}$ of dissolved particles. Since osmotic pressure is proportional to particle count, the NaCl solution exerts twice the osmotic pressure of the glucose solution.NaCl 溶液渗透压更大。渗透压 $\pi = iMRT$ 取决于总溶解颗粒浓度，而非仅仅是化学式单元摩尔浓度。两种溶液均为 $0.500\ \mathrm{mol/L}$，但 NaCl 每个化学式单元产生 $i = 2$ 个离子（有效颗粒浓度 $= 1.00\ \mathrm{mol/L}$），而葡萄糖（$i = 1$）只贡献 $0.500\ \mathrm{mol/L}$ 溶解颗粒。由于渗透压与颗粒数成正比，NaCl 溶液的渗透压是葡萄糖溶液的两倍。
Two routes to the same solution: from solid or from dilution. Both give identical results.两种方法配制同一溶液：固体溶解或稀释。两种方法结果完全一致。Method 1 (solid) uses $n = cV$ then $m = nM_r$. Method 2 (dilution) uses $c_1V_1 = c_2V_2$ to find the stock volume. The two answers $(7.31\ \mathrm{g}$ vs $31.3\ \mathrm{mL})$ are equivalent because $7.31\ \mathrm{g} \div 58.44\ \mathrm{g/mol} = 0.125\ \mathrm{mol}$ and $4.00\ \mathrm{mol/L} \times 0.0313\ \mathrm{L} = 0.125\ \mathrm{mol}$, confirming both produce $0.125\ \mathrm{mol}$ NaCl in $250\ \mathrm{mL}$. Part (d) links to the biological context of the question: physiological saline ($\approx 0.154\ \mathrm{mol/L}$ NaCl, or $0.9\%$ w/v) is used intravenously because its osmotic pressure matches blood plasma. Using glucose at the same molarity would give half the osmotic pressure, causing cells to swell. This integration of chemistry with biology is tested on BC Chemistry 11 and AP Biology.方法 1（固体）用 $n = cV$，再 $m = nM_r$。方法 2（稀释）用 $c_1V_1 = c_2V_2$ 求储备液体积。两个答案（$7.31\ \mathrm{g}$ 与 $31.3\ \mathrm{mL}$）等价：$7.31\ \mathrm{g} \div 58.44\ \mathrm{g/mol} = 0.125\ \mathrm{mol}$，$4.00\ \mathrm{mol/L} \times 0.0313\ \mathrm{L} = 0.125\ \mathrm{mol}$，证明两种方法均在 $250\ \mathrm{mL}$ 中产生 $0.125\ \mathrm{mol}$ NaCl。第 (d) 小问联系题目的生物背景：生理盐水（约 $0.154\ \mathrm{mol/L}$ NaCl，即 $0.9\%$ w/v）用于静脉注射，因为其渗透压与血浆匹配。若用同摩尔浓度的葡萄糖，渗透压只有一半，会导致细胞膨胀。这种化学与生物的整合在卑诗省化学 11 和 AP 生物中均有考查。