PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 25 marksAP 风格选择题 + 安/卑省考短答 · 共 25 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. For short-answer items, state units in every answer. Use $M_r(\mathrm{NaCl})=58.44\ \mathrm{g/mol}$ where needed. No calculator on Q1-Q2; calculator permitted on Q3-Q5.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足以让阅卷人核对的过程。短答题每道都要写出单位。需要时取 $M_r(\mathrm{NaCl})=58.44\ \mathrm{g/mol}$。Q1-Q2 不可使用计算器；Q3-Q5 可用计算器。
A student dissolves copper sulfate ($\mathrm{CuSO_4}$) in water to make a blue solution. Which statement correctly identifies the components?一位同学将硫酸铜（$\mathrm{CuSO_4}$）溶于水，配制出蓝色溶液。以下哪项正确识别了各组成部分？
(A)$\mathrm{CuSO_4}$ is the solvent; water is the solute$\mathrm{CuSO_4}$ 是溶剂；水是溶质
(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
(C)Both $\mathrm{CuSO_4}$ and water are solvents$\mathrm{CuSO_4}$ 和水都是溶剂
(D)The blue solution is a heterogeneous mixture蓝色溶液是非均匀混合物
A student dissolves $11.70\ \mathrm{g}$ of NaCl ($M_r = 58.44\ \mathrm{g/mol}$) in enough water to prepare 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。
(a)Calculate the number of moles of NaCl.计算 NaCl 的摩尔数。[2]
(b)Calculate the molar concentration of the solution, with units.计算溶液的摩尔浓度，写出单位。[2]
(c)State one laboratory procedure that ensures the volume is exactly $400.0\ \mathrm{mL}$.写出一种确保体积恰好为 $400.0\ \mathrm{mL}$ 的实验操作。[1]
(d)State the assumption about the solute-solvent interaction that allows $c = n/V$ to give a reliable result.写出允许 $c = n/V$ 给出可靠结果的溶质-溶剂相互作用假设。[1]
Q4MEDIUM中🇨🇦 BC卑BC Provincial-style卑诗省考风格§4 Dilution稀释 · BC Chem 11[6 marks][6 分]
A laboratory has a stock solution of $\mathrm{HCl}$ at $8.00\ \mathrm{mol/L}$. A student needs to prepare $500\ \mathrm{mL}$ of a $0.400\ \mathrm{mol/L}$ $\mathrm{HCl}$ solution for a titration.实验室有 $8.00\ \mathrm{mol/L}$ 的盐酸（$\mathrm{HCl}$）储备液。一位同学需要配制 $500\ \mathrm{mL}$ 的 $0.400\ \mathrm{mol/L}$ $\mathrm{HCl}$ 溶液用于滴定。
(a)State the dilution equation and identify the two conserved quantities it encodes.写出稀释方程，并说明它所表达的两个守恒量。[2]
(b)Calculate the volume of stock solution required. Give the answer in mL.计算所需储备液的体积，结果以 mL 表示。[2]
(c)Describe the safe laboratory procedure for preparing the diluted solution. Warn of one safety hazard.描述配制稀溶液的安全实验操作，并指出一个安全隐患。[2]
At $60\ ^\circ\mathrm{C}$, the solubility of $\mathrm{KNO_3}$ is $110\ \mathrm{g}$ per $100\ \mathrm{g}$ of water. A student dissolves $80\ \mathrm{g}$ of $\mathrm{KNO_3}$ in $100\ \mathrm{g}$ of water at $60\ ^\circ\mathrm{C}$. The solution is then slowly cooled to $20\ ^\circ\mathrm{C}$, at which temperature the solubility of $\mathrm{KNO_3}$ is $32\ \mathrm{g}$ per $100\ \mathrm{g}$ of water.在 $60\ ^\circ\mathrm{C}$ 时，$\mathrm{KNO_3}$ 的溶解度为每 $100\ \mathrm{g}$ 水 $110\ \mathrm{g}$。一位同学在 $60\ ^\circ\mathrm{C}$ 下将 $80\ \mathrm{g}$ $\mathrm{KNO_3}$ 溶于 $100\ \mathrm{g}$ 水中，然后缓慢冷却至 $20\ ^\circ\mathrm{C}$，此温度下 $\mathrm{KNO_3}$ 的溶解度为每 $100\ \mathrm{g}$ 水 $32\ \mathrm{g}$。
(a)Classify the solution at $60\ ^\circ\mathrm{C}$ as saturated, unsaturated, or supersaturated. Justify.将 $60\ ^\circ\mathrm{C}$ 时的溶液分类为饱和、不饱和或过饱和溶液，并说明理由。[2]
(b)Calculate the mass of $\mathrm{KNO_3}$ that crystallises when the solution is cooled to $20\ ^\circ\mathrm{C}$.计算溶液冷却至 $20\ ^\circ\mathrm{C}$ 时析出的 $\mathrm{KNO_3}$ 晶体质量。[2]
(c)A gas is dissolved in the same water. State what happens to its solubility as the temperature rises from $20\ ^\circ\mathrm{C}$ to $60\ ^\circ\mathrm{C}$, and give one real-world consequence.同一水中溶有一种气体。说明从 $20\ ^\circ\mathrm{C}$ 升温至 $60\ ^\circ\mathrm{C}$ 时该气体溶解度的变化，并给出一个实际生活中的后果。[2]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 30 marksAP 衔接简答题 + 荣誉级 · 共 30 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. State the formula used before substituting. Include units in every final answer. Calculator permitted on Q6-Q9. For Q9, state assumptions explicitly.每一步推理都要写出。代入数值前先写出所用公式。每个最终答案都要写单位。Q6-Q9 可用计算器。Q9 须明确写出假设条件。
A solution of magnesium chloride $\mathrm{MgCl_2}$ is prepared by dissolving $23.8\ \mathrm{g}$ of $\mathrm{MgCl_2}$ ($M_r = 95.21\ \mathrm{g/mol}$) in enough water to make $1.000\ \mathrm{L}$ of solution. Assume complete dissociation.将 $23.8\ \mathrm{g}$ 氯化镁 $\mathrm{MgCl_2}$（$M_r = 95.21\ \mathrm{g/mol}$）溶于足量水中，配制 $1.000\ \mathrm{L}$ 溶液。假设完全解离。
(a)Calculate the molar concentration of $\mathrm{MgCl_2}$.计算 $\mathrm{MgCl_2}$ 的摩尔浓度。[2]
(b)Write the equation for the complete dissociation of $\mathrm{MgCl_2}$ in water.写出 $\mathrm{MgCl_2}$ 在水中完全解离的方程式。[1]
(c)Calculate $[\mathrm{Mg^{2+}}]$ and $[\mathrm{Cl^-}]$ in the solution.计算溶液中 $[\mathrm{Mg^{2+}}]$ 和 $[\mathrm{Cl^-}]$。[2]
(d)Calculate the total ion concentration in the solution.计算溶液中总离子浓度。[1]
(e)Classify $\mathrm{MgCl_2}$ as a strong electrolyte, weak electrolyte, or nonelectrolyte. Justify briefly.将 $\mathrm{MgCl_2}$ 分类为强电解质、弱电解质或非电解质，并简要说明理由。[1]
A student has a $2.40\ \mathrm{mol/L}$ stock solution of $\mathrm{H_2SO_4}$. She pipettes $75.0\ \mathrm{mL}$ of this stock solution into a $500\ \mathrm{mL}$ volumetric flask and dilutes to the mark with distilled water to make Solution A.一位同学有 $2.40\ \mathrm{mol/L}$ 的硫酸（$\mathrm{H_2SO_4}$）储备液。她用移液管吸取 $75.0\ \mathrm{mL}$ 储备液，转移至 $500\ \mathrm{mL}$ 容量瓶中，加蒸馏水稀释至刻度线，配制成溶液 A。
(a)Calculate the concentration of Solution A.计算溶液 A 的浓度。[2]
(b)What volume (mL) of Solution A contains exactly $0.0720\ \mathrm{mol}$ of $\mathrm{H_2SO_4}$?溶液 A 中哪个体积（mL）恰好含有 $0.0720\ \mathrm{mol}$ 的 $\mathrm{H_2SO_4}$？[2]
(c)The student takes $200\ \mathrm{mL}$ of Solution A and further dilutes it to $600\ \mathrm{mL}$ to make Solution B. Calculate the concentration of Solution B.该同学取 $200\ \mathrm{mL}$ 溶液 A，进一步稀释至 $600\ \mathrm{mL}$，配制成溶液 B。计算溶液 B 的浓度。[2]
(d)In both dilution steps, explain what quantity is conserved and why $c_1V_1 = c_2V_2$ holds.在两步稀释中，说明哪个量保持守恒，以及 $c_1V_1 = c_2V_2$ 成立的原因。[2]
A student dissolves $9.96\ \mathrm{g}$ of anhydrous copper sulfate $\mathrm{CuSO_4}$ ($M_r = 159.6\ \mathrm{g/mol}$) in water and transfers the solution to a $250.0\ \mathrm{mL}$ volumetric flask, adding water to the mark. She labels this Solution X.一位同学将 $9.96\ \mathrm{g}$ 无水硫酸铜 $\mathrm{CuSO_4}$（$M_r = 159.6\ \mathrm{g/mol}$）溶于水，转移至 $250.0\ \mathrm{mL}$ 容量瓶并加水至刻度线，标记为溶液 X。
(a)Calculate the concentration of Solution X. Show the full conversion from mass to moles to molarity.计算溶液 X 的浓度，展示从质量到摩尔数再到摩尔浓度的完整换算过程。[3]
(b)The student pipettes exactly $50.0\ \mathrm{mL}$ of Solution X into a $200.0\ \mathrm{mL}$ volumetric flask and dilutes to the mark to make Solution Y. Calculate the concentration of Solution Y.该同学用移液管精确吸取 $50.0\ \mathrm{mL}$ 溶液 X，转移至 $200.0\ \mathrm{mL}$ 容量瓶，加水至刻度线配制溶液 Y。计算溶液 Y 的浓度。[2]
(c)Calculate the mass of $\mathrm{CuSO_4}$ present in $150.0\ \mathrm{mL}$ of Solution Y.计算 $150.0\ \mathrm{mL}$ 溶液 Y 中 $\mathrm{CuSO_4}$ 的质量。[2]
(d)Without calculation, state whether Solution Y is more or less concentrated than Solution X, and explain why.无需计算，说明溶液 Y 比溶液 X 浓还是稀，并解释原因。[1]
Three aqueous solutions are each prepared at a concentration of $0.10\ \mathrm{mol/L}$: (A) glucose $\mathrm{C_6H_{12}O_6}$ (a nonelectrolyte), (B) sodium chloride $\mathrm{NaCl}$ (a strong electrolyte), (C) calcium chloride $\mathrm{CaCl_2}$ (a strong electrolyte).以 $0.10\ \mathrm{mol/L}$ 浓度分别配制三种水溶液： (A) 葡萄糖 $\mathrm{C_6H_{12}O_6}$（非电解质）， (B) 氯化钠 $\mathrm{NaCl}$（强电解质）， (C) 氯化钙 $\mathrm{CaCl_2}$（强电解质）。
(a)Write the dissociation equations for NaCl and $\mathrm{CaCl_2}$ in water.写出 NaCl 和 $\mathrm{CaCl_2}$ 在水中的解离方程式。[2]
(b)Calculate the total dissolved particle concentration (in $\mathrm{mol/L}$) for each of the three solutions. Show your reasoning.计算三种溶液各自的总溶解颗粒浓度（$\mathrm{mol/L}$），写出推理过程。[3]
(c)Rank the three solutions in order of increasing boiling-point elevation (smallest first). Explain the trend using your particle counts from (b).按沸点升高从小到大排列三种溶液（最小在前），并用 (b) 中的颗粒浓度解释这一趋势。[2]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 26 marks阿省毕业考 + 通用题型 · 共 26 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define symbols (with units) at the start of each question. State the equation used before substituting. Conclude each question with a one-sentence contextual answer. Calculator permitted throughout Part III.每题开始时定义符号（含单位）。代入数值前先写出所用方程。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。
A chemistry technician prepares a solution by dissolving $55\ \mathrm{g}$ of potassium nitrate ($\mathrm{KNO_3}$) in $100\ \mathrm{g}$ of water at $40\ ^\circ\mathrm{C}$. At $40\ ^\circ\mathrm{C}$ the solubility of $\mathrm{KNO_3}$ is $65\ \mathrm{g}$ per $100\ \mathrm{g}$ water. The solution is then heated to $70\ ^\circ\mathrm{C}$ (solubility $= 138\ \mathrm{g}/100\ \mathrm{g}$ water) and more $\mathrm{KNO_3}$ is added until the solution is just saturated at $70\ ^\circ\mathrm{C}$. The saturated solution is then rapidly cooled to $10\ ^\circ\mathrm{C}$ (solubility $= 21\ \mathrm{g}/100\ \mathrm{g}$ water).一名化学技术员在 $40\ ^\circ\mathrm{C}$ 时将 $55\ \mathrm{g}$ 硝酸钾（$\mathrm{KNO_3}$）溶于 $100\ \mathrm{g}$ 水中。$40\ ^\circ\mathrm{C}$ 时 $\mathrm{KNO_3}$ 的溶解度为每 $100\ \mathrm{g}$ 水 $65\ \mathrm{g}$。溶液加热至 $70\ ^\circ\mathrm{C}$（溶解度 $= 138\ \mathrm{g}/100\ \mathrm{g}$ 水），继续添加 $\mathrm{KNO_3}$ 直至在 $70\ ^\circ\mathrm{C}$ 时恰好饱和。将饱和溶液快速冷却至 $10\ ^\circ\mathrm{C}$（溶解度 $= 21\ \mathrm{g}/100\ \mathrm{g}$ 水）。
(a)Classify the solution at $40\ ^\circ\mathrm{C}$ (before heating) as saturated, unsaturated, or supersaturated. Justify.将 $40\ ^\circ\mathrm{C}$ 时（加热前）的溶液分类为饱和、不饱和或过饱和，并说明理由。[2]
(b)Calculate the additional mass of $\mathrm{KNO_3}$ that must be added at $70\ ^\circ\mathrm{C}$ to just saturate the solution.计算在 $70\ ^\circ\mathrm{C}$ 时使溶液恰好饱和需额外加入的 $\mathrm{KNO_3}$ 质量。[3]
(c)Calculate the mass of $\mathrm{KNO_3}$ that crystallises when the saturated solution (from b) is rapidly cooled to $10\ ^\circ\mathrm{C}$.计算将（b）中的饱和溶液快速冷却至 $10\ ^\circ\mathrm{C}$ 时析出的 $\mathrm{KNO_3}$ 晶体质量。[2]
(d)Rapid cooling can produce a supersaturated solution. Describe what this means at the molecular level and what event triggers crystallisation.快速冷却可产生过饱和溶液。从分子层面描述其含义，以及什么事件会触发结晶。[2]
A student adds crystals of aluminium sulfate $\mathrm{Al_2(SO_4)_3}$ ($M_r = 342.2\ \mathrm{g/mol}$) to water. The aluminium sulfate dissolves completely, and the solution conducts electricity well.一位同学将硫酸铝 $\mathrm{Al_2(SO_4)_3}$（$M_r = 342.2\ \mathrm{g/mol}$）晶体加入水中，硫酸铝完全溶解，溶液导电性良好。
(a)Describe the role of water molecules in dissolving the $\mathrm{Al_2(SO_4)_3}$ crystal. Reference ion-dipole forces in your answer.描述水分子在溶解 $\mathrm{Al_2(SO_4)_3}$ 晶体中的作用，在回答中引用离子-偶极力。[3]
(b)Write the complete dissociation equation for $\mathrm{Al_2(SO_4)_3}$ in water, including state symbols.写出 $\mathrm{Al_2(SO_4)_3}$ 在水中的完整解离方程式，包括状态符号。[2]
(c)A $0.150\ \mathrm{mol/L}$ solution of $\mathrm{Al_2(SO_4)_3}$ is prepared. Calculate $[\mathrm{Al^{3+}}]$ and $[\mathrm{SO_4^{2-}}]$ in this solution.配制 $0.150\ \mathrm{mol/L}$ 的 $\mathrm{Al_2(SO_4)_3}$ 溶液。计算该溶液中 $[\mathrm{Al^{3+}}]$ 和 $[\mathrm{SO_4^{2-}}]$。[2]
(d)Explain why the solution conducts electricity, whereas a glucose solution of the same concentration does not.解释为什么该溶液能导电，而相同浓度的葡萄糖溶液却不能。[1]
A lab technician needs to prepare $250\ \mathrm{mL}$ of a $0.500\ \mathrm{mol/L}$ NaCl solution to simulate physiological saline. She has solid NaCl ($M_r = 58.44\ \mathrm{g/mol}$) and also a $4.00\ \mathrm{mol/L}$ NaCl stock solution available.一名实验技术员需要配制 $250\ \mathrm{mL}$ 的 $0.500\ \mathrm{mol/L}$ NaCl 溶液，用于模拟生理盐水。她有固体 NaCl（$M_r = 58.44\ \mathrm{g/mol}$）和 $4.00\ \mathrm{mol/L}$ NaCl 储备液两种原料。
(a)Method 1 (from solid): Calculate the mass of NaCl required to prepare $250\ \mathrm{mL}$ of $0.500\ \mathrm{mol/L}$ NaCl solution directly from the solid.方法 1（固体溶解）：计算直接用固体 NaCl 配制 $250\ \mathrm{mL}$ 的 $0.500\ \mathrm{mol/L}$ NaCl 溶液所需的 NaCl 质量。[3]
(b)Method 2 (dilution): Calculate the volume (mL) of the $4.00\ \mathrm{mol/L}$ stock solution needed to prepare $250\ \mathrm{mL}$ of $0.500\ \mathrm{mol/L}$ NaCl by dilution.方法 2（稀释）：计算通过稀释 $4.00\ \mathrm{mol/L}$ 储备液配制 $250\ \mathrm{mL}$ 的 $0.500\ \mathrm{mol/L}$ NaCl 溶液所需的储备液体积（mL）。[2]
(c)NaCl is a strong electrolyte. Write the dissociation equation and calculate the concentration of each ion in the final $0.500\ \mathrm{mol/L}$ NaCl solution.NaCl 是强电解质。写出解离方程式，并计算最终 $0.500\ \mathrm{mol/L}$ NaCl 溶液中每种离子的浓度。[2]
(d)A glucose solution and the NaCl solution are both at $0.500\ \mathrm{mol/L}$. Which has the greater osmotic pressure, and why?葡萄糖溶液和 NaCl 溶液均为 $0.500\ \mathrm{mol/L}$。哪种溶液的渗透压更大？为什么？[2]
🇨🇦 Alberta阿尔伯塔Chem 20 Unit C · C-GO1 · Chem 30 diploma-style calculations化学 30 毕业考计算
Full Syllabus Map in ../Study Guides/Unit_8_Solutions_and_Solubility.html. Colligative properties (Q9) exceed the NGSS/SCH3U floor but are core for IB/AP feeders; Q9 is Honors-flagged accordingly.完整大纲对照见 ../Study Guides/Unit_8_Solutions_and_Solubility.html。依数性（Q9）超出 NGSS/SCH3U 基准，但为 IB/AP 衔接核心内容，Q9 已标注荣誉级。