📚 Year 12 CCEA Science: Case Study Practice | Year 12 CCEA 科学：案例分析实战演练
Case study questions are a key part of the CCEA Year 12 Science assessment. They test your ability to apply scientific knowledge, analyse experimental data, and evaluate procedures. This article guides you through a complete case study example, showing you step by step how to tackle each section with confidence. The model investigation explores how different concentrations of sodium chloride solution affect the mass of potato chips. You will practise calculating percentage change, plotting graphs, interpreting trends using osmosis, and evaluating the experiment — all essential skills for your exam.
案例分析题是 CCEA 12 年级科学考试的重要组成部分，考查你运用科学知识、分析实验数据和评价实验方法的能力。本文将带你完整演练一个案例，逐步展示如何从容应对每个部分。示范研究探究不同浓度的氯化钠溶液对土豆条质量的影响。你将练习计算质量变化百分比、绘制图表、借助渗透作用解释趋势，以及评价实验——这些都是你考试必备的技能。
1. Understanding the Case Study Background | 理解案例背景
A student wanted to find out how changing the concentration of sodium chloride solution influences the mass of potato chips. Five beakers were set up, each containing a different concentration of NaCl solution: 0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 mol dm⁻³. Potato chips of similar size were cut, blotted dry, and their initial mass was measured. After 30 minutes, the chips were removed, blotted again, and their final mass was recorded. The student then calculated the percentage change in mass for each chip.
一名学生想探究改变氯化钠溶液浓度会如何影响土豆条的质量。他准备了六个烧杯，分别盛有不同浓度的 NaCl 溶液：0.0、0.2、0.4、0.6、0.8 和 1.0 mol dm⁻³。切出大小相近的土豆条，吸干表面水分，测量初始质量。30 分钟后取出土豆条，再次吸干水分，记录最终质量。该学生随后计算了每条土豆条的质量变化百分比。
Reading the question carefully is vital. Identify what is being measured and what is being changed. In this case, the student is investigating the effect of salt concentration on mass change in potato tissue, which points toward osmosis as the underlying process.
仔细阅读题目至关重要。要明确测量的对象和改变的因素。在此案例中，学生研究盐浓度对土豆组织质量变化的影响，这提示背后的原理是渗透作用。
2. Identifying Variables | 识别变量
The independent variable is the concentration of sodium chloride solution. It is deliberately changed by the experimenter across the range 0.0 to 1.0 mol dm⁻³. Always state the variable and give the units where appropriate.
自变量是氯化钠溶液的浓度。实验者有目的地将其在 0.0 到 1.0 mol dm⁻³ 范围内改变。记得陈述变量并标注单位。
The dependent variable is the percentage change in mass of the potato chips. This is the outcome that responds to changes in the independent variable. Using percentage change rather than absolute change allows fairer comparison because chips may have slightly different starting masses.
因变量是土豆条质量的百分比变化。这是随自变量变化而响应的结果。采用百分比变化而非绝对变化，能让比较更为公平，因为薯条初始质量可能略有差异。
Several control variables must be kept constant: the volume of solution in each beaker (e.g. 100 cm³), the temperature (room temperature), the time of immersion (30 minutes), the surface area and shape of the chips, and the type of potato. Controlling these ensures that any mass change is due only to the salt concentration.
必须保持恒定的控制变量包括：每个烧杯中溶液的体积（如 100 cm³）、温度（室温）、浸泡时间（30 分钟）、薯条的表面积和形状，以及土豆品种。控制这些变量可确保任何质量变化仅由盐浓度引起。
3. Experimental Setup and Control | 实验设置与对照
A proper experimental setup requires a clear plan. The 0.0 mol dm⁻³ beaker contains distilled water and acts as a control. This provides a baseline for comparison, showing the mass change when no solute is present. The chip in pure water is expected to gain mass because water potential in the cells is lower than that of pure water.
合理的实验设计需要有清晰的方案。0.0 mol dm⁻³ 的烧杯盛装蒸馏水，充当对照。这提供了比较的基线，显示无溶质存在时的质量变化。预计纯水中的薯条质量会增加，因为细胞内的水势低于纯水。
To blot the chips, the student should use the same number of dabs with absorbent paper each time. This reduces variation in surface water, a common source of random error. All chips should be fully submerged in their respective solutions.
学生每次应用吸水纸轻压薯条相同的次数以吸干水分。这可以减少表面水带来的变异，这是一种常见的随机误差来源。所有薯条都应完全浸没在各自的溶液中。
The experiment should be repeated at least three times for each concentration. Replicates allow calculation of a mean percentage change, improving reliability and helping identify anomalous results.
每一浓度至少应重复实验三次。重复允许计算平均百分比变化，提高可靠性，并有助于识别异常结果。
4. Data Collection and Presentation | 数据收集与呈现
The results table below shows typical data collected by the student. Note how the table includes initial mass, final mass, mass change, and percentage change. Units are given in the headings.
下表展示了学生收集的典型数据。注意表格中包含了初始质量、最终质量、质量变化和百分比变化，并在表头标注单位。
| NaCl concentration / mol dm⁻³ | Initial mass / g | Final mass / g | Mass change / g | Percentage change / % |
|---|---|---|---|---|
| 0.0 | 2.50 | 2.72 | +0.22 | +8.8 |
| 0.2 | 2.45 | 2.56 | +0.11 | +4.5 |
| 0.4 | 2.52 | 2.51 | -0.01 | -0.4 |
| 0.6 | 2.48 | 2.33 | -0.15 | -6.0 |
| 0.8 | 2.51 | 2.22 | -0.29 | -11.6 |
| 1.0 | 2.47 | 2.15 | -0.32 | -13.0 |
When presenting your own data, always use clear headings and record values to the same number of decimal places. The mass change is calculated by subtracting initial mass from final mass; a negative value indicates a loss in mass.
在呈现你的数据时，务必使用清晰的表头，并将数值记录到相同的小数位数。质量变化由最终质量减去初始质量得出；负值表示质量减少。
5. Performing Calculations: Percentage Change | 进行计算：质量变化百分比
The percentage change in mass is found using the formula:
质量变化百分比用下列公式计算：
Percentage change = (final mass – initial mass) / initial mass × 100%
Let us verify the calculation for 0.4 mol dm⁻³: (-0.01 g / 2.52 g) × 100% = -0.4%. For 1.0 mol dm⁻³: (-0.32 g / 2.47 g) × 100% = -13.0%. Always show one example working in your answer.
我们来验证 0.4 mol dm⁻³ 的计算：（-0.01 g / 2.52 g）× 100% = -0.4%。对于 1.0 mol dm⁻³：（-0.32 g / 2.47 g）× 100% = -13.0%。作答时务必展示一个计算示例。
A positive percentage change means the chip gained water by osmosis, while a negative percentage change means it lost water. If repeated readings exist, calculate the mean percentage change for each concentration and use the mean for plotting.
正百分比变化表示薯条通过渗透吸收了水分，负百分比变化则表示失去了水分。若存在重复读数，应计算每个浓度的平均百分比变化，并用平均值作图。
6. Drawing an Accurate Graph | 绘制精准图表
To visualise the relationship, plot a line graph with sodium chloride concentration on the x-axis (independent variable) and percentage change in mass on the y-axis (dependent variable). Choose scales that use more than half of the graph paper in both directions.
为直观显示关系，应绘制折线图，x 轴为氯化钠浓度（自变量），y 轴为质量变化百分比（因变量）。坐标轴刻度要利用图表纸的一半以上区域。
ne of best fit or connect the points with straight line segments, depending on the question' r biological data, a smooth curve is often most appropriate. Add a descriptive title, e.g. 'Graph showing the effect of NaCl concentration on potato chip mass change'.
用叉号（×）小心标记每个数据点，然后根据题目要求画出最佳拟合线或用直线连接各点。对于生物学数据，平滑曲线通常最为合适。添加描述性标题，如"显示 NaCl 浓度对土豆条质量变化影响的图"。
The graph must show where the line crosses the x-axis — this is the concentration at which there is no net mass change, indicating that the solution is isotonic to the potato cells' cytoplasm.
图表必须显示线与 x 轴的交点——即无净质量变化时的浓度，表明该溶液与土豆细胞细胞质等渗。
7. Interpreting the Graph and Trends | 解释图表与趋势
From the data, we see a clear trend: as NaCl concentration increases, the percentage change in mass becomes more negative. The chip gains mass in pure water and low salt concentrations, loses mass slightly at 0.4 mol dm⁻³, and loses significantly more at higher concentrations.
从数据可看出明显趋势：随着 NaCl 浓度升高，质量变化百分比越发呈负值。薯条在纯水和低盐浓度中质量增加，在 0.4 mol dm⁻³ 时质量略微减少，在更高浓度下质量显著减少。
The point where the plotted line crosses the x-axis represents the water potential equilibrium. At this concentration (around 0.4 mol dm⁻³), the water potential inside the potato cells equals that of the surrounding solution, so there is no net movement of water.
图中线与 x 轴的交点代表水势平衡。在该浓度（大约 0.4 mol dm⁻³）下，土豆细胞内部的水势与周围溶液的水势相等，因此没有净水分移动。
Below the equilbrium concentration, water enters the cells by osmosis, causing the mass to increase and the cells to become turgid. Above it, water leaves the cells, the cytoplasm shrinks away from the cell wall (plasmolysis), and the mass decreases.
低于平衡浓度时，水分通过渗透进入细胞，使质量增加，细胞变得饱满。高于平衡浓度时，水分离开细胞，细胞质从细胞壁收缩分离（质壁分离），质量下降。
8. Explaining the Science: Osmosis | 解释科学原理：渗透作用
Osmosis is the net movement of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane. The potato cell membrane acts as the partially permeable membrane.
渗透是水分子通过部分透性膜从水势较高区域向水势较低区域的净移动。土豆细胞膜充当了部分透性膜。
Pure water has a water potential of zero; adding solute lowers water potential (makes it more negative). Therefore, the 0.0 mol dm⁻³ solution has the highest water potential, and the 1.0 mol dm⁻³ solution has the lowest (most negative).
纯水的水势为零；加入溶质会降低水势（使之更负）。因此，0.0 mol dm⁻³ 溶液水势最高，1.0 mol dm⁻³ 溶液水势最低（最负）。
In hypotonic solutions (lower solute concentration than the cell contents), water enters the cells, increasing mass. In hypertonic solutions, water leaves the cells, decreasing mass. When the solutions are isotonic, there is no net change in mass. These terms should be used in exam explanations.
在低渗溶液中（溶质浓度低于细胞内含物），水进入细胞，质量增加。在高渗溶液中，水离开细胞，质量减少。当溶液等渗时，质量没有净变化。考试解释时应使用这些术语。
9. Evaluating Anomalous Data | 评估异常数据
In any experiment, some data points may not fit the overall pattern. For instance, if the 0.4 mol dm⁻³ result were much higher than expected, it would be an anomaly. In our table, the values follow a smooth trend, but consider what to do if an outlier is present.
在任何实验中，某些数据点可能不符合整体规律。例如，如果 0.4 mol dm⁻³ 的结果远高于预期，那就是异常值。在我们的表格中，数值遵循平滑趋势，但仍要考虑若出现离群值该如何处理。
Possible causes of anomalies include: failure to blot consistently, chips with different surface areas, evaporation during weighing, or incorrect reading of the balance. Always identify the most likely source and suggest a practical remedy, such as using a template to cut chips to uniform dimensions.
异常值可能的原因包括：吸干水分不一致、薯条表面积不同、称重过程中蒸发，或读数错误。务必指出最可能的原因并提出实际改进方法，例如用模板将薯条切成统一尺寸。
If an anomaly is identified, it should be circled on the graph, excluded from the line of best fit, and repeated readings should be considered. State that exclusion helps prevent skewed conclusions.
如果识别出异常值，应在图上圈出，不纳入最佳拟合线计算，并考虑重复实验数据。明确说明排除异常值可以防止结论偏离。
10. Identifying Limitations and Improvements | 识别局限与改进措施
Every experiment has limitations. In this case study, the main limitation is difficulty in controlling all variables perfectly — potato chips may vary in shape, exact surface area, and tissue type (e.g. more or less vascular tissue).
每个实验都有局限性。在本案例中，主要局限是难以完美控制所有变量——土豆条在形状、确切表面积和组织类型（如维管组织多少）上可能存在差异。
To improve accuracy, use a cork borer to produce chips of identical diameter and cut them to exactly the same length using a scalpel and ruler. Measure mass to 0.01 g or better, and control temperature with a water bath to avoid fluctuations that affect the rate of osmosis.
为提高准确性，用打孔器制作直径相同的薯条，并用手术刀和直尺切成完全相同的长度。质量测量精确到 0.01 g 或更佳，使用水浴控制温度以避免影响渗透速率的波动。
Extend the investigation by measuring the change in length as well as mass, or by testing other varieties of potato to confirm the pattern. Also, more closely spaced concentrations near the isotonic point would refine the estimate of the cell's solute potential.
可延伸的研究包括同时测量长度变化与质量变化，或测试其他土豆品种以确认规律。此外，在等渗点附近设置更密集的浓度可以更精确地估计细胞的溶质势。
11. Conclusion and Exam Tips | 结论与考试技巧
The case study demonstrates that increasing NaCl concentration reduces the water potential gradient between potato cells and the solution, leading to a shift from water gain to water loss. The isotonic point is found where the line of best fit crosses zero percentage change.
本案例表明，提高 NaCl 浓度会减小土豆细胞与溶液之间的水势梯度，导致从吸水转为失水。等渗点可以在最佳拟合线与零百分比变化相交处找到。
Exam tip: always use precise scientific vocabulary — 'water potential', 'partially permeable membrane', 'plasmolysis'. Support your answers with numerical evidence from the table or graph. For evaluation questions, structure your response: identify a specific limitation, explain its effect on results, and propose a realistic improvement.
考试技巧：始终使用精确的科学词汇——"水势""部分透性膜""质壁分离"。用表格或图表中的数值证据支撑答案。面对评价题，构建你的回答：指出具体的局限，解释它对结果的影响，提出切实可行的改进措施。
Practice drawing graphs quickly and accurately, and always check that your plotted points match the table values. By repeatedly applying this structured approach to different case studies, you will develop the confidence and skill needed to excel in the CCEA Year 12 Science exam.
练习快速准确地绘图，并始终检查所描点是否与表格数据一致。通过反复将这种结构化方法应用于不同的案例分析，你将培养出在 CCEA 12 年级科学考试中脱颖而出所需的信心与技能。
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