Osmosis: mass changes and rates use different denominators
導入済み| English | 意味 |
|---|---|
| osmosis/ɒzˈməʊsɪs/ | 浸透 |
| percentage mass change/pəˈsentɪdʒ mæs tʃeɪndʒ/ | 質量変化率 |
What would explain this observation?
- Two plant samples can gain the same mass but show different percentage changes. Starting mass and exposure time both matter when comparing uptake.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Osmosis 渗透 is net movement of water from a dilute solution to a more concentrated solution through a partially permeable membrane. Plant tissue can gain water in a more dilute surrounding solution and lose water in a more concentrated one. A zero net change suggests balanced water exchange under the measured conditions.
- osmosis: Net water movement through a partially permeable membrane from dilute to more concentrated solution; percentage mass change 质量变化百分比: Mass change divided by initial mass, multiplied by one hundred.
Which denominator is used for percentage mass change?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Percentage change compares the change with where the sample STARTED, so the denominator is the initial mass. Using the final mass or the time gives a different quantity with a different meaning.
Match each technical term to its precise meaning.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- Percentage change = (final−initial)/initial×100. Mean water-uptake rate compares an estimated water amount with elapsed time. These are different quantities: percentage change adjusts for starting mass, while a rate describes change per unit time. Plot a concentration series and estimate where the trend crosses zero rather than expecting every individual repeat to lie on one line.
- For required practical 3, use repeated equal-size pieces from comparable tissue in teacher-approved sugar or salt solutions. Keep solution volume, temperature and duration fixed; blot consistently before weighing. Record all raw masses and units. A range of concentrations allows a meaningful zero-change estimate and a check for anomalies.
Which two habits make the investigation or model in this case more defensible?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Stating the conditions and assumptions lets others repeat the work exactly, and recording the supporting evidence lets them check the conclusion. Hiding disagreeing results removes the evidence that could challenge a wrong conclusion, and one measurement can never establish a wider claim.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: initial mass 3.00 g and final mass 3.30 g after 40 min. Gain=0.30 g. Percentage gain=0.30/3.00×100=10%. Mean gain rate=0.30/40=0.0075 g/min under the model that the mass change represents water. The two answers have different meanings and units.
A 2.00 g sample gains 0.20 g in 20 min. Calculate mean mass-gain rate. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
A 2.00 g sample gains 0.20 g in 20 min. Calculate mean mass-gain rate.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
The result is 0.01 g/min. Known: 2.00 g sample gains 0.20 g in 20 min. Use mean rate = gain / time. Rate = 0.20/20 = 0.01 g per minute. This rate is distinct from the percentage change, which would be 0.20/2.00 x 100 = 10%.
Check the conclusion and its limits
- Surface liquid adds mass without entering the cells, so blotting is essential. A negative percentage indicates loss. An estimated zero-change concentration does not mean water molecules stop moving or that every cell is undamaged and identical.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Percentage mass change and mass-gain rate always have the same units. This claim is false: Percentage change is dimensionless (change / initial mass); rate carries units of mass per time. The two answer different questions - how much relative to the start versus how fast.
Osmosis: mass changes and rates use different denominators: Percentage change = (final−initial)/initial×100. Mean water-uptake rate compares an estimated water amount with elapsed time. These are different quantities: percentage change adjusts for starting mass, while a rate describes change per unit time. Plot a concentration series and estimate where the trend crosses zero rather than expecting every individual repeat to lie on one line.
Percentage mass change and mass-gain rate always have the same units.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Percentage change is dimensionless (change / initial mass); rate carries units of mass per time. The two answer different questions - how much relative to the start versus how fast.
Net water movement through a partially permeable membrane from dilute to more concentrated solution: write the technical term.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
osmosis means Net water movement through a partially permeable membrane from dilute to more concentrated solution.