Osmosis: mass changes and rates use different denominators
| English | 中文 | Pinyin |
|---|---|---|
| osmosis/ɒzˈməʊsɪs/ | 渗透作用 | shèn tòu zuò yòng |
| percentage mass change/pəˈsentɪdʒ mæs tʃeɪndʒ/ | 质量变化百分比 | zhì liàng biàn huà bǎi fēn bǐ |
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 渗透作用 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?
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.
Match each technical term to its precise meaning.
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?
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.
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.
The result is 0.01 g/min. 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.
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: 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.
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.
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.
Net water movement through a partially permeable membrane from dilute to more concentrated solution: write the technical term.
osmosis means Net water movement through a partially permeable membrane from dilute to more concentrated solution.