Separate insoluble solids and recover dissolved solutes
| English | 中文 | Pinyin |
|---|---|---|
| residue/ˈresɪdjuː/ | 滤渣 | lǜ zhā |
| filtrate/ˈfɪltreɪt/ | 滤液 | lǜ yè |
What would explain this observation?
- A filter can remove sand from salt water, but the salt solution passes through. Recovering both solids requires two different physical processes.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- A mixture contains elements or compounds that are not chemically combined with one another; each retains its chemical properties. Filtration separates an insoluble solid from a liquid: sand stays as the residue 滤渣 and salt solution passes through as the filtrate 滤液. Dissolved ions pass through ordinary filter paper with the solvent. Crystallisation recovers a dissolved solid by concentrating its solution and allowing crystals to form as it cools.
- residue: The insoluble solid retained by a filter; filtrate: The liquid or solution that passes through a filter.
After filtering sand mixed with salt water, where is the dissolved salt?
For sand mixed with salt, add water and stir to dissolve the salt, filter the insoluble sand, then gently evaporate some water from the filtrate. Stop concentration before boiling dry and allow the solution to cool; filter and dry the crystals. Washing the sand removes adhering solution, and washing crystals with a little cold suitable solvent removes some surface impurity while limiting dissolution.
Match each technical term to its precise meaning.
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- For sand mixed with salt, add water and stir to dissolve the salt, filter the insoluble sand, then gently evaporate some water from the filtrate. Stop concentration before boiling dry and allow the solution to cool; filter and dry the crystals. Washing the sand removes adhering solution, and washing crystals with a little cold suitable solvent removes some surface impurity while limiting dissolution.
- Use actual teacher-supervised school separation equipment, eye protection and a risk-assessed heat source. Fold and support the paper in a funnel, collect filtrate in a clean vessel and transfer solution to an evaporating basin. Use a safe end-point such as crystals forming in a cooled test drop. Hot glass can appear cold. Record mass only after the recovered material is dry.
Which two habits make the investigation or model in this case more defensible?
Use actual teacher-supervised school separation equipment, eye protection and a risk-assessed heat source. Fold and support the paper in a funnel, collect filtrate in a clean vessel and transfer solution to an evaporating basin. Use a safe end-point such as crystals forming in a cooled test drop. Hot glass can appear cold. Record mass only after the recovered material is dry.
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: a dry starting mixture contains 6.0 g sand and 10.0 g salt. A separation recovers 5.4 g dry sand and 8.0 g dry salt. Sand recovery = 5.4/6.0×100 = 90%; salt recovery = 8.0/10.0×100 = 80%. Lower recovery can reflect transfer losses or solute left in the remaining solution; it does not show that salt atoms vanished.
A mixture initially contains 12.0 g salt; 9.0 g dry salt is recovered. Calculate percentage recovery. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
A mixture initially contains 12.0 g salt; 9.0 g dry salt is recovered. Calculate percentage recovery.
The result is 75 %. Known: a dry starting mixture contains 6.0 g sand and 10.0 g salt. A separation recovers 5.4 g dry sand and 8.0 g dry salt. Sand recovery = 5.4/6.0×100 = 90%; salt recovery = 8.0/10.0×100 = 80%. Lower recovery can reflect transfer losses or solute left in the remaining solution; it does not show that salt atoms vanished.
Check the conclusion and its limits
- Filtering salt solution does not remove dissolved salt. Crystallisation changes physical arrangement, not the chemical identity of the solute. A wet recovered mass overestimates dry solid mass. Heating every solution to dryness is not a universal purification method because some substances decompose.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Ordinary filter paper removes dissolved salt from water. This claim is false: Filtering salt solution does not remove dissolved salt. Crystallisation changes physical arrangement, not the chemical identity of the solute. A wet recovered mass overestimates dry solid mass. Heating every solution to dryness is not a universal purification method because some substances decompose.
Separate insoluble solids and recover dissolved solutes: For sand mixed with salt, add water and stir to dissolve the salt, filter the insoluble sand, then gently evaporate some water from the filtrate. Stop concentration before boiling dry and allow the solution to cool; filter and dry the crystals. Washing the sand removes adhering solution, and washing crystals with a little cold suitable solvent removes some surface impurity while limiting dissolution.
Ordinary filter paper removes dissolved salt from water.
Filtering salt solution does not remove dissolved salt. Crystallisation changes physical arrangement, not the chemical identity of the solute. A wet recovered mass overestimates dry solid mass. Heating every solution to dryness is not a universal purification method because some substances decompose.
The insoluble solid retained by a filter: write the technical term.
residue means The insoluble solid retained by a filter.