Higher Tier: explain how mass and volume change concentration
| English | 中文 | Pinyin |
|---|---|---|
| final solution volume | 溶液最终体积 | róng yè zuì zhōng tǐ jī |
| dilution | 稀释 | xī shì |
What would explain this observation?
- Higher Tier: Adding water to a solution can lower concentration while leaving the dissolved solute mass unchanged. Concentration describes a ratio, not whether solute has disappeared.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Mass concentration c=m/V compares solute mass with final solution volume 溶液最终体积. At fixed volume, doubling dissolved mass doubles concentration. At fixed dissolved mass, doubling final solution volume halves concentration. If both change, calculate the new ratio: multiplying mass and volume by the same factor leaves concentration unchanged. This explanatory relationship is an embedded Higher-only requirement in 4.3.2.5.
- dilution 稀释: Lowering concentration by adding solvent while retaining the stated solute amount; final solution volume: The total volume of solution after the stated preparation or dilution.
At fixed solute mass, what happens when final solution volume doubles?
During simple dilution with pure solvent and no losses or reaction, the solute mass stays constant. Therefore c₁V₁=c₂V₂ when both concentrations use the same mass units and both volumes the same volume units. The final volume includes the original sample and added solvent; it is not the volume of water added alone. Dissolution and dilution should not be confused with a chemical transformation of solute.
Match each technical term to its precise meaning.
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- During simple dilution with pure solvent and no losses or reaction, the solute mass stays constant. Therefore c₁V₁=c₂V₂ when both concentrations use the same mass units and both volumes the same volume units. The final volume includes the original sample and added solvent; it is not the volume of water added alone. Dissolution and dilution should not be confused with a chemical transformation of solute.
- Identify what is held constant before stating a direction of change. Rearrange c=m/V to V=m/c or m=cV. For a planned school dilution, use appropriate measured transfer and make up to the required final volume under teacher supervision. Concentrated acids require the school’s specific safe procedure; the generic arithmetic is not an instruction to mix hazardous liquids.
Which two habits make the investigation or model in this case more defensible?
Identify what is held constant before stating a direction of change. Rearrange c=m/V to V=m/c or m=cV. For a planned school dilution, use appropriate measured transfer and make up to the required final volume under teacher supervision. Concentrated acids require the school’s specific safe procedure; the generic arithmetic is not an instruction to mix hazardous liquids.
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: 5.0 g solute in 0.250 dm³ solution gives 20 $\dfrac{\text{g}}{\text{dm}^3}$. Diluting to final volume 0.500 dm³ gives 10 $\dfrac{\text{g}}{\text{dm}^3}$. Alternatively, keeping final volume 0.250 dm³ and increasing dissolved solute to 10.0 g gives 40 $\dfrac{\text{g}}{\text{dm}^3}$. The two operations change different parts of the ratio.
A solution contains 6 g solute and is diluted to 0.75 dm³. Find its mass concentration. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
A solution contains 6 g solute and is diluted to 0.75 dm³. Find its mass concentration.
The result is 8 g per dm³. Known: 5.0 g solute in 0.250 dm³ solution gives 20 grams per dm³. Diluting to final volume 0.500 dm³ gives 10 grams per dm³. Alternatively, keeping final volume 0.250 dm³ and increasing dissolved solute to 10.0 g gives 40 grams per dm³. The two operations change different parts of the ratio.
Check the conclusion and its limits
- Do not say concentration always rises when volume rises: the held quantity matters. The equation assumes a uniform solution and that the stated mass is dissolved. Evaporation may also remove volatile solute, so constant solute mass must be a stated condition rather than automatically inferred.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Adding pure water removes half the solute atoms during dilution. This claim is false: Do not say concentration always rises when volume rises: the held quantity matters. The equation assumes a uniform solution and that the stated mass is dissolved. Evaporation may also remove volatile solute, so constant solute mass must be a stated condition rather than automatically inferred.
Higher Tier: explain how mass and volume change concentration: During simple dilution with pure solvent and no losses or reaction, the solute mass stays constant. Therefore c₁V₁=c₂V₂ when both concentrations use the same mass units and both volumes the same volume units. The final volume includes the original sample and added solvent; it is not the volume of water added alone. Dissolution and dilution should not be confused with a chemical transformation of solute.
Adding pure water removes half the solute atoms during dilution.
Do not say concentration always rises when volume rises: the held quantity matters. The equation assumes a uniform solution and that the stated mass is dissolved. Evaporation may also remove volatile solute, so constant solute mass must be a stated condition rather than automatically inferred.
Lowering concentration by adding solvent while retaining the stated solute amount: write the technical term.
dilution means Lowering concentration by adding solvent while retaining the stated solute amount.