Solution concentration by mass: convert volume before calculating
導入済み| English | 意味 |
|---|---|
| solute/ˈsɒljuːt/ | 溶質 |
| mass concentration/mæs ˌkɒnsənˈtreɪʃn/ | 質量濃度 |
What would explain this observation?
- A bottle labelled 8 grams per cubic decimetre contains 8 g solute 溶质 in each 1 dm³ of solution. A quarter of that solution volume contains a quarter of the solute mass.
- 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 质量浓度 states the mass of dissolved solute in a given volume of solution. With concentration c in $\dfrac{\text{g}}{\text{dm}^3}$, mass m in grams and solution volume V in dm³, m=cV. One dm³ equals 1,000 cm³. Thus 250 cm³ equals 0.250 dm³, not 250 dm³. Use final solution volume rather than assuming the solvent volume is identical.
- solute: The substance dissolved in a solvent to form a solution; mass concentration: The mass of solute per stated volume of solution.
Which volume should be used with a concentration in grams per dm³?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The concentration is per CUBIC DECIMETRE, so the volume must be in dm^3: 250 cm^3 = 0.250 dm^3. Using 250 with a per-dm^3 concentration is a thousandfold error, and a mass is never a volume.
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
- For a uniformly mixed solution, the amount of solute taken is proportional to the sample volume. A 0.250 dm³ sample of an 8.0 $\dfrac{\text{g}}{\text{dm}^3}$ solution contains 2.0 g solute. The unit calculation cancels dm³ and leaves grams. This shared-tier numerical task is distinct from the embedded Higher-only explanation of how changing mass or solution volume changes concentration.
- Write the concentration unit beside the value and convert volume before multiplying. Mark whether a number refers to solute, solvent or solution. For an actual teacher-approved preparation, weigh the selected solute, dissolve it and make the solution up to the required final volume using appropriate school apparatus. Check complete dissolution and safe handling; the written calculation does not replace actual supervised technique.
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: 30 $\dfrac{\text{g}}{\text{dm}^3}$ solution is sampled at 150 cm³. Volume=150/1,000=0.150 dm³, so solute mass=30×0.150=4.5 g. A 300 cm³ portion contains 9.0 g under the same uniform-concentration condition. The sample’s total solution mass is not given by this solute calculation.
Find solute mass in 200 cm³ of a 15 $\dfrac{\text{g}}{\text{dm}^3}$ solution. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
Find solute mass in 200 cm³ of a 15 grams per dm³ solution.
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 3 g. Known: 200 cm^3 = 0.200 dm^3; concentration 15 g per dm^3. Use mass = concentration x volume. Mass = 15 x 0.200 = 3 g of solute. This is the solute mass, not the mass of the whole solution.
Check the conclusion and its limits
- Do not multiply concentration by a cm³ number when the concentration uses dm³. Dissolved solute is still present even if invisible. A concentration is not the total mass of the bottle, and equal solution volumes need not contain equal solute masses when concentrations differ.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Solute mass calculated from concentration is the total mass of the solution. This claim is false: Concentration x volume gives the mass of SOLUTE only. The solution also contains the solvent - roughly a kilogram of water per litre - so total solution mass is far larger.
Solution concentration by mass: convert volume before calculating: For a uniformly mixed solution, the amount of solute taken is proportional to the sample volume. A 0.250 dm³ sample of an 8.0 $\dfrac{\text{g}}{\text{dm}^3}$ solution contains 2.0 g solute. The unit calculation cancels dm³ and leaves grams. This shared-tier numerical task is distinct from the embedded Higher-only explanation of how changing mass or solution volume changes concentration.
Solute mass calculated from concentration is the total mass of the solution.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Concentration x volume gives the mass of SOLUTE only. The solution also contains the solvent - roughly a kilogram of water per litre - so total solution mass is far larger.
The substance dissolved in a solvent to form a 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
solute means The substance dissolved in a solvent to form a solution.