Higher Tier, Chemistry-only: concentration in moles per solution volume
| English | 中文 | Pinyin |
|---|---|---|
| molar concentration | 物质的量浓度 | wù zhì dì liàng nóng dù |
| aliquot | 定量分取液 | dìng liàng fēn qǔ yè |
What would explain this observation?
- Higher Tier: A concentration of 0.20 moles per cubic decimetre tells the amount of dissolved solute, not its gram mass. The molar mass is needed to connect amount and mass.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Molar concentration 物质的量浓度 c=n/V uses solute amount n in mol and final solution volume V in dm³. Rearrange to n=cV or V=n/c. To find solute mass use m=nM, where M is molar mass in grams per mol. Combining gives c=m/(MV). The same gram mass of two different solutes can have different mole amounts because their molar masses differ.
- molar concentration: The amount in moles of stated solute per volume of solution; aliquot 定量分取液: A measured portion of a larger sample of solution.
Which sequence finds solute mass from molar concentration?
Convert cm³ to dm³ by dividing by 1,000. A 250 cm³ portion of 0.20 moles per dm³ solution contains 0.20×0.250=0.050 mol solute. If solute molar mass is 40 g per mol, that amount weighs 2.0 g. At fixed amount, greater final solution volume means lower molar concentration. At fixed final volume, more solute moles means higher concentration.
Match each technical term to its precise meaning.
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- Convert cm³ to dm³ by dividing by 1,000. A 250 cm³ portion of 0.20 $\dfrac{\text{mol}}{\text{dm}^3}$ solution contains 0.20×0.250=0.050 mol solute. If solute molar mass is 40 $\dfrac{\text{g}}{\text{mol}}$, that amount weighs 2.0 g. At fixed amount, greater final solution volume means lower molar concentration. At fixed final volume, more solute moles means higher concentration.
- Write units beside concentration and volume before multiplying. Identify which chemical species concentration describes. A bottle’s molar concentration normally refers to the stated solute formula amounts, not automatically the total amount of every ion after dissociation. Actual solution preparation requires a measured final volume, complete dissolution and the teacher-approved apparatus and procedure.
Which two habits make the investigation or model in this case more defensible?
Write units beside concentration and volume before multiplying. Identify which chemical species concentration describes. A bottle’s molar concentration normally refers to the stated solute formula amounts, not automatically the total amount of every ion after dissociation. Actual solution preparation requires a measured final volume, complete dissolution and the teacher-approved apparatus and procedure.
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: 4.0 g NaOH with supplied molar mass 40 $\dfrac{\text{g}}{\text{mol}}$ is 0.100 mol. In final volume 0.500 dm³ its concentration is 0.100/0.500=0.200 $\dfrac{\text{mol}}{\text{dm}^3}$. A 100 cm³ aliquot contains 0.0200 mol and 0.800 g NaOH. This is solute content rather than total solution mass.
A 200 cm³ solution contains 0.30 $\dfrac{\text{mol}}{\text{dm}^3}$ solute. Calculate its amount. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
A 200 cm³ solution contains 0.30 mol per dm³ solute. Calculate its amount.
The result is 0.06 mol. Known: 4.0 g NaOH with supplied molar mass 40 g per mol is 0.100 mol. In final volume 0.500 dm³ its concentration is 0.100/0.500=0.200 moles per dm³. A 100 cm³ aliquot contains 0.0200 mol and 0.800 g NaOH. This is solute content rather than total solution mass.
Check the conclusion and its limits
- Do not substitute grams directly into c=n/V. Do not use the volume of solvent added when a final solution volume is specified. Molar concentration and mass concentration are related through molar mass, but their numerical values and units are different.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
A molar concentration states grams of solute per dm³. This claim is false: Do not substitute grams directly into c=n/V. Do not use the volume of solvent added when a final solution volume is specified. Molar concentration and mass concentration are related through molar mass, but their numerical values and units are different.
Higher Tier, Chemistry-only: concentration in moles per solution volume: Convert cm³ to dm³ by dividing by 1,000. A 250 cm³ portion of 0.20 $\dfrac{\text{mol}}{\text{dm}^3}$ solution contains 0.20×0.250=0.050 mol solute. If solute molar mass is 40 $\dfrac{\text{g}}{\text{mol}}$, that amount weighs 2.0 g. At fixed amount, greater final solution volume means lower molar concentration. At fixed final volume, more solute moles means higher concentration.
A molar concentration states grams of solute per dm³.
Do not substitute grams directly into c=n/V. Do not use the volume of solvent added when a final solution volume is specified. Molar concentration and mass concentration are related through molar mass, but their numerical values and units are different.
The amount in moles of stated solute per volume of solution: write the technical term.
molar concentration means The amount in moles of stated solute per volume of solution.