Learning objective

Calculate amounts using concentration, volume and moles in solution.

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At a glance

5

Flashcards

7

Questions

Topic

Amount of substance

Subtopic

The mole and the Avogadro constant

AQA A Level ChemistryPhysical chemistry

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Short explanation

In the subtopic The mole and the Avogadro constant, this AQA A-Level Chemistry 7405 learning objective focuses on calculate amounts using concentration, volume and moles in solution. It belongs to Amount of substance, so revision should stay anchored to this exact subtopic rather than drifting into a generic GCSE-level chemistry summary. Approved keywords to use include solution, mole, concentration. Concentration. means the amount of solute present in a given volume of solution, typically expressed in mol/dm³ or g/dm³ Avoid the mistake of students often forget to convert volume from cm³ to dm³ when calculating concentration, leading to incorrect results; instead, to fix this, remember that 1 dm³ equals 1000 cm³. Use the formula for concentration: concentration (c) = mass (m) / volume (V). If the volume is given in cm³, convert it to dm³ by dividing by 1000 before substituting into the formula. For example, if you have 50 g of solute in 250 cm³ of solution, first convert 250 cm³ to dm³: 250 cm³ = 0.250 dm³. Then, substitute: c = 50 g / 0.250 dm³ = 200 g/dm³ For exam answers, use the formula for concentration to calculate moles in solution: c = n / V

Key concepts

concentrationmoles

Why it matters

This objective helps connect The mole and the Avogadro constant to exam-style questions, flashcards, and revision notes for Amount of substance.

Common mistakes

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  • Common Mistake in Concentration Calculations: To fix this, remember that 1 dm³ equals 1000 cm³. Use the formula for concentration: concentration (c) = mass (m) / volume (V). If the volume is given in cm³, convert it to dm³ by dividing by 1000 before substituting into the formula. For example, if you have 50 g of solute in 250 cm³ of solution, first convert 250 cm³ to dm³: 250 cm³ = 0.250 dm³. Then, substitute: c = 50 g / 0.250 dm³ = 200 g/dm³.

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