Question detail
For Concentrations in mol/dm3 and titration calculations, which ratio reasoning answer best supports this Unit 4.3 objective: (chemistry only) (HT only) Rearrange concentration equations. (MS 3b)?
Try the question, check the answer, then read the explanation to understand the curriculum point.
At a glance
MCQ
Type
practice
Style
Topic
Using concentrations of solutions in mol/dm3 (chemistry only) (HT only)
Question
- A. 0.34 mol/dm3 - correct ratio reasoning for chemistry only
- B. Wrong ratio reasoning: confuses chemistry only with a nearby Unit 4.3 idea
- C. Wrong unit check: uses the wrong unit for Concentrations in mol/dm3 and titration calculations
- D. Wrong calculation link: does not support (chemistry only) (HT only) Rearrange concentration equations. (MS 3b)
Answer
The correct option is 0.34 mol/dm3 - correct ratio reasoning for chemistry only. The calculated answer is 0.34 mol/dm3.
Explanation
The correct option is 0.34 mol/dm3 - correct ratio reasoning for chemistry only. This uses Concentration in mol/dm3 because the objective is about (chemistry only) (HT only) Rearrange concentration equations. (MS 3b). This ratio reasoning variant asks students to separate chemistry only from similar Unit 4.3 calculation steps. The reasoning belongs to Concentrations in mol/dm3 and titration calculations within Using concentrations of solutions in mol/dm3 (chemistry only) (HT only), so it should not be confused with nearby quantitative ideas such as mass, moles, concentration, yield, atom economy, or gas volume unless those are named in the objective. Use the focus term chemistry only to keep the answer aligned with AQA GCSE Chemistry 8462 Unit 4.3. The other options are weaker because they either use the wrong formula, the wrong unit, a vague relationship, or the wrong quantitative context.
Common mistake
Misunderstanding Rearrangement
Students often forget to isolate the variable they need when rearranging concentration equations, leading to incorrect formulas.
Practice isolating the variable step-by-step, ensuring to perform the same operation on both sides of the equation.
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