OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2017: Question 12

1 mark · Medium difficulty · Multiple Choice

Calculate the number of molecules of water formed when 0.1 mol of HOOCCH2COOH is reacted with 0.1 mol of aqueous NaOH.

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Question

Multiple choice question 12 asking how many molecules of water are formed when 0.1 mol of HOOCCH2COOH is reacted with 0.1 mol of aqueous NaOH. Four options are provided: A (6.02 x 10^22), B (3.01 x 10^22), C (6.02 x 10^23), and D (3.01 x 10^23), with an answer box.
Question text

12 0.1 mol of HOOCCH2COOH are reacted with 0.1 mol of aqueous NaOH.

How many molecules of water are formed?

A 6.02 × 1022

B 3.01 × 1022

C 6.02 × 1023

D 3.01 × 1023

Your answer

[1]

Mark scheme

Show the mark scheme The mark scheme table shows question number 12 with the correct answer option A and 1 mark.

12 A 1

How to answer it

Neutralisation Stoichiometry & Avogadro's Constant

OCR A-Level Chemistry • Multiple Choice Question

What this question tests

This question assesses your understanding of acid-base stoichiometry, specifically dealing with a dibasic (diprotic) acid reacting with a limited amount of a strong base (sodium hydroxide). It also tests your ability to apply the molar ratio to determine moles of a product and use Avogadro's constant to convert moles into actual numbers of molecules.

Question 12

Full Worked Solution & Examiner Breakdown

✅ Correct Answer

A ( 6.02 × 10²² )

Mark Allocation: 1 mark for selecting option A.

💡 Key Knowledge

  • Propanedioic acid ( HOOCCH₂COOH ) is a dicarboxylic acid containing two acidic carboxylic acid (-COOH) groups per molecule.
  • In a neutralisation reaction, 1 mole of -COOH reacts with 1 mole of OH⁻ ions to form 1 mole of H₂O .
  • Avogadro's constant ( N_A ≈ 6.02 × 10²³ mol⁻¹ ) links moles to the number of particles (molecules/atoms/ions).

📐 Step-by-Step Calculation

  1. Identify limiting reagent: We have 0.1 mol of HOOCCH₂COOH and 0.1 mol of NaOH . Because the acid is diprotic, it has twice as many acidic protons available as moles of NaOH added. Therefore, NaOH is the limiting reagent.
  2. Use mole ratios: Each mole of OH⁻ reacted produces 1 mole of H₂O . Since 0.1 mol of NaOH reacts completely, exactly 0.1 mol of water is formed.
  3. Convert moles to molecules:
    Number of molecules = Moles × Avogadro's constant
    = 0.1 × 6.02 × 10²³
    = 6.02 × 10²² molecules.

🧠 Exam Technique

Don't fall into the trap of assuming 1 mole of acid always reacts with 1 mole of base! Always check the structure of the acid to see if it is mono-, di-, or tribasic. For quick multiple-choice questions, writing out a quick mental or scratchpad mole balance saves you from silly magnitude errors.

❌ Common Errors & Traps

  • Assuming complete neutralisation: Students often see "acid + alkali" and assume both carboxylic acid groups react, leading them to think 0.2 mol of acid requires 0.2 mol of NaOH . Here, only half-neutralisation occurs because NaOH is limited.
  • Stoichiometry confusion: Mistakenly thinking 1 mole of diprotic acid produces 2 moles of water per mole of acid reacted, regardless of the limiting reagent.
  • Power of 10 slips: Forgetting that multiplying by 0.1 ( 10⁻¹ ) shifts the standard Avogadro index ( 10²³ ) down to 10²² , leading candidates to incorrectly choose option C.

Topics

Module 2: Foundations in chemistry · Module 6: Organic chemistry and analysis · 6.1 Aromatic compounds, carbonyls and acids · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.