Chemistry 9701/35 — May/June 2025

Background Concept

At room temperature and pressure (r.t.p.), one mole of any gas occupies approximately 24 dm³, which is 24,000 cm³. This is the molar gas volume. The amount of a gas in moles is therefore its volume in cm³ divided by 24,000.

Understanding the Question

You measured the volume of carbon dioxide collected in part (a). Now convert that volume to an amount in mol using the molar gas volume at room conditions.

Approach

Use the relationship amount = volume ÷ molar gas volume. Since the volume is in cm³, divide by 24,000 cm³ mol⁻¹.

Step-by-Step Reasoning

Taking the representative volume of 200 cm³:

The answer should be given to 2–4 significant figures. 0.00833 has three significant figures, which is acceptable.

Key Takeaways

• The molar gas volume at r.t.p. is 24 dm³ mol⁻¹ = 24,000 cm³ mol⁻¹.
• Always match the units of the volume to the units of the molar volume.

Common Mistakes

• Using 24 dm³ instead of 24,000 cm³, giving an answer 1000 times too large.
• Giving the answer to too many significant figures.

Things to Be Careful About

• The mark scheme requires the answer to 2–4 significant figures.
• Use the volume you actually recorded in part (a), not a textbook value.

Background Concept

A metal carbonate reacts with an acid to give a salt, water and carbon dioxide. For a group 1 carbonate M₂CO₃ reacting with sulfuric acid, the salt is the sulfate M₂SO₄. The general pattern is: carbonate + acid → salt + water + carbon dioxide.

Understanding the Question

Write the balanced symbol equation, with state symbols, for the reaction between the metal carbonate M₂CO₃ and sulfuric acid.

Approach

Identify the three products — the sulfate salt, water and carbon dioxide — then balance the equation and add state symbols.

Step-by-Step Reasoning

The reactants are M₂CO₃(s) and H₂SO₄(aq). The products are M₂SO₄(aq), H₂O(l) and CO₂(g). Balancing:

Each element balances: two M, one C, three O (plus four from the acid), two H, one S. The equation is already balanced as written.

Key Takeaways

• Acid + carbonate → salt + water + carbon dioxide.
• State symbols: solid carbonate, aqueous acid and salt, liquid water, gaseous carbon dioxide.

Common Mistakes

• Forgetting state symbols — the mark scheme requires them all.
• Writing the sulfate salt incorrectly (e.g. MSO₄ instead of M₂SO₄).

Things to Be Careful About

• The mark scheme requires all four state symbols.
• The metal is monovalent (group 1), so the sulfate is M₂SO₄.

Background Concept

From the balanced equation in (b)(ii), one mole of M₂CO₃ produces one mole of CO₂, so the amounts are equal. The relative formula mass of a compound is its mass divided by the amount in mol: M_r = m ÷ n.

Understanding the Question

Use the amount of CO₂ from (b)(i) and the 1:1 stoichiometry from (b)(ii) to find the amount of M₂CO₃, then divide the mass of FA 1 (from part (a)) by that amount to obtain the relative formula mass.

Approach

First, set the amount of M₂CO₃ equal to the amount of CO₂. Then divide the mass of FA 1 by this amount.

Step-by-Step Reasoning

From the equation, 1 mol M₂CO₃ gives 1 mol CO₂, so:

The answer is given to 3 significant figures, within the required 2–4 sf range.

Key Takeaways

• The 1:1 stoichiometry between carbonate and CO₂ is the key link.
• M_r = mass ÷ amount.

Common Mistakes

• Using the wrong stoichiometric ratio.
• Confusing mass and amount when substituting into the formula.

Things to Be Careful About

• The answer must be to 2–4 significant figures.
• Use the mass of FA 1 you calculated in part (a).

Background Concept

The relative formula mass of M₂CO₃ is the sum of the relative atomic masses: 2×A_r(M) + A_r(C) + 3×A_r(O) = 2A_r(M) + 12 + 48 = 2A_r(M) + 60. Rearranging gives A_r(M) = (M_r − 60) ÷ 2.

Understanding the Question

Deduce the relative atomic mass of metal M from the M_r calculated in (b)(iii), then identify which group 1 metal it is by comparing with known A_r values.

Approach

Subtract 60 (the mass of the CO₃ group) from M_r, divide by 2, then match the result to the periodic table.

Step-by-Step Reasoning

The group 1 metals have A_r values: Li 6.9, Na 23.0, K 39.1, Rb 85.5, Cs 132.9. The value 22.8 is closest to sodium (23.0).

The mark scheme uses boundary values between consecutive metals: Li ≤ 14.95 ≤ Na ≤ 31.05 ≤ K ≤ 62.30 ≤ Rb ≤ 109.20 ≤ Cs ≤ 250. Since 22.8 lies between 14.95 and 31.05, the metal is sodium.

Key Takeaways

• A_r(M) = (M_r(M₂CO₃) − 60) ÷ 2.
• Identify the metal by matching the calculated A_r to the periodic table.

Common Mistakes

• Forgetting to divide by 2.
• Not subtracting the mass of the carbonate group.

Things to Be Careful About

• The mark scheme provides boundary values for assigning the metal.
• Use the periodic table A_r values, not rounded mass numbers.

Background Concept

The amount of substance (in mol) is calculated from the relationship:

The concentration of FA 4 is given as 0.0500 mol dm$^{-3}$. The volume used is the mean titre from part (b), 34.20 cm$^3$. Since the concentration is in mol dm$^{-3}$, the volume must be converted from cm$^3$ to dm$^3$ by dividing by 1000.

Understanding the Question

You are asked to calculate the amount, in mol, of sulfuric acid present in the mean titre volume of FA 4 from part (b). This is the first step in a chain of calculations that will eventually identify the metal Z.

Approach

Apply the formula amount = concentration × volume. Convert the titre from cm$^3$ to dm$^3$ by dividing by 1000, then multiply by the concentration.

Step-by-Step Reasoning

The answer is given to 3 significant figures, which is consistent with the data (0.0500 has 3 sf, 34.20 has 4 sf; 3 sf is appropriate).

Key Takeaways

• amount (mol) = concentration × volume (in dm$^3$).
• Always convert cm$^3$ to dm$^3$ by dividing by 1000.
• Give the answer to 3 or 4 significant figures.

Common Mistakes

• Forgetting to convert cm$^3$ to dm$^3$ (would give 1.71 mol instead of $1.71 \times 10^{-3}$ mol).
• Using the wrong volume (e.g., the rough titre instead of the mean titre).
• Giving the answer to too few significant figures.

Things to Be Careful About

• The volume used must be the mean titre from part (b) — this is the value you selected as most reliable.
• The answer must be given to 3 or 4 significant figures per the mark scheme.
• Keep the units consistent: mol dm$^{-3}$ × dm$^3$ = mol.

Background Concept

The reaction between a metal carbonate and sulfuric acid is a neutralisation (acid–carbonate) reaction:

For every 1 mol of carbonate, 1 mol of acid is consumed. This 1:1 stoichiometric ratio is the key to converting the moles of acid (from part (c)(i)) into moles of carbonate.

Understanding the Question

You are asked to use your answer to (c)(i) — the moles of H$_2$SO$_4$ in the titre — to find the amount of Z$_2$CO$_3$ in 1.0 dm$^3$ of FA 3. The 25.0 cm$^3$ pipetted into the flask contains the same number of moles of Z$_2$CO$_3$ as the moles of acid that neutralised it (1:1 ratio). You then scale this up from 25.0 cm$^3$ to 1 dm$^3$ (1000 cm$^3$).

Approach

1. Use the 1:1 stoichiometry: moles of Z$_2$CO$_3$ in 25.0 cm$^3$ = moles of H$_2$SO$_4$ from (c)(i).
2. Scale to 1 dm$^3$: multiply by 1000/25 = 40.

Step-by-Step Reasoning

From (c)(i), the amount of H$_2$SO$_4$ is $1.71 \times 10^{-3}$ mol. Since the reaction is 1:1, the amount of Z$_2$CO$_3$ in the 25.0 cm$^3$ sample is also $1.71 \times 10^{-3}$ mol.

To find the amount in 1 dm$^3$, scale up:

This is the concentration of Z$_2$CO$_3$ in FA 3 in mol dm$^{-3}$.

Key Takeaways

• The stoichiometric ratio from the balanced equation converts moles of one reactant to moles of another.
• Scaling from a sample volume to 1 dm$^3$ uses the factor 1000/sample volume.
• This result (moles per dm$^3$) is the molar concentration of Z$_2$CO$_3$ in FA 3.

Common Mistakes

• Using the wrong stoichiometric ratio (e.g., assuming 2:1 because of the subscript 2 in Z$_2$CO$_3$). The balanced equation shows 1:1.
• Forgetting to scale from 25 cm$^3$ to 1 dm$^3$.
• Using the rough titre instead of the mean titre.

Things to Be Careful About

• The answer must be given to 3 or 4 significant figures.
• This value (moles of Z$_2$CO$_3$ per dm$^3$) is used in part (c)(iii) to find the molar mass, so accuracy matters.
• The balanced equation must be correct: Z$_2$CO$_3$ + H$_2$SO$_4$ → Z$_2$SO$_4$ + CO$_2$ + H$_2$O.

Background Concept

The concentration of FA 3 is given as 7.26 g dm$^{-3}$, meaning 1 dm$^3$ of solution contains 7.26 g of Z$_2$CO$_3$. From part (c)(ii), 1 dm$^3$ contains $6.84 \times 10^{-2}$ mol of Z$_2$CO$_3$. The molar mass is therefore:

The formula Z$_2$CO$_3$ contains two atoms of Z, one carbon atom (Ar = 12), and three oxygen atoms (Ar = 16 each). So:

Rearranging gives the relative atomic mass of Z. Comparing this with the periodic table identifies the metal.

Understanding the Question

You are asked to identify the Group 1 metal Z. You have the mass per dm$^3$ (7.26 g) and the moles per dm$^3$ (from (c)(ii)). The molar mass links these. From the molar mass, subtract the contribution of CO$_3$ (60) and divide by 2 to get the atomic mass of Z. Then identify the metal.

Approach

1. Calculate $M_r$(Z$_2$CO$_3$) = mass per dm$^3$ ÷ moles per dm$^3$.
2. Write $M_r$(Z$_2$CO$_3$) = $2A_r$(Z) + 60.
3. Solve for $A_r$(Z).
4. Compare with known Group 1 atomic masses: Li = 7, Na = 23, K = 39, Rb = 85, Cs = 133.

Step-by-Step Reasoning

Step 1 — Molar mass:

Step 2 — Set up the formula:

Step 3 — Solve:

Step 4 — Identify: The Group 1 metal with $A_r \approx 23$ is sodium (Na, $A_r = 23.0$).

The mark scheme provides ranges to allow for experimental error: Li ≤ 14.95 ≤ Na ≤ 31.05 ≤ K ≤ 62.30 ≤ Rb ≤ 109.20 ≤ Cs ≤ 250. Since 23.05 falls within the Na range (14.95–31.05), Z is sodium.

Key Takeaways

• Molar mass = mass ÷ moles.
• The formula of a compound relates its molar mass to the atomic masses of its elements.
• Experimental values may differ slightly from theoretical; use the mark scheme ranges to identify the metal.
• Group 1 carbonates have the general formula Z$_2$CO$_3$.

Common Mistakes

• Forgetting to subtract the CO$_3$ contribution (60) before dividing by 2.
• Forgetting to divide by 2 (the subscript in Z$_2$CO$_3$).
• Misidentifying the metal (e.g., choosing K because 23 is "close" to K's Ar of 39 — it is not; 23 is clearly Na).
• Using the wrong value for moles per dm$^3$ (e.g., using moles in 25 cm$^3$ instead of 1 dm$^3$).

Things to Be Careful About

• The answer must be given to 3 or 4 significant figures throughout.
• The mark scheme's ranges are: Li ≤ 14.95 ≤ Na ≤ 31.05 ≤ K ≤ 62.30 ≤ Rb ≤ 109.20 ≤ Cs ≤ 250. Your calculated $A_r$(Z) must fall within one of these ranges.
• If your titre were different, the metal identified could change — the calculation must be carried through with YOUR mean titre.

Background Concept

Qualitative analysis of aqueous cations relies on the formation of characteristic precipitates when aqueous sodium hydroxide (NaOH) or aqueous ammonia (NH₃) is added. The colour of the precipitate and its behaviour in excess reagent are diagnostic.

• Iron(II), Fe²⁺: Forms a green precipitate of iron(II) hydroxide, $\text{Fe(OH)}_2$, with both NaOH and NH₃. It is insoluble in excess. However, $\text{Fe}^{2+}$ is easily oxidised by atmospheric oxygen to $\text{Fe}^{3+}$, so the green precipitate gradually turns brown on standing as $\text{Fe(OH)}_3$ forms.
• Chromium(III), Cr³⁺: Forms a grey-green precipitate of chromium(III) hydroxide, $\text{Cr(OH)}_3$, with both NaOH and NH₃. With NaOH, the amphoteric hydroxide dissolves in excess to form the dark green tetrahydroxochromate(III) ion, $[\text{Cr(OH)}_4]^-$. With NH₃, it remains insoluble. On standing, $\text{Cr}^{3+}$ can be oxidised to $\text{CrO}_4^{2-}$ (chromate(VI)), turning the solution pale purple / mauve / lilac.
• Hydrogen peroxide test: $\text{H}_2\text{O}_2$ oxidises $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ (solution turns yellow/brown). When NaOH is then added, a dark brown $\text{Fe(OH)}_3$ precipitate forms. $\text{H}_2\text{O}_2$ also decomposes spontaneously (especially catalysed by metal ions) to give oxygen gas: $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$. Oxygen is identified by relighting a glowing splint.

Understanding the Question

You are given two unknown aqueous solutions, FA 6 and FA 7, each containing one cation and one anion. You must carry out three specific tests using a 1 cm depth of each solution and record your observations in Table 3.1. Note that Test 2 is only required for FA 7 (the FA 6 box is crossed out), and Test 3 is only required for FA 6 (the FA 7 box is crossed out).

Approach

Recall the standard observations for Fe²⁺ and Cr³⁺ with NaOH and NH₃. For Test 3, recognise that hydrogen peroxide will oxidise Fe²⁺ to Fe³⁺ and also decompose to release oxygen gas.

Step-by-Step Reasoning

Test 1: Add aqueous sodium hydroxide.

• FA 6 (Fe²⁺): $\text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_2(\text{s})$. Observation: green precipitate, insoluble in excess. On standing, it turns brown due to oxidation to $\text{Fe(OH)}_3$.
• FA 7 (Cr³⁺): $\text{Cr}^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{Cr(OH)}_3(\text{s})$. Observation: grey-green precipitate. In excess NaOH, it dissolves: $\text{Cr(OH)}_3(\text{s}) + \text{OH}^-(\text{aq}) \rightarrow [\text{Cr(OH)}_4]^-(\text{aq})$, forming a dark green solution.

Test 2: Add aqueous ammonia, let stand 5 minutes.

• Only FA 7 is tested. Cr³⁺ forms $\text{Cr(OH)}_3$, which is insoluble in excess NH₃ (unlike Cu²⁺ or Zn²⁺). On standing, aerial oxidation of Cr³⁺ to chromate(VI) turns the supernatant pale purple / mauve / lilac.

Test 3: Add hydrogen peroxide, then NaOH.

• Only FA 6 is tested. $\text{H}_2\text{O}_2$ oxidises Fe²⁺ to Fe³⁺ (solution may turn yellow). Decomposition of $\text{H}_2\text{O}_2$ produces $\text{O}_2$ gas (bubbling/effervescence; relights glowing splint). Adding NaOH then gives a dark brown precipitate of $\text{Fe(OH)}_3$.

Key Takeaways

• Fe²⁺ gives a green, air-sensitive precipitate with OH⁻; Cr³⁺ gives a grey-green precipitate that is soluble in excess NaOH but not excess NH₃, and oxidises to purple chromate on standing.
• Hydrogen peroxide acts as both an oxidising agent (Fe²⁺ → Fe³⁺) and a source of oxygen gas via decomposition.

Common Mistakes

• Writing "brown precipitate" for Fe²⁺ with NaOH directly; it is initially green and only turns brown on standing.
• Saying Cr³⁺ precipitate is soluble in excess NH₃; it is not (unlike Zn²⁺ or Cu²⁺).
• Forgetting to mention the gas test (relights glowing splint) for Test 3, which is a specific mark point.

Things to Be Careful About

• Ensure you only record observations for the uncrossed boxes in Table 3.1.
• Use precise colour descriptions: "green" (not just "coloured"), "grey-green", "dark green", "pale purple / mauve / lilac", "dark brown".

Background Concept

Anion identification in aqueous solutions uses specific precipitation reactions:

• Sulfate, $\text{SO}_4^{2-}$: Added to barium chloride or barium nitrate solution in the presence of dilute nitric acid. A white precipitate of barium sulfate, $\text{BaSO}_4$, forms, which is insoluble in dilute acids. The acid is added first to remove carbonate ions that could also form a white precipitate ($\text{BaCO}_3$) but would dissolve in acid.
• Chloride, $\text{Cl}^-$: Added to silver nitrate solution acidified with dilute nitric acid. A white precipitate of silver chloride, $\text{AgCl}$, forms. This precipitate is soluble in dilute aqueous ammonia.
• Bromide, $\text{Br}^-$: Cream precipitate of $\text{AgBr}$, sparingly soluble in dilute $\text{NH}_3$(aq).
• Iodide, $\text{I}^-$: Yellow precipitate of $\text{AgI}$, insoluble in dilute $\text{NH}_3$(aq).

Understanding the Question

You must identify the anions in FA 6 and FA 7. The marking scheme requires you to present your tests and observations clearly, typically in a table with headings. You must test both solutions with at least one common reagent (like $\text{BaCl}_2$ or $\text{AgNO}_3$), or test one solution with two different reagents.

Approach

Construct a simple table with columns for the test, reagent, and observations for each solution. Perform the standard sulfate test on FA 6 and the halide test on FA 7.

Step-by-Step Reasoning

M1: Table format. Create a table with clear headings: "Test", "Reagent", "Observation (FA 6)", "Observation (FA 7)". Ensure at least one reagent and one observation are listed.

M2: Reagent choice. Test both FA 6 and FA 7 with one reagent (e.g., $\text{BaCl}_2(\text{aq})$ or $\text{AgNO}_3(\text{aq})$), or test one with two reagents.

M3: Sulfate test for FA 6. Add $\text{BaCl}_2(\text{aq})$ (or $\text{Ba(NO}_3)_2(\text{aq})$) to FA 6, followed by dilute $\text{HNO}_3$. Observation: white precipitate, insoluble in dilute $\text{HNO}_3$. This confirms $\text{SO}_4^{2-}$.

M4: Halide test for FA 7. Add $\text{AgNO}_3(\text{aq})$ to FA 7. Observation: white precipitate. (Optionally add dilute $\text{NH}_3$(aq) to confirm it dissolves, proving $\text{Cl}^-$). This confirms $\text{Cl}^-$.

Key Takeaways

• Always acidify barium and silver nitrate tests with dilute $\text{HNO}_3$ to prevent false positives from carbonates.
• Present qualitative data in a clear, headed table to earn the presentation mark.

Common Mistakes

• Forgetting to include headings in the results table.
• Not testing both solutions with at least one common reagent, or not using enough reagents to identify both anions.
• Writing "white precipitate" for silver chloride without noting it is soluble in ammonia (though just "white ppt" with $\text{AgNO}_3$ often scores if the conclusion is correct).

Things to Be Careful About

• The mark scheme requires a table with headings (M1). A loose list of observations may not score this mark.
• Use dilute $\text{HNO}_3$ for acidification, not $\text{HCl}$ (which would add chloride ions and interfere with the silver nitrate test).
• Ensure your observations are specific: "white precipitate" is better than "a precipitate forms".

Background Concept

An ionic compound is formed from cations and anions in a ratio that ensures overall electrical neutrality. Once the individual ions are identified through qualitative analysis, the formula is written by balancing the charges.

• Iron(II) is $\text{Fe}^{2+}$, sulfate is $\text{SO}_4^{2-}$ → $\text{FeSO}_4$.
• Chromium(III) is $\text{Cr}^{3+}$, chloride is $\text{Cl}^-$ → $\text{CrCl}_3$.

Understanding the Question

Using the cation tests from (a)(i) and anion tests from (a)(ii), deduce the complete formulae for FA 6 and FA 7.

Approach

Pair the identified cation with the identified anion for each solution and ensure the charges balance.

Step-by-Step Reasoning

For FA 6:

• Cation test: Green precipitate with NaOH, insoluble in excess, turns brown on standing → $\text{Fe}^{2+}$.
• Anion test: White precipitate with $\text{BaCl}_2$, insoluble in dilute $\text{HNO}_3$ → $\text{SO}_4^{2-}$.
• Combine: $\text{Fe}^{2+}$ and $\text{SO}_4^{2-}$ balance in a 1:1 ratio → $\text{FeSO}_4$.

For FA 7:

• Cation test: Grey-green precipitate with NaOH, soluble in excess (dark green solution); grey-green ppt insoluble in excess NH₃, turns purple on standing → $\text{Cr}^{3+}$.
• Anion test: White precipitate with $\text{AgNO}_3$ → $\text{Cl}^-$.
• Combine: $\text{Cr}^{3+}$ and $\text{Cl}^-$ require a 1:3 ratio to balance charges → $\text{CrCl}_3$.

Key Takeaways

• Always verify that the final formula is electrically neutral by checking the charges of the identified ions.
• Roman numerals or oxidation states must match the qualitative behaviour (e.g., Fe²⁺ not Fe³⁺ for the initial green precipitate).

Common Mistakes

• Writing $\text{Fe}_2(\text{SO}_4)_3$ for FA 6; the cation test clearly indicates Fe(II), not Fe(III).
• Writing $\text{CrCl}$ or $\text{Cr}_2\text{Cl}_3$ instead of the correct $\text{CrCl}_3$.
• Forgetting that the question states "none of the ions contains nitrogen", which rules out nitrate or ammonium.

Things to Be Careful About

• The question asks for the formulae, not the names. Write $\text{FeSO}_4$ and $\text{CrCl}_3$, not "iron(II) sulfate".
• Ensure state symbols are not required unless specified; here, just the formulae are needed.

Background Concept

Acidified aqueous potassium manganate(VII), $\text{KMnO}_4$, is a strong oxidising agent. In acidic solution, the purple $\text{MnO}_4^-$ ion is reduced to the pale pink / colourless $\text{Mn}^{2+}$ ion:

Organic compounds that can be oxidised by $\text{KMnO}_4$ include:

• Primary (1º) alcohols: oxidised to aldehydes, then to carboxylic acids.
• Secondary (2º) alcohols: oxidised to ketones.
• Aldehydes: oxidised to carboxylic acids.

Compounds that are not oxidised by $\text{KMnO}_4$ under these mild conditions:

• Tertiary (3º) alcohols: no hydrogen on the carbon bearing the -OH group.
• Ketones: already at a high oxidation state and resist further oxidation.
• Alkanes, alkenes (slow), aromatic rings: generally unreactive or require harsher conditions.

Understanding the Question

You are testing an organic liquid FA 8 (containing only C, H, O and one functional group) with acidified $\text{KMnO}_4$ in a hot water bath. You must record the observation and draw a conclusion about what type of compound FA 8 is or is not.

Approach

Recall that $\text{KMnO}_4$ is purple and turns colourless (or pale yellow/brown if concentrated) when reduced. Since the test is positive (it reacts), FA 8 must be a compound that is easily oxidised: a 1º alcohol, 2º alcohol, or aldehyde. Therefore, it cannot be a ketone or 3º alcohol.

Step-by-Step Reasoning

M1: Observation. The purple $\text{KMnO}_4$ solution decolourises, turning colourless, pale yellow, or pale brown as $\text{MnO}_4^-$ is reduced to $\text{Mn}^{2+}$.

M2: Conclusion. Since the reaction occurs, FA 8 contains a functional group that is oxidisable. Therefore, FA 8 could be a 1º alcohol, 2º alcohol, or aldehyde. Conversely, FA 8 cannot be a ketone or a 3º alcohol, as these do not react with acidified $\text{KMnO}_4$.

Key Takeaways

• Acidified $\text{KMnO}_4$ is a positive test for 1º/2º alcohols and aldehydes; a negative test (purple remains) indicates a ketone, 3º alcohol, or other unoxidisable group.
• Always state what the compound could be or cannot be based on the result.

Common Mistakes

• Saying the solution turns "blue" or "green"; those are for alkaline $\text{KMnO}_4$ or $\text{Cu}^{2+}$ reactions.
• Concluding FA 8 is definitely an alcohol; it could also be an aldehyde. The mark scheme allows either framing (could be X or cannot be Y).
• Forgetting to mention that the solution is heated; the reaction may be slow at room temperature.

Things to Be Careful About

• The colour change is from purple to colourless/pale yellow. Do not just write "decolourises" without mentioning the original purple colour.
• The conclusion must be logically linked to the observation: positive oxidation test → oxidisable functional group present.

Background Concept

Carboxylic acids are weak acids that react with carbonates and hydrogen carbonates to produce carbon dioxide gas, water, and a salt:

The production of $\text{CO}_2$ is observed as effervescence (bubbling/fizzing). This is a standard test to distinguish carboxylic acids from other oxygen-containing organic compounds like alcohols and aldehydes, which do not react with carbonates.

Understanding the Question

You add solid sodium carbonate to FA 8. You must record the observation and conclude what this tells you about the identity of FA 8.

Approach

If FA 8 were a carboxylic acid, $\text{CO}_2$ would be produced (effervescence). Since it is not, FA 8 cannot be a carboxylic acid.

Step-by-Step Reasoning

M1: Observation. Adding sodium carbonate to FA 8 results in no visible change: no bubbles, no fizzing, no effervescence. The solid may remain unchanged or simply dissolve without gas evolution.

M2: Conclusion. Since no $\text{CO}_2$ is produced, FA 8 does not react as an acid with a carbonate. Therefore, FA 8 is not a carboxylic acid.

Key Takeaways

• Sodium carbonate test is a quick way to rule out carboxylic acids. No effervescence = not a carboxylic acid.
• Combine this with the $\text{KMnO}_4$ test: positive oxidation + no carbonate reaction = alcohol or aldehyde (not carboxylic acid).

Common Mistakes

• Writing "bubbles form" when no reaction occurs.
• Concluding FA 8 is an alcohol; the test only rules out carboxylic acids, it does not positively identify an alcohol.
• Forgetting to link the observation (no gas) to the conclusion (not an acid).

Things to Be Careful About

• Use precise language: "no effervescence" or "no change" is better than "nothing happens".
• The conclusion must specifically state that FA 8 is not a carboxylic acid; saying "it is an alcohol" is not fully supported by this test alone.

Background Concept

From parts (b)(i) and (b)(ii), we know FA 8 is oxidisable by $\text{KMnO}_4$ (so it is a 1º alcohol, 2º alcohol, or aldehyde) but does not react with $\text{Na}_2\text{CO}_3$ (so it is not a carboxylic acid). To distinguish between these remaining possibilities, we use specific functional group tests:

• Aldehydes vs Alcohols:

  • Tollens' reagent (ammoniacal $\text{AgNO}_3$): Aldehydes reduce $\text{Ag}^+$ to metallic silver, forming a silver mirror. Alcohols do not react.
  • Fehling's / Benedict's solution: Aldehydes reduce $\text{Cu}^{2+}$ (blue) to $\text{Cu}_2\text{O}$ (brick-red precipitate). Alcohols do not react.
  • 2,4-DNPH (Brady's reagent): Reacts with carbonyl compounds (aldehydes and ketones) to form an orange/yellow precipitate (2,4-dinitrophenylhydrazone). Since ketones are ruled out, a positive test confirms an aldehyde.

• Alcohols:

  • Sodium metal: Reacts with alcohols to produce hydrogen gas (effervescence). $2\text{ROH} + 2\text{Na} \rightarrow 2\text{RO}^-\text{Na}^+ + \text{H}_2$. Aldehydes do not react with Na.
  • Phosphorus pentachloride ($\text{PCl}_5$): Reacts with alcohols to produce steamy white fumes of $\text{HCl}$. Aldehydes do not react.

Understanding the Question

You must suggest one further test to identify one of the compound types that FA 8 could be (alcohol or aldehyde). State the reagent and what a positive result would indicate.

Approach

Choose a test that gives a clear positive result for either an aldehyde or an alcohol, and describe the observation and its meaning.

Step-by-Step Reasoning

Option 1: Test for aldehyde.

• Reagent: Tollens' reagent (warm gently in a water bath).
• Positive result: Formation of a silver mirror on the inside of the test tube.
• Conclusion: FA 8 is an aldehyde.

Option 2: Test for aldehyde (alternative).

• Reagent: 2,4-DNPH (Brady's reagent).
• Positive result: Formation of an orange / yellow precipitate.
• Conclusion: FA 8 is an aldehyde (or ketone, but ketone is already ruled out).

Option 3: Test for alcohol.

• Reagent: Sodium metal (add a small piece to FA 8).
• Positive result: Effervescence / bubbling; gas relights a glowing splint ($\text{H}_2$).
• Conclusion: FA 8 is an alcohol.

Any one of these valid options scores the mark.

Key Takeaways

• Use functional group tests to narrow down possibilities: Tollens'/Fehling's/2,4-DNPH for aldehydes; Na/$\text{PCl}_5$ for alcohols.
• Always link the specific observation to the specific functional group identified.

Common Mistakes

• Suggesting a test for carboxylic acids (e.g., pH paper, $\text{NaHCO}_3$); these are already ruled out.
• Suggesting a test for ketones (e.g., 2,4-DNPH) and concluding it is a ketone; ketones are ruled out by the $\text{KMnO}_4$ test.
• Forgetting to state what the positive result indicates about the identity of FA 8.

Things to Be Careful About

• The question says "Do not carry out your test." You only need to describe it.
• Ensure the reagent and the positive result match: Tollens' → silver mirror; 2,4-DNPH → orange ppt; Na → effervescence.
• State the conclusion clearly: "indicating FA 8 is an aldehyde" or "indicating FA 8 is an alcohol".