2024 Cambridge IAS-Level Chemistry (9701) 模擬試題連答案詳解

First, write the balanced chemical equation: \(\text{MgCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MgCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\). Calculate the moles of hydrochloric acid: \(n(\text{HCl}) = 50.0 \times 10^{-3}\text{ dm}^3 \times 1.20\text{ mol dm}^{-3} = 0.0600\text{ mol}\). According to the stoichiometry, 2 moles of \(\text{HCl}\) react to produce 1 mole of \(\text{CO}_2\). Therefore, the moles of \(\text{CO}_2\) produced is \(0.0600 / 2 = 0.0300\text{ mol}\). Finally, convert this to volume at r.t.p.: \(\text{Volume} = 0.0300\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.720\text{ dm}^3 = 720\text{ cm}^3\).

1 mark: Correct calculation of HCl moles (0.0600 mol), application of the 1:2 stoichiometric ratio to find CO2 moles (0.0300 mol), and conversion to volume (720 cm³).

Let the percentage abundance of \(^{25}X\) be \(a\%\) and the percentage abundance of \(^{26}X\) be \(b\%\). Since the total abundance is \(100\%\), we have \(78.99 + a + b = 100\), which gives \(a = 21.01 - b\). The relative atomic mass is the weighted average of the isotopic masses: \(A_r = \frac{(24 \times 78.99) + (25 \times a) + (26 \times b)}{100} = 24.31\). Substituting the expression for \(a\): \(1895.76 + 25(21.01 - b) + 26b = 2431\). Solving for \(b\): \(1895.76 + 525.25 - 25b + 26b = 2431 \implies 2421.01 + b = 2431 \implies b = 9.99\%\).

1 mark: Correct setup of simultaneous equations based on total percentage abundance and relative atomic mass, leading to the correct percentage of 9.99%.

Concentrated sulfuric acid is a strong oxidising agent and solid potassium iodide is a strong reducing agent. Iodide ions reduce sulfuric acid to several species, including sulfur dioxide (\(\text{SO}_2\)), sulfur (\(\text{S}\)), and hydrogen sulfide (\(\text{H}_2\text{S}\)), while iodide is oxidised to molecular iodine (\(\text{I}_2\)). Iodine easily vaporises due to the heat generated by the reaction, forming a characteristic purple vapour. Hydrogen sulfide is an acidic gas (not basic, so it doesn't turn red litmus paper blue), sulfur dioxide is a colourless gas (not a yellow solid, which is sulfur), and hydrogen iodide appears as misty white fumes (not a brown gas).

1 mark: Correctly identifying iodine as the purple vapour product resulting from the oxidation of iodide ions by concentrated sulfuric acid.

Reaction with 2,4-DNPH to form an orange precipitate indicates that compound \(Y\) is a carbonyl compound (either an aldehyde or a ketone). The lack of reaction with Tollens' reagent shows that \(Y\) is not an aldehyde, so it must be a ketone. The only ketone with four carbon atoms is butanone, \(\text{CH}_3\text{COCH}_2\text{CH}_3\). Butanone contains a methyl carbonyl group (\(\text{CH}_3\text{CO}-\)), which allows it to undergo the triiodomethane (iodoform) reaction with alkaline aqueous iodine to produce a yellow precipitate of triiodomethane (\(\text{CHI}_3\)).

1 mark: Correct deduction of the ketone functional group from 2,4-DNPH and Tollens' test, and confirmation of methyl ketone structure from the iodoform reaction.

The reaction is: \(\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O}\). Calculate moles of \(\text{HCl}\): \(n(\text{HCl}) = 0.200\text{ mol dm}^{-3} \times 0.03125\text{ dm}^3 = 0.00625\text{ mol}\). According to the equation, 1 mole of \(\text{Na}_2\text{CO}_3\) reacts with 2 moles of \(\text{HCl}\). Therefore, moles of pure \(\text{Na}_2\text{CO}_3\) = \(0.00625 / 2 = 0.003125\text{ mol}\). Mass of pure \(\text{Na}_2\text{CO}_3\) = \(0.003125\text{ mol} \times 106.0\text{ g mol}^{-1} = 0.33125\text{ g}\). Percentage purity = \(\frac{0.33125\text{ g}}{2.00\text{ g}} \times 100 = 16.56\% \approx 16.6\%\).

1 mark: Correctly calculates moles of HCl, applies the 1:2 molar ratio to find moles of Na2CO3, calculates mass of pure Na2CO3, and correctly determines the percentage purity of 16.6%.

The cream precipitate is silver bromide, \(\text{AgBr}\). Silver chloride, \(\text{AgCl}\) (white), is soluble in dilute aqueous ammonia. Silver bromide, \(\text{AgBr}\) (cream), is insoluble in dilute aqueous ammonia but dissolves in concentrated aqueous ammonia. Silver iodide, \(\text{AgI}\) (yellow), is insoluble in both dilute and concentrated aqueous ammonia. Thus, demonstrating that the precipitate is insoluble in dilute ammonia but soluble in concentrated ammonia confirms it is silver bromide.

1 mark: Correctly identifies the differential solubility of silver halides in dilute and concentrated aqueous ammonia as the distinguishing test for silver bromide.

Fehling's reagent is a mild oxidising agent containing copper(II) complexes. It oxidises aldehydes to carboxylic acids, and in the process, blue copper(II) ions are reduced to a brick-red precipitate of copper(I) oxide, \(\text{Cu}_2\text{O}\). Ketones (such as butanone, option A) and alcohols (such as butan-1-ol, option C, and butan-2-ol, option D) are not oxidised by Fehling's reagent. Therefore, only the aldehyde, butanal (option B), will react to form the brick-red precipitate.

1 mark: Correctly identifies that only aldehydes (such as butanal) are oxidised by Fehling's reagent to yield a brick-red precipitate.

First, find the number of neutrons in a single molecule of \(^2\text{H}_2\text{O}\): Deuterium (\(^2\text{H}\)) has 1 proton and 1 neutron. There are two deuterium atoms per molecule, contributing \(2 \times 1 = 2\) neutrons. Oxygen (\(^{16}\text{O}\)) has 8 protons and 8 neutrons. Total number of neutrons in one molecule of \(^2\text{H}_2\text{O} = 2 + 8 = 10\) neutrons. Now, calculate the total number of neutrons in \(0.100\text{ mol}\) of heavy water: Moles of neutrons = \(0.100\text{ mol} \times 10 = 1.00\text{ mol}\). Number of neutrons = \(1.00 \times L = 1.00 L\).

1 mark: Correctly determines the number of neutrons in a molecule of heavy water (10) and multiplies by the given mole value to express the answer in terms of the Avogadro constant L.

First, the absorption of gas by concentrated aqueous KOH represents the volume of \(CO_2\) produced, which is \(110 - 50 = 60\text{ cm}^3\). Since 20 cm³ of \(C_xH_y\) produced 60 cm³ of \(CO_2\), we find \(x = 3\). Next, the remaining 50 cm³ of gas is the excess, unreacted oxygen. The volume of oxygen that reacted is \(150 - 50 = 100\text{ cm}^3\). The ratio of reacted oxygen to hydrocarbon is \(100 / 20 = 5\). Using the general combustion equation, \(x + \frac{y}{4} = 5\). Substituting \(x = 3\) gives \(3 + \frac{y}{4} = 5\), which yields \(y = 8\). Therefore, the formula is \(C_3H_8\).

Award 1 mark for the correct choice D. Deduct 0 marks for incorrect choices.

Concentrated sulfuric acid acts as an oxidising agent with halide ions, except fluoride and chloride which only undergo acid-base reactions. Bromide ions reduce sulfuric acid to sulfur dioxide, \(SO_2\) (a choking gas that reduces acidified dichromate(VI) from orange to green). Iodide ions, being even stronger reducing agents, reduce sulfuric acid to sulfur dioxide, elemental sulfur (a yellow solid), and hydrogen sulfide. Since a yellow solid is deposited, the halide ion must be iodide, \(I^-\).

Award 1 mark for the correct choice D. Deduct 0 marks for incorrect choices.

The reaction with 2,4-DNPH shows that \(W\) is a carbonyl compound (aldehyde or ketone). The negative test with Tollens' reagent shows that \(W\) is a ketone rather than an aldehyde. Since \(W\) has four carbon atoms, it must be butanone, \(CH_3COCH_2CH_3\). Reduction of butanone with \(NaBH_4\) produces butan-2-ol, \(CH_3CH(OH)CH_2CH_3\), which contains a chiral carbon atom. Therefore, compound \(Z\) exhibits optical isomerism.

Award 1 mark for the correct choice C. Deduct 0 marks for incorrect choices.

The solubility of Group 2 sulfates decreases down the group, so magnesium sulfate is highly soluble while barium sulfate is highly insoluble. Adding dilute sulfuric acid will selectively precipitate barium ions as white barium sulfate, \(BaSO_4(s)\), leaving magnesium ions in solution. The other options (sodium chloride, sodium nitrate, and hydrochloric acid) do not form precipitates with either of these metal ions.

Award 1 mark for the correct choice C. Deduct 0 marks for incorrect choices.

First, calculate the mass of carbon: \(1.98 \times (12.0 / 44.0) = 0.54\text{ g}\). Next, calculate the mass of hydrogen: \(0.81 \times (2.0 / 18.0) = 0.09\text{ g}\). The mass of oxygen is \(1.35 - 0.54 - 0.09 = 0.72\text{ g}\). Now find the moles of each element: \(n(C) = 0.54 / 12.0 = 0.045\text{ mol}\), \(n(H) = 0.09 / 1.0 = 0.090\text{ mol}\), and \(n(O) = 0.72 / 16.0 = 0.045\text{ mol}\). The mole ratio of \(C : H : O\) is \(0.045 : 0.090 : 0.045\), which simplifies to \(1 : 2 : 1\). Therefore, the empirical formula is \(CH_2O\).

Award 1 mark for the correct choice B. Deduct 0 marks for incorrect choices.

The molar mass of \(Na_2CO_3\) is 106.0 g/mol. Moles of \(Na_2CO_3\) in 2.65 g is \(2.65 / 106.0 = 0.025\text{ mol}\). The concentration of the 250.0 cm³ solution is \(0.025 / 0.250 = 0.100\text{ mol dm}^{-3}\). A 25.0 cm³ sample contains \(0.100 \times 0.025 = 0.0025\text{ mol}\) of \(Na_2CO_3\). The reaction equation is \(Na_2CO_3 + 2HCl \rightarrow 2NaCl + CO_2 + H_2O\). Thus, moles of \(HCl\) required is \(2 \times 0.0025 = 0.0050\text{ mol}\). Since this is in 20.0 cm³, the concentration of \(HCl\) is \(0.0050 / 0.0200 = 0.250\text{ mol dm}^{-3}\).

Award 1 mark for the correct choice C. Deduct 0 marks for incorrect choices.

The thermal stability of hydrogen halides decreases down Group 17. Because the atomic radius of iodine is much larger than that of chlorine, the orbital overlap in the \(H-I\) covalent bond is less effective than in the \(H-Cl\) bond. This results in a longer and weaker \(H-I\) bond (lower bond enthalpy), which requires less energy to break upon heating.

Award 1 mark for the correct choice A. Deduct 0 marks for incorrect choices.

Propanone, \(CH_3COCH_3\), undergoes nucleophilic addition with HCN to form the hydroxynitrile, 2-hydroxy-2-methylpropanenitrile, \((CH_3)_2C(OH)CN\) (product \(Y\)). Heating \(Y\) under reflux with dilute hydrochloric acid hydrolyses the nitrile group (\(-CN\)) into a carboxylic acid group (\(-COOH\)), yielding 2-hydroxy-2-methylpropanoic acid, \((CH_3)_2C(OH)COOH\) (compound \(Z\)).

Award 1 mark for the correct choice B. Deduct 0 marks for incorrect choices.

First, calculate the number of moles of \(\text{CO}_2\) gas collected: moles of \(\text{CO}_2 = \frac{270 \times 10^{-3}\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.01125\text{ mol}\). From the balanced equation: \(\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\), the mole ratio is 1:1. Therefore, the moles of pure \(\text{CaCO}_3\) is also \(0.01125\text{ mol}\). Molar mass of \(\text{CaCO}_3 = 40.1 + 12.0 + 3(16.0) = 100.1\text{ g mol}^{-1}\). Mass of pure \(\text{CaCO}_3 = 0.01125\text{ mol} \times 100.1\text{ g mol}^{-1} = 1.126\text{ g}\). Percentage purity = \(\frac{1.126\text{ g}}{1.25\text{ g}} \times 100\% = 90.1\%\).

1 mark for the correct calculation leading to option C.

Use the ideal gas equation: \(pV = nRT = \frac{m}{M_{\text{r}}}RT\). Rearranging for \(M_{\text{r}}\): \(M_{\text{r}} = \frac{mRT}{pV}\). Convert units to SI: volume \(V = 88.7\text{ cm}^3 = 8.87 \times 10^{-5}\text{ m}^3\). Substitute the values: \(M_{\text{r}} = \frac{0.222 \times 8.31 \times 360}{1.01 \times 10^5 \times 8.87 \times 10^{-5}} = \frac{664.135}{8.9587} = 74.1\text{ g mol}^{-1}\).

1 mark for the correct calculation leading to option C.

By Avogadro's hypothesis, the volume ratios of reacting gases are equal to their mole ratios. Let the hydrocarbon formula be \(\text{C}_x\text{H}_y\). The equation for complete combustion is: \(\text{C}_x\text{H}_y + (x + \frac{y}{4})\text{O}_2 \rightarrow x\text{CO}_2 + \frac{y}{2}\text{H}_2\text{O}\). Volume of \(\text{CO}_2\) / Volume of \(\text{C}_x\text{H}_y = x = \frac{40}{10} = 4\). Therefore, \(x = 4\). Volume of \(\text{O}_2\) / Volume of \(\text{C}_x\text{H}_y = x + \frac{y}{4} = \frac{65}{10} = 6.5\). Substitute \(x = 4\) into the equation: \(4 + \frac{y}{4} = 6.5 \implies \frac{y}{4} = 2.5 \implies y = 10\). The molecular formula of \(\text{X}\) is \(\text{C}_4\text{H}_{10}\).

1 mark for identifying the correct molecular formula as C4H10 (option C).

Down Group 17, the number of electrons per molecule increases, which increases the size and polarisability of the electron cloud. This leads to stronger instantaneous dipole-induced dipole forces (London dispersion forces) between molecules, so more energy is required to separate the molecules, resulting in higher boiling points. Electronegativity and covalent bond strength do not determine the boiling point of these simple molecular elements.

1 mark for selecting the correct statement (option C).

Concentrated sulfuric acid reacts with sodium iodide to produce hydrogen iodide (\(\text{HI}\)), which is a very strong reducing agent. \(\text{HI}\) reduces the sulfuric acid to various products, including iodine (\(\text{I}_2\), purple fumes), sulfur (\(\text{S}\), a yellow solid), and hydrogen sulfide (\(\text{H}_2\text{S}\), an acidic gas). Fluoride and chloride cannot reduce sulfuric acid, and bromide is only strong enough to reduce sulfuric acid to sulfur dioxide (\(\text{SO}_2\), a colourless gas) and bromine (\(\text{Br}_2\), orange/brown vapour), but not to elemental sulfur.

1 mark for identifying iodide as the correct halide ion (option D).

The formation of an orange precipitate with 2,4-DNPH shows that \(\text{Y}\) contains a carbonyl group (it is an aldehyde or a ketone). The lack of reaction with Tollens' and Fehling's reagents shows that \(\text{Y}\) is not an aldehyde. Therefore, \(\text{Y}\) must be a ketone. With four carbon atoms, the only possible ketone is butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)). The reduction of butanone with \(\text{NaBH}_4\) yields the secondary alcohol, butan-2-ol.

1 mark for identifying butan-2-ol as the correct reduction product (option B).

The reaction begins with the nucleophilic attack of the cyanide ion (\(\text{CN}^-\)) on the electron-deficient carbon of the \(\text{C}=\text{O}\) group in propanone. This forms an intermediate alkoxide ion, which subsequently abstracts a proton from \(\text{HCN}\) to form the hydroxynitrile product and regenerate the \(\text{CN}^-\cap\) catalyst.

1 mark for choosing the correct mechanism description (option B).

Observation 1 indicates the cation forms an amphoteric hydroxide (\(\text{Al}^{3+}\), \(\text{Zn}^{2+}\), or \(\text{Pb}^{2+}\)). Observation 2 shows that the white hydroxide precipitate does not dissolve in excess ammonia, which is characteristic of \(\text{Al}^{3+}\) (\(\text{Zn}^{2+}\) forms a soluble complex \([\text{Zn}(\text{NH}_3)_4]^{2+}\) and its precipitate dissolves in excess ammonia). Observation 3 shows the presence of the sulfate ion (\(\text{SO}_4^{2-}\)), which forms insoluble barium sulfate (\(\text{BaSO}_4\)) with barium ions in acidic conditions. Therefore, the ions present are \(\text{Al}^{3+}\) and \(\text{SO}_4^{2-}\).

1 mark for identifying the correct combination of \(\text{Al}^{3+}\) and \(\text{SO}_4^{2-}\) (option A).

1. The volume of
\text{CO}_2
produced is equal to the decrease in volume when passed through aqueous
\text{NaOH}
:
\text{Volume of CO}_2 = 85\text{ cm}^3 - 55\text{ cm}^3 = 30\text{ cm}^3
. Since
10\text{ cm}^3
of
\text{C}_x\text{H}_y
was used, we get
x = 30 / 10 = 3
. 2. The volume of unreacted oxygen remaining is
55\text{ cm}^3
. Therefore, the volume of oxygen reacted is:
\text{Volume of O}_2\text{ reacted} = 100\text{ cm}^3 - 55\text{ cm}^3 = 45\text{ cm}^3
. 3. The general equation for combustion is
\text{C}_3\text{H}_y + (3 + y/4)\text{O}_2 \rightarrow 3\text{CO}_2 + (y/2)\text{H}_2\text{O}
. The ratio of
\text{O}_2
reacted to hydrocarbon reacted is
3 + y/4 = 45 / 10 = 4.5
. Solving for
y
gives
y/4 = 1.5
, hence
y = 6
. Thus, the hydrocarbon is
\text{C}_3\text{H}_6
.

1 mark for correct option B. 1 mark is awarded for identifying that
30\text{ cm}^3
of
\text{CO}_2
is produced, establishing
x=3
. 1 mark is awarded for calculating that
45\text{ cm}^3
of
\text{O}_2
reacted, establishing
y=6
.

1. Calculate the amount of
\text{Fe}_2(\text{SO}_4)_3
dissolved:
\text{moles} = \text{concentration} \times \text{volume} = 0.120\text{ mol dm}^{-3} \times (25.0 / 1000)\text{ dm}^3 = 0.00300\text{ mol}
. 2. Determine the number of ions per formula unit:
\text{Fe}_2(\text{SO}_4)_3
dissociates to form two
\text{Fe}^{3+}
ions and three
\text{SO}_4^{2-}
ions, giving a total of 5 ions per formula unit. 3. Calculate the total moles of ions:
\text{total moles of ions} = 0.00300\text{ mol} \times 5 = 0.0150\text{ mol}
. 4. Calculate the total number of ions:
\text{number of ions} = 0.0150\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 9.03 \times 10^{21}
.

1 mark for correct option D. Award marks for calculating the moles of iron(III) sulfate as
0.00300\text{ mol}
and multiplying by 5 to find total ion moles, then scaling by the Avogadro constant.

Boiling points depend on intermolecular forces.
\text{HF}
has hydrogen bonding because hydrogen is bonded to the highly electronegative fluorine atom, whereas
\text{HCl}
has weaker permanent dipole-dipole attractions. Hence,
\text{HF}
has a higher boiling point than
\text{HCl}
. Both
\text{F}_2
and
\text{Cl}_2
are non-polar molecules held together only by instantaneous dipole-induced dipole forces. Since
\text{Cl}_2
has more electrons than
\text{F}_2
, its electron cloud is more polarisable, leading to stronger dispersion forces and a higher boiling point than
\text{F}_2
.

1 mark for correct option A. Award mark for correctly identifying the intermolecular forces in both pairs (hydrogen bonding vs dipole-dipole for the hydrogen halides, and instantaneous-induced dipole forces for the halogens).

The cream precipitate that is sparingly soluble in dilute ammonia but soluble in concentrated ammonia is silver bromide (
\text{AgBr}
), so the halide is
\text{NaBr}
. When solid
\text{NaBr}
reacts with concentrated
\text{H}_2\text{SO}_4
,
\text{HBr}
gas is produced which fumes in moist air. Because bromide is a strong enough reducing agent, it reduces the concentrated
\text{H}_2\text{SO}_4
to sulfur dioxide (
\text{SO}_2
) while being oxidised to bromine (
\text{Br}_2
). Thus, redox products are formed.

1 mark for correct option B. Award mark for identifying bromide from the silver halide solubility tests and correctly stating that bromide reduces sulfuric acid to sulfur dioxide while being oxidized to bromine.

The positive Tollens' test indicates the presence of an aldehyde group (
-\text{CHO}
). Among the options, only option A (
\text{CH}_3\text{CH(OH)CH}_2\text{CH}_2\text{CHO}
) contains an aldehyde group. Options B and D are ketones, and option C is an ester. The reduction of the aldehyde group in
\mathbf{Y}
by
\text{NaBH}_4
yields the diol
\mathbf{Z}
, which is
\text{CH}_3\text{CH(OH)CH}_2\text{CH}_2\text{CH}_2\text{OH}
.

1 mark for correct option A. Award mark for identifying that only the aldehyde in option A can react with Tollens' reagent and undergo reduction with
\text{NaBH}_4
to form a diol.

1. Find the mass of water lost:
4.99\text{ g} - 3.19\text{ g} = 1.80\text{ g}
. 2. Find the moles of water:
1.80\text{ g} / 18.0\text{ g mol}^{-1} = 0.100\text{ mol}
. 3. Determine
x
by finding the ratio of moles of water to moles of salt:
x = 0.100\text{ mol} / 0.0200\text{ mol} = 5
. 4. Calculate the molar mass of the anhydrous residue:
M_r = 3.19\text{ g} / 0.0200\text{ mol} = 159.5\text{ g mol}^{-1}
. 5. Identify
\text{M}
:
A_r(\text{M}) = 159.5 - (32.1 + 64.0) = 63.4
, which is copper (
\text{Cu}
).

1 mark for correct option A. Award marks for calculating the moles of water lost, finding the mole ratio
x = 5
, and using the molar mass of the residue to identify the metal as copper.

The orange precipitate with 2,4-DNPH indicates a carbonyl group (aldehyde or ketone), which eliminates propan-2-ol. The lack of reaction with Fehling's solution indicates it is a ketone, which eliminates propanal. The yellow precipitate with alkaline aqueous iodine indicates a methyl carbonyl group (
\text{CH}_3\text{CO}-
). Butanone (
\text{CH}_3\text{COCH}_2\text{CH}_3
) has this group and reacts positively, while pentan-3-one (
\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3
) does not. Therefore,
\mathbf{W}
is butanone.

1 mark for correct option C. Award mark for using the three test results to systematically eliminate other options and identify butanone as the methyl ketone.

The addition of silver nitrate forms precipitates of the halides. Silver chloride (
\text{AgCl}
) is soluble in dilute ammonia, so the portion of the precipitate that dissolves indicates the presence of
\text{Cl}^-
. Silver bromide (
\text{AgBr}
) is insoluble in dilute ammonia but soluble in concentrated ammonia, while silver iodide (
\text{AgI}
) is insoluble in both. Since the remaining residue does not dissolve in concentrated ammonia, it must be
\text{AgI}
, indicating the presence of
\text{I}^-
. Thus, the mixture contains
\text{Cl}^-
and
\text{I}^-
.

1 mark for correct option B. Award mark for linking the solubility of silver halide precipitates in dilute and concentrated ammonia to the correct halide ions.

First, determine the volume of carbon dioxide produced:
- The aqueous sodium hydroxide absorbs the carbon dioxide, \(CO_2\).
- Therefore, the decrease in volume represents the volume of \(CO_2\) produced: \(55\text{ cm}^3 - 25\text{ cm}^3 = 30\text{ cm}^3\).
- The remaining volume of \(25\text{ cm}^3\) is the unreacted oxygen.

Next, find the volume of oxygen reacted:
- Volume of oxygen reacted = \(70\text{ cm}^3 - 25\text{ cm}^3 = 45\text{ cm}^3\).

Now, analyze the mole ratios:
- \(10\text{ cm}^3\) of \(C_xH_y\) reacts with \(45\text{ cm}^3\) of \(O_2\) to form \(30\text{ cm}^3\) of \(CO_2\).
- This gives a volume ratio of \(C_xH_y : O_2 : CO_2 = 1 : 4.5 : 3\).
- Since 1 mole of \(C_xH_y\) forms 3 moles of \(CO_2\), \(x = 3\).
- The general equation for combustion is:
\(C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\)
- Therefore, \(x + \frac{y}{4} = 4.5\).
- Substituting \(x = 3\): \(3 + \frac{y}{4} = 4.5 \Rightarrow \frac{y}{4} = 1.5 \Rightarrow y = 6\).

The hydrocarbon is \(C_3H_6\).

- Correct identification of volume of carbon dioxide and oxygen reacted: 1 mark
- Calculation of the molecular formula from the stoichiometric ratios: 1 mark
- Select Option B: 1 mark

When concentrated sulfuric acid acts as an oxidising agent on iodide ions, the sulfur in \(H_2SO_4\) (oxidation state +6) is reduced to various species:
- In sulfur dioxide (\(SO_2\)), the oxidation state of sulfur is +4 (a decrease from +6).
- In sulfur (\(S\)), the oxidation state of sulfur is 0 (a decrease from +6).
- In hydrogen sulfide (\(H_2S\)), the oxidation state of sulfur is -2 (a decrease from +6).

Let's evaluate the options:
- \(HI\) and \(KHSO_4\) are products of the acid-base reaction of \(KI\) and \(H_2SO_4\), in which no redox occurs (oxidation state of S remains +6).
- \(I_2\) is formed by the oxidation of iodide ions (oxidation state increases from -1 to 0), not by the reduction of sulfur.

Therefore, only the list in option C contains solely products where the oxidation state of sulfur has decreased.

- Correct identification of the reduction products of sulfuric acid: 1 mark
- Select Option C: 1 mark

1. \(Y\) is a carbonyl compound because it reacts with 2,4-DNPH.
2. \(Y\) does not react with Tollens' reagent, indicating it is a ketone, not an aldehyde. The only ketone with the molecular formula \(C_4H_8O\) is butanone, \(CH_3COCH_2CH_3\).
3. When butanone (\(Y\)) reacts with HCN, nucleophilic addition occurs to form a hydroxynitrile (\(Z\)):
\(CH_3COCH_2CH_3 + HCN \rightarrow CH_3C(OH)(CN)CH_2CH_3\)
4. When \(Z\) is heated under reflux with dilute hydrochloric acid, the nitrile group (\(-CN\)) undergoes acid hydrolysis to form a carboxylic acid group (\(-COOH\)):
\(CH_3C(OH)(CN)CH_2CH_3 + 2H_2O + H^+ \rightarrow CH_3C(OH)(COOH)CH_2CH_3 + NH_4^+\)

This structure can be rewritten as \(CH_3CH_2C(OH)(CH_3)COOH\), which corresponds to option B.

- Deduction of the identity of butanone: 1 mark
- Identification of the structure of the nitrile and its hydrolysis product: 1 mark
- Select Option B: 1 mark

1. Calculate the number of moles of \(CO_2\) gas collected:
\(n(CO_2) = \frac{600\text{ cm}^3}{24000\text{ cm}^3\text{ mol}^{-1}} = 0.0250\text{ mol}\)

2. Use the stoichiometric ratio from the equation:
Since 1 mol of \(MCO_3\) produces 1 mol of \(CO_2\):
\(n(MCO_3) = n(CO_2) = 0.0250\text{ mol}\)

3. Calculate the molar mass (\(M_r\)) of \(MCO_3\):
\(M_r(MCO_3) = \frac{\text{mass}}{\text{moles}} = \frac{2.11\text{ g}}{0.0250\text{ mol}} = 84.4\text{ g mol}^{-1}\)

4. Determine the relative atomic mass (\(A_r\)) of \(M\):
\(M_r(MCO_3) = A_r(M) + A_r(C) + 3 \times A_r(O)\)
\(84.4 = A_r(M) + 12.0 + 3 \times 16.0 = A_r(M) + 60.0\)
\(A_r(M) = 84.4 - 60.0 = 24.4\)

Comparing this to the periodic table, the relative atomic mass of magnesium is \(24.3\), which is the closest value. Therefore, the metal \(M\) is magnesium.

- Calculation of the moles of gas and carbonate: 1 mark
- Calculation of the relative atomic mass of M and identification of magnesium: 1 mark
- Select Option A: 1 mark

- Addition of aqueous silver nitrate to halide ions produces silver halide precipitates: \(AgCl\) (white), \(AgBr\) (cream), and \(AgI\) (yellow).
- Dilute aqueous ammonia dissolves \(AgCl\) to form a colourless solution containing the complex ion \([Ag(NH_3)_2]^+\):
\(AgCl(s) + 2NH_3(aq) \rightarrow [Ag(NH_3)_2]^+(aq) + Cl^-(aq)\)
- Dilute aqueous ammonia does not dissolve \(AgBr\) or \(AgI\).
- Therefore, the part of the precipitate that dissolved in dilute ammonia must be silver chloride, which shows that chloride ions (\(Cl^-\)) were present in the solution.
- Concentrated aqueous ammonia dissolves \(AgBr\) but does not dissolve \(AgI\).
- Since the pale yellow residue does not dissolve even in concentrated ammonia, the residue must be silver iodide, which shows that iodide ions (\(I^-\)) were present in the solution.

Thus, the two halide ions present are chloride and iodide.

- Analysis of the solubility of silver halide precipitates in dilute and concentrated ammonia: 1 mark
- Correct deduction of chloride and iodide ions: 1 mark
- Select Option C: 1 mark

The relative atomic mass, \(A_r\), is the weighted average mass of the isotopes compared to one-twelfth of the mass of a carbon-12 atom. It is calculated using the formula:

\(A_r = \frac{\sum (\text{isotopic mass} \times \text{abundance})}{\text{total abundance}}\)

\(A_r = \frac{(24 \times 78.99) + (25 \times 10.00) + (26 \times 11.01)}{100}\)

\(A_r = \frac{1895.76 + 250.00 + 286.26}{100}\)

\(A_r = \frac{2432.02}{100} = 24.3202\)

To two decimal places, this value is \(24.32\).

- Calculation of the weighted average: 1 mark
- Rounding to two decimal places: 1 mark
- Select Option C: 1 mark

Let's evaluate each option:
- **A is incorrect**: The boiling point *increases* down the group because the number of electrons in each molecule increases, which leads to stronger instantaneous dipole-induced dipole (van der Waals') forces, requiring more thermal energy to overcome.
- **B is incorrect**: The electronegativity *decreases* down the group because the atomic radius and shielding increase, meaning the nucleus has a weaker attraction for the shared pair of electrons in a covalent bond.
- **C is correct**: The first ionisation energy *decreases* down the group because the outer electrons are in shells further from the nucleus and experience greater shielding by inner electrons, making them easier to remove.
- **D is incorrect**: The volatility *decreases* down the group (as shown by the physical states changing from gas to liquid to solid) due to the increasing strength of intermolecular van der Waals' forces.

- Evaluation of the trends in Group 17 physical properties: 1 mark
- Select Option C: 1 mark

- The molecular formula \(C_5H_{10}O\) with a carbonyl group represents an aldehyde or a ketone.
- Warming with alkaline aqueous iodine (the iodoform reaction) tests for the presence of a \(CH_3CO-\) group (or a \(CH_3CH(OH)-\) group, but \(W\) is a carbonyl compound). Only methyl ketones, \(CH_3COR\), give a positive result (yellow precipitate of \(CHI_3\)). This rules out pentan-3-one (option A), pentanal (option C), and 3-methylbutanal (option D).
- Reduction of \(W\) with \(NaBH_4\) produces a secondary alcohol, confirming that \(W\) is a ketone rather than an aldehyde (aldehydes are reduced to primary alcohols).
- Therefore, \(W\) must be pentan-2-one, \(CH_3COCH_2CH_2CH_3\).

- Deducing that \(W\) is a methyl ketone from the iodoform and reduction reactions: 1 mark
- Correct structure selection: 1 mark
- Select Option B: 1 mark

(a) Equations:
(i) \(2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g)\) [1]
(ii) \(\text{NaHCO}_3(s) + \text{HCl}(aq) \rightarrow \text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)\) [1]
(iii) \(\text{Na}_2\text{CO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{CO}_2(g) + \text{H}_2\text{O}(l)\) [1]

(b) Effervescence / bubbling / gas produced AND the solid dissolves to form a colorless solution. [1]

(c) \(\text{CO}_2\) and \(\text{H}_2\text{O}\) are produced as gases and escape. [1]

(d) Only \(\text{NaHCO}_3\) decomposes on heating.
From the equation: \(2\text{ mol}\) of \(\text{NaHCO}_3\) (\(2 \times 84.0\text{ g} = 168.0\text{ g}\)) produces \(1\text{ mol } \text{CO}_2\) (\(44.0\text{ g}\)) and \(1\text{ mol } \text{H}_2\text{O}\) (\(18.0\text{ g}\)).
Total mass loss of volatile products per 168.0 g of reactants = \(44.0 + 18.0 = 62.0\text{ g}\). [1]
Let \(x\) be the mass of \(\text{NaHCO}_3\).
\(\text{Mass loss} = x \times \frac{62.0}{168.0} = 0.111\text{ g}\). [1]
\(x = 0.111 \times \frac{168.0}{62.0} = 0.301\text{ g}\). [1]

(e) Mass of \(\text{Na}_2\text{CO}_3\) in the sample = \(0.500\text{ g} - 0.301\text{ g} = 0.199\text{ g}\). [1]
Percentage of \(\text{Na}_2\text{CO}_3 = \frac{0.199}{0.500} \times 100\% = 39.8\%\). [1]

(f) Moles of \(\text{NaHCO}_3 = \frac{0.301}{84.0} = 0.00358\text{ mol}\).
Moles of \(\text{Na}_2\text{CO}_3 = \frac{0.199}{106.0} = 0.00188\text{ mol}\). [1]
Total moles of \(\text{HCl}\) required:
Each mole of \(\text{NaHCO}_3\) reacts with \(1\text{ mol}\) of \(\text{HCl}\); each mole of \(\text{Na}_2\text{CO}_3\) reacts with \(2\text{ mol}\) of \(\text{HCl}\) (using methyl orange, which indicates complete reaction).
Total moles of \(\text{HCl} = 0.00358 + (2 \times 0.00188) = 0.00734\text{ mol}\). [1]
Volume of \(\text{HCl} = \frac{\text{moles}}{\text{concentration}} = \frac{0.00734}{0.200} = 0.0367\text{ dm}^3 = 36.7\text{ cm}^3\). [1]

(g) Phenolphthalein [1]. The color changes from pink to colorless at the first equivalence point (where \(\text{CO}_3^{2-}\) is converted only to \(\text{HCO}_3^-\)). [1]

(a)(i) 1 mark for correct balanced equation including states.
(a)(ii) 1 mark for correct balanced equation.
(a)(iii) 1 mark for correct balanced equation.
(b) 1 mark for identifying both effervescence and solid dissolving.
(c) 1 mark for explaining that gaseous products (carbon dioxide and water) escape.
(d) 1 mark for stating the mass loss of 62.0 g corresponds to 168.0 g of NaHCO3.
1 mark for setting up the correct mathematical ratio.
1 mark for showing the calculation that leads to 0.301 g.
(e) 1 mark for finding the correct mass of Na2CO3 (0.199 g).
1 mark for calculating the correct percentage (39.8%).
(f) 1 mark for calculating moles of both compounds.
1 mark for calculating the total moles of HCl (0.00734 mol).
1 mark for calculating the correct volume of HCl (36.7 cm3).
(g) 1 mark for phenolphthalein.
1 mark for color change (pink to colorless).

(a)(i) Hydrogen chloride / \(\text{HCl}\). [1]
(ii) \(\text{KCl}(s) + \text{H}_2\text{SO}_4(l) \rightarrow \text{KHSO}_4(s) + \text{HCl}(g)\)
(Note: \(2\text{KCl} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{HCl}\) is also acceptable). [1]
(iii) Not redox [1], because the oxidation states of all elements remain unchanged (e.g., \(\text{H} = +1\), \(\text{Cl} = -1\), \(\text{S} = +6\), \(\text{O} = -2\)). [1]

(b)(i) Dark gray solid: \(\text{I}_2\) (iodine) [1]
Gas with rotten egg smell: \(\text{H}_2\text{S}\) (hydrogen sulfide) [1]
(ii) Oxidation state of \(\text{S}\) in \(\text{H}_2\text{SO}_4 = +6\) [1]
Oxidation state of \(\text{S}\) in \(\text{H}_2\text{S} = -2\) [1]
(iii) \(\text{H}_2\text{SO}_4 + 8\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\)
(Accept ionic form: \(\text{SO}_4^{2-} + 10\text{H}^+ + 8\text{e}^- \rightarrow \text{H}_2\text{S} + 4\text{H}_2\text{O}\)). [2]
(iv) Iodide ions are larger and have more shielding [1]. The outer electrons are further from the nucleus and less strongly attracted, making iodide ions stronger reducing agents than chloride ions (which hold their valence electrons too tightly to reduce sulfuric acid). [1]

(c)(i) \(\text{Cl}_2 + 2\text{NaOH} \rightarrow \text{NaCl} + \text{NaClO} + \text{H}_2\text{O}\) [2]
(1 mark for correct products, 1 mark for correct balancing).
(ii) Disproportionation is a reaction in which the same element is simultaneously oxidized and reduced. [1]

(a)(i) 1 mark for correct identification of HCl.
(a)(ii) 1 mark for correct equation.
(a)(iii) 1 mark for stating "not a redox reaction" and 1 mark for justification (no changes in oxidation numbers).
(b)(i) 1 mark for I2 and 1 mark for H2S.
(b)(ii) 1 mark for +6 and 1 mark for -2.
(b)(iii) 2 marks for fully balanced half-equation (1 mark if species correct but unbalanced).
(b)(iv) 1 mark for stating iodide is a stronger reducing agent than chloride due to larger size/more shielding.
1 mark for explaining that outer electrons are lost more easily.
(c)(i) 2 marks for balanced equation (1 mark for species, 1 mark for balancing).
(c)(ii) 1 mark for definition of disproportionation.

(a) The three carbonyl isomers are:
1. Butanal: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO}\) (Skeletal: zigzag line of 4 carbons ending in double-bonded O).
2. 2-methylpropanal: \((\text{CH}_3)_2\text{CHCHO}\) (Skeletal: Y-shaped carbon backbone with aldehyde group).
3. Butanone: \(\text{CH}_3\text{COCH}_2\text{CH}_3\) (Skeletal: 4-carbon chain with double-bonded O on carbon-2).
[3 marks: 1 mark for each correct pair of skeletal structure and IUPAC name].

(b)(i) Butanal and 2-methylpropanal (since both are aldehydes). [1]
(ii) Oxidation (or redox). [1]
(iii) No change / solution remains colorless (since ketones cannot be oxidized by Tollens' reagent). [1]

(c)(i) Reagents: \(\text{NaCN}\) (or \(\text{KCN}\)) and a trace of acid (to provide both \(\text{CN}^-\)\ and \(\text{H}^+\)\ ions / buffer pH 7–8). [1]
Conditions: Room temperature (or cold). [1]

(ii) Mechanism:
- Step 1: Carbonyl carbon is \(\delta+\) and oxygen is \(\delta-\). A curly arrow starts from the lone pair on the carbon of the cyanide ion (\(^{-}\text{C}\equiv\text{N}\)) and points to the carbonyl carbon of butanal. [1]
- Step 2: A curly arrow starts from one of the bonds in the \(\text{C}=\text{O}\) double bond and points to the oxygen atom. [1]
- Step 3: An intermediate alkoxide ion is formed: \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}(\text{O}^-)\text{CN}\) with a lone pair on the negatively charged oxygen. [1]
- Step 4: A curly arrow starts from the lone pair on \(\text{O}^-\)\ pointing to a hydrogen ion (\(\text{H}^+\)\ or the \(\text{H}\)\ in \(\text{HCN}\)) to form the final hydroxynitrile product. [1]

(iii) 2-hydroxypentanenitrile. [1]

(iv) The carbonyl group in butanal is planar [1]. The \(\text{CN}^-\) nucleophile can attack the planar carbon with equal probability from either side, resulting in an equimolar (racemic) mixture of both enantiomers, which has no net optical rotation. [1]

(a) 3 marks: 1 mark per isomer structure paired with correct name (butanal, 2-methylpropanal, and butanone).
(b)(i) 1 mark for naming both aldehydes.
(b)(ii) 1 mark for oxidation.
(b)(iii) 1 mark for stating no change/no precipitate.
(c)(i) 1 mark for NaCN/KCN and trace acid (or HCN with NaCN catalyst).
1 mark for room temperature.
(c)(ii) 4 marks for mechanism:
- 1 mark for correct dipoles on C=O and curly arrow from lone pair on CN- to carbon.
- 1 mark for curly arrow from C=O bond to oxygen.
- 1 mark for drawing the correct intermediate structure with O-.
- 1 mark for curly arrow from O- lone pair to H+.
(c)(iii) 1 mark for the correct IUPAC name (2-hydroxypentanenitrile).
(c)(iv) 1 mark for noting the carbonyl carbon is planar.
1 mark for explaining that attack is equally likely from either side, forming a racemic mixture.

(a)(i) Cation: \(\text{Ca}^{2+}\) (calcium) [1].
Reason: Calcium hydroxide is slightly soluble and precipitates with a strong alkali (NaOH), but because the concentration of hydroxide ions in weak ammonia solution is very low, the ionic product does not exceed the solubility product, so no precipitate forms. [1]

(ii) Halide: \(\text{Br}^-\). [1]
Observations: The cream precipitate of silver bromide is insoluble in dilute aqueous ammonia [1] but dissolves in concentrated aqueous ammonia to form a colorless solution. [1]

(b) Mass of water lost = \(4.620\text{ g} - 3.000\text{ g} = 1.620\text{ g}\).
Moles of water lost = \(\frac{1.620\text{ g}}{18.0\text{ g mol}^{-1}} = 0.0900\text{ mol}\). [1]
Molar mass of anhydrous \(\text{CaBr}_2 = 40.1 + (2 \times 79.9) = 199.9\text{ g mol}^{-1}\). [1]
Moles of anhydrous \(\text{CaBr}_2 = \frac{3.000\text{ g}}{199.9\text{ g mol}^{-1}} = 0.0150\text{ mol}\). [1]
Ratio of \(\text{H}_2\text{O} : \text{CaBr}_2 = \frac{0.0900}{0.0150} = 6\).
So, \(x = 6\). [1]

(c)(i) \(\text{Ag}^+(aq) + \text{Br}^-(aq) \rightarrow \text{AgBr}(s)\) [1]
(ii) Moles of \(\text{CaBr}_2\) in \(250.0\text{ cm}^3 = 0.0150\text{ mol}\).
Concentration = \(\frac{0.0150\text{ mol}}{0.250\text{ dm}^3} = 0.0600\text{ mol dm}^{-3}\). [2] (1 mark for formula, 1 mark for correct value).
(iii) In \(25.0\text{ cm}^3\) of this solution:
Moles of \(\text{CaBr}_2 = 0.0600\text{ mol dm}^{-3} \times 0.0250\text{ dm}^3 = 1.50 \times 10^{-3}\text{ mol}\). [1]
Moles of \(\text{Br}^-\) ions = \(2 \times (1.50 \times 10^{-3}) = 3.00 \times 10^{-3}\text{ mol}\). [1]
Reaction is 1:1 between \(\text{Ag}^+\) and \(\text{Br}^-\), so moles of \(\text{AgNO}_3\) needed = \(3.00 \times 10^{-3}\text{ mol}\).
Volume of \(\text{AgNO}_3 = \frac{3.00 \times 10^{-3}\text{ mol}}{0.100\text{ mol dm}^{-3}} = 0.0300\text{ dm}^3 = 30.0\text{ cm}^3\). [1]

(a)(i) 1 mark for Ca2+.
1 mark for explaining that calcium hydroxide is slightly soluble and does not precipitate with weak ammonia.
(a)(ii) 1 mark for bromide (Br-).
1 mark for stating that the cream precipitate is insoluble in dilute ammonia.
1 mark for stating that it dissolves in concentrated ammonia.
(b) 1 mark for calculating mass and moles of water (0.0900 mol).
1 mark for calculating molar mass of CaBr2 (199.9 g/mol).
1 mark for calculating moles of anhydrous CaBr2 (0.0150 mol).
1 mark for finding x = 6.
(c)(i) 1 mark for the correct balanced equation with state symbols.
(c)(ii) 1 mark for moles/volume setup.
1 mark for 0.0600 mol dm-3.
(c)(iii) 1 mark for calculating moles of CaBr2 in 25.0 cm3 (1.50 x 10^-3 mol).
1 mark for multiplying by 2 to get moles of Br- (3.00 x 10^-3 mol).
1 mark for calculating volume of AgNO3 (30.0 cm3).

Assuming a sample mean titre of \(25.00\text{ cm}^3\) of FA 2:

**(a)** The candidate's table should include initial and final burette readings, and the volume of FA 2 added for rough and accurate runs, recorded to 2 decimal places.

**(b)** Appropriate concordant titres (within \(0.10\text{ cm}^3\)) are selected and averaged:
\(\text{Mean titre} = 25.00\text{ cm}^3\).

**(c)** \(\text{Moles of HCl} = 0.100\text{ mol dm}^{-3} \times \left(\frac{25.00}{1000}\right)\text{ dm}^3 = 2.50 \times 10^{-3}\text{ mol}\).

**(d)** Balanced chemical equation:
\(\text{Na}_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})\)
\(\text{Moles of Na}_2\text{CO}_3 = \frac{1}{2} \times \text{Moles of HCl} = 1.25 \times 10^{-3}\text{ mol}\).

**(e)** \(\text{Concentration of FA 1} = 1.25 \times 10^{-3}\text{ mol} \times \left(\frac{1000}{25.0}\right)\text{ dm}^{-3} = 0.0500\text{ mol dm}^{-3}\).

**(f)** \(M_r\text{ of } \text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O} = \frac{\text{mass per dm}^3}{\text{concentration}} = \frac{14.30\text{ g dm}^{-3}}{0.0500\text{ mol dm}^{-3}} = 286.0\text{ g mol}^{-1}\).

**(g)** \(M_r\text{ of anhydrous } \text{Na}_2\text{CO}_3 = 2(23.0) + 12.0 + 3(16.0) = 106.0\text{ g mol}^{-1}\).
\(\text{Mass of water of crystallisation} = 286.0 - 106.0 = 180.0\text{ g mol}^{-1}\).
\(x = \frac{180.0}{18.0} = 10\).

**(a) Titration Table [4 marks]**
- 1 mark: Table is clearly presented with appropriate headings and units.
- 1 mark: All burette readings (initial and final) and volume added are recorded to 2 d.p. (e.g., \(0.00\) or \(0.05\)).
- 1 mark: At least two accurate titres are concordant within \(0.10\text{ cm}^3\).
- 1 mark: Accuracy mark based on how close the student's titre is to the supervisor's value.

**(b) Selection and Mean [1 mark]**
- 1 mark: Selecting at least two concordant titres and correctly calculating the mean value.

**(c) Moles of HCl [1 mark]**
- 1 mark: Correct calculation of moles of \(\text{HCl}\) to 3 sig fig (e.g., \(2.50 \times 10^{-3}\text{ mol}\) for a \(25.00\text{ cm}^3\) titre).

**(d) Equation and Moles of Salt [1 mark]**
- 1 mark: Balanced equation provided and calculation of moles of \(\text{Na}_2\text{CO}_3\) is half the moles of \(\text{HCl}\) from (c).

**(e) Concentration of FA 1 [1.5 marks]**
- 1 mark: Correctly scales moles in \(25.0\text{ cm}^3\) to \(1000\text{ cm}^3\).
- 0.5 marks: Final answer given to 3 significant figures.

**(f) Relative Formula Mass [2 marks]**
- 1 mark: Correct formula used: \(M_r = \frac{\text{mass}}{\text{moles}}\).
- 1 mark: Calculated value matches the concentration calculated in (e).

**(g) Determination of x [2 marks]**
- 1 mark: Correctly subtracts \(106.0\) from the calculated \(M_r\).
- 1 mark: Divides the difference by \(18.0\) to obtain a value for \(x\) to the nearest integer.

Assuming a mean titre of \(24.50\text{ cm}^3\) of FA 4:

**(a)** Endpoint color change: blue-black to colorless.
Starch is not added at the start because high concentrations of iodine bind strongly/irreversibly to starch, preventing some of it from releasing and causing an inaccurate (overestimated) endpoint.

**(b)**
**(i)** \(\text{Moles of } \text{S}_2\text{O}_3^{2-} = 0.100\text{ mol dm}^{-3} \times \left(\frac{24.50}{1000}\right)\text{ dm}^3 = 2.45 \times 10^{-3}\text{ mol}\).
ewline
**(ii)** From the equations:
\(1\text{ mol of } \text{I}_2\) reacts with \(2\text{ mol of } \text{S}_2\text{O}_3^{2-}\).
\(1\text{ mol of } \text{ClO}^-\rightleftharpoons\) produces \(1\text{ mol of } \text{I}_2\).
Therefore, \(\text{Moles of } \text{ClO}^- = \frac{1}{2} \times \text{Moles of } \text{S}_2\text{O}_3^{2-} = 1.225 \times 10^{-3}\text{ mol}\).
ewline
**(iii)** \(\text{Concentration of } \text{ClO}^- \text{ in FA 3} = \frac{1.225 \times 10^{-3}\text{ mol}}{0.0250\text{ dm}^3} = 0.0490\text{ mol dm}^{-3}\).
ewline
**(iv)** Since the original bleach was diluted by a factor of \(10\) (from \(10.0\text{ cm}^3\) to \(100.0\text{ cm}^3\)):
\(\text{Concentration of } \text{ClO}^- \text{ in original bleach} = 0.0490\text{ mol dm}^{-3} \times 10 = 0.490\text{ mol dm}^{-3}\).
ewline
**(v)** \(M_r(\text{NaClO}) = 23.0 + 35.5 + 16.0 = 74.5\text{ g mol}^{-1}\).
\(\text{Mass concentration of } \text{NaClO} = 0.490\text{ mol dm}^{-3} \times 74.5\text{ g mol}^{-1} = 36.5\text{ g dm}^{-3}\).

**(c)**
- *Source of error*: Atmospheric/aerial oxidation of iodide ions (\(4\text{I}^-(\text{aq}) + \text{O}_2(\text{g}) + 4\text{H}^+(\text{aq}) \rightarrow 2\text{I}_2(\text{aq}) + 2\text{H}_2\text{O}(\text{l})\)) which generates extra iodine and increases the titre of sodium thiosulfate.
- *Modification*: Titrate immediately after adding acid and potassium iodide / keep the flask stoppered until titration commences / add sodium carbonate to produce an inert \(\text{CO}_2\) atmosphere.

**(a) Endpoint & Starch [1.5 marks]**
- 0.5 marks: Correctly states endpoint color change as blue-black to colorless (or loss of blue color).
- 1 mark: Explains that starch binds strongly/irreversibly with high concentrations of iodine, making the endpoint inaccurate/sluggish if added too early.

**(b) Calculations [8 marks]**
- **(i)** 1 mark: Correct calculation of moles of \(\text{S}_2\text{O}_3^{2-}\) to 3 sig fig.
- **(ii)** 2 marks: Deduces 1:2 ratio of \(\text{ClO}^- : \text{S}_2\text{O}_3^{2-}\) and calculates \(1.225 \times 10^{-3}\text{ mol}\) (or equivalent tracking from student's (i)).
- **(iii)** 1.5 marks: Correctly calculates concentration of FA 3 (1 mark) and gives answer to 3 sig fig (0.5 marks).
- **(iv)** 1.5 marks: Correctly multiplies concentration of FA 3 by 10 (1 mark) and gives answer to 3 sig fig (0.5 marks).
- **(v)** 2 marks: Calculates \(M_r\) of \(\text{NaClO}\) as \(74.5\) (1 mark) and correctly multiplies to get concentration in \(\text{g dm}^{-3}\) (1 mark).

**(c) Error and Modification [3 marks]**
- 1 mark: Identifies aerial oxidation of iodide to iodine as the systematic error.
- 1 mark: Explains that this leads to extra iodine being produced, increasing the titre (and overestimating bleach concentration).
- 1 mark: Proposes an appropriate experimental modification (e.g., titrating rapidly, covering the conical flask, or executing in a carbon dioxide atmosphere).

### Expected Observations and Identification

#### Part (a): Observations Table
* **Test 1**: On adding \(\text{NaOH}(aq)\) dropwise to **FA 3**, a **green precipitate** (or dirty green precipitate) is formed which is **insoluble** in excess of the reagent.
* **Test 2**: On adding \(\text{NH}_3(aq)\) dropwise to **FA 3**, a **green precipitate** (or dirty green precipitate) is formed which is **insoluble** in excess of the reagent.
* **Test 3**: On adding aqueous barium chloride (or barium nitrate) to **FA 3**, a **white precipitate** is formed. On addition of dilute nitric acid, the precipitate **does not dissolve** (remains insoluble).
* **Test 4**: On adding acidified \(\text{KMnO}_4(aq)\) to **FA 3**, the **purple solution turns colourless** (decolourised), indicating a redox reaction where \(\text{Fe}^{2+}\) is oxidised to \(\text{Fe}^{3+}\) and \(\text{MnO}_4^-\rightarrow\text{Mn}^{2+}\) is reduced.
* **Test 5**: On adding \(\text{NaOH}(aq)\) dropwise to **FA 4**, a **white precipitate** is formed which **dissolves** in excess of the reagent to form a **colourless solution**.
* **Test 6**: On adding \(\text{NH}_3(aq)\) dropwise to **FA 4**, a **white precipitate** is formed which **dissolves** in excess of the reagent to form a **colourless solution**.
* **Test 7**: On adding \(\text{AgNO}_3(aq)\) to **FA 4**, a **white precipitate** is formed. On addition of dilute aqueous ammonia, this white precipitate **dissolves** to form a **colourless solution**.

#### Part (b): Identification of Ions
* **FA 3**:
* **Cation**: \(\text{Fe}^{2+}\) (Iron(II))
* **Anion**: \(\text{SO}_4^{2-}\) (Sulfate)
* **Evidence**: The green precipitate with both NaOH and ammonia that is insoluble in excess confirms the presence of \(\text{Fe}^{2+}\). The formation of a white precipitate with barium ions that is insoluble in dilute nitric acid confirms the presence of \(\text{SO}_4^{2-}\).
* **FA 4**:
* **Cation**: \(\text{Zn}^{2+}\) (Zinc)
* **Anion**: \(\text{Cl}^-\rightleftharpoons\) (Chloride)
* **Evidence**: The formation of a white precipitate with both NaOH and ammonia that dissolves in excess of both reagents uniquely confirms the presence of \(\text{Zn}^{2+}\). The formation of a white precipitate with silver nitrate that dissolves in dilute ammonia confirms the presence of \(\text{Cl}^-\).

#### Part (c): Ionic Equations
* *(i)* Reaction of \(\text{Fe}^{2+}\) with \(\text{OH}^-\):
$$\text{Fe}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Fe(OH)}_2(s)$$
* *(ii)* Reaction of \(\text{Cl}^-\rightleftharpoons\) with \(\text{Ag}^+\):
$$\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)$$

#### Part (d): Further Test for Iron(III)
* To confirm the presence of \(\text{Fe}^{3+}\) ions in the oxidised solution, add aqueous sodium hydroxide (or aqueous ammonia) to a sample of the mixture. A **red-brown precipitate** of iron(III) hydroxide, \(\text{Fe(OH)}_3\), will be formed, which is insoluble in excess.
* (Alternatively, add a few drops of aqueous potassium thiocyanate, \(\text{KSCN}\), which will form a characteristic **blood-red solution** due to the formation of the \([\text{Fe(H}_2\text{O)}_5(\text{SCN})]^{2+}\) complex).

### Marking Scheme

#### Part (a): Observations [8 marks]
* **Test 1**: Award 1 mark for: **Green / dirty green precipitate** which is **insoluble in excess** NaOH.
* **Test 2**: Award 1 mark for: **Green / dirty green precipitate** which is **insoluble in excess** ammonia.
* **Test 3**: Award 1 mark for: **White precipitate** which **remains insoluble / does not dissolve** in dilute nitric acid.
* **Test 4**: Award 1 mark for: **Purple solution turns colourless** (or purple color discharged / decolourised). *Reject 'clear' instead of 'colourless'.*
* **Test 5**: Award 1 mark for: **White precipitate** which **dissolves / is soluble** in excess NaOH to give a colourless solution.
* **Test 6**: Award 1 mark for: **White precipitate** which **dissolves / is soluble** in excess ammonia to give a colourless solution.
* **Test 7**: Award 2 marks:
* 1 mark for **white precipitate** on adding \(\text{AgNO}_3(aq)\).
* 1 mark for precipitate **dissolving / soluble** on adding dilute \(\text{NH}_3(aq)\).

#### Part (b): Ion Identification and Evidence [4 marks]
* Award 1 mark for identifying **FA 3** containing **\(\text{Fe}^{2+}\)** (iron(II)) and **\(\text{SO}_4^{2-}\)** (sulfate).
* Award 1 mark for identifying **FA 4** containing **\(\text{Zn}^{2+}\)** (zinc) and **\(\text{Cl}^-\)** (chloride).
* Award 1 mark for supporting evidence for **FA 3**:
* Explicitly links the green precipitate insoluble in excess NaOH/ammonia to \(\text{Fe}^{2+}\) AND the white precipitate insoluble in acid with barium ions to \(\text{SO}_4^{2-}\).
* Award 1 mark for supporting evidence for **FA 4**:
* Explicitly links the white precipitate soluble in excess NaOH and ammonia to \(\text{Zn}^{2+}\) AND the white precipitate with silver nitrate soluble in dilute ammonia to \(\text{Cl}^-\).

#### Part (c): Ionic Equations [2 marks]
* *(i)* Award 1 mark for:
$$\text{Fe}^{2+}(aq) + 2\text{OH}^-(aq) \rightarrow \text{Fe(OH)}_2(s)$$
*Both correct formulas and state symbols required.*
* *(ii)* Award 1 mark for:
$$\text{Ag}^+(aq) + \text{Cl}^-(aq) \rightarrow \text{AgCl}(s)$$
*Both correct formulas and state symbols required.*

#### Part (d): Test for Iron(III) [1 mark]
* Award 1 mark for:
* Reagent: Add **aqueous sodium hydroxide** (or **aqueous ammonia**), AND observation: **red-brown / rust-brown precipitate**.
* OR Reagent: Add **potassium thiocyanate / \(\text{KSCN}\)** solution, AND observation: **blood-red solution** formed.

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