Skill Align TASC Chemistry Level 4 - Free Online Pack 0
An independently authored four-section CHM415115 practice examination themed around Derwent water quality, marine infrastructure and renewable industry.
- Paper
- Written practice examination
- Reading
- 15 minutes preparation
- Writing
- 3 hours
- Assessment
- 180 marks
The current TASC CHM415115 Chemistry External Exam Information Sheet is approved for use throughout the examination and is linked as a separate official resource.
Section A - Electrochemistry
Five compulsory questions, 45 marks. Suggested working time: 45 minutes. This section assesses Criterion 5.
Question 1
6 marksA Derwent monitoring laboratory uses acidified permanganate to determine dissolved iron(II).
Question 2
7 marksA river sensor is powered by a standard Zn/Zn²⁺ and Ag⁺/Ag galvanic cell. E⁰(Zn²⁺/Zn) = -0.76 V and E⁰(Ag⁺/Ag) = +0.80 V.
Question 3
9 marksA workshop electrorefines copper for a river-conductivity probe. A current of 2.40 A flows for 15.0 min through a cell with an impure Cu anode, pure Cu cathode and a solution containing Cu²⁺. Use F = 96 500 C mol⁻¹.
Question 4
11 marksA steel river pontoon is electrically connected to magnesium blocks in aerated water. Use E⁰(Fe²⁺/Fe) = -0.41 V and E⁰(Mg²⁺/Mg) = -2.38 V as comparative evidence.
Question 5
12 marksA copper half-cell is connected to a standard hydrogen electrode (SHE). The copper electrode is positive and the voltmeter reads 0.34 V under standard conditions.
Section B - Thermochemistry, kinetics and equilibrium
Five compulsory questions, 45 marks. Suggested working time: 45 minutes. This section assesses Criterion 6.
Question 6
7 marksA Derwent field class burns ethanol beneath a simple water calorimeter. 180.0 g of water warms by 8.4 degrees C when 0.01200 mol of ethanol reacts. Use c(water) = 4.184 J g⁻¹ K⁻¹ and initially neglect calorimeter heat capacity.
Question 7
9 marksA process model considers CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). DeltaHf: CH₄(g) -74.8, CO₂(g) -393.5, H₂O(l) -285.8 kJ mol⁻¹; DeltaHf[O₂(g)] = 0.
Question 8
11 marksA water laboratory records CO₂ evolved when a carbonate sample reacts with acid. The graph shows one run under fixed conditions.
Question 9
12 marksA closed reactor reaches equilibrium for H₂(g) + I₂(g) ⇌ 2HI(g). At one temperature, [H₂] = 0.200, [I₂] = 0.150 and [HI] = 0.800 mol L⁻¹.
Question 10
6 marksA river-quality model uses ethanoic acid, CH₃COOH. At 25 degrees C, Ka = 1.8 x 10⁻⁵ and the analytical acid concentration is 0.100 mol L⁻¹. Treat water autoionisation as negligible for the initial estimate.
Section C - Properties and reactions of organic and inorganic matter
Five compulsory questions, 45 marks. Suggested working time: 45 minutes. This section assesses Criterion 7.
Question 11
9 marksA dissolved-oxygen study collects an oxygen sample in a flexible vessel. The oxygen changes from 101.3 kPa, 1.200 L and 298 K to 95.0 kPa and 310 K without loss of gas.
Question 12
11 marksA river laboratory converts propene to a cleaning solvent through hydration and oxidation.
Question 13
12 marksAn unknown has M⁺ at m/z 60, a very broad IR absorption from 2500-3300 cm⁻¹ and a strong 1710 cm⁻¹ band. A fragment at m/z 43 is prominent.
Question 14
6 marksSpecies Na has atomic number 11 and charge 0; species Mg has atomic number 12 and charge 0. The graph gives successive ionisation energies for neutral Na.
Question 15
7 marksBarium chloride and sodium sulfate solutions are mixed during a sulfate test.
Section D - Quantitative chemical problem solving
Five compulsory questions, 45 marks. Suggested working time: 45 minutes. This section assesses Criterion 8.
Question 16
11 marksA production trial uses 2H₂ + O₂ → 2H₂O. It combines 4.00 g H₂ with 40.00 g O₂. The limiting reactant reacts completely; any shortfall in isolated product arises solely during product recovery. Use M(H₂) = 2.00, M(O₂) = 32.00 and M(H₂O) = 18.00 g mol⁻¹.
Question 17
12 marksA sample of CuSO₄.xH₂O has mass 4.992 g. After repeated heat-cool-weigh cycles to constant mass, 3.192 g of anhydrous CuSO₄ remains. Use M(CuSO₄) = 159.6 and M(H₂O) = 18.0 g mol⁻¹.
Question 18
9 marksA 25.00 mL aliquot of diluted HCl requires 21.40 mL of 0.0850 mol L⁻¹ NaOH. The original sample was diluted by a factor of 5. The reaction is HCl + NaOH → NaCl + H₂O. The nominated indicator is bromothymol blue.
Question 19
7 marksOxygen from a water-quality reaction is collected over water. Total pressure is 101.30 kPa, water-vapour pressure is 3.17 kPa, gas volume is 0.425 L and temperature is 298 K. Use R = 8.31 L kPa K⁻¹ mol⁻¹.
Question 20
6 marksA health-products pilot converts salicylic acid to aspirin with the other reagent in excess. 6.00 g of salicylic acid is used. The stoichiometric product-to-reactant amount ratio is 1:1. M(salicylic acid) = 138.12 and M(aspirin) = 180.16 g mol⁻¹. The isolated product mass is 6.80 g at 96.0% purity.
Worked Solutions And Marking Guide
General marking principles
- Apply each question-specific criterion to the evidence shown and accept chemically equivalent equations, structures and reasoning.
- Carry forward an earlier numerical error when later chemistry is internally consistent, unless a later criterion independently assesses the same step.
- Award every numerical mark against the mapped external criterion and the evidence supplied for that question.
- Do not penalise a missing unit more than once in a linked calculation unless units are separately assessed.
Section A Question 1
(a) MnO₄⁻ is reduced from Mn(+7) to Mn(+2) and is the oxidiser. Fe²⁺ is oxidised to Fe³⁺ and is the reducer.
(b) Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Oxidation: Fe²⁺ → Fe³⁺ + e⁻.
Detailed marking criteria
Part a (1 mark)
MnO₄⁻ is reduced from Mn(+7) to Mn(+2) and is the oxidiser.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Fe²⁺ is oxidised to Fe³⁺ and is the reducer.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Links oxidation to electron loss and reduction to electron gain.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Writes the balanced reduction half-equation: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Writes the balanced oxidation half-equation: Fe²⁺ → Fe³⁺ + e⁻.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Verifies that atoms and net charge are conserved in both acidic half-equations.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section A Question 2
(a) Anode: Zn(s) → Zn²⁺(aq) + 2e⁻. Cathode: Ag⁺(aq) + e⁻ → Ag(s).
(b) E⁰_cell = E⁰_cathode - E⁰_anode = +0.80 - (-0.76) = 1.56 V. The positive value predicts a spontaneous cell reaction under standard conditions.
(c) States that tabulated standard potentials apply to standard-state conditions.
Detailed marking criteria
Part a (1 mark)
Identifies Zn as the anode because its reduction potential is lower.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Writes Zn(s) → Zn²⁺(aq) + 2e⁻.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Writes Ag⁺(aq) + e⁻ → Ag(s).
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Uses E⁰_cell = E⁰_cathode - E⁰_anode.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates E⁰_cell = 1.56 V.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Uses the positive standard cell potential to justify spontaneity in the written direction.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
States that tabulated standard potentials apply to standard-state conditions.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section A Question 3
(a) Anode: Cu(s) → Cu²⁺(aq) + 2e⁻. Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s). Ideally the same amount of Cu²⁺ is made at the anode as is consumed at the cathode.
(b) Q = It = (2.40)(900) = 2160 C. n(e⁻) = Q/F = 2.24 × 10⁻² mol.
(c) n(Cu) = n(e⁻)/2 = 1.12 × 10⁻² mol, so m = nM = 0.711 g. Competing cathodic reactions, loss of deposited metal during rinsing, or lower actual charge than assumed can lower the measured gain. Incomplete drying leaves retained liquid and can make the apparent gain too high.
Detailed marking criteria
Part a (1 mark)
Writes anode oxidation of Cu to Cu²⁺ with 2 electron(s).
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Writes cathode reduction of Cu²⁺ to Cu with 2 electron(s).
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Explains that equal electron transfer couples Cu²⁺ production and consumption.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Converts the operating time to 900 s.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates Q = 2160 C.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates n(e⁻) = 2.24 × 10⁻² mol using Faraday's constant.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Uses the 2:1 electron-to-Cu stoichiometric ratio.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Calculates a theoretical cathode gain of 0.711 g.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Identifies one valid lower-mass cause, such as competing cathodic reactions, loss of deposited metal or lower actual charge than assumed. Retained rinse liquid biases mass high and is not an acceptable low-mass cause.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section A Question 4
(a) At the operating anode Mg(s) → Mg²⁺(aq) + 2e⁻. At an oxygen-rich cathodic site on the steel, O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq). Mobile ions complete the internal circuit and enable separated redox sites.
(b) E⁰(Mg) - E⁰(Fe) = -1.97 V. Mg has the more negative reduction potential and is preferentially oxidised, so the steel is cathodically protected.
(c) A suitable strategy is to install and periodically replace magnesium sacrificial anodes. Rust is an ionic, porous solid rather than a coherent metallic layer; it occupies a greater volume than the consumed iron and can crack or lift a coating, exposing fresh metal and electrolyte pathways.
(d) States one valid controlled variable, such as exposed steel area, electrolyte composition or oxygen access. Uses replicated mass loss, corrosion current or rust-area measurements over a stated time.
Detailed marking criteria
Part a (1 mark)
Writes Mg(s) → Mg²⁺(aq) + 2e⁻ at the operating anode.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Writes O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq) at an oxygen-rich cathodic site.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Explains that dissolved ions provide an internal ionic pathway between separated sites.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Compares -2.38 V with -0.41 V rather than relying on coating appearance.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Identifies Mg as the sacrificial anode.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Explains that electron supply suppresses oxidation of the steel.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Recommends that the operator install and periodically replace magnesium sacrificial anodes.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Contrasts porous ionic rust with coherent metallic steel or a continuous barrier.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Explains that the larger rust volume can crack or lift protection and expose fresh steel.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
States one valid controlled variable, such as exposed steel area, electrolyte composition or oxygen access.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Uses replicated mass loss, corrosion current or rust-area measurements over a stated time.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section A Question 5
(a) Specifies H₂(g) at standard pressure in contact with standard-concentration H⁺(aq). Specifies 25 degrees C for the tabulated standard measurement. Explains that Pt conducts and catalyses without being a net reactant.
(b) Writes H₂(g) → 2H⁺(aq) + 2e⁻ at the hydrogen anode. Writes Cu²⁺(aq) + 2e⁻ → Cu(s) at the copper cathode. States that electrons travel from the SHE to the copper electrode.
(c) Uses E⁰_cell = E⁰_cathode - E⁰_anode. Obtains E⁰(Cu²⁺/Cu) = +0.34 V. Explains that Cu²⁺ is reduced more readily than H⁺ under standard conditions.
(d) Controls temperature OR the relevant dissolved-ion concentrations or activities in both half-cells. Either valid single control earns this mark; stating both is also accepted. Uses voltmeter polarity or electron direction to check the sign assigned to the unknown half-cell. Repeats the measurement after renewing electrode surfaces or solutions to detect drift.
Detailed marking criteria
Part a (1 mark)
Specifies H₂(g) at standard pressure in contact with standard-concentration H⁺(aq).
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Specifies 25 degrees C for the tabulated standard measurement.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Explains that Pt conducts and catalyses without being a net reactant.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Writes H₂(g) → 2H⁺(aq) + 2e⁻ at the hydrogen anode.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Writes Cu²⁺(aq) + 2e⁻ → Cu(s) at the copper cathode.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
States that electrons travel from the SHE to the copper electrode.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Uses E⁰_cell = E⁰_cathode - E⁰_anode.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Obtains E⁰(Cu²⁺/Cu) = +0.34 V.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Explains that Cu²⁺ is reduced more readily than H⁺ under standard conditions.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Controls temperature OR the relevant dissolved-ion concentrations or activities in both half-cells. Either valid single control earns this mark; stating both is also accepted.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Uses voltmeter polarity or electron direction to check the sign assigned to the unknown half-cell.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Repeats the measurement after renewing electrode surfaces or solutions to detect drift.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section B Question 6
(a) q_water = mcDeltaT = (180.0)(4.184)(8.4) = 6.326 kJ. Therefore DeltaH = -q_water/n = -527.2 kJ mol⁻¹.
(b) Heat lost to the wider surroundings makes the measured water temperature rise too small, so the calculated enthalpy is insufficiently exothermic. A calibration factor accounts for energy absorbed by the calorimeter: q = C_f DeltaT, combined with any separately modelled water term.
(c) Identifies heat absorbed by the vessel or escaping to air as energy omitted from the simple water model.
Detailed marking criteria
Part a (1 mark)
Calculates q_water = 6.326 kJ.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Uses q_reaction = -q_surroundings for the measured process.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Calculates DeltaH = -527.2 kJ mol⁻¹ with a negative sign.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Predicts that external heat loss makes the magnitude of measured DeltaH too small.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Explains that some reaction energy bypasses the measured water temperature change.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Uses a calibration factor with the measured temperature change to include calorimeter energy.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Identifies heat absorbed by the vessel or escaping to air as energy omitted from the simple water model.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section B Question 7
(a) Hess' law states that enthalpy change depends only on initial and final states, not the route. For the written reaction, DeltaH = sum[stoichiometric coefficient x DeltaHf(products)] - sum[stoichiometric coefficient x DeltaHf(reactants)].
(b) Every formation enthalpy is multiplied by its stoichiometric coefficient. The coefficient-weighted product total minus the coefficient-weighted reactant total gives DeltaH = -890.3 kJ for the reaction as written.
(c) Reversing changes the sign and halving the equation halves DeltaH, so the result is 445.15 kJ. Enthalpy is an extensive quantity tied to the stoichiometric amount represented.
Detailed marking criteria
Part a (1 mark)
States that reaction enthalpy is path independent for fixed initial and final states.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Uses products minus reactants.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Includes stoichiometric coefficients in the formation-enthalpy sum.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Substitutes every supplied formation enthalpy on the correct side.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Applies all coefficients before summing.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Obtains DeltaH = -890.3 kJ for the equation as written.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Changes the sign from -890.3 kJ to 890.3 kJ for reversal.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Halves the magnitude to 445.15 kJ.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Explains that reaction enthalpy scales with the amount represented by the equation.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section B Question 8
(a) Average rate = (24.8 - 0)/(30 - 0) = 0.827 mL s⁻¹. It is a secant average over a finite interval; the instantaneous initial rate is the tangent gradient at t = 0.
(b) Predicts an increased initial reaction rate. More acid particles per unit volume increase the frequency of collisions with the carbonate. The fraction of collisions with sufficient energy remains unchanged because the temperature, energy distribution and activation energy are unchanged.
(c) Gas volume increases from 47.5 to 54.0 mL between 90 and 150 s, a further 6.5 mL, so gas production continues; the recorded curve does not establish complete consumption of the limiting reactant. Reactants are progressively consumed, so declining acid concentration and/or declining accessible carbonate surface can reduce effective collision frequency and slow gas production. Either defensible reactant-based explanation is accepted; a lower rate alone does not prove the reaction has stopped. Compares a mean with spread from independent repeats, using the same time interval or defined initial-gradient method for every tested condition.
(d) Subtracts 2.0 mL from each absolute gas-volume reading to correct the positive zero offset. The constant offset cancels when the two volume readings are subtracted, so it does not bias the volume-difference rate if the time readings are correct; an offset in absolute volume must not be confused with a changing offset or scale error.
Detailed marking criteria
Part a (1 mark)
Reads 24.8 mL at 30 s from the graph.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Calculates the average rate as 0.827 mL s⁻¹.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Distinguishes a secant average from the tangent gradient at t = 0.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Predicts an increased initial reaction rate.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
More acid particles per unit volume increase the frequency of collisions with the carbonate.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
The fraction of collisions with sufficient energy remains unchanged because the temperature, energy distribution and activation energy are unchanged.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Gas volume increases from 47.5 to 54.0 mL between 90 and 150 s, a further 6.5 mL, so gas production continues; the recorded curve does not establish complete consumption of the limiting reactant.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Reactants are progressively consumed, so declining acid concentration and/or declining accessible carbonate surface can reduce effective collision frequency and slow gas production. Either defensible reactant-based explanation is accepted; a lower rate alone does not prove the reaction has stopped.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Compares a mean with spread from independent repeats, using the same time interval or defined initial-gradient method for every tested condition.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Subtracts 2.0 mL from each absolute gas-volume reading to correct the positive zero offset.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
The constant offset cancels when the two volume readings are subtracted, so it does not bias the volume-difference rate if the time readings are correct; an offset in absolute volume must not be confused with a changing offset or scale error.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section B Question 9
(a) Kc = [HI]²/([H₂][I₂]). Substitution gives Kc = 21.3 using the concentration values as written.
(b) At unchanged temperature, the forward rate immediately rises and the system shifts toward HI until the rates are equal again. The value of Kc is unchanged because only temperature changes an equilibrium constant.
(c) At dynamic equilibrium the forward and reverse reaction rates are equal while both reactions continue, so macroscopic concentrations remain constant. Le Chatelier's principle predicts direction but not the size or speed of change; K, kinetics, energy cost and catalyst behaviour are also needed for process design.
(d) The first pass converts 0.0400 mol H₂, leaving 0.0600 mol; the second converts a further 0.0600 × 0.400 = 0.0240 mol. Total H₂ conversion is 0.0400 + 0.0240 = 0.0640 mol, or 64.0% of the original feed. Kc is unchanged at the same temperature: separating product and recycling reactant improves overall process conversion without changing the equilibrium constant.
Detailed marking criteria
Part a (1 mark)
Writes Kc = [HI]²/([H₂][I₂]).
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Substitutes equilibrium rather than initial concentrations with coefficients used as powers.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Calculates Kc = 21.3.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
the forward rate immediately rises and the system shifts toward HI until the rates are equal again
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Explains the prediction using competing forward and reverse rates or a reaction quotient.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
States that Kc is unchanged at the same temperature.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
States that forward and reverse rates are equal and non-zero at dynamic equilibrium.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Links equal rates to constant macroscopic concentrations in a closed system.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Identifies that Le Chatelier's principle alone gives neither the magnitude nor the rate of a response.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
The first pass converts 0.0400 mol H₂, leaving 0.0600 mol; the second converts a further 0.0600 × 0.400 = 0.0240 mol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Total H₂ conversion is 0.0400 + 0.0240 = 0.0640 mol, or 64.0% of the original feed.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Kc is unchanged at the same temperature: separating product and recycling reactant improves overall process conversion without changing the equilibrium constant.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section B Question 10
(a) Ka approximately equals x²/c, so [H⁺] = x = sqrt((1.8 × 10⁻⁵)(0.100)) = 1.34 × 10⁻³ mol L⁻¹ and pH = -log[H⁺] = 2.87.
(b) Percentage ionisation is about 100x/c = 1.34%. This is at or below the usual 5% check, so the approximation is reasonable. A strong acid at 0.100 mol L⁻¹ would have pH 1.00, lower because it ionises essentially completely.
Detailed marking criteria
Part a (1 mark)
Uses Ka approximately equal to x²/c for a weak monoprotic acid.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Calculates [H⁺] = 1.34 × 10⁻³ mol L⁻¹.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Calculates pH = 2.87.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates percentage ionisation as approximately 1.34%.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Judges the small-ionisation approximation reasonable using the 5% check.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Predicts the strong-acid pH as 1.00 and explains the lower value by near-complete ionisation.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section C Question 11
(a) Uses P₁V₁/T₁ = P₂V₂/T₂ with temperatures in kelvin. Calculates V₂ = (101.3)(1.200)(310)/[(298)(95.0)] = 1.331 L. States that the amount of gas and its identity remain constant.
(b) At fixed volume, higher absolute temperature raises mean translational kinetic energy and collision impulse. At fixed temperature, compression increases collision frequency per unit wall area. Connects macroscopic pressure to particle collisions with the vessel walls.
(c) Predicts greatest departure at high pressure and low temperature. Explains that intermolecular attractions are no longer negligible. Explains that molecular volume is no longer negligible compared with container volume.
Detailed marking criteria
Part a (1 mark)
Uses P₁V₁/T₁ = P₂V₂/T₂ with temperatures in kelvin.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Calculates V₂ = (101.3)(1.200)(310)/[(298)(95.0)] = 1.331 L.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
States that the amount of gas and its identity remain constant.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
At fixed volume, higher absolute temperature raises mean translational kinetic energy and collision impulse.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
At fixed temperature, compression increases collision frequency per unit wall area.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Connects macroscopic pressure to particle collisions with the vessel walls.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Predicts greatest departure at high pressure and low temperature.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Explains that intermolecular attractions are no longer negligible.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Explains that molecular volume is no longer negligible compared with container volume.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section C Question 12
(a) Identifies propene as an alkene and propan-2-ol as a secondary alcohol. Uses steam with an acid catalyst for hydration. Writes CH₃CH=CH₂ + H₂O → CH₃CH(OH)CH₃.
(b) Identifies propanone as a ketone. Uses acidified dichromate or another suitable controlled oxidant. Writes CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O.
(c) Links alcohol hydrogen bonding to greater water solubility than propene. Explains that propanone accepts hydrogen bonds from water through its carbonyl oxygen. Distinguishes intermolecular forces from breaking covalent bonds during dissolving.
(d) Identifies one plausible competing reaction or incomplete-conversion risk for the stated functional group. Names a separation or purification step suited to the represented product class.
Detailed marking criteria
Part a (1 mark)
Identifies propene as an alkene and propan-2-ol as a secondary alcohol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Uses steam with an acid catalyst for hydration.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Writes CH₃CH=CH₂ + H₂O → CH₃CH(OH)CH₃.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Identifies propanone as a ketone.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Uses acidified dichromate or another suitable controlled oxidant.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Writes CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Links alcohol hydrogen bonding to greater water solubility than propene.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Explains that propanone accepts hydrogen bonds from water through its carbonyl oxygen.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Distinguishes intermolecular forces from breaking covalent bonds during dissolving.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Identifies one plausible competing reaction or incomplete-conversion risk for the stated functional group.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Names a separation or purification step suited to the represented product class.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section C Question 13
(a) Uses M⁺ = 60 as relative molecular-mass evidence. Uses the broad O-H and carbonyl absorptions to identify a carboxylic acid. Identifies C₂H₄O₂ as a formula consistent with Mr 60 and a carboxylic acid.
(b) Associates m/z 43 with a plausible CH₃CO⁺ fragment. Recognises that fragmentation assignments require charge and mass consistency. States that a shared fragment cannot by itself establish complete connectivity.
(c) Combines molecular-ion, IR and fragmentation evidence rather than using one peak alone. Identifies structural isomerism or impurity as a remaining limitation. Nominates suitable crystallographic or additional spectroscopic evidence for confirmation.
(d) Writes CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l), conserving atoms and charge; accepts the equivalent hydronium-based net ionic representation with the coupled acid equilibrium. Neutralisation removes the carboxyl proton and forms ethanoate, which no longer contains the acid O-H bond responsible for the original broad 2500-3300 cm⁻¹ absorption. Ethanoate anions and sodium cations form ion-dipole attractions with polar water molecules; hydration of the charged particles supports the salt's water solubility.
Detailed marking criteria
Part a (1 mark)
Uses M⁺ = 60 as relative molecular-mass evidence.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Uses the broad O-H and carbonyl absorptions to identify a carboxylic acid.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Identifies C₂H₄O₂ as a formula consistent with Mr 60 and a carboxylic acid.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Associates m/z 43 with a plausible CH₃CO⁺ fragment.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Recognises that fragmentation assignments require charge and mass consistency.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
States that a shared fragment cannot by itself establish complete connectivity.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Combines molecular-ion, IR and fragmentation evidence rather than using one peak alone.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Identifies structural isomerism or impurity as a remaining limitation.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Nominates suitable crystallographic or additional spectroscopic evidence for confirmation.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Writes CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l), conserving atoms and charge; accepts the equivalent hydronium-based net ionic representation with the coupled acid equilibrium.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Neutralisation removes the carboxyl proton and forms ethanoate, which no longer contains the acid O-H bond responsible for the original broad 2500-3300 cm⁻¹ absorption.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Ethanoate anions and sodium cations form ion-dipole attractions with polar water molecules; hydration of the charged particles supports the salt's water solubility.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section C Question 14
(a) Constructs Na as 1s2 2s2 2p6 3s1. Constructs Mg as 1s2 2s2 2p6 3s2. Explains that the different electron counts mean the pair is not isoelectronic.
(b) Identifies the largest displayed jump for neutral Na from the supplied graph. Explains that the jump begins when ionisation removes an electron from a lower, more strongly bound shell. Explains that discrete absorption or emission lines arise from transitions between quantised electron energies.
Detailed marking criteria
Part a (1 mark)
Constructs Na as 1s2 2s2 2p6 3s1.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Constructs Mg as 1s2 2s2 2p6 3s2.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Explains that the different electron counts mean the pair is not isoelectronic.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Identifies the largest displayed jump for neutral Na from the supplied graph.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Explains that the jump begins when ionisation removes an electron from a lower, more strongly bound shell.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Explains that discrete absorption or emission lines arise from transitions between quantised electron energies.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section C Question 15
(a) Predicts that a white barium sulfate precipitate forms. Writes Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). Identifies Na⁺ and Cl⁻ as spectator ions.
(b) Applies the relevant solubility exception to predict the solid product. Explains that precipitation removes the selected ion pair from the aqueous phase. Recognises that other ions or mixtures can coexist, so one precipitate does not establish complete sample purity.
(c) Identifies one ion that could interfere with the selected precipitation observation.
Detailed marking criteria
Part a (1 mark)
Predicts that a white barium sulfate precipitate forms.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Writes Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Identifies Na⁺ and Cl⁻ as spectator ions.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Applies the relevant solubility exception to predict the solid product.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Explains that precipitation removes the selected ion pair from the aqueous phase.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Recognises that other ions or mixtures can coexist, so one precipitate does not establish complete sample purity.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Identifies one ion that could interfere with the selected precipitation observation.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section D Question 16
(a) Calculates n(H₂) = 2.0000 mol. Calculates n(O₂) = 1.2500 mol. Compares n/coefficient and identifies H₂ as limiting and O₂ as excess.
(b) Uses reaction extent 1.0000 mol of reaction. Calculates n(H₂O) = 2.0000 mol. Calculates theoretical m(H₂O) = 36.000 g.
(c) Calculates percentage yield = 29.520/36.000 × 100 = 82.0%. Calculates the amount of O₂ consumed from the balanced-equation ratio. Subtracts consumed from initial amount to obtain remaining O₂: 0.2500 mol.
(d) The unrecovered water mass is 36.000 - 29.520 = 6.480 g, using the theoretical mass formed and the isolated mass. Rejects the interpretation: all limiting H₂ reacts, so its conversion is 100%, while the isolated yield is 82.0%. The stated shortfall occurs only during product recovery and does not imply unreacted H₂.
Detailed marking criteria
Part a (1 mark)
Calculates n(H₂) = 2.0000 mol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Calculates n(O₂) = 1.2500 mol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Compares n/coefficient and identifies H₂ as limiting and O₂ as excess.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Uses reaction extent 1.0000 mol of reaction.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates n(H₂O) = 2.0000 mol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates theoretical m(H₂O) = 36.000 g.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Calculates percentage yield = 29.520/36.000 × 100 = 82.0%.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Calculates the amount of O₂ consumed from the balanced-equation ratio.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Subtracts consumed from initial amount to obtain remaining O₂: 0.2500 mol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
The unrecovered water mass is 36.000 - 29.520 = 6.480 g, using the theoretical mass formed and the isolated mass.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Rejects the interpretation: all limiting H₂ reacts, so its conversion is 100%, while the isolated yield is 82.0%. The stated shortfall occurs only during product recovery and does not imply unreacted H₂.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section D Question 17
(a) Calculates water mass = 4.992 - 3.192 = 1.800 g. Calculates n(H₂O) = 0.1000 mol. Calculates n(CuSO₄) = 0.0200 mol.
(b) Calculates n(H₂O)/n(CuSO₄) = 5:1 and therefore x = 5. Writes the hydrate as CuSO₄.5H₂O. Calculates water percentage = 36.06%.
(c) Predicts incomplete dehydration leaves the final mass too high and makes calculated x too low. Explains that repeated constant mass provides evidence that further heating no longer removes water. Explains that covered cooling limits reabsorption of atmospheric moisture before weighing.
(d) The measured residue is 3.092 g. Apparent n(H₂O) = (4.992 - 3.092)/18.0 = 0.1056 mol, and apparent n(CuSO₄) = 3.092/159.6 = 0.01937 mol. Apparent x = n(H₂O)/n(CuSO₄) = 5.45; accepts equivalent working without premature rounding. Calculated x is too high relative to the true x = 5: lost salt is falsely counted as water loss while the calculated salt amount is reduced. Constant mass cannot restore the missing CuSO₄ or remove this systematic bias.
Detailed marking criteria
Part a (1 mark)
Calculates water mass = 4.992 - 3.192 = 1.800 g.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Calculates n(H₂O) = 0.1000 mol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Calculates n(CuSO₄) = 0.0200 mol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates n(H₂O)/n(CuSO₄) = 5:1 and therefore x = 5.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Writes the hydrate as CuSO₄.5H₂O.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates water percentage = 36.06%.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Predicts incomplete dehydration leaves the final mass too high and makes calculated x too low.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Explains that repeated constant mass provides evidence that further heating no longer removes water.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Explains that covered cooling limits reabsorption of atmospheric moisture before weighing.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
The measured residue is 3.092 g. Apparent n(H₂O) = (4.992 - 3.092)/18.0 = 0.1056 mol, and apparent n(CuSO₄) = 3.092/159.6 = 0.01937 mol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Apparent x = n(H₂O)/n(CuSO₄) = 5.45; accepts equivalent working without premature rounding.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part d (1 mark)
Calculated x is too high relative to the true x = 5: lost salt is falsely counted as water loss while the calculated salt amount is reduced. Constant mass cannot restore the missing CuSO₄ or remove this systematic bias.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section D Question 18
(a) Uses the balanced ratio of 1 mol NaOH per mol HCl. Selects bromothymol blue because its transition is within the steep pH change near the nominated endpoint. Distinguishes the stoichiometric equivalence point from the observed indicator endpoint.
(b) Calculates n(NaOH) = (0.0850)(0.02140) = 0.001819 mol. Uses the equation to obtain n(HCl) = 0.001819 mol in the aliquot. Calculates diluted c(HCl) = 0.0728 mol L⁻¹.
(c) Applies the dilution factor to obtain original c(HCl) = 0.3638 mol L⁻¹. States one valid procedural control, such as conditioning the burette with NaOH, conditioning the pipette with diluted analyte, rinsing the volumetric flask with deionised water, removing a burette-tip air bubble, or reading the meniscus at eye level. States a second distinct, scientifically valid procedural control. Any two valid controls earn the two control marks; the same action cannot earn both marks and the listed examples are not all compulsory.
Detailed marking criteria
Part a (1 mark)
Uses the balanced ratio of 1 mol NaOH per mol HCl.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Selects bromothymol blue because its transition is within the steep pH change near the nominated endpoint.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Distinguishes the stoichiometric equivalence point from the observed indicator endpoint.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates n(NaOH) = (0.0850)(0.02140) = 0.001819 mol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Uses the equation to obtain n(HCl) = 0.001819 mol in the aliquot.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates diluted c(HCl) = 0.0728 mol L⁻¹.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Applies the dilution factor to obtain original c(HCl) = 0.3638 mol L⁻¹.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
States one valid procedural control, such as conditioning the burette with NaOH, conditioning the pipette with diluted analyte, rinsing the volumetric flask with deionised water, removing a burette-tip air bubble, or reading the meniscus at eye level.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
States a second distinct, scientifically valid procedural control. Any two valid controls earn the two control marks; the same action cannot earn both marks and the listed examples are not all compulsory.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section D Question 19
(a) Uses Dalton's law: P_total = P(O₂) + P(H₂O). Calculates P(O₂) = 101.30 - 3.17 = 98.13 kPa. Explains that the collected sample contains both product gas and equilibrium water vapour.
(b) Uses PV = nRT with pressure in kPa, volume in litres and temperature in kelvin. Substitutes n = (98.13)(0.425)/[(8.31)(298)]. Calculates n(O₂) = 1.68 × 10⁻² mol.
(c) Check that the water levels inside the cylinder and in the surrounding bath are equal, so no hydrostatic pressure difference separates the collected gas pressure from atmospheric pressure.
Detailed marking criteria
Part a (1 mark)
Uses Dalton's law: P_total = P(O₂) + P(H₂O).
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Calculates P(O₂) = 101.30 - 3.17 = 98.13 kPa.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Explains that the collected sample contains both product gas and equilibrium water vapour.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Uses PV = nRT with pressure in kPa, volume in litres and temperature in kelvin.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Substitutes n = (98.13)(0.425)/[(8.31)(298)].
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates n(O₂) = 1.68 × 10⁻² mol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part c (1 mark)
Check that the water levels inside the cylinder and in the surrounding bath are equal, so no hydrostatic pressure difference separates the collected gas pressure from atmospheric pressure.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Section D Question 20
(a) Calculates n(salicylic acid) = 0.04344 mol. Uses the 1:1 ratio to obtain n(aspirin) = 0.04344 mol. Calculates theoretical m(aspirin) = 7.826 g.
(b) Calculates pure product mass = 6.80 × 0.960 = 6.528 g. Uses purity-corrected product mass rather than total wet or impure isolate mass. Calculates percentage yield = 83.4%.
Detailed marking criteria
Part a (1 mark)
Calculates n(salicylic acid) = 0.04344 mol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Uses the 1:1 ratio to obtain n(aspirin) = 0.04344 mol.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part a (1 mark)
Calculates theoretical m(aspirin) = 7.826 g.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates pure product mass = 6.80 × 0.960 = 6.528 g.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Uses purity-corrected product mass rather than total wet or impure isolate mass.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Part b (1 mark)
Calculates percentage yield = 83.4%.
Acceptable alternatives: Accept a chemically equivalent representation that preserves the assessed proposition.
Diagnostic Checklist
| Topic | Questions | Marks | Marks Lost | Action |
|---|---|---|---|---|
| Criterion 5 - Electrochemistry | Q1, Q2, Q3, Q4, Q5 | 45 | ___ | Review redox equations, electrode potentials, galvanic and electrolytic cells, Faraday's laws and corrosion. |
| Criterion 6 - Thermochemistry, kinetics and equilibrium | Q6, Q7, Q8, Q9, Q10 | 45 | ___ | Review calorimetry, bond-energy ledgers, thermochemical-equation cycles, constructed energy diagrams, rate evidence, dynamic equilibrium, K, Q, acids and pH. |
| Criterion 7 - Properties and reactions of organic and inorganic matter | Q11, Q12, Q13, Q14, Q15 | 45 | ___ | Review gas mixtures and experimental models, functional groups and reactions, isotope spectra, benzene bonding, spectroscopy, periodicity and inorganic reactions. |
| Criterion 8 - Quantitative chemical problem solving | Q16, Q17, Q18, Q19, Q20 | 45 | ___ | Review stoichiometry, limiting reagents, gravimetry, titration, gas laws, synthesis yield and multi-stage calculations. |