2023 HSC Chemistry Exam Paper Solutions

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2023 HSC Chemistry Exam Paper Worked Solutions

These are the responses to the 2023 Chemistry paper, which can be found here on the NESA website.

Section 1: Multiple Choice Questions

Question	Answer	Explanation
1	D	Hydrocarbons cannot be poured down the sink as many are toxic to aquatic organisms. It is impractical to dispose of a liquid in a garbage bin. It is unsafe to burn hydrocarbons as a means of disposal as many are highly flammable and this poses a needless fire risk.
2	C	In atomic absorption spectroscopy (AAS) a sample is aspirated into a flame and light of characteristic wavelengths is passed through the sample. A monochromator separates these wavelengths of light and focuses them separately onto a detector, which measures the amount of light absorbed by the sample at each wavelength.
3	B	The longest continuous carbon chain consists of 5 carbons, so the stem name is ‘pent’. There is a triple bond between two carbon atoms which appends the suffix ‘yne’. The lowest locant number that can be assigned to this triple bond is 2, counting from right to left as drawn. Hence the name is pent-2-yne.
4	D	Ksp = [Na+][Cl–]  = 6.132 = 37.6 (3 sig. figs.)
5	D	A weak acid only partially ionises in solution, so particles of intact molecular acid, hydrogen ions, and anions must all be present. This disqualifies options B and C. Of the remaining two, there are more particles present in the same volume in option D than option A. Option D therefore represents the most concentrated weak acid.
6	C	pH = -log10⁡[H+]. Thus, if pH decreases by 3 units, the concentration of hydrogen ions will increase by a factor of 103 = 1000.
7	A	From the information given, the following ICE table can be constructed: Conc. (M) CO(g) + 2H2(g) ⇋ CH3OH(g) Initial 0.8 0.8 0 Change −0.3 −0.6 +0.3 Equilibrium 0.5 0.2 0.3
8	C	With three carbon atoms and eight monovalent atoms, only a linear three carbon chain is possible. There are then four unique ways of arranging two fluorine atoms on three available carbon positions:
9	C	The initial pH of the solution is 11, which indicates that a basic substance began in the conical flask. This disqualifies options A and B. There is a buffering region between the start and equivalence point, indicating that the initial base was weak, and that addition of a strong acid produces a weak base/conjugate acid mixture during the titration that can resist changes in pH until the buffer capacity is exceeded.
10	A	All four molecules have the same length of carbon chain and similar molar masses, and hence similar strength dispersion forces. Heptane is non-polar and can therefore only form dispersion forces, so it has the weakest overall intermolecular forces that will require the least thermal energy to overcome, and hence the lowest boiling point. Heptan-2-one is polar and can form dipole-dipole forces but not hydrogen bonds, giving it stronger overall intermolecular forces than heptane but weaker than heptan‑1-ol or heptanoic acid. Heptan-1-ol and heptanoic acid are both polar and can form hydrogen bonds, however heptan-1-ol can only form one hydrogen bond per molecule on average whereas heptanoic acid can form two. Heptanoic acid therefore has the strongest overall intermolecular forces and thus the highest boiling point.
11	C	H2SO4 is diprotic. The first ionisation is complete, but the second ionisation is not. This means the concentration of H+ will be between 1 x 10−5 M and 2 × 10−5 M. Since pH=-log10⁡[H+], the pH will be between 4.70 and 5.00. A solution with pH 4.70 – 5.00 will be purple in this indicator. The [OH−] in a 5 × 10−5 M solution of NaOH is 5 × 10−5 M. pOH= -log10⁡[OH–] = 4.3.  pH = 12 – pOH = 14 – 4.3 = 9.7.  A solution with pH 9.7 will be blue-green in this indicator.
12	B	Increasing pressure disturbs the equilibrium. According to Le Châtelier’s Principle, the system will favour the forward reaction to reduce the number of gas moles and reduce the pressure, thus minimising the disturbance. Therefore [NH3] will increase, resulting in a higher yield. Lower temperatures will favour the exothermic forward reaction and result in a higher yield, not a lower yield. Higher temperatures will favour the endothermic reverse reaction and result in a lower yield, not a higher yield.
13	A	Acetate and carbonate ions are basic and would be expected to turn red litmus blue if sufficiently dissolved in solution. This makes options B and D very unlikely. Bromide ions do not form a white precipitate with barium ions, whereas sulfate ions will form a white precipitate with barium ions (BaSO4). The remaining observations match for silver sulfate; silver forms a brown precipitate with hydroxide (AgOH) and a white precipitate with chloride (AgCl).
14	D	2HCl(aq) + Na2CO3(s) ⟶ 2NaCl(aq) + CO2(g) + H2O(l) n(Na2CO3) = m/MM = 1.34/(2 × 22.99 + 12.01 + 3 × 16.00) = 0.0126…mol n(HCl) = n(Na2CO3) × 2 = 0.025285…mol V(HCl) = n/c = (0.025285…)/0.540 = 0.04682… L
15	B	Propan-1-ol has one more CH2 unit than ethanol and one less than butan-1-ol. Its molar heat of combustion should therefore be approximately midway between the molar heats of combustion of ethanol and butan-1-ol. Mathematically, this can be expressed: (molar heat of combustion of ethanol + molar heat of combustion of butan-1-ol)/2.
16	B	Ksp(AgI) = 8.52 × 10−17 and Ksp(AgCl) = 1.77 × 10−10. Since both salts are highly insoluble but silver iodide is 107 times less soluble than silver chloride, it is reasonable to expect that adding silver ions will precipitate all the iodide (first equivalence point) before precipitating all the chloride (second equivalence point).
17	A	PbI2(s) ⇌ Pb2+(aq) + 2I−(aq)          Ksp = [Pb2+][I−]2 = 9.8 x 10−9 Let the number of moles of PbI­2 that can dissolve in 1 L be x. Therefore, [Pb2+] = 2x M, and [I−] = 2x M. Therefore: 9.8×10-9=x(2x)2 x=∛9.8×10-9/4 = 1.348…×10-3…mol L-1 Therefore, in 375 mL: n(PbI2) = c × V = 1.348…×10-3 × 0.375 = 5.055… × 10-4 mol m(PbI2)=n × MM = 5.055…×10-4 × 461 = 0.233 g
18	A	Since the rates of both forward and reverse reaction decreased at time t, the temperature must have decreased, as decreasing temperature reduces the amount of kinetic energy available to particles and reduces the number of collisions overcoming activation energy. This disqualifies options C and D. Since the activation energy of the endothermic reaction is larger than the activation energy of the exothermic reaction, there will be a greater percentage decrease in successful collision for the endothermic reaction, resulting in a larger decrease in rate of the endothermic reaction. The rate of the forward reaction decreases more, so the forward reaction is endothermic (ΔH > 0).
19	B	CH3CH2CH­2 is not a valid fragment that can be obtained from fragmenting the radical cation of butan-2-one. This disqualifies options C and D. Only charged species can be accelerated in mass spectrometry, so for a fragment to be detected at m/z 43 there needs to be a fragment with MM = 43 g mol−1 and charge = +1. This is the CH3CO+
20	D	Let the change in concentration of NO2 as the system approaches equilibrium be 2x M. Since there are initially only products present, this change must be a decrease (i.e., –2x). [NO2 ]= 1.80/2.00 = 0.90 M Then the following ICE table can be constructed: Conc. (M) 2NO(g) + O2(g) ⇋ 2NO2(g) Initial 0 0 0.90 Change +2x +x –2x Equilibrium 2x x 0.90 – 2x Keq = 2.47 × 1012, so equilibrium lies very far to the right and the change –2x will be insignificant in comparison the initial [NO2]. We can therefore assume that the equilibrium [NO2]  0.90 M. Therefore: Keq = [NO2]2 / [NO]2 [O2] 2.47 × 1012 ≈ (0.90)2/(2x)2(x) 4x3 ≈ 0.81/2.47×1012 x ≈ ∛0.81/4×2.47×1012 ≈ 4.344…×10-5 Therefore, the equilibrium [NO2] = 2 × 4.344… × 10−5 = 8.69 × 10−5 M.

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Section 2: Long Response Questions

Question 21

Functional group isomers: butan-2-one and butanal. Both structures have the same molecular formula (C₄H₈O) but different functional groups (ketone vs aldehyde).
Chain isomers: butanal and 2-methylpropanal. Both structures have the same molecular formula (C₄H₈O) but different parent chains (4 carbons vs 3 carbons)

Question 22

HCl would be classified as an acid under the Arrhenius definition as it produces hydrogen ions when it ionises in solution:
HCl_(aq) ⟶ H⁺_(aq) + Cl⁻_(aq)

HCl would be classified as an acid under the Brønsted-Lowry definition as it donates protons:
HCl_(aq) + H₂O_(l) ⟶ H₃O⁺_(aq) + Cl⁻_(aq)

NH₄Cl would be classified as neutral under the Arrhenius definition as it produces neither H⁺ nor OH⁻ when it dissociates in solution:
NH₄Cl_(aq) ⟶ NH₄⁺_(aq) + Cl⁻_(aq)

NH₄Cl would be classified as an acid under the Brønsted-Lowry definition as the ammonium ion donates protons:
NH₄⁺_(aq) + H₂O_(l) ⇌ NH_3(aq) + H₃O⁺_(aq)

Question 23

Solution X is a neutral solution (initial pH = 7) and is not a buffer, as it demonstrates no capability to resist changes in pH when a small amount of base is added. Concentrated NaOH solution contains a high concentration of OH⁻ ions that neutralise H⁺ ions in solution X when the two are mixed, thus reducing the [H⁺] in solution X. Since pH = −log₁₀[H⁺], this causes the pH of solution X to increase.

Solution Y is also a neutral solution (initial pH = 7) but is a buffer, as it demonstrates an ability to resist changes in pH when a small amount of base is added. This solution contains equimolar amounts of a weak acid (HA) and its conjugate base (A⁻) in equilibrium:

HA_(aq) + H₂O_(l) ⇌ H₃O⁺_(aq) + A⁻_(aq)

When concentrated NaOH is added, it neutralises with the weak acid (the most prevalent acid in the system), forming a salt and water:

HA_(aq) + NaOH_(aq) ⟶  NaA_(aq) + H₂O_(l)

The buffer equilibrium will shift slightly to reduce [A⁻] and increase [HA], causing a small decrease in the [H⁺] and thus a small increase in pH. However, this increase is much less than that observed in solution X.

Question 24

An amphiprotic ion is one that can either donate or accept a proton depending on what it reacts with (i.e., it can react as either a Brønsted-Lowry (B/L) acid or base). The hydrogen oxalate ion meets these criteria:

Reacting as B/L acid: HC₂O₄⁻_(aq) + OH⁻_{(aq) ⟶}  C₂O₄²⁻_(aq) + H₂O_(l)
Reacting as B/L base: HC₂O₄⁻_(aq) + H₃O⁺_(aq) ⟶  H₂C₂O_4(aq) + H₂O_(l)

Question 25 (a)

q = mcΔT = 205 × 4.18 × (60.4-23.7) = 31448.23 J = 31.4…kJ

n = -q/ΔH = -31.4…/-5294 = 0.00594…mol

m = n × MM = 0.00594… × 130.23 = 0.774 g (3 sig figs)

Question 25 (b)

Biofuels are derived from biomass, a renewable resource, whereas fossil fuels are non-renewable resources. Since the biomass is renewable, it can be replenished in a short timeframe, enabling more biofuel to be produced to replace what is consumed. In contrast, fossil fuels take millions of years to regenerate, so they are effectively a finite resource. This makes using biofuels more sustainable than using fossil fuels.

Question 26(a)

NO_2(g) and N₂O_4(g) are in equilibrium at the end of Reactor 2: 2NO_2(g) ⇋ N₂O_4(g). When NO₂ is consumed by the reaction with water in Reactor 3, this decreases the [NO₂]. According to Le Châtelier’s Principle, the NO₂/N₂O₄ system will shift left as written above to increase the [NO₂] and minimise the disturbance. Consequently, the [N₂O₄] will decrease. Since the reaction in Reactor 3 goes to completion, it will consume all NO₂ available. This is effectively an open system that will continue to shift to replenish the lost NO₂ until all the N₂O₄ has reacted.

Question 26(b)

Improvement 1: Rather than disposing of the water that is removed in Separator 1, this water could instead be used as a reactant in Reactor 3 (after purification), or it could be used in the cooler/condenser to reduce the temperature of the nitric oxide before it enters Reactor 2. This minimises material wastage and reduces the need to acquire additional water from elsewhere, which is more economically and environmentally sustainable.

Improvement 2: Rather than catalytically reducing the nitric oxide that is removed in Separator 2, which requires heat and an expensive catalyst, and releasing the nitrogen and oxygen gas to the atmosphere, this leftover nitric oxide could be collected and returned to Reactor 2, where it can go on to become additional nitric acid product. Again, this minimises material wastage and eliminates the need for a difficult, dangerous, and expensive waste processing procedure, which is also more economically and environmentally sustainable.
Other answers possible.

Question 27

Calculation:

m(ethanol) = D × V = 0.789 × 185 = 145.965 g

n(ethanol) = m/MM = 145.965/46.068 = 3.16…mol = n(CO₂)

V(CO₂ ) = nRT/P = 3.16… × 8.314 × 310/100 = 81.7 L (3 sig figs)

Question 28 (a)

5 mL of substance Q was added to a clean test tube in a fume hood. 10 drops of liquid bromine were added to the test tube and the colour of the mixture inspected carefully as the bromine was added. If immediate decolourisation of the brown bromine leaving a colourless solution was observed, this confirmed that substance Q was an alkene.

Question 28 (b)

Structure:

Since R is the addition product of an alkene (Q) and Cl_2(g), it must contain an even number of chlorine atoms; 2 per double bond in the alkene. Since the highest m/z peak in the mass spectrum occurs at m/z 114, there can only be two chlorine atoms in the molecule since these will account for a minimum of 70 g mol⁻¹.

This means that there must be three molecular ion peaks due to different possible combinations of two naturally abundant chlorine isotopes, ³⁵Cl and ³⁷C, which occur in roughly 3:1 ratio. The first molecular ion peak (2 × ³⁵Cl) is therefore the peak at m/z 112, the next (³⁵Cl + ³⁷Cl) is the peak at m/z 114, and the last (2 × ³⁷Cl) is of too low abundance to be observed. Subtracting 70 from 112 leaves 42 g mol⁻¹, which can consist of only carbon and hydrogen atoms. Three carbon atoms (3 × 12) and six hydrogen atoms (6 × 1) reasonably accounts for this mass and yields a reasonable structure when paired with two chlorine atoms. Finally, the proposed structure matches the quoted carbon percentage:

%C=MM(C)/MM(molecule) ×100

%C= (3×12.01)/(3×12.01+6×1.008+2×35.45) ×100=31.89%

Question 29

The hydroxyl group in alkan-1-ols can form strong adhesive hydrogen bonds with water, while the hydrocarbon chain can only form weak adhesive dispersion forces with water. Thus, cohesive hydrogen bonds in water can be disrupted by the adhesive hydrogen bonds, but not the adhesive dispersion forces. This allows water to solvate the hydroxyl group but not the hydrocarbon chain. When the chain is short (and MM is small), solvation is not disrupted, hence solubility is high. However, as chain length increases (and MM increases), a larger and larger proportion of the molecule becomes unsolvated, resulting in an exponential decrease in solubility.

Question 30

1. To the water sample, add 1.0 mol L⁻¹ nitric acid (HNO₃). Effervescence (appearance of bubbles) indicates the presence of carbonate ions, as carbonate ions react with acid to form carbon dioxide:

2HNO_3(aq) + CO₃²⁻_(aq) ⟶ CO_2(g) + H₂O_(l) + 2NO₃⁻_(aq)

The gas can be confirmed as carbon dioxide by collecting it and bubbling it through a beaker of limewater. The limewater (Ca(OH₂)) turns cloudy as carbon dioxide reacts with calcium hydroxide to form insoluble calcium carbonate:

CO_2(g) + Ca(OH)_2(aq) ⟶  CaCO_3(s) + H₂O_(l)

2. If there was effervescence, continue adding nitric acid to the same sample until effervescence ceases. Then add 1.0 mol L⁻¹ silver nitrate (AgNO₃) to the same water sample. Formation of a pale-yellow precipitate indicates the presence of bromide ions, as bromide ions react with silver ions to form insoluble silver bromide:

Ag⁺_(aq) + Br⁻_(aq) ⟶  AgBr_(s)

The precipitate can be confirmed as silver bromide by exposing it to sunlight. The solid will darken as it decomposes into silver metal and bromine gas.

Question 31

From graph:

c([Cu(C₃H₆O₃)₂]²⁺) at equilibrium = 0.046 mol L⁻¹

K= [[Cu(C₃H₆O₃)₂]²⁺]]/[Cu²⁺][C₃H₆O₃]² = 0.046 /(0.010 × 0.019² ) = 12742.38227 = 13000 (2 s.f.)

Question 32

HCl_(aq) + NaOH_(aq) ⟶ NaCl_(aq) + H₂O_(l)

V(HCl)average = 22.05+22.00+21.95/3 = 22.00 mL = 0.022 L

n(HCl) = 0.1102 × 0.022 = 2.4244 ×10^-3 mol = n(NaOH) in 20.00 mL

n(NaOH) in 250.0 mL (excess) = (2.4244 × 10^-3mol)/(20) × 250 = 0.030305 mol

n(NaOH)total = 0.050 × 1.124 = 0.0562 mol

n(NaOH) reacted = 0.0562 – 0.030305 = 0.025895 mol = n(NH₄⁺)

m(NH₄⁺) = 0.025895 × (14.01 + 1.008 × 4) = 0.4672 g (4 s.f.)

Question 33(a)

Between 6-8 minutes, the system is at dynamic equilibrium. At dynamic equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, hence the moles of all reaction components remain constant.

Question 33(b)

The disturbance at 8 minutes does not have any immediate effect on the moles of A₂, B₂ and AB₂. Hence, the disturbance must be a change in temperature or volume. After the disturbance at 8 minutes, the moles of A₂ and B₂ increases and the moles of AB₂ decreases. This means a shift in the reverse direction occurs to counteract the disturbance.

The reverse direction is endothermic, hence an increase in temperature could have occurred at 8 minutes. According to Le Chatelier’s principle, an increase in temperature shifts the equilibrium in the endothermic direction to remove added heat and minimise the disturbance.

The reverse direction has more moles of gas, hence an increase in volume or decrease in pressure could have occurred at 8 minutes. According to Le Chatelier’s principle, a decrease in pressure shifts the equilibrium in the direction with more moles of gas to increase pressure and minimise the disturbance.

Question 34

MgF_2(s) ⇌ Mg²⁺_(aq) + 2F⁻_(aq)                               K_sp = [Mg²⁺][F⁻]² = 5.16 ×10^-11

n(NaF) = 1.50 × 0.175 = 0.2625 mol = n(F^–) total

n(MgF₂ ) = 0.6231/62.31 = 0.01 mol

n(F⁻)used = 0.02 mol

n(F⁻)remaining = 0.2625 – 0.02 = 0.2425 mol

[F⁻] = 0.2425/0.125 +0.175 = 0.808333 mol L^-1

[Mg²⁺] = K_sp /[F⁻]² = (5.16 ×10^-11)/0.808333² =  7.90 × 10^-11 mol L^-1 (3 s.f.)

Question 35 (a)

CHCl₂COOH_(aq) ⇌ CHCl₂COO⁻_(aq) + H⁺_(aq)

[H⁺] at equilibrium = 10^−pH = 10^−1.107

K_a = [CHCl₂COO^–][H⁺]/[CHCl₂COOH] = (−10^−1.107)/(0.2000 – 10^−1.107) = 0.0501 (3 s.f.)

Question 35 (b)

Ionisation of acetic acid:

CH₃COOH_(aq) ⇌ CH₃COO⁻_(aq) + H⁺_(aq)                          K_a = [CH₃COO^–][H⁺]/[CH3COOH]

Ionisation of trichloroacetic acid:

CCl₃COOH_(aq) ⇌ CCl₃COO⁻_(aq) + H⁺_(aq)                        K_a = [CCl₃COO^–][H⁺]/[CCl₃COOH]

A stronger acid is more ionised; hence the equilibrium position lies further towards the right.

Since the acid dissociation constant K_a is given by [products]/[reactant], and pK_a = −log₁₀K_a, a stronger acid will have a larger K_a value and a smaller pK_a. Since trichloroacetic acid has a smaller pK_a than acetic acid, it is a stronger acid.

The change in Gibbs free energy indicates the equilibrium position for the ionisation of each acid. A larger positive ΔG value indicates the Gibbs free energy of the products (the ions) is much greater than the Gibbs free energy of the reactants (the intact acid). Since a system will move towards a state of lower Gibbs free energy, the system with a larger positive ΔG value will have more reactants in the system, and thus the equilibrium lies further to the left.

Since acetic acid has a larger positive ΔG value than trichloroacetic acid, its equilibrium lies further to the left and thus is a weaker acid

Question 36

Compound A – Functional group: carbon-carbon double bond

Compound B – Functional group: hydroxyl group

Compound C – Functional group: carbonyl group (ketone)

Compound A is an alkene, given that it undergoes a hydration reaction with dilute sulfuric acid to form alcohol B. The molar mass of A matches that of C₆H₁₂.

The ¹³C NMR spectrum indicates that only three unique carbon environments are present. The only structure with this molecular formula and 3 unique carbon environments is hex-3-ene, as shown in the diagram below, with unique environments labelled A, B and C.

The alkyl carbon environment A corresponds to the peak at ~15 ppm, and the alkyl carbon environment B corresponds to the peak at ~25 ppm (slightly more downfield as it is slightly closer to the deshielding alkene). The alkene carbon environment C corresponds to the much more downfield peak at ~130-135 ppm.

Compound B must be hexan-3-ol since it is produced from hydration of hex-3-ene. The presence of the hydroxyl group is further supported by the broad absorption between 3230-3550 cm⁻¹ in the IR spectrum, which is characteristic of O-H bonds of alcohols.

Compound C must be hexan-3-one since it is the oxidation product of hexan-3-ol. This is further supported by the ¹H NMR data. Hexan-3-one has 5 unique hydrogen environments, labelled a, b, c, d, and e.

Signal at 2.46 corresponds to the 2 hydrogens in environment b. The 2 hydrogens give a peak area of 2 and the signal will be split by 3 hydrogens in environment a to give a quartet. The chemical shift is typical of hydrogens on a carbon adjacent to a carbonyl carbon.

Signal at 2.42 corresponds to the 2 hydrogens in environment c. The 2 hydrogens give a peak area of 2 and the signal will be split by 2 hydrogens in environment d to give a triplet. The chemical shift is typical of hydrogens on a carbon adjacent to a carbonyl carbon.

Signal at 1.65 corresponds to the 2 hydrogens in environment d. The 2 hydrogens give a peak area of 2 and the signal is split by 2 hydrogens in environment c and 2 hydrogens in environment c, making it a multiplet. The chemical shift is typical of hydrogens on an alkyl carbon.

Signal at 1.05 corresponds to the 3 hydrogens in environment a. The 3 hydrogens give a peak area of 3 and the signal will be split by 2 hydrogens in environment b to give a triplet. The chemical shift is typical of hydrogens on an alkyl carbon.

Signal at 1.01 corresponds to the 3 hydrogens in environment e. The 3 hydrogens give a peak area of 3 and the signal will be split by 2 hydrogens in environment d to give a triplet. The chemical shift is typical of hydrogens on an alkyl carbon.

Question 37 (a)

Q = [CO₂]/[CO]² = (1.21 × 10^-3)/(1.10 × 10^-2)²) = 10

Since Q = K, the system is at equilibrium.

Question 37 (b)

When CO_2(g) is added, equilibrium shifts left according to Le Châtelier’s Principle to reduce [CO₂] and minimise the disturbance.

Let additional [CO₂] = Y M

[CO₂]initial = 1.21 × 10^-3 + Y

At equilibrium [CO] = [CO₂] = 1.10 × 10⁻²  + 2x

K = [CO₂]/[CO]² = (1.10 × 10^-2+ 2x)/((1.10 × 10^-2+ 2x)² ) = 10.00

1/1.10 × 10^-2+ 2x = 10.00

2x = 1/10.00 -1.10 × 10^-2 =0.089

x = 0.0445

[CO] at equilibrium = 1.10 × 10^-2 +0.089 = 0.10 M

∴ [CO₂] at equilibrium = 0.10 M

[CO₂]initial = 0.10 + 0.0445 = 0.1445mol L^-1

[CO₂]initial = 1.21 × 10⁻³ + Y = 0.1445mol L^-1

Y = 0.1445 – 1.21 × 10⁻³ = 0.14329 = additional CO₂

Since volume is 1L:

∴ moles of CO₂ added = 0.143mol (3 s.f)