Deduce the pH range in which glycine is an effective buffer in basic solution.

Enzymes are biological catalysts.

The data shows the effect of substrate concentration, [S], on the rate, v, of an enzyme-catalysed reaction.

Determine the value of the Michaelis constant (K_m) from the data. A graph is not required.

Outline the action of a non-competitive inhibitor on the enzyme-catalysed reaction.

The sequence of nitrogenous bases in DNA determines hereditary characteristics.

Calculate the mole percentages of cytosine, guanine and thymine in a double helical DNA structure if it contains 17% adenine by mole.

«pH range» 8.6–10.6

Accept any value between 8.2 and 11.0.

[1 mark]

«K_m =» 0.67 «mmol dm^–3»

Do not penalize if a graph is drawn to determine the value.

[1 mark]

does not compete for active site
OR
binds to allosteric site/away from «enzyme» active site
OR
alters shape of enzyme

reduces rate/V_max

[2 marks]

«% cytosine + % guanine = 100% – 17% – 17% = 66%»

Cytosine: 33 «%» AND Guanine: 33 «%»

Thymine: 17 «%»

[2 marks]

1.40 × 10⁻³ g of NaOH (s) are dissolved in 250.0 cm³ of 1.00 × 10⁻¹¹ mol dm⁻³ Pb(OH)₂ (aq) solution.

Determine the change in lead ion concentration in the solution, using section 32 of the data booklet.

«[OH⁻] =$\frac{1.40\times {10}^{-3}\text{g}}{40.00\text{ g mo}{\text{l}}^{-1}\times 0.2500\text{ d}{\text{m}}^3}$ =» 1.40 × 10⁻⁴ «mol dm⁻³» ✔

«[OH⁻] from dissolved Pb(OH)₂ is negligible»

NOTE: Accept «ratio $\frac{{\left[P{b}^{2+}\right]}_{initial}}{{\left[P{b}^{2+}\right]}_{final}}$ =» 13.7 OR  «ratio $\frac{{\left[P{b}^{2+}\right]}_{final}}{{\left[P{b}^{2+}\right]}_{initial}}$ =» 0.0730 for M4.

K_sp = [Pb²⁺][OH⁻]²
OR
1.43 × 10⁻²⁰ = [Pb²⁺] × (1.40 × 10⁻⁴)² ✔

[Pb²⁺]_final = 7.30 × 10⁻¹³ «mol dm⁻³» ✔

NOTE: Award [4] for correct final answer.

«change in [Pb²⁺] = 1.00 × 10⁻¹¹ − 7.30 × 10⁻¹³ =» 9.27 × 10⁻¹² «mol dm⁻³» ✔

NOTE: Award [3] for correct [Pb²⁺]_final.

Show that the mass of the ²³⁸U isotope in the rock is $0.639 \mathrm{kg}$.

The half-life of ²³⁸U is $4.46\times {10}^9$ years. Calculate the mass of ²³⁸U that remains after $0.639 \mathrm{kg}$ has decayed for $2.23\times {10}^{10}$ years.

Outline a health risk produced by exposure to radioactive decay.

Deduce the nuclear equation for the decay of uranium-238 to thorium-234.

Thorium-234 has a higher binding energy per nucleon than uranium-238. Outline what is meant by the binding energy of a nucleus.

Determine the nuclear binding energy, in $\mathrm{J}$, of ${}_{238}\mathrm{U}$ using sections 2 and 4 of the data booklet.

The mass of the ${}_{238}\mathrm{U}$ nucleus is $238.050786 \mathrm{amu}$.

$«\frac{\mathrm{mass} \%}{\mathrm{fraction} \mathrm{of} \mathrm{U} \mathrm{in} {\mathrm{UO}}_2}=»\frac{238.03}{238.03+2\times 16}$ /$0.881$/$88.1 \%$ ✔

$46.5«\mathrm{kg}»\times 0.0157\times 0.881\times 0.9928«=0.639 \mathrm{kg}»$ ✔

Award [1 max] for omitting mass composition (giving $\mathit{0}\mathit{.}\mathit{725}\mathit{ }kg$).

M2 is for numerical setup, not for final value of $\mathit{0}\mathit{.}\mathit{639}\mathit{ }kg$.

Alternative 1
$«\frac{2.23\times {10}^{10} \mathrm{year}}{4.46\times {10}^9 \mathrm{year}}=»5.00«\mathrm{half}-\mathrm{lives}»$ ✔

$«m=0.639 \mathrm{kg}\times (0.5{)}^5=»0.0200«\mathrm{kg}»$ ✔

Alternative 2
$«𝜆=\frac{\ln 2}{4.46\times {10}^9 \mathrm{year}}=»1.554\times {10}^{-10}«{\mathrm{year}}^{-1}»$ ✔

$«m=0.639 \mathrm{kg}\times {𝑒}^{–1.554\times {10}^{–10 {\mathrm{year}}^{–1}}\times 2.23\times {10}^{10 \mathrm{year}}}=»0.0200«\mathrm{kg}»$ ✔

Award [2] for correct final answer.

Any one:

«genetic» mutations ✔

«could cause» cancer ✔
Accept specific named types of cancer.

cells «in body» altered ✔

cells «in body» cannot function ✔

damaged DNA/proteins/enzymes/organs/tissue ✔

«radiation» burns ✔

hair loss ✔

damage in foetuses ✔

damages/weakens immune system ✔

${}_{92}{}^{238}\mathrm{U}\to {}_{90}{}^{234}\mathrm{Th}+{}_2{}^4\mathrm{He}$ ✔

Do not penalize missing atomic numbers in the equation.

Accept “$\alpha$” for "$He$”.

energy required to separate a nucleus into protons and neutrons/nucleons
OR
energy released when nucleus was formed from «individual/free/isolated» protons and neutrons/nucleons ✔

Do not accept “energy released when atom was formed”.

$238.050786«\mathrm{amu}»\times 1.66\times {10}^{-27}«\mathrm{kg} {\mathrm{amu}}^{-1}»$
OR

 ✔

$\left(92\times 1.672622\times {10}^{-27}\right)+\left(146\times 1.674927\times {10}^{-27}\right)-3.95\times {10}^{-25}$
OR
$3.42\times {10}^{-27}/3\times {10}^{-27}«\mathrm{kg}»$ ✔

$«E=m{c}^2=3.42\times {10}^{-27}\times (3.00\times {10}^8{)}^2=»3.08\times {10}^{-10}«\mathrm{J}»$ ✔

Accept answers in the range “$\mathit{2}\mathit{.}\mathit{7}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{10}}\mathit{–}\mathit{3}\mathit{.}\mathit{1}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{10}}\mathit{«}J\mathit{»}$”.

Award [3] for correct final answer.

This question was generally well answered. Many candidates approached this question using amount, in mol, and converting to mass at the end. Some candidates omitted the mass composition however, resulting in a mass of 0.725 kg, which yielded [1 max].

The question involving half-life was very well answered and most scored both marks giving a final answer of 0.0200 kg, via different methods of calculation.

This question on outlining a health risk produced by exposure to radioactive decay posed no difficulty and the most common, correct answers included "could cause cancer" and "can damage DNA".

Most scored the one mark for the nuclear equation for the decay of uranium-238 to thorium-234. The most common error was including a neutron on the product side

Although many scored the one mark for outlining what is meant by the binding energy of a nucleus, the question was often answered as energy involved in the formation of an atom. Some candidates did not understand the difference between nucleus and nucleons.

This demanding question on the determination of the nuclear binding energy was well executed and many scored all three marks. Even the weaker candidates still managed to gain an ECF mark for M3 by using the E = mc² equation. The general performance on this type of calculation was much better than in previous sessions.

Outline why the initial reaction should be carried out under a fume hood.

Deduce the equation for the relationship between absorbance and concentration.

Copper(II) ion solutions are blue. Suggest, giving your reason, a suitable wavelength of light for the analysis.

Outline how a solution of 0.0100 mol dm⁻³ is obtained from a standard 1.000 mol dm⁻³ copper(II) sulfate solution, including two essential pieces of glassware you would need.

The original piece of brass weighed 0.200 g. The absorbance was 0.32.

Calculate, showing your working, the percentage of copper by mass in the brass.

Deduce the appropriate number of significant figures for your answer in (e)(i).

Comment on the suitability of using brass of this composition for door handles in hospitals.

If you did not obtain an answer to (e)(i), use 70 % but this is not the correct answer.

Suggest another property of brass that makes it suitable for door handles.

Titration is another method for analysing the solution obtained from adding brass to nitric acid.

Copper(II) ions are reduced to copper(I) iodide by the addition of potassium iodide solution, releasing iodine that can be titrated with sodium thiosulfate solution, Na₂S₂O₃ (aq). Copper(I) iodide is a white solid.

4I⁻ (aq) + 2Cu²⁺ (aq) → 2CuI (s) + I₂ (aq)

I₂ (aq) + 2S₂O₃²⁻ (aq) → 2I⁻ (aq) + S₄O₆²⁻ (aq)

Suggest why the end point of the titration is difficult to determine, even with the addition of starch to turn the remaining free iodine black.

NO₂/NO/NO_x/HNO₃/gas is poisonous/toxic/irritant ✔

Accept formula or name.

Accept “HNO₃ is corrosive” OR “poisonous/toxic gases produced”.

Accept “reaction is harmful/hazardous”.

Slope (gradient):

40 ✔

Equation:

absorbance = 40 × concentration

OR

y = 40x ✔

Accept any correct relationship for slope such as $\frac{1.00}{0.025}$.

Award [2] if equation in M2 is correct.

orange is opposite blue «in the colour wheel»

OR

the complementary colour «blue» is seen/transmitted ✔

585–647 «nm would be absorbed» ✔

Accept any value or range within 550–680 «nm» for M2.

dilute 1.00 cm³ «of the standard solution with water» to 100 cm³

OR

dilute sample of standard solution «with water» 100 times ✔

«graduated/volumetric» pipette/pipet ✔

volumetric flask ✔

Accept any 1 : 100 ratio for M1.

Accept “mix 1 cm³ of the standard solution with 99 cm³ of water” for M1.

Do not accept “add 100 cm³ of water to 1.00 cm³ of standard solution” for M1.

Accept “burette/buret” for M2.

Accept “graduated/measuring flask” for M3 but not “graduated/measuring cylinder” or “conical/Erlenmeyer flask”.

concentration of copper = 0.0080 «mol dm^–3» ✔

mass of copper in 250.0 cm³ = «0.0080 mol dm^–3 × 0.2500 dm³ × 63.55 g mol^–1 =» 0.127 «g»

OR

mass of brass in 1 dm³ = «4 × 0.200 g =» 0.800 g AND [Cu2+] = «0.0080 mol dm^–3 × 63.55 g mol^–1 =» 0.5084 g dm^–3 ✔

«% copper in this sample of brass $=\frac{0.127}{0.200}\times 100=$» 64 «%»

OR

«% copper in this sample of brass $=\frac{0.5084}{0.800}\times 100=$» 64 «%» ✔

Accept any value in range 0.0075–0.0085 «mol dm^–3» for M1.

Accept annotation on graph for M1.

Award [3] for correct final answer.

Accept “65 «%»”.

two ✔

Do not apply ECF from 1(e)(i).

«since it is greater than 60%» it will reduce the presence of bacteria «on door handles» ✔

resistant to corrosion/oxidation/rusting

OR

low friction surface «so ideal for connected moving components» ✔

Accept “hard/durable”, “«high tensile» strength”, “unreactive”, “malleable” or any reference to the appearance/colour of brass (eg “gold-like”, “looks nice” etc.).

Do not accept irrelevant properties, such as “high melting/boiling point”, “non-magnetic”, “good heat/electrical conductor”, “low volatility”, etc.

Do not accept “ductile”.

precipitate/copper(I) iodide/CuI makes colour change difficult to see

OR

release of I₂/iodine from starch-I₂ complex is slow so titration must be done slowly ✔

Calculate the energy released, in $\mathrm{kJ}$, from the complete combustion of $5.00 {\mathrm{dm}}^3$ of ethanol.

State a class of organic compounds found in gasoline.

Outline the advantages and disadvantages of using biodiesel instead of gasoline as fuel for a car. Exclude any discussion of cost.

A mixture of gasoline and ethanol is often used as a fuel. Suggest an advantage of such a mixture over the use of pure gasoline. Exclude any discussion of cost.

When combusted, all three fuels can release carbon dioxide, a greenhouse gas, as well as particulates. Contrast how carbon dioxide and particulates interact with sunlight.

Methane is another greenhouse gas. Contrast the reasons why methane and carbon dioxide are considered significant greenhouse gases.

Suggest a wavenumber absorbed by methane gas.

Determine the relative rate of effusion of methane ($M_{\mathrm{r}}=16.05$) to carbon dioxide ($M_{\mathrm{r}}=44.01$), under the same conditions of temperature and pressure. Use section 1 of the data booklet.

$«21 200 \mathrm{kJ} {\mathrm{dm}}^{-3}\times 5.00 {\mathrm{dm}}^3=»106000/1.06\times {10}^5«\mathrm{kJ}»$ ✔

alkane
OR
cycloalkane
OR
arene ✔

Accept “alkene”.
Do not accept just “hydrocarbon”, since given in stem.
Do not accept “benzene/aromatic” for “arene”.

Advantages: [2 max]

renewable ✔

uses up waste «such as used cooking oil» ✔

lower carbon footprint/carbon neutral ✔

higher flashpoint ✔

produces less ${\mathrm{SO}}_{\mathrm{x}}$/${\mathrm{SO}}_2$
OR
less polluting emissions ✔

has lubricating properties
OR
preserves/increases lifespan of engine ✔

increases the life of the catalytic converter ✔

eliminates dependence on foreign suppliers ✔

does not require pipelines/infrastructure «to produce» ✔

relatively less destruction of habitat compared to obtaining petrochemicals ✔

Accept “higher energy density” OR “biodegradable” for advantage.

Disadvantages: [2 max]

needs conversion/transesterification ✔

takes time to produce/grow plants ✔

takes up land
OR
deforestation ✔

fertilizers/pesticides/phosphates/nitrates «used in production of crops» have negative environmental effects ✔

biodiversity affected
OR
loss of habitats «due to energy crop plantations» ✔

cannot be used at low temperatures ✔

variable quality «in production» ✔

high viscosity/can clog/damage engines ✔

Accept “lower specific energy” as disadvantage.

Do not accept “lower octane number” as disadvantage”.

Any one:

uses up fossil fuels more slowly ✔

lower carbon footprint/CO2 emissions ✔

undergoes more complete combustion ✔

produces fewer particulates ✔

higher octane number/rating
OR
less knocking ✔

prevents fuel injection system build up
OR
helps keep engine clean ✔

Accept an example of a suitable advantage even if repeated from 11c.

carbon dioxide allows sunlight/short wavelength radiation to pass through AND particulates reflect/scatter/absorb sunlight ✔

Accept “particulates reflect/scatter/absorb sunlight AND carbon dioxide does not”.
Accept “$C{O}_{\mathit{2}}$ absorbs $IR$ «radiation» AND particulates reflect/scatter/absorb sunlight”.

Do not accept “traps” for “absorbs”.

carbon dioxide is highly/more abundant «in the atmosphere» ✔

methane is more effective/potent «as a greenhouse gas»
OR
methane/better/more effective at absorbing $\mathrm{IR}$ «radiation»
OR
methane has greater greenhouse factor
OR
methane has greater global warming potential/GWP✔

Accept “carbon dioxide contributes more to global warming” for M1.

any value or range within $2850–3090«{\mathrm{cm}}^{-1}»$ ✔

«rate of effusion of $\frac{{\mathrm{CH}}_4}{{\mathrm{CO}}_2}=\sqrt{\frac{44.01}{16.05}}=»1.656$ ✔

Almost all were able to calculate the energy released from the complete combustion of ethanol.

The majority cited correctly that alkanes are a class of organic compounds found in gasoline.

Most gained at least one mark for an advantage of using biodiesel instead of gasoline as fuel for a car and most scored one mark at least for a disadvantage of biodiesel. Many conveyed solid understanding, though the disadvantages were not as well articulated as the advantages. Some incorrectly based their responses on cost factors which were excluded as outlined in the stem of the question.

Most scored the one mark for this question, with "less knocking or higher octane number/rating" the most common correct answer seen.

The wording of this question was critical which involved contrasting how carbon dioxide and particulates interact with sunlight. Some missed the "Contrast" command term as the action verb. Loose, non-scientific syntax was often seen such as stating "traps" instead of "absorbs".

This was another "Contrast-type" question, which was better answered compared to (e)(i). Many scored both marks by stating that carbon dioxide is more abundant in the atmosphere whereas methane is more effective at absorbing IR radiation.

The main issue with this question was that a high percentage of candidates did not realise that wavenumber is the reciprocal of wavelength and hence wavenumber has typical units of cm^-1. Many incorrectly gave wavelength values, in nm, which did not answer the question posed.

The determination of the relative rate of effusion of methane to carbon dioxide was almost universally correctly computed as 1.656.

Deduce, giving a reason, the group of elements in the periodic table most likely to undergo sublimation.

Describe the density trend across periods 4 and 5 of the periodic table.

Suggest, with a reason, whether the lanthanoids or actinoids of the f-block would have the higher density.

Compare the ease of oxidation of s-block and d-block metals to their melting points and densities. Use section 25 of the data booklet.

Sketch how the first ionization energies of elements vary with their atomic radius.

group 18/noble gases     [✔]

smallest difference between melting and boiling points
OR
weakest intermolecular forces «in that period»     [✔]

Note: Accept “group 17/halogens”.

density increases «to a maximum in the transition elements» AND then decreases     [✔]

actinoids AND density increases down all groups «due to large increase in atomic mass for small increase in atomic volume»
OR
actinoids AND «much» greater atomic mass with similar type of bonding
OR
actinoids AND density «of actinoids» atomic number 90 to 95 is greater than corresponding lanthanoids    [✔]

Note: Accept “actinoids AND on graph actinoids have «much» greater density than lanthanoids”.

Alternative 1:
«metals with» low densities oxidize easier    [✔]

«metals with» low melting points oxidize easier     [✔]

Alternative 2:
in s-block «metals with» high densities oxidize easier
OR
in s-block «metals with» low melting points oxidize easier     [✔]

in d-block «metals with» low densities oxidize easier
OR
in d-block «metals with» low melting points oxidize easier     [✔]

Note: Award [1 max] for “s-block metals more easily oxidized” OR “s-block metals have lower melting points” OR “s-block metals have lower densities”.

Accept “have greater activity” for “oxidize easier”.

    [✔]

Note: Accept any negative sloping line.

Do not award mark if line touches either axis.

Most candidates correctly identified the group of elements most likely to undergo sublimation but did not score for the reason as they referred to low melting and boiling points, rather than the smallest difference between these temperatures. There were several G2 comments that “the group of elements” was a confusing requirement as elements could be grouped in many ways, including for instance, from B to Ne. The Chemistry Guide clearly states that a group on the periodic table refers to a vertical column of elements. A few complaints were received about the inclusion of a question on sublimation, but the question was designed to make candidates think, and did not require knowledge of phase diagrams.

Required candidates to consider density trends. Most candidates correctly described trends across periods 4 and 5 but had difficulty predicting and explaining whether lanthanoids or actinoids would have the higher density. Many said that actinoids would have higher density because they have more protons and neutrons / greater atomic number / greater mass with no further detail about having similar bonding and hence similar volume. Some G2 comments complained about the inclusion of lanthanoids and actinoids in this question. However, the Chemistry Guide clearly states that these terms should be known.

Most candidates scored at least 1 mark for comparing the s-block and d-block metals and most drew a line with a negative slope.

Although many here failed to score because the line crossed or touched one of the axes. A few sketched a graph reminiscent of first ionization energy against atomic number.

Calculate the mole fraction of ethanal in the mixture.

The vapour pressure of pure ethanal at $20°\mathrm{C}$ is $101 \mathrm{kPa}$.

Calculate the vapour pressure of ethanal above the liquid mixture at $20°\mathrm{C}$.

Describe how this mixture is separated by fractional distillation.

$«{\mathrm{\chi }}_{\mathrm{ethanal}}=\frac{0.100}{0.100+0.100+0.200}=»0.250$ ✔

Accept “$\mathit{25}\mathit{\%}$”.

$«{\mathrm{\rho }}_{\mathrm{ethanal}}=0.250\times 101=»25.3«\mathrm{kPa}»$ ✔

Any two of:
continuous evaporation and condensation
OR
increased surface area in column helps condensation ✔
Accept “glass «beads» aid condensation «in fractionating column»”.

temperature decreases up the fractionating column ✔

liquids condense at different heights
OR
liquid of lowest boiling point collected first
OR
liquid with weakest intermolecular forces collected first
OR
most volatile component collected first
OR
fractions/liquids collected in order of boiling point/volatility ✔
Accept “liquids collected in order of molar mass”.

This question involving Raoult's Law was very well answered and most were able to calculate the mole fraction of ethanal in the mixture (0.250) and the corresponding vapour pressure of ethanal above the liquid mixture at 20 °C (25.3 kPa). There was one G2 comment on this question. One teacher stated that the diagram shows four fractions but the stem of the question specifically states only three components and hence the fourth test tube is not required. The teacher commented that some students may have been distracted by this. 

This question involving Raoult's Law was very well answered and most were able to calculate the mole fraction of ethanal in the mixture (0.250) and the corresponding vapour pressure of ethanal above the liquid mixture at 20 °C (25.3 kPa). There was one G2 comment on this question. One teacher stated that the diagram shows four fractions but the stem of the question specifically states only three components and hence the fourth test tube is not required. The teacher commented that some students may have been distracted by this. 

In this question candidates were required to describe how the mixture can be separated by fractional distillation. Only the better candidates scored both marks, though most gained at least one mark, usually for stating that the most volatile component is collected first. Many did not convey the idea that there is continuous evaporation and condensation in the process or the fact that the temperature decreases up the fractionating column.

Identify the experiment with the highest rate of lead dissolving.

Suggest why the relationship between time and lead concentration for Cola at 16 °C is not linear.

Examine, giving a reason, whether the rate of lead dissolving increases with acidity at 18 °C.

Lead(II) chloride, PbCl₂, has very low solubility in water.

PbCl₂ (s) $\rightleftharpoons$ Pb²⁺ (aq) + 2Cl⁻ (aq)

Explain why the presence of chloride ions in beverages affects lead concentrations.

A mean daily lead intake of greater than 5.0 × 10⁻⁶ g per kg of body weight results in increased lead levels in the body.

Calculate the volume, in dm³, of tap water from experiment 8 which would exceed this daily lead intake for an 80.0 kg man.

6   [✔]

Note: Accept “orange juice”.

equilibrium is being established «between lead in solution and in mug»
OR
solution becoming saturated
OR
concentration of lead ions/[Pb²⁺] in the solution has increased «over time»
OR
acid concentration has decreased «as reacted with lead»
OR
surface lead has decreased/formed a compound/forms insoluble layer on surface
OR
acid reacts with other metals «because it is an alloy»   [✔]

Note: Do not accept “concentration of cola, orange juice, etc… has decreased”

Do not accept a response that only discusses mathematical or proportional relationships.

no AND experiment 7/beer has lowest rate and intermediate acidity/pH
OR
no AND experiment 6/orange juice has fastest rate but lower acidity/higher pH than lemonade
OR
no AND experiment 6/orange juice has highest rate and intermediate acidity/pH  [✔]

Note: Accept no AND any comparison, with experimental support, that concludes no pattern/increase with acidity

eg: “rate of Pb/lead dissolving generally decreases with acidity as tap water has highest rate (after orange juice) while lemonade (lower pH) has lower rate”.

equilibrium shifts to the left/towards reactants  [✔]

lead «compounds/ions» precipitate
OR
concentration of lead «ions»/[Pb²⁺] decreases  [✔]

Note: Award [2] for “equilibrium shifts to the left/towards reactants due to common ion effect”.

Accept “lead ions/[Pb²⁺] removed from solution” for M2.

«daily limit = 5.0 × 10^–6 g kg^–1 × 80.0 kg =» 4.0 × 10^–4 «g of lead»  [✔]

«volume $=\frac{4.0\times {10}^{-4}\text{ g}}{1.5\times {10}^{-2}\text{g d}{\text{m}}^{-3}}=$» 2.7 × 10^–2/0.027 «dm³»  [✔]

Note: Award [2] for correct final answer

This part was correctly answered by the majority of the candidates.

A surprising number of candidates gave evidence for the non-linearity but then did not go on to explain why, giving no reasons or causes rooted in chemical theory. The command term "suggest" involves proposing a solution or hypothesis. Here the instruction "suggest why" indicates that the reason has to be explained.

The candidates who examined the data and quoted it in their answer generally scored the mark but several candidates did not refer to the data table.

Several candidates missed that this question was based on the equilibrium and it will shift to the left in presence of chloride ions.

Majority of the candidates scored two marks but some struggled with the conversion of grams to milligrams.