HKDSE Chemistry & Combined Science (Chemistry)
Report on the 2012 examination
Dr. LI Tak-man
Senior Manager (Science), HKEAA
11 & 12 Dec 2012
1
Overview
Paper
Chemistry
CS(Chem)
1A (MC)
Mean : 25 out of 36 (i.e. 69%)
Mean : 13 out of 24 (i.e. 54%)
1
~53%
~42%
2
~37%
N.A.
SBA
~69%
~65%
Subject
~53%
~47%
Candidature
17 208
6 272
2
Marking & Grading
On­screen marking (OSM) panels
Chemistry
CS(Chem)
1B­1: Q.1­6
Q.1­6
1B­2: Q.7­10
Q.7­10
1B­3: Q.11­16
­­­
2A:
Industrial Chemistry (48%) ­­­
2B:
Materials Chemistry (5%) ­­­
2C:
Analytical Chemistry (47%)
­­­
SBA marks stat. moderated with both Mean and SD
adjusted (outlining cases reviewed by Supervisors)
3
Marking & Grading
Expert Panel (7 persons including examiners etc) determines
level boundaries/cut scores with reference to
Level descriptors
Group Ability Indicator (GAI)
Actual candidate samples
CS(Chem) graded by common items and viewing candidate
samples
Endorsement by Senior Management/Public Exam Board
4
Results
Chemistry
Level
5**
5*+
5+
4+
3+
2+
1+
Percentage
2.1%
8.3%
20.9%
46.1%
73.5%
86.8%
95.0%
2+
1+
CS(Chem)
Level
5**
5*+
5+
4+
3+
Percentage
1%
3.7%
9.2%
31.3%
59.3%
69.9% 86.8%
5
Paper 1A
Chemistry (36 MC)
% Correct
>70%
50%­70%
<50%
No. of items
17
17
2
Easy
Difficult
CS (Chem) (24 MC)
% Correct
>70%
50%­70%
<50%
No. of items
7
7
10
Easy
Difficult
6
Paper 2
Analytical Industrial
Candidature
16124
16401
Max Mark (20) Analytical
> Industrial
Materials
1676
Industrial > Materials
Mean Mark difference
0.79
2.6
% Mark difference
4%
13%
7
Rough Performance Comparison
Chemistry 1B Part I VS CS(Chem) Section B
Max Mark (56)
Chemistry 1B Part I > CS(Chem) Section B
Mean Mark difference
7.58
% Mark difference
13.5%
8
Observations
Candidates were generally weak in answering questions involving calculation and data analysis. These include mass/mole/concentration calculations for a titration experiment, calculation of the enthalpy changes of reactions, and calculations on chemical equilibrium.
Many candidates were weak in redox chemistry. They were confused about the concepts of oxidation, reduction, oxidizing power, reducing power, position of chemical species in the electrochemical series, and chemical reactions that occur at the electrodes.
Many candidates were not able to state the expected colour changes/observations in chemical tests, or the difference in results of positive and negative tests. Many candidates confused the types of chemical bonding with intermolecular forces in different types of chemical species.
9
THANK YOU
10
2012 HKDSE
Chemistry and
Combined Science
(Chemistry)
Paper 1B (Part 1)
Dr Kendrew MAK
What is meant by the term “isotope”? (1)
Atoms with the same number of protons but different numbers of neutrons.
Many candidates failed to mention “isotopes are atoms of an element ……” in their answer.
Give one daily application of neon. (1)
Gas for filling luminous advertisement tubes / neon tubes / neon signs / neon lights.
Many candidates wrongly wrote that neon is
used in light bulbs or potato chips packages.
Some candidates gave vague answers such as “ffor making advertising boards”.
Explain why the boiling point of neon is lower than that of oxygen. (2)
Neon is monoatomic whereas oxygen is diatomic. O2 molecules has larger molecular size than Ne molecule.
Thus stronger van der Waals’ forces / stronger intermolecular forces among O2 molecules.
Did not point out larger molecular size of O2
今 strong van der Waals’ forces between O2
molecules
Wrongly answered that energy is needed to
break the covalent bonds of O2 molecules.
Suggest a chemical test to show how to distinguish
between ethenyl ethanoate and ethyl ethanoate. (2)
Bromine test – ethenyl ethanoate can decolourise
orange bromine solution immediately while ethyl
ethanoate cannot. (also accept acidified KMnO4)
No complete description for the expected colour
change.
Did not mention the expected observation of the negative test.
Arrange metal X, metal Y and copper in increasing order of reducing power. (1)
Some confused “reducing power” with “oxidising
power”, and reversed the order.
Write the half equation for the change that occurs at:
(i) metal X strip (1)
(ii) Copper strip (1)
Some candidates answered for (i)
Zn 今 Zn2+ + 2e­
Most candidates wrongly answered for (ii):
Cu2+ + 2e­ 今 Cu
With the aid of a diagram, explain the formation of hydrogen bonding in hydrogen fluoride. (2)
Fluoride / F is a highly electronegative element.
The H­
H­F covalent bond is very polar / The HF molecule is highly polarised.
polarised.
Most candidates gave the correct drawing.
Many candidates only mentioned:
“F is electronegative”
“F is more electronegative than H”
They failed to mention:
“F is a highly electronegative element”, or
“the H­F bond is very polar”
With the aid of a chemical equation, explain why the concentration of the CuSO4(aq) prepared was lower than 0.1 M.
Displacement reaction occurs when the iron rod is dipped into the copper(II) sulphate solution. / Copper(II) ions are reduced to copper metal and deposited on the iron rod.
CuSO4(aq) + Fe(s) 今 Cu(s) + FeSO4(aq)
(also accept ionic equation)
Common wrong answer:
­ The error is due to the dilution caused by the water of crystallisation in the copper(II) sulphate
crystals.
­ Wrongly put “CuSO4.5H2O(aq)” in the equation.
Explain why copper can be coated on the metallic object by electrolysis. Copper is lower than hydrogen in the electrochemical series, or
Cu2+ is discharged preferentially than H+ when a current is applied.
Most candidate didn’t explain the formation of Cu in terms of “preferential discharge of Cu2+ ions”.
Suggest what the bubbles were, and explain why the copper layer easily flaked off.
Hydrogen gas / H2
The hydrogen gas bubbles hinder the deposition of copper on the surface of the metallic object.
Outline the steps in preparing solid lead(II) sulphate from solid lead(II) nitrate. You have to state the additional chemical agents that are required, but need not mention the apparatus involved. (3+1)
• Dissolve solid lead(II) nitrate in water.
• Mix with excess sulphuric acid (or soluble sulphate
salt)
• Filter the mixture to obtain the solid residue (PbSO4), wash it with deionised water and dry in oven.
• Communication mark
Some candidates:
­Missed the crucial step of dissolving Pb(NO3)2(s) in water.
­Wrongly treated PbSO4(s) as a soluble salt.
­Misinterpreted the question as the preparation of Pb(NO3)2(s) from PbSO4(s) Question 7 (c)
Given that ammonia is very soluble in water, state the advantage of using an inverted funnel. (1)
Prevent sucking back as NH3(g) is very soluble.
Most candidates wrongly described that:
­The function of the inverted funnel was to prevent the ammonia from escaping to the surrounding by providing a larger surface area for dissolution.
Question 7 (d) (iii)
Calculate the percentage by mass of NH4NO3 in this fertiliser. (3)
Number of moles of HCl(aq) remained in the beaker:
=0.100 X 0.04100 X (100.00/25.00) = 0.0164
Number of moles of NH3(aq) produced
= 0.0485 – 0.0164 = 0.0321
Percentage by mass of NH4NO3 in the fertiliser
= (0.0321 x 80) / 3.150 x 100% = 81.5%
• Only a few candidates were able to perform the calculation.
• Most candidates failed to recognise which compounds are involved in each step of the expt.
Question 8 (c) (i)
Assuming that the heat capacity of the cup used is negligible, calculate the enthalpy change of Reaction(1) from the above data. Question 8 (c) (i)
Question 8 (c) (i)
• Many candidates failed to recognise that reaction (1) is endothermic.
• They omitted the “+” sign in the answer.
• Some candidates used a wrong mass of the reacting mixture in their calculations.
Question 8 (c) (ii)
In another experiment performed under the same conditions, the enthalpy change of Reaction (2) was found to be ­49.1 KJ mol­1. Calculate the enthalpy change of decomposition of KHCO3(s) under the experimental conditions. Question 8 (c) (i)
• Many candidates failed to recognise the relation between the enthalpy changes of Reaction (1) and Reaction (2). • Many candidates did not express the enthalpy change of decomposition as per mole of KHCO3.
Question 8 (d) (i)
Question 8 (d) (i)
• Many candidates were not able to construct an enthalpy change cycle based on Hess’s law for the calculation.
Question 10
Suggest THREE measures for reducing the emission of air pollutants upon using fossil fuels.
Any three
• Install catalytic converters in cars
• Use unleaded petrol
• Replace diesel with LPG for vehicles
• Install scrubbers in power plants
• Using ultra low sulphur diesel
• Use electrostatic precipitator
• Some candidates wrongly suggested measures which are not related to fossil fuel usage (e.g. use nuclear energy, plant more trees, etc)
• Some give answers with mis­matched applications (e.g., installing scrubbers in vehicles, or installing catalytic converters in factories)
The End
2012 HKDSE Chemistry
Paper 1B, Part II
Y L AU
Question 11(b)
Explain qualitatively the effect on the initial rate of the reaction of replacing the 2.0 M HCl(aq) with 2.0 M H2SO4(aq).
HCl is a monobasic acid, while H2SO4 is a dibasic acid. Initial rate increases as the concentration of H+
increases in 2.0 M H2SO4.
Remarks:
Many candidates did not know that 2.0 M H2SO4 has a higher concentration of H+(aq) than 2.0 M HCl(aq). They were not able to state that the increase in initial rate is due to an increase in the concentration of H+(aq).
Question 12 (b)
Explain why ethyl ethanoate is a better solvent than water for dissolving cinnamaldehyde.
Both cinnamaldehyde and ethyl ethanoate are relatively non­polar compounds. Their molecules are attracted by weak intermolecular forces / weak van der Waals’ forces.
Remarks:
Some candidates were not able to mention that both ethyl ethanoate and cinnamaldehyde are non­polar, so they are attracted to each other by weak intermolecular forces.
Question 12 (c)
.. In order to separate these two compounds, simple distillation can be carried out. Draw a diagram for the set­up involved, and label the name of the distillate collected.
Common mistakes:
• Drawing of set­up for fractional distill
lla
ation
• Set­up of a closed system
• Wrong position of the thermometer
Question 13 (a)
In an experiment, 25.0 cm3 of 0.010 M Fe2(SO4)3(aq) and 25.0 cm3 of 0.01 M KSCN(aq) were mixed in a conical flask at room temperature, and equilibrium was attained.
The concentration of Fe(SCN)2+(aq) in the mixture was 0.0043 M when equilibrium was attained. Calculate the equilibrium constant Kc for the above reaction at room temperature.
initial [Fe3+(aq)] after mixing = 0.010 × 2 × 0.5 = 0.01 M
initial [SCN­(aq)]
aq)] after mixing = 0.010 × 0.5 = 0.005 M
Remarks:
• Only a few candidates were able to calculate the concentration of the reacting species ( in particular, the concentration of Fe3+) in the reaction mixture right after mixing.
Question 14 (a)
(i) Given that all alkyl groups in both X and Y are straight chains, label the chiral carbon(s) by using ‘＊’
in the above diagram.
(ii) With reference to (i), explain whether a change in optical activity is involved in the above conversion.
mo
n m
s
akes
:
Com
i
st
ak
• Some candi
dates wrongly ment
i
oned
t
hat Y has
a chi
ral c
arbon.
co
mpound
Question 14 (b)
One of the products in the alkaline hydrolysis of Y has a cleansing property. Explain the cleansing property of this product.
• C17H35COO– ion has an ionic head (COO­) and a hydrocarbon tail (C17H35).
• The hydrocarbon tail dissolves in grease droplets / is hydrophobic, while the ionic head dissolves in water / is hydrophilic.
• The ionic heads of the grease droplets repel from each other and the dirts inside these droplets are then removed.
Common mistakes:
•Some of them mis­spelt the words ‘hydrophilic’ and ‘hydrophobic’.
•Candidates were not able to point out that a detergent molecule consists of an ionic ‘head’ and a hydrocarbon ‘tail’.
Question 15
Use electron diagrams to illustrate, step by step, how CH4 reacts with Br2 under sunlight to form CH3Br. (Show electrons in the outermost shell
llss only.)
Common mistakes:
•Candidates did not state clearly the steps involved in the formation of CH3Br from CH4 and Br2. •Some candidates omitted some electrons in their lly
drawings, especiall
y the lone­pairs of electrons.
Question 16 (b)
Explain why SiO2 has the highest melting point among the covalent oxides listed above.
• SiO2 has a giant covalent structure, and the Si and O atoms are linked by strong covalent bonds.
• Other covalent oxides are discrete molecules attracted by weak van der Waals’ forces.
Common mistakes:
•Some candidates wrongly considered that SiO2 exists as molecules.
The End
HKDSE 2012 Chemistry Paper II
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Marking Criteria and Candidates’ Performance
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Dr Raymond WH FONG
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Question 1
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1 (a) (i) (1) and (2)
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What is the catalyst used in …?
(1 mark, well answered)
✔ Finely divided iron / Fe / iron oxide / …
✖ Nickel / Platinum / …
Explain the effect of a catalyst on a chemical reaction
(2 marks, well answered)
✔ Increase / decrease the rate of a reaction)
✔ Alternative pathway with lower / higher activation
energy (Concept, not well answered)
✖ Lower activation energy
1 (a) (ii)
Suggest how hydrogen can be obtained …
�
(1 mark, fairly answered)
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✔ Steam reforming of natural gas / methane/ CH4 …
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✖ Na & water / Mg & acid / Electrolysis of sea water /
電解海水 – not feasible in industry
Give a chemical equation
(1 mark)
�
3H2(g) + CO(g) OR
�
✔ CH4(g) + H2O(g)
✔ CO (g) + H2O(g)
� H2(g) + CO2(g)
(Need to be balanced, state symbols are not considered)
�
1 (a) (iii)
�
… explain why such operation conditions are
chosen …
(2 marks, fairy answered)
✔ The higher temperature – speed up the reaction …
✔ A not very high pressure – safety concerns /
mechanical design / construction cost / maintenance
cost / …
✖ cost / 負出高成本 / …
✖ High temp and lower pressure can increase the
reaction rate
Need to have a matched operating condition(s) and
explanation(s)
1 (a) (iv)
�
… explain why the product mixture is removed from
the reaction chamber before the yield of about
20% …
(2 marks, poorly answered)
✔ Only a few candidates read accurately: “Removal of
the product mixture”
✖ Many candidates regarded the question as asking
“removal of product”
✔ It takes a long time to attain the equilibrium …
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✔ It can increase the total amount of ammonia
�
produced per unit time
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1 (b) (i)
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Methanol is …produced from syngas made from
methane. Why is methanol an important … industry?
(1 mark, satisfactorily answered)
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✔ starting material to make organic compounds with
�
large carbon numbers / methanal / formaldehyde /
�
ethanoic acid / acetic acid / ether / as a solvent
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✖ as a fuel / “vinegar”/ ethene (Not feasible in industry)
1 (b) (ii)
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Write the chemical equation for the reaction in the
production of methanol from syngas, and state the
conditions required.
(1 mark for the equation, satisfactorily answered)
CH3OH(g) (Accept irreversible
✔ CO(g) + 2H2(g)
sign)
(2 marks for the conditions)
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✔ Catalyst: Cu / ZnO / Al2O3 ✖ Pt / Pd/ Ni …
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✔ Temperature: 200-300oC (Require correct unit)
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✔ Pressure: 50-100 atm (Require correct unit)
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1 (b) (iii)
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State an advancement of the methanol production
technology. Explain why it is considered as an advancement.
(1 mark for the specific advancement related to methanol
�
production, and 1 mark for the reason, poorly answered)
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✔ Direct conversion of methane … metal oxide
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catalyst …catalytic reagent / high atom economy OR
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✔ Oxidation of methane … by microbial reaction … higher
energy efficiency OR
✔ Conversion of biomass to syngas/biogas … renewable feed
stocks OR
✔ Carbon dioxide in flue gas … reduce the release of CO2 OR
✔ Hydrogen left over … uses up the unconsumed raw
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materials
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1 (c) (i)
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Explain why ‘initial rate’ is commonly used in the study of the
kinetics of a reaction.
(1 mark, poorly answered)
✔ … because the initial concentrations of reactants are known
(Many students failed to answer this question)
1 (c) (ii)
�
Deduce the order of reaction with respect to NO(g) and that
to H2(g).
(2 marks, well answered)
✔ Many candidates were able to use ‘deduction method’ or an
‘arithmetic approach’ to find the order required
1 (c) (iii)
�
State the rate equation for the reaction, and calculate its rate
constant at the temperature of the experiment
(1+1 marks, satisfactorily answered)
✔ rate = k [NO(g)]2 [H2(g)]
✔ R = k [NO(g)]2 [H2(g)]
✔ r = k [NO(g)]2 [H2(g)]
✖ initial rate = k [NO(g)]2 [H2(g)]
Poor performance in finding the answer, especially the correct
unit
✔ k = 0.384 dm6 mol-2 s-1
✖ Candidates were very often failed to write s-1.
Question 2
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2 (a) (i) (1) and (2)
�
Kevlar and nylon-6.6 ... Draw the structures of the
monomers for making Kevlar.
(2 marks, fairly answered)
✔ Give two different monomers
✖ Structures of monomers – missing double bonds,
missing atoms, incorrect orders of atoms, ...
Name* the type of reaction for the formation of
Kevlar from its monomers
(1 mark, correct spelling, fairly answered)
✔ Condensation polymerisation / polymerization
2 (a) (ii) (1) and (2)
�
(1) Structure of “A”
(1 mark, fairly answered)
✔ HOOC(CH2)4COOH
✖ Missing atoms, incorrect total number of bonds, ...
(2) Give three reasons for consider why reaction (1) is
“Greener” ...
(3 marks, not well answered)
✔ Must be in comparison style
H2O2 in reaction (1) is less corrosive when compared
with concentrated nitric acid in reaction (2)
By-product H2O in reaction (1) is ... when compared with
the by-product N2O in reaction (2)
A catalyst is used in reaction (1) but not in reaction (2).
2 (a) (ii) (3)
(3) In what aspect are both reactions considered NOT
green.
(1 mark, fairly answered)
✔ Both reactions consumer starting materials that
possible obtained from a non-renewable resource /
petroleum
2 (a) (iii)
�
With reference to the structures of Kevlar and nylon6.6, account for their difference in mechanical
strength.
(2 marks, poorly answered)
✔ Benzene groups in Kevlar molecules have a more
rigid structure (fairly answered)
✔ Aromatic stacking / Intermolecular interactions
between benzene groups of adjacent polymer
�
molecules (fairly answered)
�
2 (b) (i)
�
A part of the structure of natural rubber ... Name the
treatment process, state its purpose, and explain the
principle behind
(3 marks, fairly answered)
�
✔ Vulcanisation / vulcanization
�
✖ Wrong spelling, wrong concept, ...
�
✔ To make the material strong / tough / high strength
�
/ elastic or flexible
�
✖ Durable / do not deform (wrong concept)
�
✔ Forming sulphur / disulphur / S-S cross links
�
between polymer chains
�
✖ Sulphide bonds / “cross links”
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2 (b) (ii)
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A technician wore a pair of gloves ... natural rubber
and performed an experiment involving bromine. ...
brittle ... Explain the phenomenon.
(2 marks, fairly answered)
✔ Addition of bromine to the C=C bonds of the
materials
✔ The brominated polymer have weaker
intramolecular structure / weaker intermolecular
attraction, leading to a different mechanical property.
✖ Substitution reaction, ... (wrong concept)
2 (c) (i) (ii)
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Iron – body-centred cubic structure.
(2 marks, satisfactorily answered)
✔ Draw a unit cell of iron
✖ Missing one atom (Careless)
✔ Finding the number of iron atoms in the unit cell
Why stainless steel is suitable for making knives
�
(3 marks, fairly answered)
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✔ Carbon and chromium / nickel / manganese
�
✖ Coke (Wrong concept)
�
✔ Hardness and size of carbon / chromium / nickel /
manganese ... (Fairly answered)
✔ Corrosion resistant and chromium / nickel /
manganese ... (Fairly answered)
Question 3
�
3 (a) (i)
�
Compound X (molar mass < 118 g) contains a -C6H5
group. Two chemical tests are performed on X …
Test (1): X turns acidified potassium dichromate
�
solution green
�
Test (2): X gives a negative result when tested with 2,4dinitrophenylhydrazine
(2 marks, well answered)
(i) … TWO functional groups
✔ hydroxyl group / -OH / alcohol
✔ aldehyde group / -CHO / aldehyde
3 (a) (ii)
�
(2 marks, well answered)
(1) … purpose of 2,4-dinitrophenylhydrazine
✔ Test for carbonyl / aldehyde or ketone / aldehyde
and ketone
✔ RCHO and RCOR’
(2) … expected observation if X gives a positive
result in Test (2)
✔ give yellow, orange or red precipitate
✖ brown colour
3 (a) (iii)
�
(1 mark, well answered)
(iii) With reference to the results of both Test (1) and
Test (2), suggest one functional group that may be
present in X.
✔ hydroxyl (group) / -OH
3 (a) (iv)
�
(2 marks, fairly answered)
(iv) Suggest one chemical species corresponding to
each of the signals at m/z = 91 and 108.
✔ C7H7+ / C6H5CH2+ (ion)
✔ C7H8O+ / C6H5CH2OH+ (ion)
Must shown plus sign (concept)
3 (a) (v)
(1 mark, fairly answered)
Draw a possible structure of X.
✔ C6H5CH2OH (no charge)
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✔ Drawings OK
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3 (b) (i)
�
(1 mark, poorly answered)
The dioxin levels in air are generally measured
through instrumental analysis but not …
(i) Suggest a source of dioxins in air
✔ Combustion of materials containing chlorine / PVC
✔ Incineration of materials containing chlorine / PVC
✔ Emission from incinerators
✔ Burning / combustion of plastic waste
✔ Burning / combustion of waste
✖ Car exhaust
3 (b) (ii) (iii)
�
(1 mark, poorly answered)
(ii) Explain why there is a need to measure the dioxin
levels in air (1 mark)
✔ Dioxin is carcinogenic / can cause cancer
(2 marks, poorly answered)
(iii) Suggest an instrumental analytical method for
measuring the dioxin levels in air, and state why this
method, rather than methods based on … is to be used.
✔ Gas chromatography-mass spectrometry / GC-MS / GC
MS / GC,MS / GC/MS
✔ It can measure more accurately the low level of dioxin
than using …
✖ “Faster …”, “Dioxins – gaseous”, …
3 (c) (i)
�
(2 marks, satisfactorily answered)
An aqueous ... HCl(aq) and HI(aq). Based on the fact
that AgCl(s), but not AgI(s), can dissolved in excess
NH3(aq), you are required to plan a gravimetric
analysis to determine the mole ratio of Cl−(aq) to I−
(aq) ...
Suggest TWO reagents, other than deionised water, that
should be used in the analysis
✔ AgNO3(aq) and NH3 (aq)
✔ Acidified AgNO3(aq) and NH3 (aq)
3 (c) (ii)
(4 marks, satisfactorily answered)
Outline the experimental steps involved in the analysis.
�
✔ Step 1: Add excess AgNO3(aq) to the solution ... AgCl(s) and
AgI(s)
✔ Step 2: Filter the mixture, wash with deionised water and dry
the residue
✔ Step 3: Determine / weigh the total mass of AgCl(s) and AgI(s)
✔ Step 4: Wash the solid residue with excess ammonia solution
to dissolve / remove AgCl(s), filter and dry the residue, and
determine the mass of AgI(s) remains
3 (c) (iii)
�
(2 marks, satisfactorily answered)
Outline the steps in the calculation of the mole ratio
of Cl−(aq) to I− (aq) in the solution using the data
obtained from (ii) above.
✔ Subtracting the total mass of AgCl(s) and AgI(s)
determined in Step 3 by mass of AgI(s) determined in Step
4 to get the mass of AgCl(s).
✔ Number of mole of AgCl and AgI can be obtained by
dividing their respective mass by the corresponding molar
mass. Mole ratio of Cl−(aq) to I− (aq) can then be
determined.
The End
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