2021

HIGHER SCHOOL CERTIFICATE

MOCK HSC EXAMINATION
SOLUTIONS

Chemistry

General Instructions Total marks − 100

• Reading time − 5 minutes
• Working time − 3 hours Section I Pages 2 − 11
• Write using blue or black pen
20 marks
Black pen is preferred
• Attempt Questions 1 − 20
• Draw diagrams using pencil
• Allow about 35 minutes for this part
• Board-approved calculators may be used
• A data sheet and a Periodic Table are Section II Pages 12 − 31
provided at the back of this paper
• Write your name and class at the top of 80 marks
this page • Attempt Questions 21 − 32
• Allow about 2 hours and 25 minutes for this
section

−1−
Section I: Multiple Choice Questions (20 marks)
Attempt Questions 1 − 20
Allow about 35 minutes for this section

Use the multiple choice answer sheet for Questions 1 − 20.

1. A A C D
2. A B C A

3. A B C D
4. A B C D
5. A B C A

6. A A C D
7. A B A D
8. A B C A

9. A B C A

10. A B C D
11. A A C D
12. A B C A

13. A A C D
14. A B C D
15. A B A D
16. A B A D
17. A A C D
18. A B A D
19. A B A D
20. A B C D

−2−
1. Which of the following is a pair of chain isomers where the boiling point of the first compound
is higher than that of the second compound?
(A) 2,2-dimethylbutane and hexane
(B) Butan-1-ol and 2-methylpropan-1-ol
(C) Hexan-1-amine and pentan-1-amine
(D) Butanoic acid and ethyl ethanoate

2. The graph below shows how the acid ionisation constant (Ka ) of acetic acid varies with
temperature.

1.76

1.74
Ka (× 10–5)

1.72

1.70

1.68

20 25 30 35 40
Temperature (°C)

From the graph, it can be concluded that the ionisation of acetic acid in water is:
(A) Endothermic and the pH of acetic acid will decrease as temperature increases
(B) Endothermic and the pH of acetic acid will increase as temperature increases
(C) Exothermic and the pH of acetic acid will decrease as temperature increases
(D) Exothermic and the pH of acetic acid will increase as temperature increases

3. An unknown salt dissolves in water to give a solution that causes red litmus to turn blue.
When sulfuric acid is added to this solution, no bubbling is observed but a white precipitate
does form. Furthermore, when this solution is vaporised into a flame, an apple green flame
colour is observed.

What is a possible identity for the unknown salt?

(A) Barium acetate
(B) Barium chloride
(C) Barium carbonate
(D) Calcium carbonate

−3−
4. When solid calcium hydroxide is added to water, only some of the calcium hydroxide dissolves.

What is the solubility of calcium hydroxide in water?

(A) 0.0800 g/100 mL
(B) 0.800 g/100 mL
(C) 0.0108 g L–1
(D) 0.108 g L–1

5. Consider the following compound:

Cl H F H O H
Cl C C C C C N

Cl H H H H

What is the systematic name of this compound?

(A) 1-amino-5,5,5-trichloro-3-fluoropentanal
(B) 3-fluoro-5,5,5-trichloropentanamide
(C) 5,5,5-trichloro-3-fluoropentanamine
(D) 5,5,5-trichloro-3-fluoropentanamide

6. A laboratory contains solutions of hydrochloric acid and methanoic acid, both at the same
volume and concentration.

Which of the following statements regarding these two solutions is correct?

(A) Both acids have the same pH
(B) Both solutions would be neutralised by the same volume of 0.10 mol L–1 NaOH
(C) The hydrochloric acid has a lower electrical conductivity than the methanoic acid
(D) The methanoic acid contains a higher number of anion particles than the hydrochloric acid

7. Ammonium nitrate is a highly water soluble salt. When it is dissolved in water, the temperature
of the water decreases.

Which row of the following table correctly describes the thermodynamics of the dissolution
of ammonium nitrate in water under standard conditions?

Enthalpy change Entropy change Gibbs free energy change

(A) ΔH −◦− < 0 ΔS −◦− < 0 ΔG−◦− < 0 since ΔH −◦− < T ΔS −◦−
(B) ΔH −◦− > 0 ΔS −◦− < 0 ΔG−◦− < 0 since ΔH −◦− > T ΔS −◦−
(C) ΔH −◦− > 0 ΔS −◦− > 0 ΔG−◦− < 0 since ΔH −◦− < T ΔS −◦−
(C) ΔH −◦− > 0 ΔS −◦− > 0 ΔG−◦− < 0 since ΔH −◦− > T ΔS −◦−

−4−
8. The mass spectrum of a compound is shown below.

100

80
Relaltive Intensity

60

40

20

0
10 20 30 40 50
m/z

What is the most likely identity of the compound?

(A) NO2
(B) CO2
(C) C3H8
(D) C2H5OH

9. The conjugate base of a Brønsted-Lowry acid would always have:

(A) A hydroxide ion
(B) A negative charge
(C) An unpaired electron
(D) A lone pair of electrons

−5−
10. The following compound can be produced from a reaction between two reagents, P and Q.

H H O
H C C C H H H

H H O C C C H

H H
CH3

If the boiling points of P and Q are 99◦ C and 141◦ C respectively, which row of the following
table correctly identifies the reagents?

P Q
(A) Butan-2-ol Propanoic acid
(B) 2-methylpropan-1-ol Propanoic acid
(C) Propanoic acid Butan-2-ol
(D) Propanoic acid 2-methylpropan-1-ol

11. Ultraviolet-visible (UV-vis) spectroscopy was used to measure the absorbance of three dyes over
a range of wavelengths. The absorbance spectra produced are shown below.

1.0

0.8 Blue dye

Absorbance

0.6
Yellow dye
Red dye
0.4

0.2

0
400 500 600 700
Wavelength (nm)

Which of the following is the most suitable wavelength to use to determine the concentration
of the red dye in a mixture containing all three dyes?
(A) 430 nm
(B) 500 nm
(C) 540 nm
(D) 620 nm

−6−
12. Consider the following reaction:

2 SO2(g) + O2(g) 2 SO3(g) ΔH < 0

The following graph shows how the rate of the forward and reverse reactions for this system
varies with time.

Rate of reaction

t
Time
Forward reaction
Reverse reaction

Which of the following disturbances may have been applied at time t?

(A) SO2 was removed from the reaction vessel
(B) SO3 was removed into the reaction vessel
(C) The temperature of the reaction vessel was decreased
(D) The volume of the reaction vessel was increased

13. The monomers below combine together to form nylon-6,6 and hydrogen chloride.

H H O O
N (CH2 )6 N Cl C (CH2 )4 C Cl
H H
Hexane-1,6-diamine Hexanedioyl dichloride
The table below shows the approximate molar mass of these monomers and hydrogen chloride.

Compound Approximate molar mass (g mol−1 )

Hexane-1,6-diamine 116
Hexanedioyl dichloride 183
Hydrogen chloride 36

What is the approximate molar mass of the nylon-6,6 molecule formed when 20 molecules of
hexane-1,6-diamine combines with 20 molecules of hexanedioyl dichloride?
(A) 4540 g mol–1
(B) 4576 g mol–1
(C) 4612 g mol–1
(D) 5980 g mol–1

−7−
14. The diagram below shows a spectrometer that is being used to analyse a sample solution for a
particular metal cation.

Prism
(or monochromator)

Lamp
Flame Lens

Sample solution Detector

aspirated into flame

A group of high school students made the following statements regarding this instrument:

(i) The intensity of light emitted by the lamp is higher than the intensity of light that exits
from the flame
(ii) The lamp produces a continuous spectrum and only some wavelengths are absorbed by
the element being tested
(iii) The flame vaporises and atomises the sample being tested
(iv) The flame vaporises and ionises the sample being tested

Which of the above statements are correct?

(A) (i) and (iii)
(B) (i) and (iv)
(C) (ii) and (iii)
(D) (ii) and (iv)

15. A 25.0 mL solution of 0.15 mol L–1 hydrochloric acid was mixed with a 20.0 mL solution of
0.10 mol L–1 potassium hydroxide.

What is the pH of the resultant solution?

(A) 1.06
(B) 1.15
(C) 1.41
(D) 2.76

−8−
16. Consider the following series of reactions where A, B and C are different organic compounds.

Dilute KOH K2Cr2O7/H2SO4

A (C4H9Cl) B C

The infrared (IR) spectrum and 13 C nuclear magnetic resonance (NMR) spectrum of compound
C is shown below:

Infrared spectrum

0.8

0.6
Transmitance

0.4

0.2

3000 2000 1000

Wavenumber (cm–1)

13C NMR Spectrum

200 150 100 50 0

Chemical shift (ppm)

Based on the above information, what is the identity of compound A?

(A) 1-chlorobutane
(B) 2-chlorobutane
(C) 1-chloro-2-methylpropane
(D) 2-chloro-2-methylpropane

−9−
17. Consider the following reactions:

ClCH2COO–(aq) + H2O(l) ClCH2COOH(aq) + OH–(aq) pKb = 11.13

NO2–(aq) + H2O(l) HNO2(aq) + OH–(aq) pKb = 10.84

Equimolar solutions of ClCH2COONa, ClCH2COOH, NaNO2 and HNO2 were prepared.

Which row of the following table identifies the solutions that would have the lowest and highest
pH value?

Lowest pH Highest pH
(A) ClCH2COOH ClCH2COONa
(B) ClCH2COOH NaNO2
(C) HNO2 ClCH2COONa
(D) HNO2 NaNO2

18. The structural formula of an organic compound is shown below:

H O H
Br C C N

H H

79 81
The common isotopes of bromine are Br and Br which both have a natural abundance of
approximately 50%.

When this organic compound is analysed with a mass spectrometer, the following processes
occur:
•
BrCH2CONH2(g) + e– → BrCH2CONH2 +(g) + 2 e–
• •
BrCH2CONH2 +(g) → BrCH2+(g) + CONH2 (g)
• •
BrCH2CONH2 +(g) → CONH2+(g) + BrCH2 (g)

Which row of the following table shows the number of appreciable peaks that these processes
would produce on the mass spectrum and the species directly responsible for these peaks?

Number of peaks Species directly responsible

•
(A) 3 BrCH2CONH2 +, BrCH2+, CONH2+
• • •
(B) 3 BrCH2CONH2 +, CONH2 , BrCH2
•
(C) 5 BrCH2CONH2 +, BrCH2+, CONH2+
• • •
(D) 5 BrCH2CONH2 +, BrCH2+, CONH2 , CONH2+, BrCH2

−10−
19. Which of the following combinations of acid and base would produce the largest temperature
change for the resultant mixture?
(A) 100 mL of 0.05 mol L–1 nitric acid and 100 mL of 0.50 mol L–1 sodium hydroxide
(B) 100 mL of 0.20 mol L–1 sulfuric acid and 100 mL of 0.10 mol L–1 potassium hydroxide
(C) 100 mL of 0.20 mol L–1 hydrochloric acid and 100 mL of 0.10 mol L–1 barium hydroxide
(D) 200 mL of 0.10 mol L–1 nitric acid and 200 mL of 0.40 mol L–1 potassium hydroxide

20. An aqueous solution of bromocresol green contains a yellow molecule (HInd) and a blue anion
(Ind–). The pKa of bromocresol green is 4.90.

A few drops of bromocresol green is added to a base and the mixture is then titrated against
an acid. The corresponding titration curve is shown below:

14

12

10

8
pH

0
0 5 10 15 20 25 30 35 40 45 50
Volume of acid added (mL)

Which row of the following table shows the colour change that will be observed and the relative
concentration ratio of HInd to Ind– at the equivalence point?

Colour change at [HInd]

at equivalence point
equivalence point [Ind− ]
(A) Blue to green 0.79
(B) Yellow to green 0.79
(C) Blue to green 1.26
(D) Yellow to green 1.26

−11−
Section II: Short Answer Questions (80 marks)

Question 21 (4 marks)
The following equilibrium mixture was set up in a laboratory. 4

Lid

Beaker Saturated CuSO4 solution

Solid CuSO4

Some solid sodium sulfate is then added into the beaker and the mixture is maintained at a
constant temperature for several days.

With reference to collision theory and the solubility product (Ksp ), explain how the final
equilibrium mixture would compare with the initial equilibrium mixture with respect to:

• The mass of solid CuSO4 present

• The colour of the solution
• The concentration of Cu2+ and SO42– ions in the solution

Include a relevant chemical equation in your answer.

CuSO4(s) Cu2+(aq) + SO42–(aq)

When Na2SO4 is added to this system, it dissociates and causes [SO42–] to increase. Collision
theory predicts that the rate of the reverse reaction will increase since there will be more
successful collisions between Cu2+ and SO42– ions. As such, the reverse reaction rate will
be higher than the forward reaction rate, causing the equilibrium to shift left due to the
net reverse reaction that occurs. This causes more solid CuSO4 to form which is consistent
with the common ion effect and thus the final equilibrium mixture will have a higher mass
of solid CuSO4 present.

Ksp will remain constant since temperature is kept constant. Since the equilibrium shifts
left, [Cu2+] will decrease which would cause the blue colour of the final solution to be less
intense than the initial solution. However, since Ksp = [Cu2+][SO42–], if [Cu2+] is lower, then
[SO42–] must be higher in the final equilibrium mixture for Ksp to remain constant. The
addition of Na2SO4 results in a net increase in [SO42–].

1 mark − Includes a relevant chemical equation

1 mark − Explains that the rate of the reverse reaction is increased with collision theory

1 mark − Explains that the final mixture will have a higher mass of solid CuSO4 present
and a less intense blue colour

1 mark − Explains that [Cu2+] is lower but [SO42–] must be higher in the final mixture with
reference to Ksp

−12−
Question 22 (9 marks)
The structure of two monomers which can be used to make different polymers are shown below.
H OH O O
H C C C H2 C C C

H H O H H O H
Lactic acid Acrylic acid
A polymer chemist carried out two different polymerisation reactions to convert lactic acid into
poly(lactic acid) and acrylic acid into poly(acrylic acid).

(a) Compare the process of polymerisation that the monomers above undergo. Include relevant 4
chemical equations using structural formulae in your answer.

Both polymerisation processes involve many monomer units chemically joining together
to form a large polymer. Lactic acid undergoes condensation polymerisation which
involves the monomers joining together by eliminating a small molecule (water in this
case) through a reaction between two functional groups on both ends of the monomer.

H O  H O
nH O C C O H O

C C
 H
+ (n − 1) O
H C H H C H n H
H H
In contrast, acrylic acid undergoes addition polymerisation which involves the monomers
joining together without the formation of any by-products through breaking the reactive
C=C double bond.

H H H H

nC C C C
 

H H n
C O C O

O O

H H
2 marks − Identifies the type of polymerisation each monomer undergoes and describes
the main differences between the two in terms of by-products (1 mark each)

2 marks − Includes TWO chemical equations using structural formulae showing the
polymerisation processes (1 mark each)

Note:

• The similarity wasn’t marked for but it is recommended to include it.

• Answers may also show repeating units for the polymers rather than the bracket
notation (can accept equations that show just 2 monomers combining but it is
recommended to show 3 monomer units with open ends).

−13−
(b) Describe a simple chemical test that could be used to confirm if all of the acrylic acid has 2
been completely converted into poly(acrylic acid). Include a relevant chemical equation
using structural formulae in your answer.

Bromine water can be added to the reaction mixture. If the bromine water remains
brown, then this indicates all of the acrylic acid has been converted into the polymer
which doesn’t contain a reactive C=C double bond. However, if the bromine water is
decolourised from brown to colourless, then there is still some unreacted acrylic acid.

H H H H

C C + Br Br Br C C Br

H C O H C O

O O

H H
1 mark − Describes a chemical test that could be confirm if all the acrylic acid has
been converted into its polymer including any observations

1 mark − Includes a relevant chemical equation using structural formulae

(c) Polyethylene is a commonly used commercial polymer that is very versatile. 3

Explain why polyethylene is such a versatile polymer with reference to the structure,
properties and uses of TWO different forms of polyethylene.

Low density polyethylene (LDPE) is an amorphous polymer with a high degree of

chain branching that prevents its polymer chains from packing closely together. The
dispersion forces between the polymer chains will therefore be quite weak, so LDPE is
a soft and flexible polymer which can be used as glad wrap and plastic bags.

High density polyethylene (HDPE) is a crystalline polymer with very minimal chain
branching, so its polymer chains can pack closely together. This results in stronger
dispersion forces between the polymer chains, so HDPE is a hard, rigid and strong
polymer which can be used as buckets, rubbish bins and chemical containers.

As such, polyethylene is an extremely versatile polymer as it can be produced in two

different forms with very different properties and uses.

3 marks − Explains the properties and specific uses of LDPE and HDPE with reference
to chain branching and dispersion forces

−14−
Question 23 (10 marks)
A student carried out a conductometric titration between a standardised hydrochloric acid
solution and an unknown ethanamine solution. The table below shows how the electrical
conductivity of the reaction mixture varied as ethanamine was added from a burette to a
conical flask containing 25.0 mL of 0.12 mol L–1 hydrochloric acid.

Volume of ethanamine added (mL) Conductivity (mS cm–1)

0.00 24.00
2.00 23.47
4.00 23.00
6.00 22.80
8.00 22.00
10.00 21.55
12.00 21.00
14.00 20.70
16.00 20.69
18.00 20.71
20.00 20.70

(a) Construct an appropriate graph of the data. 3

Conductivity curve for the titration of HCl and C2H5NH2

24.5

24.0 x

23.5 x
Conductivity (mS cm–1)

23.0 x
x
22.5

22.0 x

21.5 x

21.0 x
x x x x
20.5

0 2 4 6 8 10 12 14 16 18 20 22
Volume of ethanamine added (mL)

−15−
 1 mark − Labels the y-axis with conductivity and the x-axis with volume of ethanamine
added including units

1 mark − Plots all the data points correctly AND uses an appropriate scale

1 mark − Draws TWO lines of best fit that extends to a point of intersection (must
NOT directly connect the points at (0, 24) and (14, 20.70))

(b) Calculate an estimate for the pH of the unknown ethanamine solution given that the pKb 4
of ethanamine is 3.35.

From the graph, the equivalence point is reached when 13.2 mL of ethanamine is added.
HCl(aq) + C2H5NH2(aq) → C2H5NH3Cl(aq)

n(HCl) = 0.12 mol L–1 × 0.02500 L

= 3.0 × 10−3 mol
n(C2H5NH2) = 3.0 × 10−3 mol
3.0 × 10−3 mol
[C2H5NH2] =
0.0132 L
= 0.23 mol L–1

C2H5NH2(aq) + H2O(l) C2H5NH3+(aq) + OH–(aq)

C2H5NH2 C2H5NH3+ OH–

Initial 0.23 0 0
Change −x +x +x
Equilibrium 0.23 − x x x

Kb = 10−3.35
= 4.5 × 10−4
[C2H5NH3+][OH–]
Kb =
[C2H5NH2]
x2
4.5 × 10−4 =
0.23 − x

Assume that 0.23 − x + 0.23 since Kb has a small value.

x2
4.5 × 10−4 =
0.23
=⇒ x = 0.010 mol L–1
[OH–] = 0.010 mol L–1
pOH = −log10 (0.010)
= 2.00
pH = 14.00 − 2.00
= 12.00

−16−
 1 mark − Calculates the correct moles of HCl titrated (values may vary slightly)

1 mark − Calculates the correct concentration of ethanamine (values may vary slightly)

1 mark − Calculates the correct concentration of OH– (values may vary slightly)

1 mark − Calculates the correct pH (values may vary slightly)

Note that the assumption does introduce some error into the calculation but this error
is relatively small and would still be acceptable:

0.0102
Kb =
0.23 − 0.010
= 4.7 × 10−4
≈ 4.5 × 10−4

(c) Prior to the conductometric titration, the hydrochloric acid was standardised by titrating 3
it against a sodium carbonate primary standard solution. During the preparation of the
primary standard, the student filled a volumetric flask to the level shown below:

Gradation mark

Sodium carbonate solution

A burette was then rinsed and filled with the sodium carbonate solution while a conical
flask was rinsed and filled with the hydrochloric acid solution. The student then carried
out the titration and calculated the concentration of hydrochloric acid to be 0.12 mol L–1.

Explain the effect that each individual mistake made by the student would have on the
calculated concentration of hydrochloric acid.

The volumetric flask should have been filled until the base of the meniscus was on the
engraved mark. By filling it below the mark, the Na2CO3 primary standard solution
would be more concentrated than expected. This would result in a smaller volume of
Na2CO3 being required to reach the end point, leading to an underestimation of the
moles and concentration of HCl.

The conical flask should have been rinsed with only distilled water. By rinsing it
with HCl, the moles of HCl in the flask would be higher than expected. This would
result in a larger volume of Na2CO3 being required to reach the end point, leading to
an overestimation of the moles and concentration of HCl.

The net effect on the calculated concentration of HCl would depending on the relative
magnitude of these errors.
1 mark − Identifies at least ONE mistake made by the student
2 marks − Explains the effect of BOTH mistakes on the calculated concentration of
HCl with reference to the concentration of Na2CO3 and the moles of HCl

−17−
Question 24 (5 marks)
A camp stove using butane as a fuel is used to heat water in an aluminium pot as shown below.

Water in aluminium pot

Flame

Butane

The table below shows some information regarding this set up.

Mass of the aluminium pot 250.00 g

Mass of the aluminium pot and water 650.00 g
Initial temperature of the aluminium pot and water 18.5◦ C
Specific heat capacity of aluminium 0.900 J g−1 K−1
Molar heat of combustion of butane 2877 kJ mol−1
Total heat lost to the surroundings 60.0%

(a) Calculate the minimum mass of butane that needs to be combusted to heat the aluminium 4
pot and water to 100.0◦ C.

mw = 650.00 g − 250.00 g
= 400.00 g
qw = (mcΔT )w
= 400.00 g × 4.18 J g–1 K–1 × (100.0 − 18.5) K
= 136268 J
qAl = (mcΔT )Al
= 250.00 g × 0.900 J g–1 K–1 × (100.0 − 18.5) K
= 18337.5 J
qtotal = qw + qAl
= 136268 J + 18337.5 J
= 154605.5 J

If there is 60.0% heat loss to the surroundings, then only 40.0% is absorbed by the
system.

qtotal = −0.400q(C4H10)
qtotal
=⇒ q(C4H10) = −
0.400
154605.5 J
=−
0.400
= −386513.75 J

−18−
 q(C4H10)
ΔHc =
n(C4H10)
−386513.75 J
−2877 × 103 J mol–1 =
n(C4H10)
=⇒ n(C4H10) = 0.134 mol
m(C4H10) = 0.134 mol × (4 × 12.01 + 10 × 1.008) g mol–1
= 7.81 g

1 mark − Calculates the correct energy absorbed by water OR aluminium or equivalent

merit

1 mark − Calculates the correct total energy absorbed by water AND aluminium or
equivalent merit

1 mark − Calculates the correct energy that needs to be released by the combustion

1 mark − Calculates the correct mass of C4H10 that needs to be combusted

(b) It was found that to achieve the required temperature change for the aluminium pot and 1
water, a larger mass of butane was required than the value calculated in part (a).

Propose a chemical reason, other than heat loss, for this difference.

Incomplete combustion may have occurred and this releases less energy than complete
combustion which the molar heat of combustion of a fuel is based upon. More C4H10
would therefore need to be combusted to achieve the required temperature change.

1 mark − Proposes a suitable chemical reason for the difference

−19−
Question 25 (9 marks)
The graphs below shows how the boiling point and water solubility of straight chain alkanes
and primary alcohols varies with the number of carbon atoms in the molecule.

200

150

100
Boiling point (oC)

50

0
Primary Amines
−50

−100

−150

−200
0 1 2 3 4 5 6 7
Number of carbon atoms

∞ (miscible)

80

70
Water solubility (g L–1)

60

50

40

30

20

10

0
0 1 2 3 4 5 6 7
Number of carbon atoms

Primary Alcohols
Alkanes

−20−
(a) Explain the trends shown on the graphs. Include relevant diagrams in your answer. 7

From the graphs, it is evident that alcohols have a higher BP and water solubility than
alkanes of similar size. This is because:

• Alcohols are polar molecules that contain a polar hydroxyl (−OH) group and
are capable of forming strong hydrogen bonds which require significant energy to
overcome, resulting in a high BP.

• Alkanes are non-polar molecules that are only held together by weak dispersion
forces which requires little energy to overcome, resulting in much lower BPs.

• The polar alcohols can form strong hydrogen bonds with water which is also
polar, making them water soluble (especially the short chain alcohols that are
fully miscible).

• The non-polar alkanes cannot form any strong intermolecular forces with water
which makes them all insoluble in water.

As the carbon chain length increases, it can be seen that:

• The BP of substances in both homologous series increases. This is because larger

molecules have stronger dispersion forces, since they contain more electrons and
there are more instances of uneven electron distribution.

• The BPs of the series also start to converge since the longer hydrocarbon chain
results in stronger dispersion forces that begin to outweigh the effects of the
hydrogen bonds formed by the alcohols.

• The water solubility of the alcohols progressively decrease. This is because as the
non-polar hydrocarbon chain increases in length, it’s non-polar and hydrophobic
nature begins to outweigh the effects of the polar and hydrophilic −OH group.

3 marks − Explains why alcohols have higher BPs and water solubilities than alkanes
of similar size with reference to polarity and intermolecular forces

1 mark − Explains why the BP of both series increases with carbon chain length with
reference to dispersion forces

1 mark − Explains why the BP of the series start to converge as the carbon chain
length increases

−21−
 1 mark − Explains why the water solubility of alcohols decrease as carbon chain length
increases

1 mark − Provides at least TWO relevant labelled diagrams showing intermolecular

forces

(b) On the boiling point graph, draw a line to roughly show how the boiling points of straight 2
chain primary amines will compare with the other compounds and explain your answer.

Amines are polar molecules that can form hydrogen bonds due to their polar N−H
groups. However, the N−H bond is less polar than the O−H bond since N is less
electronegative than O, so amines have weaker hydrogen bonds than alcohols. This
results in the amines having lower BPs than alcohols, but still higher than the non-polar
alkanes of similar size.

1 mark − Draws a line to show the BP of primary amines relative to alcohols and
alkanes

1 mark − Explains why the amines have lower BPs than alcohols of similar size with
reference to the electronegativity of N and O

−22−
Question 26 (7 marks)
The graph shows the changes in pH for the titrations of equal volumes of two monoprotic acids,
HX and HY, against a potassium hydroxide solution.

14
12
HX
10
8
pH
6
HY
4
2
0
0 10 20 30 40 50 60
Volume of KOH (mL)

(a) A chemistry student made the following statement: 4

“Since a greater volume of potassium hydroxide is required to completely neutralise HY

compared to HX, HY is a stronger acid than HX.”

Critically assess this statement with reference to TWO pieces of evidence from the graph.

The statement is incorrect. HX is a strong acid, as evident by the equivalence point

being at pH = 7 which is characteristic of a strong acid-strong base titration which
produces a neutral salt. HY is a weak acid, as evident by the equivalence point being
at pH > 7 which is characteristic of a weak acid-strong base titration which produces a
basic salt. This is also supported by HX having a lower initial pH than HY, indicating
that HX is more ionised and has a higher [H+] than HY.

The volume of KOH needed to neutralise a monoprotic acid does not depend on the
strength of the acid because neutralisation reactions with strong bases like KOH go to
completion regardless of the strength of the acid. The reason why a greater volume of
KOH is needed to neutralise HY is because HY is more concentrated than HX (in this
case, 1.5× more concentrated).

1 mark − Provides a judgement on the statement

2 marks − Explains that HX is a strong acid and HY is a weak acid with reference to
TWO pieces of evidence from the graph (pH at equivalence point, initial pH
or the buffer region)

1 mark − Explains that a larger volume of KOH is needed to neutralise HY because

HY is more concentrated than HX

−23−
(b) Account for the difference in shape between the two titration curves in the pH range of 3
4 − 6, and describe the importance of the effect responsible for this difference in a specific
natural system.

The titration curve for HY slightly plateaus in this pH range due to the formation of
a buffer which contains similar amounts of a weak acid (HY) and its conjugate base
(Y– formed as a product of the reaction). This buffer can resist changes in pH as KOH
is added to it, so the pH increases slowly. The titration curve for HX is much steeper
in this pH range because the reaction mixture contains a strong acid (HX) and its
extremely weak conjugate base (X–) which cannot act as a buffer, so the addition of
KOH causes the pH to rapidly increase, especially near the equivalence point.

The H2CO3/HCO3– buffer in human blood is important as it helps maintain a relatively

constant pH of 7.40 to ensure that enzymes that catalyse vital biochemical reactions
remain functional.

2 marks − Accounts for the difference in shape between the titration curves with
reference to buffers and their ability to resist changes in pH

1 mark − Describes the importance of a buffer in a specified natural system

−24−
Question 27 (6 marks)
Hydrogen and iodine can react to produce hydrogen iodide according to the following equation:

H2(g) + I2(g) 2 HI(g)

A 2.0 L reaction vessel initially contained 0.50 mol L–1 hydrogen gas and 0.50 mol L–1 iodine gas
which were allowed to react. The graph below shows the variation in concentration of reactants
and products as a function of time for this system.

(a) Calculate the equilibrium constant for the reaction. 1

[HI]2
K=
[H2][I2]
0.602
=
0.20 × 0.20
= 9.0

1 mark − Calculates the correct value of K

−25−
(b) At time T1 , 0.80 mol of hydrogen iodide gas was rapidly added into the 2.0 L reaction 5
vessel, and the system was then allowed to reach equilibrium at Teq . The reaction vessel
was maintained at a constant temperature throughout this time.

Calculate the equilibrium concentration of all the gases at Teq and hence sketch two curves
on the provided graph to show how the concentrations of the reactants and products change
between T1 and Tf .

n(HI)T1 = 0.60 mol L–1 × 2.0 L + 0.80 mol

= 2.0 mol
2.0 mol
[HI]T1 =
2.0 L
= 1.0 mol L–1

Addition of HI would cause the equilibrium to shift left, so [H2] and [I2] must increase
while [HI] decreases.
H2 I2 HI
Initial 0.20 0.20 1.0
Change +x +x −2x
Equilibrium 0.20 + x 0.20 + x 1.0 − 2x

[HI]2
K=
[H2][I2]

The reaction vessel is maintained at constant temperature so the value of K remains

constant.

(1.0 − 2x)2
9.0 =
(0.20 + x)2
1.0 − 2x
3.0 =
0.20 + x
=⇒ x = 0.080 mol L–1
∴ [H2]eq = [I2]eq = 0.20 mol L–1 + 0.080 mol L–1
= 0.28 mol L–1
[HI]eq = 1.0 − 2 × 0.080
= 0.84 mol L–1

1 mark − Calculates the correct concentration of HI immediately after the disturbance

1 mark − Substitutes appropriate terms into the K expression

1 mark − Calculates the correct value of x or equivalent merit

1 mark − Calculates the correct equilibrium concentration of H2, I2 and HI

1 mark − Draws two curves on the graph which are consistent with the calculation

−26−
Question 28 (7 marks)
An unknown organic compound with a molecular formula of C7H15OCl was analysed with 7
infrared (IR) spectroscopy and nuclear magnetic resonance (NMR) spectroscopy. The following
spectra were obtained from these techniques.
Infrared Spectrum

60
Transmittance (%)

40

20

0
3500 3000 2500 2000 1500 1000 500
Wavenumber (cm–1)

1H NMR Spectrum

6H

4H
2H 2H

1H

4 3 2 1 0
Chemical shift (ppm)

13C NMR Spectrum

80 70 60 50 40 30 20 10 0
Chemical shift (ppm)

file:///D:/Downloads/graph (2).svg 1/1

Analyse the spectra above to determine the structural formula of the organic compound.

Note: 1 H NMR chemical shift data has been included on your data sheet.

−27−
 IR spectrum

• Strong and broad absorption at 3230−3550 cm−1 indicates the presence of a O−H (alcohol)
group.
• No absorption at 1680 − 1750 cm−1 indicates that there is no C=O group present.

1
H NMR spectrum

There are 5 signals on the 1 H NMR spectrum, so there are 5 1 H environments and these are
summarised in the table below:
1
H Neighbours
δ (ppm) Integration Multiplicity Conclusion
(n + 1 rule)
2 symmetrical
CH3 − groups next to
0.9 6H Triplet (3) 2 a −CH2 − group

2× CH3 CH2
2 symmetrical
−CH2 − groups next to
1.5 4H Quartet (4) 3 a CH3 − group

2× CH3 CH2
A −CH2 − group next to
another −CH2 − group
1.9 2H Triplet (3) 2

CH2 CH2
The deshielded −OH
group of the alcohol
3.1 1H Singlet (1) 0
−OH
A deshielded −CH2 −
group next to a −CH2 −
group

H
CH2 C OH
3.5 2H Triplet (3) 2 H

OR

H
CH2 C Cl

−28−
 13
C NMR spectrum

There are 5 signals on the 13 C NMR spectrum, so there are 5 C environments which is less than
the 7 C atoms in the molecule. This indicates some symmetry in the structure.

• The signals at 8, 31 and 40 ppm represent typical alkyl carbons.

C C

• The signal at 42 ppm represents a C−Cl environment.

• The signal at 73 ppm represents a C−O (alcohol) environment.

From the information above, the fragments that make up the molecule are:

H H
2× CH3 CH2 −OH CH2 C OH OR CH2 C Cl

H H

In total, there are 6 C atoms in these fragments so there is still a missing C atom. The last C
atom must be in the centre of the molecule for symmetry.

The structure is therefore:

O Cl

H C H H C H

H C H H C H
H H H H H H H H

H C C C C C H OR H C C C C C H

H H Cl H H H H O H H

1 mark − Analyses the IR spectrum and identifies the presence of a O−H bond

4 marks − Analyses the 1 H NMR spectrum to identify the major structural fragments of
the organic compound

1 mark − Analyses the 13 C NMR spectra to identify the major structural fragments of the
organic compound

1 mark − Determines the correct structural formula of the organic compound (only ONE
of the above structures is required for full marks)

−29−
Question 29 (6 marks)
Ammonium oxalate is a common reagent used to determine the concentration of metal ions
in water samples since the oxalate ion (C2O22–) readily reacts with group 2 metal ions to form
insoluble precipitates.

The most common metal ions in sea water are sodium, potassium, calcium and magnesium. A
student used the following procedure to analyse a sea water sample for its calcium ion content.

• A 200.0 mL sample of sea water was placed into a beaker.

• Excess ammonium oxalate solution was added to the sample. All of the calcium ions in
the sample was completely precipitated out according to the following equation:
Ca2+(aq) + C2O22–(aq) + H2O(l) → CaC2O2 · H2O(s)
• The mixture was then filtered using a pre-weighed 13.637 g sintered glass crucible and
the residue collected. The residue was washed with distilled water and ethanol, and then
throughly dried to constant mass at room temperature.
• The mass of the sintered glass crucible with the dried residue was found to be 14.822 g.

(a) Calculate the concentration of calcium ions in the sea water sample that would be directly 3
obtained from the student’s experimental results in ppm.

m(CaC2O2 · H2O) = 14.822 g − 13.637 g

= 1.185 g
1.185 g
n(CaC2O2 · H2O) =
(40.08 + 2 × 12.01 + 3 × 16.00 + 2 × 1.008) g mol–1
= 0.01038 mol
n(Ca2+) = 0.01038 mol
m(Ca2+) = 0.01038 mol × 40.08 g mol–1
= 0.4162 g
= 416.2 mg
416.2 mg
[Ca2+] =
0.2000 L
= 2081 mg L–1
= 2081 ppm

1 mark − Calculates the correct moles of CaC2O2 · H2O

1 mark − Calculates the correct mass of Ca2+ or concentration of Ca2+ in mol L–1

1 mark − Calculates the correct concentration of Ca2+ in ppm

Note: The question has a mistake where the oxalate ion should actually be C2O42– (not
C2O22–). The solution above is what would be calculated using the incorrect chemical
equation given in the question.

−30−
(b) For comparison, the student also used atomic absorption spectroscopy (AAS) to determine 3
the concentration of calcium ions in the sea water sample.

Predict how the calcium ion concentration obtained by AAS would compare with the value
calculated in part (a), and explain which result is more accurate.

The [Ca2+] obtained from AAS will be lower and more accurate than the value obtained
from gravimetric analysis in part (a). This is because the added C2O22– ions would have
precipitated with other metal ions in sea water such as Mg2+ ions which would lead to
a larger mass of precipitate being collected and an overestimation of the [Ca2+]. AAS,
however, is very specific and would have only detected Ca2+ ions since it would have
used a Ca lamp to emit specific wavelengths of light that are only absorbed by calcium
(and not magnesium).

1 mark − Predicts that the concentration obtained by AAS will be lower

1 mark − Explains that the concentration calculated from gravimetric analysis would
be overestimated due to interference from Mg2+ ions

1 mark − Explains that the concentration calculated from AAS is more accurate since
AAS is more specific and would not have interference issues with Mg2+ ions

−31−
Question 30 (5 marks)
Fertiliser contains nitrogen in the form of ammonium sulfate. The following procedure was used 5
to determine the nitrogen content of a lawn fertiliser.

• A 1.758 g sample of the fertiliser was dissolved in 250.0 mL of distilled water.

• A 20.0 mL aliquot of the fertiliser solution was mixed with 50.0 mL of 0.160 mol L–1
sodium hydroxide. The reaction mixture was heated and all of the ammonium sulfate was
neutralised according to the following equation:
(NH4)2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 NH3(g) + 2 H2O(l)
• The reaction mixture was heated for a further 10 minutes to remove any residual ammonia.
• The excess sodium hydroxide in the final mixture was then titrated against a 0.120 mol L–1
oxalic acid primary standard solution to the phenolphthalein end point.
• The procedure was repeated three times. The results of the titration and the structural
formula of oxalic acid are given below.

Titration Volume of oxalic acid (mL)

1 26.90 O O
2 26.50 C C
3 26.45 H O O H
4 26.55 Oxalic acid

Calculate the percentage by mass of nitrogen in the lawn fertiliser.

Oxalic acid is a diprotic acid since it contains two carboxyl (−COOH) groups.

H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l)

Discarding the 1st rough titre,

(0.02650 + 0.02645 + 0.02655) L

V (H2C2O4) =
3
= 0.02650 L
n(H2C2O4) = 0.120 mol L–1 × 0.02650 L
= 3.18 × 10−3 mol
n(NaOH) leftover = 2 × 3.18 × 10−3 mol
= 6.36 × 10−3 mol
n(NaOH) added = 0.160 mol L–1 × 0.0500 L
= 8.00 × 10−3 mol
n(NaOH) reacted = n(NaOH) added − n(NaOH) leftover
= 8.00 × 10−3 mol − 6.36 × 10−3 mol
= 1.64 × 10−3 mol
1
n((NH4)2SO4) reacted = × 1.64 × 10−3 mol
2
= 8.20 × 10−4 mol

−32−
These moles of (NH4)2SO4 came from a 20.0 mL aliquot of the fertiliser solution. Therefore,
the moles of (NH4)2SO4 in 250.0 mL of the fertiliser solution is given by:

250.0 mL
n((NH4)2SO4) = × 8.20 × 10−4 mol
20.0 mL
= 0.0103 mol

Each mole of (NH4)2SO4 contains 2 moles of N.

=⇒ n(N) = 2 × 0.0103 mol

= 0.0205 mol
m(N) = 0.0205 mol × 14.01 g mol–1
= 0.287 g
0.287 g
%N = × 100%
1.758 g
= 16.3%

1 mark − Calculates the correct moles of H2C2O4

1 mark − Calculates the correct moles of NaOH added or leftover

1 mark − Calculates the correct moles of (NH4)2SO4 reacted

1 mark − Calculates the correct moles of (NH4)2SO4 in the 250.0 mL fertiliser solution or
equivalent merit

1 mark − Calculates the correct percentage by mass of N to THREE significant figures

−33−
 Mod 7 – Question 9 (7 marks)
Question 31 (8 marks)
TheThis
flowchart below
flow chart shows
shows reactions
reactions involving
involving sixdifferent
different organic
organic compounds (AtotoF).
compounds (A H). 7 8

A KMnO4 ∕ H2SO4
C5H12O No reaction

Concentrated
H2SO4

B C

H2O/dilute H2SO4 H2O/dilute H2SO4

D A E
C5H12O

KMnO4 ∕ H2SO4 KMnO4 ∕ H2SO4

F G

Na2CO3 Na2CO3

CO2 + H2O H
No reaction

Draw the structural formula of compounds A to H in the corresponding space provided.

A H B
H
H C H
H C H
H H H
H H H
H C C C C H
H C C C C H
H H O H
H H
H – 54 –

−34−
 C D H
H
H C H
H C H
H H H
H H H
H C C C C H
H C C C C H
H O H H
H H
H

E F
H H

H C H H C H
H H H H H

H C C C C O H H C C C C H

H H H H H O H H

G H
H H

H C H H C H
H H O H H O

H C C C C O H H C C C C O− Na+

H H H H H H

8 marks − Draws the correct structural formula of compounds A to H (1 mark each)

Note: H must contain an ionic bond (NOT a covalent bond) between the ions. If incorrect
structures are provided, partial marks may still be awarded if a good understanding of the
organic reactions has been demonstrated.

Explanations are given below:

• A is a tertiary alcohol since it is not oxidised by MnO4–/H+. There is only one tertiary
alcohol with a molecular formula of C5H12O that can be drawn.

• Alcohol A is dehydrated with concentrated H2SO4 into two alkenes, B and C.

• Alkenes B and C are hydrated with H2O/dilute H2SO4 into alcohols D, A and E.

−35−
• E must be a primary alcohol since it is oxidised by MnO4–/H+ to carboxylic acid G,
and this is confirmed by the neutralisation reaction between G and Na2CO3 which
produces salt H, CO2 and H2O.

• D must be a secondary alcohol which is oxidised by by MnO4–/H+ to ketone F which

does not react with Na2CO3.

• The structures of alkenes B and C can then be deduced from alcohols D and E
respectively.

−36−
Question 32 (4 marks)
The following procedure was used to determine the solubility product (Ksp ) of silver sulfate. 4
• 25.0 mL of 1.0 mol L–1 silver nitrate was mixed with 75.0 mL of 1.0 mol L–1 sodium sulfate
and a precipitate formed.
• The precipitate was filtered off and a 10.0 mL aliquot of the filtrate was diluted to 100.0 mL
with distilled water in a volumetric flask.
• The diluted filtrate was analysed with atomic absorption spectroscopy (AAS) for its silver
ion content and it was found to have an absorbance of 0.60.
• A series of standard solutionsCalibration
was used tocurve
construct
for Agthe
+ calibration curve below.

0.90

0.80

0.70

0.60
Absorbance

0.50

0.40

0.30

0.20

0.10

0.00
0 10 20 30 40 50 60 70 80
+
Concentration of Ag (ppm)

Calculate the value of Ksp for silver sulfate.

Interpolating from the calibration curve, when A = 0.60, [Ag+] = 52 ppm = 52 mg L–1. The
saturated filtrate was diluted 10-fold prior to the AAS analysis.

∴ [Ag+]eq = 10 × 52 mg L–1
= 520 mg L–1
= 0.52 g L–1
0.52 g L–1
=
107.9 g mol–1
= 4.8 × 10−3 mol L–1
n(Ag+)i = 1.0 mol L–1 × 0.0250 L
= 0.025 mol
0.025 mol
[Ag+]i =
(0.0250 + 0.0750) L
= 0.25 mol L–1

−37−
 n(SO42–)i = 1.0 mol L–1 × 0.0750 L
= 0.075 mol
0.0750 mol
[SO42–]i =
(0.0250 + 0.0750) L
= 0.75 mol L–1

Ag2SO4(s) 2 Ag+(aq) + SO42–(aq)

Ag+ SO42–
Initial 0.25 0.75
Change −0.2452 −0.1226
Equilibrium 4.8 × 10−3 0.6274

2
Ksp = [Ag+] [SO42–]
= (4.8 × 10−3 )2 × 0.6274
= 1.5 × 10−5

1 mark − Correctly interpolates the Ag+ concentration from the graph

1 mark − Determines the correct equilibrium concentration of Ag+ in mol L–1

1 mark − Calculates the correct equilibrium concentration of SO42– or equivalent merit

1 mark − Calculates the correct value of Ksp

After determining the equilibrium concentration of Ag+ ions, answers may also include:

2 AgNO3(aq) + Na2SO4(aq) → Ag2SO4(s) + 2 NaNO3(aq)

n(AgNO3)i = 1.0 mol L–1 × 0.0250 L

= 0.025 mol
n(Na2SO4)i = 1.0 mol L–1 × 0.0750 L
= 0.075 mol

AgNO3 is the limiting reagent while Na2SO4 is in excess.

1
n(Na2SO4)leftover = 0.075 mol − × 0.025 mol
2
= 0.0625 mol
n(SO42–)leftover = 0.0625 mol
0.0625 mol
[SO42–]leftover =
(0.0250 + 0.0750) L
= 0.625 mol L–1

−38−
 Ag2SO4(s) 2 Ag+(aq) + SO42–(aq)

Ag+ SO42–
Initial 0 0.625
Change +4.8 × 10−3 +2.4 × 10−3
Equilibrium 4.8 × 10−3 0.6274

2
Ksp = [Ag+] [SO42–]
= (4.8 × 10−3 )2 × 0.6274
= 1.5 × 10−5

−39−