99AL-I
1.
a)
i)
P.1
PV = nRT
PV = (m/M)RT
M = mRT/PV
ii)
(1)
=
(0.569)(8.31)(433)
(80.0 x10 3 )(96.0 x10 6 )
(1)
=
266.59 g mol-1 (accept 266.6 and 267 g mol-1)
(1)
Assume the formula of aluminium chloride is (AlCl3)n
(26.96 + 3 x 35.45) x n = 266.59
133.31 x n = 266.59
n = 2 (0.5)
Thus, aluminium chloride exists as dimer and the molecular formula is Al2Cl6. (0.5)
The structure is
b)
i)
Cl
Cl
Al
Cl
Al
Cl
Cl
Cl
(1)
Carbon dioxide exists as simple molecules and the intermolecular interaction is weak
Van der Waals' force. Therefore, it is a gas at this conditions.
(0.5)
However, silicon dioxide has a giant covalent structure.
(0.5)
The covalent bond between Si and O in silicon dioxide is much stronger than the
ii)
weak Van der Waals' force in CO2 molecules. Therefore, it is a solid.
(0.5)
The solubility of an ionic compound depends on Hsoln = Hhyd - Hlattice
(0.5)
OH- ion is relatively small compared with Ba2+ and Mg2+.
(0.5)
2+
2+
The size of Ba is larger than that of Mg and the size of Ba(OH)2 is larger than that
of Mg(OH)2. Hence, less energy is required to break the Ba(OH)2 lattice.
However, the change in hydration energy is relatively insignificant.
As a results, the difference in Hlattice between Mg(OH)2 and Ba(OH)2 is much greater
than the difference in Hhyd between Mg2+ and Ba2+ ions.
(0.5)
Therefore, barium hydroxide is more soluble.
iii)
The intermolecular interaction between H2O molecules in ice and water are both
hydrogen bond.
(0.5)
In ice, each H2O molecule forms hydrogen bond tetrahedrally with four others H2O
molecules.
(0.5)
This gives an open structure. which make the density of ice lower.
(0.5)
In water, such open structure does not exist and the H2O molecules are packed
together.
Therefore, the density of water is higher than that of ice.
(0.5)
99AL-I
2.
a)
i)
Energy of electron is quantized.
P.2
(0.5)
When energy is absorbed, an electron will be excited and be promoted from a lower
energy level to a higher energy level.
(0.5)
When the electron falls from the higher energy level to a lower energy level,
(0.5)
energy will be emitted and the amount equals to the difference in energy between the
two energy levels.
ii)
(0.5)
The emission lines converges at the lower wavelength end because the energy
difference between each succeeding energy level decreases with increase in principle
quantum number.
b)
c)
(1)
4
Rn 
 218
(or )
84 Po 2 He
i)
222
86
ii)
Rn is a noble gas.
(0.5)
It can be inhaled into the respiratory system.
(0.5)
The isotope has a very short half life
(0.5)
and it will decay to give harmful  particle.
(0.5)
i)
(1)
There are a thin film of water which forms a stationary phase on the chromatography
paper.
(0.5)
The solvent/eluent which acts as a mobile phase moves up the chromatography paper
by capillary action.
The amino acids distribute between the stationary and the mobile phases.
(1)
As the three amino acids have different distribution coefficient, they will move up at
different rate.
(0.5)
i.e. A has the highest distribution coefficient between mobile and stationary phases,
and thus it moves the fastest.
ii)
The chromatography paper can be sprayed with ninhydrin solution which reacts with
amino acids to give purple compounds.
(1)
(or use iodine vapour to detect the presence of amino acid)
iii)
3.
a)
Rf =
5 .6
 0.8
7.0
(accept 0.77 to 0.83)
The electronic configuration of Fe3+(g) is [Ar]3d5
(1)
(0.5)
According to Hund's Rule electrons must occupy each energy level singly before electron
pairing takes place,
3+
Fe should have 5 unpaired electrons.
(0.5)
(1)
99AL-I
b)
P.3
i)
2MnO4-(aq) + 5C2O42-(aq) + 16H+(aq)  2Mn2+(aq) + 10CO2(g) + 8H2O(l)
ii)
As the products are colourless, the decolourization occurs when the reaction takes
(1)
place.
Mn2+ which is one of the product acts as an autocatalyt.
(1)
At the beginning of the reaction, the concentration of Mn2+ is low and the rate of the
reaction is slow.
(0.5)
When the concentration of Mn2+ increases gradually, the rate of the reaction becomes
faster.
(0.5)
c)
Cl
H3N
NH3
Co
H3N
Cl
NH3
Cl
trans-dichlorotetraaminecobalt(III) chloride
Cl
H3N
Cl
NH3
Cl
cis-dichlorotetraaminecobalt(III) chloride
d)
i)
ii)
4.
(0.5)
NH3
Co
H3N
(0.5)
(0.5)
(0.5)
A colourless solution turns brown.
(1)
IO3-(aq) + 6H+ (aq) + 5I-(aq)  3I2(aq) + 3H2O(l)
(1)
Brown gas evolved with choking smell. (1)
2KBr(s) + H2SO4(l)  K2SO4(s) + 2HBr(g)
(0.5)
2HBr(g) + H2SO4(l)  SO2(g) + Br2(g) + 2H2O(l)
(0.5)
a)
H2O and NH4+
b)
i)
(0.5, 0.5)
99AL-I
ii)
P.4
Thymolphthalein is chosen.
(0.5)
It is because its pH range of colour change falls into the steepest region of the pH
titration curve.
c)
(1)
A known volume sample is titrated with standard HCl.
Phenolphthalein is used as the indicator to show the first end-point at the volume of HCl
used = V1.
(0.5)
Na2CO3 + HCl  NaHCO3 + NaCl
(0.5)
The titration is continued with methyl orange as the indicator to show the second end-point.
d)
The volume of HCl used is recorded as V2.
(0.5)
NaHCO3 + HCl  H2CO3 + NaCl
(0.5)
Concentration of Na2CO3 =
vol. of sample
x conc. of HCl
V1
(0.5)
Concentration of NaHCO3 =
vol. of sample
x conc.of HCl
V2  V1
(0.5)
pH = 2
-log[H+] = 2
[H+] = 0.01M
(0.5)
3
No. of mole of HCl required to prepare 1.00dm solution
=
0.01 x 1.00 = 0.01mole
(1)
The mass of constant boiling hydrochloric acid required
5.
a)
=
0.01 x (1.008  35.45)
0.202
=
1.80g
(0.5)
i)
The reaction is carried out under diffused light
(0.5)
and methane must be in excess.
(0.5)
ii)
(1)
Initiation reaction:
Cl
2 Cl .
Cl
Chain propagation reaction:
Cl . +
H CH3
Cl
+
Cl
. CH3
HCl +
. CH3
CH3Cl + Cl .
99AL-I
P.5
Termination reaction:
Cl . +
. CH3
CH3 +
iii)
CH3Cl
. CH3
CH3CH3
No, the reaction of methane with chlorine is not an appropriate method for the
preparation of dichloromethane. It is because a mixture of CH3Cl, CH2Cl2, CHCl3
and CCl4 will be formed.
5.
b)
i)
(1)
OH
D:
C
H
CN
(1)
E:
OH
C
H
ii)
CH2NH2
(1)
The type of reaction involved in the formation of D in called
Nucleophilic addition reaction.
(1)
Mechanism:
O-
(0.5) O
C
C
H
H
CN
(0.5)
O (0.5)
OH
C
H
CN
(0.5)
-:CN
C
+
H
+
H
(0.5)
+ H
O
C
H
C
H
(0.5)
(0.5)
(1.5 M for 3 arrows; 0.5M for intermediate)
OH
-:CN
H
CN
99AL-I
6.
P.6
H
a)
H
H3 N+CH2CONH C CO 2
Ar
H3 N+ C CONHCH CO +
2
2
Ar
H
H
H3 N+CH2CONH C Ar
CO2-
b)
i)
Ar C CONHCH CO +
2
2
H3 N+
(0.5 x 4)
The car exhaust contains carbon monoxide because of the incomplete combustion of
petrol which is a mixturen of hydrocarbons.
(1)
It contains nitrogen oxides because N2 and O2 are combined at high temperature in
the engine to form nitrogen monoxide.
(
N2(g) + O2(g)  2NO(g)
)
(0.5)
Besides, air oxidation of NO occurs.
(
ii)
2NO(g) + O2(g)  2NO2(g)
)
(0.5)
(0.5)
2NO(g) + 2CO(g)  N2(g) + 2CO2(g)
(0.5)
Oxygen in air is introduced to the converter and acts as an oxidant.
(0.5)
Pt/Pd catalyses the reactions
(0.5)
2CO(g) + O2(g)  2CO2(g)
C7H16(g) + 11O2(g)
 7CO2(g) + 8H2O(g)
(0.5)
(0.5)
Leaded petrol is not used in cars equipped with catalytic converters because lead
compounds can poison the Pt/Pd catalysts.
iv)
(0.5)
The catalytic converter contains two kinds of catalyst which is Rh and Pt/Pd.
Rh catalyses the reaction
iii)
(0.5)
(1)
Nitrogen dioxide breaks down to give NO and O atom upon the absorption of UV
radiation / sunlight.
uv
NO2(g)  NO(g) + O (g)
(0.5)
(0.5)
O atom reacts with O2 in air to form O3 (0.5)
O(g) + O2(g)  O3(g)
7.
a)
(0.5)
Any two of the following:
(2)
Lower the temperature of the sample of reaction mixture by immersing it in ice
(1)
Dilute the sample of reaction with water or appropriate solvent
(1)
Remove one of the reactants or catalyst by adding suitable quenching agent
(1)
99AL-I
b)
c)
P.7
Any 3 of the following:
(3)
heat loss to the surroundings
(1)
specific heat capacity of the reaction mixture does not equal to that of water
(1)
heat may be absorbed by the polystyrene foam cup and the thermometer
(1)
the density of the solution does not equal to that of water
(1)
error in the scale of the thermometer
(1)
error in the mass/volume of the acid or alkali used
(1)
i)
(1)
A platinum wire is cleaned.
The wire is dipped into concentrated HCl and then into the powder of the sample of
salt.
(1)
The wire with sample is heated in non-luminous bunsen flame to observe the colour.
(1)
ii)
d)
Sodium salt gives a golden yellow flame.
(0.5)
Potassium salt gives a lilac flame.
(0.5)
The sample is heated first.
(0.5)
The water vapour given off will turn the anhydrous cobalt chloride paper from blue to pink.
(0.5)
(The water vapour will turn the anhydrous CuSO4 from white to blue.)
8.
a)
i)
ii)
(0.5)
Any two of the following:
(2)
Remove oxygen/air
(1)
Remove fuels
(1)
Lower the temperature
(1)
I)
Use foam/ carbon dioxide/BCF/BTM type extinguisher
(1)
(cover the flask with a fire blanket)
(1)
Use powder type extinguisher
(1)
(use a fire blanket)
(1)
II)
99AL-I
b)
P.8
i)
water in
water-out
mixture of C4H9OH, KBr and conc. H2SO4
anti-bumping granules
(1M for correct diagram of the set-up; 0.5M for indicating the direction of water flow in condenser;
0.5M for labelling the anti-bumping granules.)
ii)
The reaction mixture is cleared into a separating funnel.
Water is added into the reaction mixture.
(0.5)
Ether is then added
(0.5)
and the separating funnel is shaked.
(0.5)
NaOH(aq)/NaHCO3(aq) is added to neutralize the acid.
(0.5)
The aqueous layer is removed.
(0.5)
Anhydrous Na2SO4 is used to dry the organic layer.
(0.5)
Na2SO4 is filtered off.
(0.5)
The organic layer is distillated and the distillate is collected at the boiling point of 1bromobutane.
(0.5)