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YEE 6301 SMK MERADONG Kecergasan Untuk Kemajuan
1
NAME: CLASS: PRA U S
DIAGNOSTIC TEST [AUGUST 2015]
PAPER CHEMISTRY 1
CODE 962/1
COHORT STPM 2016
DURATION 1 hour 30 minutes
SUBJECT TEACHER MS UNG HIE HUONG
INSTRUCTIONTO CANDIDATES:
This paper consists of Section A, Section B and Section C. Answer ALL questions.
Arrange and stapler your answers in numerical order.
SECTION A(15 marks)
Answer ALL the questions in Section A.
Blacken the corresponding answer on the objective answer sheet provided.
1. The compound PCl3 is formed from 31
P, 35
Cl and 37
Cl isotopes. The relative abundance of 35
Cl
to 37
Cl is 3:1. Which statement about the mass spectrum of PCl3 is true?
A The base peak corresponds to P
+
ion.
B The
𝑚
𝑒
value for the last peak is 142.
C The number of peaks for PCl3
+
ion is 5.
D The relative abundance of P35
Cl3
+
ion to P37
Cl3
+
ion is 3:1.
2. Which statement is true of the nuclides of silicon, 𝑆𝑖14
28
and 𝑆𝑖14
30
?
A The mass spectrum shows two peaks only.
B Isotope 𝑆𝑖14
28
is more reactive than isotope 𝑆𝑖14
30
.
C The tetrachlorides, 𝑆𝑖𝐶𝑙414
28
and 𝑆𝑖14
30
𝐶𝑙4 have different shape.
D 𝑆𝑖𝑂214
28
dissolves in hot concentrated solution of sodium hydroxide, 𝑆𝑖𝑂214
30
does not.
3. The radioisotope tritium, 𝑇1
3
, slowly turns into a helium isotope, 𝐻𝑒2
3
.
Which statement is true of the two isotopes?
A 𝑇1
3
and 𝐻𝑒2
3
have the same nuclear charge.
B 𝑇1
3
and 𝐻𝑒2
3
have the same number of neutral sub-atomic particles.
C 𝑇+
1
3
and 𝐻𝑒+
2
3
have the same number of charged sub-atomic particles.
D 𝑇+
1
3
and 𝐻𝑒+
2
3
are deflected to the same point in a magnetic field whose strength is not
varied.
4. 100 cm3
of liquid A and 100 cm3
of liquid B are allowed to evaporate. It is found that A
evaporates first before B does. Which statement explains this observation?
A B is more volatile than A.
B The density of A is higher than that of B.
C The vapour pressure of A is lower than that of B.
D The intermolecular forces of A is weaker than that of B.

2
5. Which hydride has the highest boiling point?
A HCl B HF C H2O D NH3
6. Gallium is a soft, silvery metal. The element gallium exists naturally as isotopes
x
Ga and
y
Ga
which have 38 and 40 neutrons respectively. What is the percentage abundance of
x
Ga?
A 35.0 B 45.5 C 54.5 D 65.0
7. A phase diagram of water is shown in the margin.
What can be deduced from the phase diagram?
A Ice sublimes at a pressure higher than
611 Pa.
B An increase in pressure will decrease the
boiling point of water.
C An increase in pressure will decrease the
freezing point of water.
D Water exists as liquid at a pressure of
611 Pa and a temperature of 298K.
Pressure/ Pa
Temperature/ K
A
O
BC
611
273.16
8. Electronic transition between energy levels in an atom will cause an absorption or emission of
light. Which energy level diagram shows the transition of electrons that emits light with the
shortest wavelength?
A Energy
n = 5
n = 4
n = 3
n = 2
n = 1
B Energy
n = 5
n = 4
n = 3
n = 2
n = 1
C Energy
n = 5
n = 4
n = 3
n = 2
n = 1
D Energy
n = 5
n = 4
n = 3
n = 2
n = 1
9. Coordinate bond does not exist in
A BeCl2 B CO C PH4
+ D [Fe(CN)6]3–

3
10. Polyatomic molecules and ions have varied geometries. Which species and geometry
correspond correctly?
Species Geometry
A NH2
–
Linear
B H3O+
Trigonal planar
C SiCl4 Square planar
D ICl3 T-shaped
11. Based on the band theory, the conductivity of metal is due to delocalized electrons in the
conduction band. How many electrons are found in the conduction band of a 10.0 g
magnesium strip?
[Avogadro’s constant is 6.02 × 1023
mol–1
]
A 2.48 × 1023 B 4.95 × 1023 C 2.48 × 1024 D 3.01 × 1024
12. An atom of element Z has nucleon number 55 and 30 fundamental uncharged particles in its
nucleus. What is the electronic configuration of a Z2+
ion?
A 1𝑠22𝑠22𝑝63𝑠23𝑝63𝑑5 B 1𝑠22𝑠22𝑝63𝑠23𝑝63𝑑34𝑠2
C 1𝑠22𝑠22𝑝63𝑠23𝑝63𝑑44𝑠1 D 1𝑠22𝑠22𝑝63𝑠23𝑝63𝑑54𝑠2
13. The boiling points of CH3OH and CH3SH are 64.5°C and 5.8°C respectively. What is the cause
of the difference in the boiling points?
A The O―H bond is stronger than the S―H bond.
B CH3SH molecule is bigger than CH3OH molecule.
C Hydrogen bonds exist between CH3OH molecules.
D The electronegativity of oxygen is higher than that of Sulphur.
14. The unit cell of an oxide of X is shown below.
= X
= O
The formula of the oxide of X is
A 𝑋2 𝑂4 B 𝑋𝑂3 C 𝑋3 𝑂4 D 𝑋2 𝑂5
15. Compounds have either ionic or covalent bonds or both. Which statement is not true of ionic
bonds?
A They involve the transfer of one or more electrons from s, p or d orbitals.
B The strength of ionic bonds is proportional to the size of the ions.
C They involve ions with stable electronic configurations.
D They result in the formation of solid compounds.

4
SECTION B (15 marks)
Answer ALL the questions in Section B.
Write your answers in the spaces provided.
16. (a) Hydrogen cyanide, HCN is a colourless and poisonous gas.
(i) Draw the Lewis diagram for the hydrogen cyanide molecule and predict its shape.
Lewis diagram: [2 marks]
Shape:
(ii) State the type of hybridization the carbon atom undergo in the molecule. [1 mark]
(iii) Draw a labelled diagram to show the overlapping of the bonding orbitals in hydrogen
cyanide. [2 marks]
(b) A sample of carbon with relative atomic mass 12.01112 consists of two isotopes 12
C
and 13
C. Calculate the percentage abundance of 12
C and 13
C isotopes to two decimal
places. [3 marks]

5
17. A real gas X behaves almost like an ideal gas. For n mol of gas X at pressure P, the graph of
volume V versus temperature T is shown below.
60 -
50 -
40 -
30 -
20 -
10 -
0 -
-300 -200 -100 0 100 200
Volume V/ cm3
Temperature T/ °C
(a) What is the most probable identity of gas X? Explain your answer. [3 marks]
Gas X:
Explanation:
(b) On the graph above, sketch and label a graph of variation in volume with temperature at a
lower pressure P’ while other conditions remain constant. [1 mark]
(c) At pressure P and temperature 0°C, a gas Y shows a negative deviation from an ideal
gas. Mark the expected volume of gas Y on the above graph.
Explain your answer. [3 marks]
Explanation:

6
SECTION C (30 marks)
Answer ALL the questions in Section C.
Write your answers in the additional answer sheets on page 7-9.
18. (a) The table below lists the temperature and pressure for the critical point and the triple point
of carbon dioxide.
Temperature/ °C Pressure/ atm
Critical point 31 73
Triple point –57 5
Carbon dioxide sublimes at –78°C under atmosphere pressure. The freezing point of
carbon dioxide increases by 2°C for every increase of 10 atm in pressure.
(i) Based on the information given above, sketch the phase diagram of carbon dioxide.
[5 marks]
(ii) Calculate the freezing point of CO2, in °C, under a pressure of 75 atm. [2 marks]
(iii) Explain why the freezing point of CO2 increases with pressure. [1 mark]
(iv) Solid carbon dioxide is known as dry ice. How can liquid CO2 be obtained from dry
ice? [1 mark]
(v) Dry ice has the advantage of being relatively cheap and non-toxic. It is commonly
used in concerts to form fog. Explain the formation of the fog. [3 marks]
(b) The boiling point of hydrogen halides are given in the following table:
Hydrogen halide HF HCl HBr HI
Boiling point/ K 293 188 206 238
Explain the variation in boiling points of the hydrogen halides. [3 marks]
19. (a) (i) Ammonium nitrate is an explosive compound and it decomposes at a high
temperature according to the following equation:
2NH4NO3(s) → 4H2O(g) + 2N2(g) + O2(g)
Calculate the total volume of gases collected from the decomposition of 100 g of
ammonium nitrate at 1.01 × 105
Pa and 25 °C.
[Gas constant, R = 8.31 J g–1
°C–1
] [3 marks]
(ii) Sketch a graph of
𝑃𝑉
𝑅𝑇
against p for 1.0 mol of ammonia gas at 0°C.
Based on the graph, explain the negative deviation of ammonia gas compared to an
ideal gas. [5 marks]
(b) (i) With the aid of a diagram, describe the structures of diamond and graphite. [5 marks]
(ii) What is the relationship between diamond and graphite in term of structure? Explain
your answer. [2 marks]

7
ANSWER SHEET
SECTION C Examiner’s
use onlyQuestion Answers
18

8

9

10
Periodic Table (Jadual Berkala)
Group (Kumpulan)
1
(I)
2
(II)
3 4 5 6 7 8 9 10 11 12 13
(III)
14
(IV)
15
(V)
16
(VI)
17
(VII)
18
(VIII)
1.0
H
1
4.0
He
2
6.9
Li
3
9.0
Be
4
a
X
b
a = relative atomic mass (jisim atom relatif)
X = atomic symbol (symbol atom)
b = atomic number (nombor atom)
10.8
B
5
12.0
C
6
14.0
N
7
16.0
O
8
19.0
F
9
20.2
Ne
10
23.0
Na
11
24.3
Mg
12
27.0
Al
13
28.1
Si
14
31.0
P
15
32.1
S
16
35.5
Cl
17
40.0
Ar
18
39.1
K
19
40.1
Ca
20
45.0
Sc
21
47.9
Ti
22
50.9
V
23
52.0
Cr
24
54.9
Mn
25
55.8
Fe
26
58.9
Co
27
58.7
Ni
28
63.5
Cu
29
65.4
Zn
30
69.7
Ga
31
72.6
Ge
32
74.9
As
33
79.0
Se
34
79.9
Br
35
83.8
Kr
36
85.5
Rb
37
87.6
Sr
38
88.9
Y
39
91.2
Zr
40
92.9
Nb
41
95.9
Mo
42
[98]
Tc
43
101
Ru
44
103
Rh
45
106
Pd
46
108
Ag
47
112
Cd
48
115
In
49
119
Sn
50
122
Sb
51
128
Te
52
127
I
53
131
Xe
54
133
Cs
55
137
Ba
56
139
La
57
178
Hf
72
181
Ta
73
184
W
74
186
Re
75
190
Os
76
192
Ir
77
195
Pt
78
197
Au
79
201
Hg
80
204
Tl
81
207
Pb
82
209
Bi
83
[209]
Po
84
[210]
At
85
[222]
Rn
86
[223]
Fr
87
[226]
Ra
88
227
Ac
89
[261]
Rf
104
[262]
Db
105
[266]
Sg
106
[264]
Bh
107
[269]
Hs
108
[268]
Mt
109
[281]
Ds
110
[272]
Rg
111
[285]
Cn
112
140
Ce
58
141
Pr
59
144
Nd
60
[145]
Pm
61
150
Sm
62
152
Eu
63
157
Gd
64
159
Tb
65
163
Dy
66
165
Ho
67
167
Er
68
169
Tm
69
173
Yb
70
175
Lu
71
232
Th
90
231
Pa
91
238
U
92
237
Np
93
[244]
Pu
94
[243]
Am
95
[247]
Cm
96
[247]
Bk
97
[251]
Cf
98
[252]
Es
99
[257]
Fm
100
[258]
Md
101
[259]
No
102
[262]
Lr
103
 The proton numbers and approximate relative atomic masses shown in the table are for use in the examination unless stated otherwise in an individual question.
 (Nombor proton dan anggaran jisim atom relatif yang ditunjukkan dalamjadual adalah untuk digunakan dalampeperiksaan kecuali yang sebaliknya dinyatakan dalamsoalan
yang tertentu.)

11
MARKING SCHEME
DIAGNOSTIC TEST [AUGUST 2015]
Q RUBRIC M
SECTION A [15 marks]
1 B 1
2 A 1
3 D 1
4 D 1
5 C 1
6 D 1
7 C 1
8 B 1
9 A 1
10 D 1
11 B 1
12 A 1
13 C 1
14 A 1
15 B 1
16 a(i)
H C NX
X
X
X
Linear
1
1
a(ii) sp 1
a(iii)  Orbitals
 Labels (Orbitals and bonds)
NOTE:
1
1
b Let the % abundance of 12
C = x% ; % abundance of 13
C = (100 – x)%
12𝑥 + 13(100−𝑥)
100
= 12.01112
𝑥 = 98.89%
𝑦 = 1.11%
1
1
1
H C Nspsp spsp
1s
py
py
py
py
pz
pz
pz
pz
bond
bond
 bond
 bond

12
Q RUBRIC M
17 a Hydrogen gas// H2 // Helium// He
Non-polar, small H2 molecule/ He atom/ gas particle has …
 negligible volume/ size of particle relative to volume of container.
 Weak, negligible intermolecular van der Waals forces.
1
1
1
b
c
60 -
50 -
40 -
30 -
20 -
10 -
0 -
-300 -200 -100 0 100 200
Volume V/ cm3
Temperature T/ °C
P
P'
(b)
(c)
B1
C1
c Negative deviation at lower pressure is due to:
 Stronger intermolecular forces between gas particles.
 Gas particles are drawn closer to each other, volume occupied by gas is smaller
than expected/ smaller than ideal gas.
1
1
18 a(i)  Drawn and labelled axes + boiling curve + sublimation curve
 Correct melting line (straight line with positive gradient)
 Mark and state coordinates for  Triple point = (-57°C, 5 atm)
 Critical point = (31°C, 73 atm)
 Normal sublimation point = (-78°C, 1 atm)
Temperature/ °C
Pressure/ atm
31-57
5
73
-78
1
Solid Liquid
Gas
NOTE:
1. If axes are not labelled  0 mark
2. Curves must have positive gradient (upwards from left to right)
3. The three phase transition lines/ curves must meet at triple point
1
1
1
1
1

13
Q RUBRIC M
18 a(ii) Under a pressure of 75 atm,
Freezing point = −57 +
(75−5)
10
× 2
= −57 + 14
= −43℃
1
1
a(iii) As external pressure increases, the melting point/ freezing point of CO2 increases.
REASON (Either one):
 The density of solid CO2 (dry ice) is higher than the density of liquid CO2.
 When solid CO2 (dry ice) melts, its volume expands.
NOTE: Accept reverse argument
1
a(iv) By increasing the temperature of dry ice to above –57°C and increasing the pressure
to above 5 atm.
1
a(v) Dry ice sublimes to form CO2 gas.
The sublimation process absorbs heat from surrounding/ is endothermic, thereby
lowering surrounding temperature.
Moisture/ water vapour in air condense into fine water droplets or “fog”.
1
1
1
(b)  Boiling point of HF is exceptionally high due to strong hydrogen bonding exist
between HF molecules
 Weak intermolecular van der Waals forces exist in HCl, HBr and HI.
 Molecular size/ mass increases from HCl to HI, therefore the strength of
intermolecular van der Waals forces increases from HCl to HI.
1
1
1
19 a(i)
Number of moles of gases formed =
3
2
×
100
80.0
= 1.875 mole
𝑝𝑉 = 𝑛𝑅𝑇 @ 𝑉 =
𝑛𝑅𝑇
𝑝
Volume of gases formed, 𝑉 =
(1.875 ) (8.31) (273+25)
1.01 ×105
m3
= 0.046 m3
1
1
1
a(ii)
 Ideal gas has no particle/ molecular volume and no intermolecular (attractive or
repulsive) forces.
 Ammonia is a non-ideal gas @ real gas because it is a polar molecule. It has a
large molecular size/ volume and strong intermolecular attractive forces (hydrogen
bonds) between the molecules.
 The attractive force causes molecules to be nearer to each other, causing the
measured volume to be smaller than expected.
 The attractive forces also lowers the speed of the molecules colliding with the
walls of the vessel. Therefore, the measured pressure is lower than expected.
Both factors cause
𝑝𝑉
𝑅𝑇
< 1, giving negative deviation.
1
1
1
1
1

14
Q RUBRIC M
19 b(i)  Structures:
 Both diamond and graphite have giant 3-dimensional covalent structures.
In diamond,
 each cabon atom undergoes sp3
hybridisation and is covalently bonded to four
other carbon atoms in a tetrahedral manner.
In graphite,
 each cabon atom undergoes sp2
hybridisation and is covalently bonded to three
other carbon atoms in a trigonal planar shape.
 This forms hexagonal rings in a layered structure. The layers are held together by
weak van der Waals forces.
NOTE: Maximum 5 marks
1
1
1
1
1
1
b(ii) Both are allotropes of carbon.
The physical properties are different due to different arrangement of carbon atoms.
1
1

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