ANSWER SCHEME
PSPM 2019/2020
No. Answer
1. (a) i.
Proton 3759 8315
Electron 35 35
Neutron
35 35
44 46
Isotopes are atoms of the same element with the same protons number but different number of
neutrons.
ii. 0.443
2 = 18 −1
−
No of electrons = 36 = 0.0246
(b) Mass of substance = 1.0 g
CO2 = 2.52g
H2O = 0.443g
2.52
2 = 44 −1
= 0.0573
1 2 ≡ 1 1 2 ≡ 2
0.0573 2 ≡ 0.0573 0.0246 2 ≡ 0.0492
= 0.0573 12 −1 = 0.0492 1 −1
= 0.06872 = 0.0492
= 1.0 − (0.6872 + 0.0492 )
= 0.2636
Element C H O
Mass (g) = 0.0492
Mole (mol) = 0.6872 0.0492 = 0.2636
0.6872 0.2636
Simplest ratio = 1 −1
= 12 −1 = 0.0492 = 16 −1
= 0.0573 = 0.0165
0.0492
0.0573 = 0.0165 0.0165
= 0.0165 = [2.98] 2 = 0.0165
= [3.47] 2 =6 = [1] 2
=7 =2
C7H6O2
Emprical formula
1
No. Answer
(c)
( 2 3 2)2 = 158.1 −1
= 250
= 1.509 −1
=?
=
= 1.509 −1 250
= 377.25
=
( )
= 0.25 0.250
= 0.0625
= 0.0625 158 −1
= 9.88
= 377.25 − 9.88
= 367.25
=
( )
0.0625
= 0.367
= 0.1703
(d) MgCl2 + 2NaOH →2NaCl + Mg(OH)2
2 = 15.1
= 9.35
15.1 9.53
2 = 95.3 −1 = 40.0 −1
= 0.158 = 0.2338
( )
1 2 ≡ 2
0.158 2 ≡ 0.316
( )
<
∴
2
No. Answer
2. (a) i. Balmer series 2 ≡ 1 ( )2
ii. 1
x 0.2338 ≡ 2 ( )2
= 0.1169 ( )2
( )2 = 0.1169 58.3 −1
= 6.815
n=5
n=4
n=3
n=2
n=1
Line X produced when an electron at ground state gain energy. It will excite to higher energy level
(n=4). At higher energy level, electron is unstable. It will fall back to lower energy level (n=2) and
produce a second line in Balmer series and emitted a light in a formed of photons.
iii.
1 11
= ( 12 − 22) , 1 < 2
1 1 1
= 1.097 107 (22 − 42)
= 486
iv.
ℎ
∆ =
(6.6256 10−34)(3.0 108)
= 486 10−9
= 4.098 10−19
3
No. Answer
(b) Anomalous in 24
24 expected configuration
1s 2s 2p 3s 3p 4s 3d
24 actual configuration
1s 2s 2p 3s 3p 4s 3d
3d electronic configuration in expected configuration is partially filled while the actual is half-filled.
Half-filled electronic configuration is more stable than partially filled.
3. (a) i.
• Both O and S are elements from group 16.
..• O is an element from period 2. It cannot expend the octet.
F – O.. – F
• While S is an element from period 3. Thus it can expand their octet.
• O can only use p orbital
• S has an empty d orbital
ii.
Both bonded to 5 fluorine. PF5 have equal repulsion according to VSEPR theory.
Thus, the shape is trigonal bipyramidal. 5 bonding pair.
IF5 has 5 bonding pair + 1 lone pair. Thus the shape become square pyramidal.
4
No. Answer
iii.
Structure III is the most plausible structure because the formal charge is almost to zero and
I II III
IStructure III is more plausible because it has low formal charge. The negative charge (-1) is on more
electronegative atom (oxygen).
(b)
S =1x6 =6
F =4x7 =28
Minus electron bond =34 – 8
Minus lone pair electron =26 – 24
Extra electron =2
1 lone pair + 4 bonding pair
4 sigma + 1 lone pair
sp3d
For the central atom:
Sground state :
3s 3p 3d
3d
Sexcited state :
3s 3p
sp3d
Shybrid state :
For the terminal atom 2p
Fground state :
2s
5
No. Answer
(c) • Group 1 have metallic bond between atoms / particles.
• Group 17 have Van der Waals between molecules.
• Metallic bond is stronger than Van der Waals forces.
• More energy required to overcome the metallic bonds.
• Thus melting point Group 1> Group 17
•
4. (a) S(s) + O2(g) →SO2(g)
2.54 103
ℎ = 32.1 −1
= 79.128
T= 30.5 + 273.15 = 303.65K
P=851.2mmHg = 1.12 atm
1 ≡ 1 2
79.128 ≡ 79.128 2
=
=
(79.128)(0.08206)(303.65)
= 1.12
= 1760.43
(b) T= 24.0 + 273.15 = 297.15 K
PT=851.2 mmHg
V=128 mL = 0.128 L
2 = 762 − 22.4
2 = 739.6
2 = 0.973
=
=
6
No. Answer
=
(0.973)(0.128)(17.0)
= (0.08206)(297.15)
= 0.0868
5. (a)
2SO3(g) SO2(g) + O2 (g)
1.0 0 0
ni/mol +2x +x
n/mol -2x
n /mol 2x X
1.0-2x =0.4 mol =0.2 mol
[ ] /M
=0.6 mol 0.4 0.2
= 2 = 2
= 0.6 = 0.2 M = 0.1
2
= 0.3 M
1 − = 0.6
= 0.2
= [ 2]2[ 2]
[ 3]2
(0.2)2(0.1)
= (0.3)2
= 0.044
(b) i.
When the temperature decrease, the system will shift to the right. Since it is an exothermic process.
Thus, Kp increase.
ii.
Adding inert gas does not affect the equilibrium position. Since the partial pressure of each gas does
not change when argon added at constant volume.
6. (a) i.
HNO2(aq) + H2O(l) H3O+(aq) + NO2-(g)
0.215M 0 0
[ ]i/M -x +x +x
[ ]/M x X
[ ] /M 0.215-x
[ ] /M = 8.6 10−3 = 8.6 10−3
= 0.215 − 8.6 10−3
= 0.2064
% = [ ] 100
]
[
4% = 0.215 100
= 8.6 10−3
7
No. Answer
ii.
= [ 3 +]2[ 2−]
[ 2]
(8.6 10−3)2
= 0.2064
= 3.58 10−4
iii.
= − log[ 3 +]
= − log (8.6 10−3)
= 2.065
(b)
Ca(OH)2(aq) Ca2+(aq) + 2OH- (aq)
0 0
[ ]i/M 0.3 0.3
0.3 x 2
[ ] /M 0 =0.6M
= − log[ −]
= − log(0.6)
= 0.22
= 14 − 0.22
= 13.78
(c) i.
HCN(aq) + KOH(aq) →KCN(aq) + H2O(l)
ii.
MKOH= 0 M
iii.
KCN(s) → K+(aq) + CN-(aq)
K+(aq) + H2O(l) → cannot hydrolysed
CN-(aq) + H2O(l) HCN(aq) + OH- (aq)
CN- is a strong conjugate base. The hydrolysis CN- with water produce OH-. Therefore KCN is a basic salt.
iii.
Phenolphtalein
Tymol blue
8