KML - SET 2

#TEAMSIRSHAZWAN2223 SK015 SEMESTER I, SESSION 2022/23

CHEMISTRY UNIT
LABUAN MATRICULATION COLLEGE

#TEAMSIRSHAZWAN2223
KIMIA

CHEMISTRY

SK015

CADANGAN SKEMA JAWAPAN
SUGGESTED ANSWER SCHEME

1

#TEAMSIRSHAZWAN2223 SK015 SEMESTER I, SESSION 2022/23

NO PART SCHEME CODE MARKS
1 (a) M1 1
= ∑
∑ M2 1
M3 1
= (203 × 29.5) + (205 × 70.5)
M4 1
(29.5 + 70.5) M5 1
M6 1
= 204.41 M7 1
M8 1
(b) Ca Cl H2O
(c)(i) 7.2 M9 1
Mass / g 4.0 7.1 0.4
4
Moles/ mol 0.0998 0.2

Simplest 1 2

Ratio

CaCl2.4H2O

3H2O + 3NO2- → 3NO3- + 6H+ + 6e-

6e- + 14 H+ + Cr2O72- → 2Cr3+ + 7 H2O
_______________________________________________

3NO2- + Cr2O72- + 8H+ → 3 NO3- + 2 Cr3+ + 4 H2O

(c)(ii) 2 = 0.52 = 7.54 × 10−3
69
M10 1
= 7.54 × 10−3 M11 1
0.1 = 0.0754

(c)(iii) 2 2 = 3 M12 1
2 2 7 2 2 7 1
M13 1
0.0754(25) 3 M14 1
2 2 7(30) = 1

2 2 7 = 0.0209

(c)(iv) 2 = (0.0754 )(25 ) = 1.885 × 10−3
1000
M15 1
3 mol NO2- produced 4 mol H2O M16 1
M17 1
1.885 ×10-3 mol NO2- produced 2.51 ×10-3 mol H2O M18 1

Mass of H2O = (2.51 ×10-3)(18) = 0.0452 g

2

#TEAMSIRSHAZWAN2223 SK015 SEMESTER I, SESSION 2022/23

NO PART SCHEME CODE MARKS
(c)(v) M19 1
% = ℎ × 100
M20 1

80% = 0.0452 × 100

Actual mass = 0.0362 g M21 1
TOTAL 21

3

#TEAMSIRSHAZWAN2223 SK015 SEMESTER I, SESSION 2022/23

NO PART SCHEME CODE MARKS
2 (a) Transition from n=6 → n=3 M1 1
M2 1
(b)(i) 11
∆ = ( 12 − 22) M3 1
M4 1
∆ = 2.18 × 10−18 1 − 1
(62 32) M5 1
M6
ΔE = -1.82 ×10-19 J M7 2

∆ = ℎ

1.82 × 10−19
= 6.6256 × 10−34

V= 2.74 ×1014 s-1

(b)(ii) The partially filled 3d-orbitals is more stable than partially M8 M9 2
(c) filled 3-d orbitals M10 1
TOTAL 10
(n,l, m,s ) = (4,0,0, ½)

4

#TEAMSIRSHAZWAN2223 SK015 SEMESTER I, SESSION 2022/23

NO PART SCHEME CODE MARKS
3 (a)(i) M1 1

M2 1

M3 1

(a)(ii) H2O = 104.5o 1
NH3 = 107o 1
CH4 = 109.5o M4 1
M5
H2O has 2 lone pair and 2 bonding pair M6 1

NH3 has 1 lone pair and 3 bonding pair M7 1

CH4 has 4 bonding pair M8

Lone pair-lone pair > lone pair-bonding pair > bonding pair-
bonding pair

(b)

M9 1
M10 1
M11 1

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#TEAMSIRSHAZWAN2223 SK015 SEMESTER I, SESSION 2022/23

NO PART SCHEME CODE MARKS

M12 1

(Shape 1
sp3d) 1

M13

(label
orbitals)

M14
(label σ)

(c) The valence band of sodium is 3s and the conduction band of M15 1
sodium is 3p

There is no band gap between the valence band and the M16 1
conduction band

The electron can transfer from the valence band to the M17 1
conduction band resulting electrical conductivity TOTAL 17

6

#TEAMSIRSHAZWAN2223 SK015 SEMESTER I, SESSION 2022/23

NO PART SCHEME CODE MARKS
4 (a) = M1 1

= M2 1


(1 )(32 −1)
= (1.428 −3)(273.15 )

R= 0.08204 atm dm-3 mol-1 K M3 1
1
(b) • Ethanol has hydrogen bond while propane has van der M4
1
Waals forces/ London Forces
1
• Hydrogen bond is stronger than London Forces M5

• Less energy needed to overcome the weak London M6
forces resulting propane easily evaporated.

(c)(i) • Crystalline solid has sharp melting point while M7 1
amorphous solid has range melting point

(c)(ii) Ionic solid: NaCl M8 1

Molecular Covalent Solid : Ice M9 1
• Any logical answers for question 4(c) TOTAL 9

7

#TEAMSIRSHAZWAN2223 SK015 SEMESTER I, SESSION 2022/23

NO PART SCHEME CODE MARKS
5 (a) [ ]2 M1 1

= [ 2][ 2] M2 1

[0.49]2
= [0.08][0.06] = .

(b) Shift to the right M3 1
M4 1
(c) H2 (g) I2 (g) 2 HI (g) M5 1
0.09 0.49
[ ]i / M 0.08 -x + 2x
0.09-x
[ ]c / M -x 0.49 + 2x

[ ]e / M 0.08-x

[0.49 + 2 ]2 M6 1
= [0.08 − ][0.09 − ] = 50.02
M7 1
46.02 x2 – 10.4634 x + 0.12 = 0 M8 1
M9 1
X1 = 0.2152 (rejected) ; X2 = 0.0121
TOTAL 9
At equilibrium

[H2] = 0.0679 M
[I2] = 0.0779 M
[HI] = 0.5142 M

8

#TEAMSIRSHAZWAN2223 SK015 SEMESTER I, SESSION 2022/23

NO PART SCHEME CODE MARKS
6 (a) C6H5COOH (aq) + H2O (l) ⇋ C6H5COO- (aq) + H3O+ (aq) M1 1

C6H5COOH H2O C6H5COO- H3O+ M2 1

[ ]i / M 0.186 - 0 0

[ ]c / M -x - +x +x

[ ]e / M 0.186-x - X x

= [ 6 5 −][ 3 +] M3 1
[ 6 5 ]
M4 1
6.8 × 10−5 = [ ][ ] ] M5 1
[0.186 − M6 1
M7 1
Assume x is very small 0.186- x= 0.186 M8 1
X=3.4 ×10-3 M M9 1
M10 1
[H3O+]= x = 3.4 ×10-3 M
pH = -log [H3O+]
=-log (3.48 ×10-3)

pH= 2.46
[ 6 5 −]
(b) = + log [ 6 5 ]

4.6 = −log (6.8 × 10−5) + log [ 6 5 −]
0.5

[ 6 5 −] = .

6 5 = 1.2938 × 0.05

6 5 = .

6 5 = 0.06469 × 144 −1

6 5 = .

When small amount of NaOH added, it will be consumed by
C6H5COOH

C6H5COOH (aq) + OH- (aq) → C6H5COO- (aq) + H2O (l)

The concentration of C6H5COOH will decrease while the
concentration of C6H5COO- will increase

9

#TEAMSIRSHAZWAN2223 SK015 SEMESTER I, SESSION 2022/23

NO PART SCHEME CODE MARKS
(c) Mg3(PO4)2 (s) ⇋ 3 Mg2+ (aq) + 2 PO43- (aq) M11 1

Ksp = [Mg2+]3 [PO43-]2 M12 1
5.2 ×10-24 = [3s]3 [2s]2
M13 1
5.2 ×10-24 = 108 s5
s = 8.64 ×10-6 mol L-1 M14 1
Solubility = 8.64 ×10-6 mol L-1 × 262 gmol-1 TOTAL 14
Solubility = 2.26 ×10-3 g L-1

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