KMK - SET 2

CHEMISTRY PRA PSPM 1 SET 3
SK 015 KOLEJ MATRIKULASI KEDAH
2022/2023

KOLEJ MATRIKULASI KEDAH

BAHAGIAN MATRIKULASI
KEMENTERIAN PENDIDIKAN MALAYSIA

PRA PSPM 1
SEMESTER 1, SESI 2022/2023

KIMIA
2 jam

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

ARAHAN KEPADA CALON:
Kertas ini mengandungi 6 soalan. Jawab semua soalan.

1

CHEMISTRY PRA PSPM 1 SET 3
SK 015 KOLEJ MATRIKULASI KEDAH
2022/2023
TABLE OF RELATIVE ATOMIC MASSES
Element
Aluminum Symbol Proton number Relative atomic mass
27.0
Silver Al 13
Argon 107.9
Arsenic Ag 47 40.0
Gold 74.9
Barium Ar 18
Beryllium 197.0
Bismuth As 33 137.3
Boron
Bromine Au 79 9.0
Iron 208.0
Fluorine Ba 56
Phosphorus 10.8
Helium Be 4 79.9
Mercury 55.9
Hydrogen Bi 83 19.0
Iodine 31.0
Cadmium B5
Potassium 4.0
Calcium Br 35 200.6
Carbon
Chlorine Fe 26 1.0
Cobalt 126.9
Cerium F9 112.4
Krypton
Chromium P 15 39.1
Copper 40.1
Lithium He 2 12.0
Magnesium 35.5
Manganese Hg 80 58.9
Sodium 140.1
Neon H1 83.8
Nickel 52.0
Nitrogen I 53 63.6
Oxygen
Platinum Cd 48 6.9
Lead 24.3
Protactinium K 19 54.9
Radium 23.0
Radon Ca 20 20.2
Rubidium 58.7
Selenium C6 14.0
Cerium 16.0
Cesium Cl 17 195.1
Silicon 207.2
Scandium Co 27 231.0
226.0
Tin Ce 58 222.0
Antimony 85.5
Strontium Kr 36 79.0
Sulphur 140.1
Uranium Cr 24 132.9
Tungsten 28.1
Cu 29 45.0
Zinc 118.7
Li 3 121.8
87.6
Mg 12 32.1
238.0
Mn 25 183.9
65.4
Na 11

Ne 10

Ni 28

N7

O8

Pt 78

Pb 82

Pa 91

Ra 88

Rn 86

Rb 37

Se 34

Ce 58

Cs 55

Si 14

Sc 21

Sn 50

Sb 51

Sr 38

S 16

U 92

W 74

Zn 30

2

CHEMISTRY PRA PSPM 1 SET 3
SK 015 KOLEJ MATRIKULASI KEDAH
2022/2023
LIST OF SELECTED CONSTANT VALUES

Ionization constant for water at 25C Kw = 1.0  1014 mol2 dm6
Molar volume of gases
Vm = 22.4 dm3 mol1 at STP
Speed of light in a vacuum = 24 dm3 mol1 at room condition
Specific heat of water
c = 3.0  108 m s1
Avogadro’s number
Faraday constant = 4.18 kJ kg1 K1
Planck constant = 4.18 J g1 K1
Rydberg constant = 4.18 J g1 C1

Molar of gases constant NA = 6.021023 mol1

Density of water F = 96500 C mol1
Freezing point of water
h = 6.631034 J s

RH = 1.097  107 m1
= 2.18  1018 J

R = 8.314 J mol-1 K1
= 0.08206 L atm mol1 K1

 = 1 g cm3

= 0.00 C
= 273.15 K

Vapour pressure of water at 25C P H2O = 23.76 torr

VOLUME UNIT AND CONVERSION FACTOR
ENERGY
1 L = 1 dm3
PRESSURE 1 mL = 1 cm3
TEMPERATURE
OTHERS 1 J = 1 kg m2 s2 = 1 N m = 1  107 erg
1 calorie = 4.184 J
1eV molecule1 = 96.7 kJ mol1

1 atm = 760 mmHg = 760 torr = 101 325 Pa = 101 325 N m-2

0C = 273.15 K

1 Faraday (F) = 96 500 Coulomb
1 Newton (N) = 1 kg m s2

3

CHEMISTRY PRA PSPM 1 SET 3
SK 015 KOLEJ MATRIKULASI KEDAH
2022/2023

Answer all questions.

1 (a) A naturally occurring chlorine isotopes, 35Cl and 37Cl has a ratio of relative
abundance 35Cl = 3.13.
37Cl
The relative mass of 35Cl and 37Cl are 34.9686 a.m.u. and 36.9659 a.m.u.

respectively. What is the relative atomic mass of chlorine.

[3 marks]

(b) Hydrated potassium carbonate has the formula 2 3. 2 . From 10.00g
of the hydrate, 7.95g of anhydrous salt was left after heating. Determine the

value of x in the formula.

[4 marks]

(c) A 50.0 mL of 2 4 solution containing 52.0% 2 4 by mass with a
density of 1.48 g/mL, is used to prepare a 0.200 M 2 4 solution. Determine
the initial molarity of the 2 4 solution and the final volume of
2 4 solution after the preparation.

[6 marks]

(d) The reaction of calcium carbonate, CaCO3 with hydroiodic acid, HI produces
calcium iodide, CaI2, water H2O and carbon dioxide, CO2.

i) Write a balance equation for the above chemical reaction

ii) If 30g of CaCO3 react with 50mL of 2.0 M HI, identify the limiting
reactant of the reaction.

iii) Calculate the mass of CaI2 produced.

[8 marks]

2. (a) Calculate the frequency when an electron falls from n=6 to n=4. Name the
series and state the region of electromagnetic spectrum for the line.
[6 marks]

(b) The proton number of element X is 15. Write the valence electronic
configuration of element X and its ion.

Give TWO possible sets of quantum number for the highest energy electrons
of the ion.

[4 marks]

3. (a) Chlorine reacts with fluorine to form ClF3 compound. For each of the
molecules given IF3, IF4+ and IF4- :
i) Draw and name the molecular shape.

ii) Predict the type of hybridisation and deduce its molecular polarity.
[12 marks]

4

CHEMISTRY PRA PSPM 1 SET 3
SK 015 KOLEJ MATRIKULASI KEDAH
2022/2023

(b) The valence orbitals of N in the nitrogen trifluoride, NF3 molecule are
hybridized. N uses the hybridized orbitals to form bonds in NF3 molecules.
Name the molecular geometry of NF3. Describe the hybridization process of
NF3.
[5 marks]

4. The relative molecular mass of carbon dioxide, CO2 gas was determined experimentally
by matriculation students. The following data was obtained:
Mass of empty vessel= 32.50g
Mass of vessel filled with CO2 gas at 200C and 75 atm = 34.72g
Mass of vessel filled with water with density 1g/mL= 465.20g

i) Calculate the relative molecular mass of CO2

ii) State the condition when a gas deviate from an ideal gas. Explain.
[9 marks]

5. When the nitrogen monoxide gas reacts with oxygen gas, nitrogen dioxide gas is formed
according to the following equation:

2NO (g) + O2(g) ⇌ 2NO2(g)

If 0.154 mol of NO gas is mixed with 0.250mol of O2 gas in a 2.00 litre evacuated flask
at 300C, the equilibrium concentration of O2 gas is found as 0.0890M. Calculate the
equilibrium constant, and degree dissociation of NO. Based on the value,
calculate for this reaction at the same temperature.

[9 Marks]

6. An experiment was carried out by using two separate beakers as in FIGURE 1. Acetic
acid, CH3COOH (20 mL, 0.1M) was added into each beaker. Calculate the pH of each
solution in the beakers. (Ka of CH3COOH= 1.8x10-5)

40mL water 18mL
0.l M NaOH

Beaker 1 Beaker 2

FIGURE 1 [14 marks]

5

CHEMISTRY PRA PSPM 1 SET 3
SK 015 KOLEJ MATRIKULASI KEDAH
2022/2023 MARKS
END OF QUESTION PAPER 1
NO
1 (a) ANSWER SCHEME 1
35Cl = 3.13 1
1 (b) 37Cl 1
1
∑
Relative atomic mass of Cl = ∑ 1
1
= 3.13 (34.9686) + 1(36.9659) 1
4.13 4

= 35.45

7.95
2 3 = 39.1(2) + 12.0 + 16.0(3)

=7.95 = 0.0575 mol

138.2= 7.95 = 0.0575 (3 )

138.2

1 2 3. 2 ≡ 1 2 3@
0.0575 2 3. 2 ≡ 0.0575 2 3

10.00
0.0575 = 138.2 + 18.0

= 1.98 ≈ 2

6

CHEMISTRY PRA PSPM 1 SET 3
SK 015 KOLEJ MATRIKULASI KEDAH
2022/2023 1
Density= mass solution
1 (c) volume solution 1

1.48g/mL= mass solution 1
50
1
Mass solution = 74 g 1
1
52% = mass solute x 100% ____
74 6
1
Mass solute = 38.48g
1
Mole solute = 38.48 1
2+ 32.1 +4(16) 1
1
= 0.392 mol 1
1
Molarity = mole of solute 1
volume solution (L) 8
21
= 0.392
0.05L

= 7.84L

M1V1=M2V2 @
(7.84) (50) = 0.2(V2)
V2 = 1960 mL

1 (d) i CaCO3 + 2HI CaI2 + H2O +CO2

Mol CaCO3 =30/ 100.1
ii = 0.30 mol

Mol HI = 2.0 x 50 / 1000
= 0.10 mol

1 mol CaCO3 ≡ 2 mol HI @
0.30 mol CaCO3 ≡ 0.60 mol HI
Mol HI available< mol HI needed

So, HI is the limiting reactant

2 mol HI ≡ 1 mol CaI2 @
iii 0.10 mol HI ≡ 0.05 mol CaI2

Mass of CaI2 produced = 0.05 x 293.9
=14.70 g

TOTAL

7

CHEMISTRY PRA PSPM 1 SET 3
SK 015 KOLEJ MATRIKULASI KEDAH
2022/2023

NO ANSWER SCHEME MARKS
2 (a) 1
∆ = [ 1 − 1 2 ] @
2 (b) 2 1

∆ = 2.18 10−18[ 1 − 412] 1
62
1
∆ = −7.57 10−20 (unit insist) 1
1
∆ = ℎ @ ______
7.57 10−20 6
1
= 6.63 10−34 1
= 1.14 1014 −1 (unit insist) 1+1
4
Brackett series (spelling must correct)
10
Infrared/ IR

X = 3s2 3p3 (choose either TWO)
X3- = 3s2 3p6 TOTAL
n=3, l=1, m= -1, s= -1/2
n=3, l=1, m=-1, s= +1/2
n=3, l=1, m=0, s=-1/2
n=3, l=1, m=0, s= +1/2
n=3, l=1, m=+1, s= -1/2
n=3, l=1, m=+1, s=+1/2

NO ANSWER SCHEME MARKS
3 (a) i
+ 1+1+1
ii
FF 1+1+1
F IF F I 1+1+1
F F 1+1+1
FF
I 12
F

F

T-shaped See-saw or distorted Square planar
tetrahedral
sp3d sp3d sp3d2
Polar molecule Non-polar molecule
Polar molecule

8

CHEMISTRY PRA PSPM 1 SET 3
SK 015 KOLEJ MATRIKULASI KEDAH
2022/2023 1
Lewis structure of NF3:
3 (b) 1
1
N 1
FF
1
Fl 5
17
Molecular shape of NF3: Trigonal pyramidal

Valence electronic configuration of N: 2s2 2p3

Ground state: ↑↓ ↑ ↑ ↑

2s 2p

Hybrid state: ↑↓ ↑ ↑ ↑
sp3

TOTAL

NO ANSWER SCHEME MARKS
4 (i) Mass of CO2 = 34.72g- 32.50g 1
1
= 2.22g 1
Mass of H2O = 465.2g-32.50g
1
= 432.70g 1
Volume H2O= Volume CO2 = 432.σ70mL
1
From Ideal gas equation: 1
Molar Mass = (2.22) (0.08206)(293.15K) 1
1
(75)(0.4327) 9
Relative molecular mass CO2= 1.65

(ii) Low Temperature
The kinetic energy of gas molecule will be lowered. Hence, the
attraction forces between gas molecules is significant.

High Pressure
The gas is compressed and the gas molecules are close to each other.
Hence, the volume of the gas molecules occupied is significant.

TOTAL

9

CHEMISTRY PRA PSPM 1 SET 3
SK 015 KOLEJ MATRIKULASI KEDAH MARKS
2022/2023
ANSWER SCHEME 1
NO 1
5 1
1
Equation 2 ( ) + 2( ) ⇌ 2 2( ) 1
Initial conc. (M) 0.154 0.250 0
1
2.00 2.00 1
= 0.0770 = 0.125 1
1
Change conc. (M) -2x -x +2x 1
Equilibrium conc. 0.077-2x 0.125-x 2x
(M) 10
9
At equilibrium, concentration of oxygen is 0.089
0.125-x =0.0890 x= 0.036 M

[NO]at equilibrium = 0.0770-2x

=0.077-2(0.036)
= 5.00 x 10-3 M

[NO2]at equilibrium = 2x
=2(0.036)
= 0.072 M

= [ 2]2
[ ]2[ 2]

(0.072)2
= (5 10−3)2(0.089)

= 2.33 103
∆ = 2 − 3 = −1

= ( )∆

= 2.33 103(0.08206 303.15)−1

= 93.66
Degree dissociation = [changes]

[Initial]
= 2(0.036) = 0.935

0.077

TOTAL
MAX

10

CHEMISTRY PRA PSPM 1 SET 3
SK 015 KOLEJ MATRIKULASI KEDAH
2022/2023

NO ANSWER SCHEME MARKS
6 (a) 1
1
Equation CH3COOH ⇌ CH3COO- + H+
[ ]i (M) 0 1
[ ]c (M) 0.033 0 +x 1
[ ]f (M) +x 1
-x +x
1
0.033-x +x 1
1
Ka ═ [CH3COO-][ H+] 1
[CH3COOH]
1
Ka<<1, (0.033 – x ) ═ 0.033 1+1

1.8 x 10-5 ═ x2 1
0.033
1
x ═ [H+] ═ 7.71 x 10-4 14

pH ═ - log [H+]

═ - log [7.71 x 10-4]

═ 3.11

Beaker 2

Equation CH3COOH + NaOH → CH3COONa + H2O
Mol (i) -
20 x 0.1 18 x 0.1 0
Mol (c) -
Mol (f) 1000 1000 -
[ ]f (M) ═ 2x10-3 ═ 1.8 x10-3 -

-1.8 x10-3 -1.8 x10-3 +1.8x10-3

2x10-4 0 1.8 x10-3

2x10-4 - 1.8 x10-3

0.038L 0.038L
═5.26 x10- ═ 0.047

3

Buffer solution
pH ═ pKa + log [conjugate base]

weak acid
═ - log( 1.8 x10-5) + log [ 0.047 ]

[ 5.26 x10-3]

═ 5.70

TOTAL

11