KMK - SET 1

CHEMISTRY PRA PSPM 1 SET 1
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 1
SK 015 KOLEJ MATRIKULASI KEDAH
2022/2023
TABLE OF RELATIVE ATOMIC MASSES
Element
Aluminum Symbol Proton number Relative atomic mass
Al 13 27.0
Silver Ag 47
Argon Ar 18 107.9
Arsenic As 33 40.0
Gold Au 79 74.9
Barium Ba 56
Beryllium Be 4 197.0
Bismuth Bi 83 137.3
Boron B 5
Bromine Br 35 9.0
Iron Fe 26 208.0
Fluorine F 9
Phosphorus P 15 10.8
Helium He 2 79.9
Mercury Hg 80 55.9
Hydrogen H 1 19.0
Iodine I 53 31.0
Cadmium Cd 48
Potassium K 19 4.0
Calcium Ca 20 200.6
Carbon C 6
Chlorine Cl 17 1.0
Cobalt Co 27 126.9
Cerium Ce 58 112.4
Krypton Kr 36
Chromium Cr 24 39.1
Copper Cu 29 40.1
Lithium Li 3 12.0
Magnesium Mg 12 35.5
Manganese Mn 25 58.9
Sodium Na 11 140.1
Neon Ne 10 83.8
Nickel Ni 28 52.0
Nitrogen N 7 63.6
Oxygen O 8
Platinum Pt 78 6.9
Lead Pb 82 24.3
Protactinium Pa 91 54.9
Radium Ra 88 23.0
Radon Rn 86 20.2
Rubidium Rb 37 58.7
Selenium Se 34 14.0
Cerium Ce 58 16.0
Cesium Cs 55 195.1
Silicon Si 14 207.2
Scandium Sc 21 231.0
Sn 50 226.0
Tin Sb 51 222.0
Antimony Sr 38 85.5
Strontium S 16 79.0
Sulphur U 92 140.1
Uranium W 74 132.9
Tungsten Zn 30 28.1
45.0
Zinc 118.7
121.8
87.6
32.1
238.0
183.9
65.4

2

CHEMISTRY PRA PSPM 1 SET 1
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
Vapour pressure of water at 25C
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

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 1
SK 015 KOLEJ MATRIKULASI KEDAH
2022/2023

Answer all questions.

1. (a) For a solution of acetic acid (CH3COOH) to be called ‘vinegar’ it must contain
5% acetic acid by mass. If a vinegar is made up only of acetic acid and water and
density of vinegar is 1.006 g/mL. Determine,

i) molarity of acetic acid

ii) molality of acetic acid

[8 marks]

(b) During the extraction of iron, the iron of hematite (Fe2O3) was reduced to iron metal
by carbon monoxide according to the equation.

Fe2O3 (s) + CO (g) Fe (s) + CO2 (g) (not balanced)

i) Determine the limiting reactant if 300g of hematite and 60g of CO are
used in the extraction.

ii) How many grams of iron are produced? If the iron extracted is 70g,
calculate the percentage yield.

iii) Determine the mass of excess reactant left over.

[13 marks]

2. (a) The wavelength of a photon emitted in the Brackett series of hydrogen emission
spectrum in 2625 nm.

i) State the region of electromagnetic spectrum of this emission.

ii) Determine the initial and final energy level corresponding to this
emission.

iii) Calculate the frequency of this emission.

[6 marks]

(b) Iron has 26 proton numbers and located at block d.
i) Write electronic configuration for 26Fe2+ and 26Fe3+.

ii) Which of these two ions are more stable? Give reason.

[4 marks]

3 (a) O

H3C C CH3

Acetone is a colourless, volatile and flammable liquid. It is miscible with water and
serves as an important solvent typically for cleaning purposes in the laboratory.

4

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

i) Draw the overlapping orbitals of acetone and label sigma () and pi ()
bond in the molecule.

ii) State the bond angle between H-C-H and C-C-O

[10 marks]

(b) Magnesium, Mg and Aluminium, Al are metals.

i) Draw the electron sea model diagram for Magnesium.

ii) Why Aluminum has higher boiling point than Magnesium?

iii) By using band theory of solid, explain how Magnesium conducts
electricity.
[7 marks]

4 (a) A certain mass of nitrogen gas is added to a vessel of 405 ml containing 0.52 g
oxygen. The gas pressure is increased from 745 mmHg to 980 mmHg.
Assuming the temperature remains constant and that the nitrogen gas does not
react with oxygen gas, calculate the mass of the added nitrogen.
[6 marks]

(b) Explain how temperature affects the vapour pressure of a liquid.
[3 marks]

5 (a) A 2.50 L flask contains 0.525 mol of CO2, 1.250 mol of CF4 and 0.750 mol of
COF2 at 1000oC. Explain in which direction will a net reaction occur to reach

equilibrium shown below.

CO2 (g) + CF4 (g) 2COF2(g) Kc = 0.50 at 1000oC
[5 marks]

(b) Sodium hydrogen carbonate decomposes at 100oC according to the equation
below:

2NaHCO3 (s) Na2CO3(s) + H2O(g) + CO2(g)

If the value of Kp is 0.231, calculate the total gas pressure (atm) at equilibrium.
[4 marks]

6 A buffer solution is prepared by mixing 100 mL 0.1M NH3 with 100 mL 1.0 NH4Cl.

Calculate the pH of the solution before and after the addition of 1 mL 0.1 M HCl.
(Kb NH3 = 1.74x10-5)

[14 marks]

END OF QUESTION PAPER
5

CHEMISTRY PRA PSPM 1 SET 1
SK 015 KOLEJ MATRIKULASI KEDAH MARKS
2022/2023 1
ANSWER SCHEME
NO Assume, mass solute = 5g 1
1 (a) i
Mass solution = 100g 1

Mole CH3COOH = 5 1
60 1
1
= 0.0833 mol 1

1.006 = 100 1
volume 8
1
Volume = 99.403mL
1
Molarity = mole @
Volume 1
1
= 0.0833 mol 1
0.0994 L 1
1
= 0.83M 1

ii Mass solvent = 100-5 = 95g

Molality = mole solute @
mass solvent (kg)

Molality = 0.0833mol
0.095kg

= 0.87 m

1 (b) i Fe2O3 (s) + 3 CO (g) 2 Fe (s) + 3 CO2 (g)

mole of hematite = 300g
159.8gmol-1

= 1.877mol

mole of CO = 60g
28 gmol-1

= 2.14 mol

1mol Fe2O3≡ 3 mol of CO
1.877 mol Fe2O3≡ 5.631 mol of CO

Thus, mol CO available less than needed
making CO the limiting reactant

ii 3 mol CO ≡ 2 mol of Fe
2.14 mol CO ≡ 1.427 mol Fe

Mass of Fe produced = 1.427 mol x 55.9gmol-1 = 79.77g

6

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

Thus, the percentage yield = actual yield x 100% 1
theoretical yield 1
= 70 x 100% @
79.77 1
= 87.75%
1
iii 3 mol CO ≡ 1 mol Fe2O3 _______
2.14 mol CO ≡ 0.713 mol Fe2O3 used up
13
Mole excess = 1.877 mol – 0.713 mol 21
= 1.164 mol

Mass Fe2O3 excess = 1.164 mol x 159.8 gmol-1
= 186.0 g

TOTAL

NO ANSWER SCHEME MARKS
2 (a) i infrared region 1

ii final energy level, n2 = 4 @ 1

1/λ = RH (1/n12 – 1/n22) 1
1/ 2625x10-9 = 1.097x107(1/42 – 1/n22)
1
n2 = 6
1
initial ni = 6 @
1
iii frequency, v = c/ λ _______
= (3.0x10 8) / 2625x10-9
= 1.14x1014 s-1 6

2 (b) 1

i Fe2+ : 1s22s22p63s23p63d6 1
Fe3+ : 1s22s22p63s23p63d5
1+1
ii Fe3+ because it has stability of half-filled 3d orbital arrangement _______

TOTAL 4

10

7

CHEMISTRY PRA PSPM 1 SET 1
SK 015 KOLEJ MATRIKULASI KEDAH MARKS
2022/2023
ANSWER SCHEME 1
NO 1
3 (a) i O 1
1
H3C C CH3 O 1
2p 1
O 1
1
 2p 1
1
σ ________
10
C sp2 1
1
2p σ 1s H

σ sp2 sp2

σ C sp3 σ sp3 sp3 σ

sp3 sp3 σC sp3 1s
sp3
1s sp3 H

Hσ 1s σ

1s H 1s
H

H

By referring to the diagram above

 orbital overlap sideways

Labelling 3 of  bond (label 9 )

Label  bond

Correct shape of s-orbital of H atom
Correct shape of p-orbital of O atom

C1 = hybrid sp3
C2 = hybrid sp2
C3 = hybrid sp3

ii Bond angle of C-C-H = 109.50
Bond angle of C-C-O = 1200

3 (b) i

Mg2+ Mg2+ charge ion Mg2+
8 sea-electrons
e e

e Mg2+
Mg2+ e
e

8

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

ii Al has more valence electron (3 valence electron) than Mg (2 valence 1
electron) 1
The strength of metallic bond is proportional to number of valence

electron

iii There is no gap between valence band and conduction band 1
The valence band dan conduction band is overlap 1
Continuous availability of electron to flow 1
_______
7

TOTAL 17

NO ANSWER SCHEME MARKS
4 (a) 1
P N2 = PT – PO2 @ 1
4 (b) = 980 – 745 1

= 235 mmHg 1

mol O2 = 0.52 @ 0.01625 mol 1
32
1
P N2 = X2 . PT @ 6
1
235 = x x 980 1
1
x + 0.01625 mol 3
9
x = 5.125 x 10-3 mol

mass N2 = 5.125 x 10-3 mol x 28
= 0.1435 g (unit insist)

When T increase, average kinetic energy/velocity of molecule increase.
More liquid molecules are able to escape.
So, vapour pressure increase.

TOTAL

9

CHEMISTRY PRA PSPM 1 SET 1
SK 015 KOLEJ MATRIKULASI KEDAH MARKS
2022/2023
ANSWER SCHEME 1
NO (all
5 (a) corrects)

[CO2] = 0.525 = 0.21 M 1
2.5
1
[CF4] = 1.25 = 0.50 M 1
2.5 1
5
[COF2] = 0.75 = 0.30 M
2.5 1

CO2 (g) + CF4 (g) 2COF2(g) Kc = 0.50 1
@ 1
Qc = [COF2]2 1
[CO2] [CF4] 4
9
= (0.30)2
(0.21)(0.50)

= 0.857

Qc > Kc

Thus the reaction shift to the left until Qc = Kc

5 (b) 2NaHCO3 (s) Na2CO3(s) + H2O(g) + CO2(g)

Kp = (PH2O) (PCO2)
xx

From stoichiometric equation, at equilibrium,

PH2O = PCO2 = x

Thus, Kp = (x)(x)

0.231 = (x)2
x = 0.48

PT = PH2O + PCO2 @
= 0.48 + 0.48
= 0.96 atm

TOTAL
10

CHEMISTRY PRA PSPM 1 SET 1
SK 015 KOLEJ MATRIKULASI KEDAH MARKS
2022/2023 1
ANSWER SCHEME 1
NO pH of the solution before addition of HCl
6 1
1
nNH3 = 0.1 x 100 = 0.01 mol 1
1000
1
nNH4+ = 1.0 x 100 = 0.1 mol 1
1000
1 (phase &
[NH3] = 0.01 mol = 0.05M @
0.2 L reversible
arrow)
[NH4+] = 0.1 mol = 0.5M
0.2L 1
1+1
pOH = pKb + log [NH4+]
[NH3] 1

= - log 1.74x10-5 + log [0.5] 1
[0.05] 1
14
= 5.76

pH = 14 – 5.76
= 8.24

pH after the addition of 1mL HCl

NH3 (aq) + H3O+ (aq) NH4+ (aq) + H2O (l)
1 x 10-4
ni 0.01 -1 x 10-4 0.1 -
∆n -1 x 10-4
0 +1 x 10-4 -
neq 0.0099
0.1001 -

[]eq 0.0099 0.1001
0.201 0.201
= 0.0493M = 0.498M

pOH = pKb + log [NH4+] @
[NH3]

= - log 1.74x10-5 + log [0.498]
[0.0493]

= 5.77

pH = 14-5.77
= 8.23

TOTAL

11