SUGGESTED ANSWER PSPM I SK015

SUGGESTED ANSWER PSPM I
20182019

[7 marks]
b) Under the same condition of temperature and density, determine which gas

behaves less ideally: CH4 or SO2
molar mass CH4 = 16 g/mol
molar mass SO2 = 64.1 g/mol
 SO2 will behave less ideally.
 SO2 has higher molecular weight compare to CH4.
 Higher molecular weight will have higher volume of particles. Thus, volume of

SO2 > CH4.
 Strength of intermolecular force of SO2 > CH4.

[3 marks]
c) In an experiment when gelatin was added to water, the water became viscous.

Explain the relationship between viscosity and intermolecular forces.
 Viscosity is the resistance of fluid to flow.
 Higher intermolecular forces will have higher viscousity.

[2 marks]

5. The equation of simulated photosynthesis reaction is represented by

6CO2 (g)  6H2O(g) C6H12O6(g)  6O2(g) ∆Ho = +ve

At 31oC, the following equilibrium concentrations were found:

[H2O] = 7.91×10-2 M, [CO2] = 9.30×10-1 M, [O2] = 2.40×10-3 M

a) Calculate the equilibrium constant, Kp for the reaction.

Kc = O2 6
CO2 6 H2O6

 2.40103 M 6
   = 9.30101M 6 7.91102 M 6

= 1.21×10-9

Kc = Kp (RT)∆n
= Kp (RT)-6

Kp = Kc (RT)-6
= (1.21×10-9) (0.08206×304.15)-6
= 5.02×10-18

8

SUGGESTED ANSWER PSPM I
20182019

b) Determine the initial mass of CO2 involved in the above reaction.

6CO2 6H2O 6O2

[ ]i a b 0

[ ]c -6x -6x +6x

[ ]e a-6x b-6x 6x

6x = 2.40×10-3 M
x = 4×10-4 M

a - 6x = 9.30×10-1 M
a - 2.40×10-3 M = 9.30×10-1 M
a = 0.9324 M
[CO2] I = 0.9324 M

Assume V = 1L
Mole of CO2 = 0.9324 mol
∴ mass of CO2 = (0.9324 mol)(44 g/mol)

= 41.03 g

c) Explain how the equilibrium position would be affected for each of the following
changes:
i. Water is added.
 Equilibrium position will shift forward to consume the added reactant.
ii. Temperature is increased.
 Equilibrium position will shift forward to absorb heat added.
[12 marks]

9

SUGGESTED ANSWER PSPM I
20182019

6. a) A sample of 0.214 g of unknown monoprotic weak acid, HA was dissolved in
25.00 mL of water and titrated with 0.1 M NaOH. The acid required 27.40 mL of the
base to reach the equivalent point.
i. Determine the molarity of the acid.

HA(aq)  NaOH (aq)  NaA(aq)  H2O(l)

Mole of NaOH = 2.74×10-3 mol

1 mol NaOH ≡ 1 mol HA
2.74×10-3 mol NaOH ≡ 2.74×10-3 mol HA

2.74103 mol

Molarity =

0.025L

= 0.1096 M

ii. Calculate the pH of the solution if 40 mL NaOH is added to the acid solution.
Mole of NaOH = 4×10-3 mol

HA NaOH NaA

ni 2.74×10-3 mol 4×10-3 mol 0

nc -2.74×10-3 mol -2.74×10-3 mol +2.74×10-3 mol

nf 0 1.26×10-3 mol 2.74×10-3 mol

1.26103 mol
Molarity of NaOH =

0.065L
= 0.0194 M

pOH = -log(0.0194 M)
= 1.71

∴pH = 12.29

iii. Explain qualitatively the pH of solution at equivalent point.
 At equivalent point, all acid is completely reacted with base.
 The only remaining particles in the solution is salt, NaA solution.
 NaA salt is from weak acid and strong base, the anion A- undergoes
hydrolysis forms OH-.

A (aq)  H2O(l) HA(aq)  OH  (aq)

 pH > 7.
[12 marks]

10

SUGGESTED ANSWER PSPM I
20182019

b) Magnesium arsenate, Mg3(AsO4)2, is used as an insecticides and it is toxic to humans.

Calculate the solubility of Mg3(AsO4)2 in water at 25oC.

[Given: Ksp Mg3(AsO4)2 = 2.2×10-20]

Mg3(AsO4 )2 (s) 3Mg2 (aq)  2AsO43 (aq)

[ ]I - 00

[ ]c - +3x +2x

[ ]e - 3x 2x

Ksp = [Mg2+]3[AsO43-]2
2.2×10-20 = (3x)3(2x)2

= 108x5
x = 4.59×10-5
∴Solubility = 4.59×10-5 M

[5 marks]

11

SUGGESTED ANSWER PSPM I 2019/2020

Answer all questions.

1. (a) Bromine has two stable isotopes, 79Br and 81Br.

(i) By comparing the number of sub-atomic particles of these isotopes, explain
what is meant by term isotopes.

(ii) Determine the number of electrons of Br – ion.

[4 marks]

(b) In a complete combustion, 1.00 g of sample W (CxHyOz) was burnt to produce
2.52 g of carbon dioxide, CO2 and 0.443 g of water vapour, H2O. Determine
the empirical formula of the compound.

[6 marks]

(c) Calcium acetate, Ca(C2H3O2)2 solution is the substance used for reducing
phosphate level in late-stage kidney failure. In an experiment, 250 ml of 0.25
M Ca(C2H3O2)2 solution was prepared. Determine the molality of the solution
with a density of 1.509 g ml-1.

[7 marks]

(d) Magnesium hydroxide, Mg(OH)2 is an antacid that is used to relieve
indigestion, sour stomach and heartburn. It can be prepared by reacting
magnesium chloride, MgCl2 and sodium hydroxide, NaOH with the by-product
of sodium chloride, NaCl. In an experiment, a students allowed 15.1 g of
MgCl2 to react with 9.35 g of NaOH. Calculate the mass (in grams) of Mg(OH)2
that could be obtained at the end of the experiment.

[9 marks]

1. (a)(i) Isotopes are atoms of the same element that have the same number of
protons but different number of neutrons.

79Br have 35 protons and 44 neutrons.

81Br have 35 protons and 46 neutrons.

(ii) No. of electrons of Br – ion is 36.

(b) Mass of C atoms = 12.0 x 2.52 g = 0.687 g

44.0

Mass of H atoms = . x 0.443 g = 0.0492 g
.

Mass of O atoms = 1.00 – (0.687 + 0.0492) = 0.264 g

C H O
0.0492 0.264
Mass (g) 0.687 0.0492 . = 0.0165
No. of moles . = 0.0573
(mol) . = 3.0
Simplest ratio
. 1
. = 3.5
. 1x2=2

Ratio 3.5 x 2 = 7 3x2=6

Thus, empirical formula of W is C7H6O2 .

(c) Given:

Volume of solution = 250 ml

Molarity of Ca(C2H3O2)2 solution = 0.25 M
Density of solution = 1.509 g ml-1

Mole of solute , Ca(C2H3O2)2 = Molarity x volume of solution

= 0.25 x 0.25

= 0.0625 mol

Mass of solute , Ca(C2H3O2)2 = 0.0625 x 158.1

= 9.881 g

Mass of solution = Density x volume of solution

= 1.509 x 250

= 377.25 g

Mass of solvent = 377.25 – 9.881

= 367.369 g

= 0.367 kg

Molality =
( )

= .

.

= 0.17 molal

(d) MgCl2 + 2 NaOH 2 NaCl + Mg(OH)2

n of MgCl2 = . = 0.158 mol
.

n of NaOH = . = 0.234 mol


1 mol of MgCl2 ≡ 2 mol of NaOH

0.158 mol of MgCl2 ≡ 0.158 x 2

= 0.316 mol of NaOH

Mol of NaOH (needed) > mol of NaOH (given)

Thus, NaOH is limiting reactant.

2 mol of NaOH ≡ 1 mol of Mg(OH)2
0.234 mol of NaOH ≡ 0.234 ÷ 2

= 0.117 mol of Mg(OH)2

Mass of Mg(OH)2 = 0.117 x 58.3

= 6.82 g

2. (a) The line spectrum of hydrogen atom in the visible region is shown in the
following diagram.

X

frequency increases →

(i) Name the series of the line spectrum.
(ii) Sketch an energy level diagram of a hydrogen atom for the formation

of line X. Explain how line X is formed.
(iii) Calculate the wavelength corresponding to line X.
(iv) Determine the energy involved in 2(a)(iii).

[10 marks]
(b) Describe the anomalous electronic configuration of chromium atoms.

[3 marks]
2. (a) (i) Balmer series

(ii) Energy

n=4
n=3

n=2
X

n=1

Line X is produced when an electron falls from n=4 to n=2
and photon/energy is emitted in a form of light.

(iii) = ( − ) , <


= . × − ( − )



λ = 4.86 x 10-7 m @ 486 nm

(iv) ∆E =



= ( . − )( . )
. −

= 4.09 x 10-19 J

(b) Expected electronic configuration of Cr : 1s22s22p63s23p64s23d4

Actual electronic configuration of Cr : 1s22s22p63s23p64s13d5

Anomalous electronic configuration of Cr occur due to stability of half-
filled 3d orbital.

3. (a) Explain each of the following statements.

(i) Upon reaction with fluorine, oxygen forms only OF2 whereas sulphur
forms SF2, SF4 and SF6 molecules.

(ii) The shape of a PF5 molecule differs form that of an IF5 molecule.

(iii) Of the three possible resonance structures for OCN- below, III is the
best structure.

[O = C= N] - [O ≡ C N ] - [O C ≡ N] -
I II III
[10 marks]

(b) Illustrate the hydribisation of the central atom in SF4 using orbital diagrams.
Show and label the overlapping of orbitals in the molecule.

[7 marks]

(c) Explain the difference in melting point between elements in Group 1 and
Group 17.

[5 marks]

3. (a) (i) Lewis Structure:

.. .. ..
: .F. .. F..: :F: ..
.. .. .. .. .. .. : .. F..:
:.F. O.. .F.: :.F. S.. .F.: .F.
S.. ..
:..F.. .F.: :..F.. S ..
.F.:
:.F.:

Oxygen, O is in period 2 while Sulphur, S is in period 3. Therefore
O can only use p orbitals (or has no d orbital) while S can use d
orbitals (or form expanded octet).

(ii)

F
F

PF
F

F

The shape of PF5 is trigonal bipyramidal because it has 5 bonding
pairs and no lone pairs.

F
FF

.I.

FF

The shape of IF5 is square pyramidal because it has 5 bonding
pairs and one lone pair.

(iii)

.. .. - O..+ C .. - .. .. -
O.. C .N. N..: :O.. C N

(0) (0) (-1) (+1) (0) (-2) (-1) (0) (0)

I II III

I and III has the lower formal charge but III is most plausible
because the negative charge is on the more electronegative atom,
O.

(b) Lewis structure:

.. ..
.F. F..:
: ..

:..F.. S.. ..
.F.:

Valence orbital diagram:

S ground state : 3p
3s

S excited state:

3s 3p 3d

S hybrid :

sp3d

F: 2p
2s

Orbital Overlapping:

(c) Elements of Group 1 have metallic bonds between the atoms. Group 17
has van der Waals forces between the molecules. Metallic bond is
stronger than van der Waals forces. More energy needed to overcome
the metallic bonds. Thus, melting point Group 1 higher than Group 17.

4. (a) When coal is burnt, the sulphur present in coal is converted to sulphur dioxide
which is responsible for the acid rain phenomenon.

S (s) + O2 (g) SO2 (g)

If 2.54 kg of sulphur is reacted with oxygen, determine the volume of sulphur
dioxide gas formed at 30.5oC and 851.2 mmHg.

[5 marks]

(b) In an experiment, NH3 gas was produced and collected using water
displacement method. At 24oC and atmospheric pressure of 762 mmHg, the
volume of the gas collected was 128 ml. Calculate the mass of the gas
obtained.

[Given the vapour pressure of water = 22.4 mmHg]

[6 marks]

4. (a) n of S = . = 79.13 mol

.

1 mol of S ≡ 1 mol of SO2
79.13 mol of S ≡ 79.13 mol of SO2

PV = nRT

( . ) = . ( . ) ( . + . )

V = 1760 L

(b) PT = PNH3 + PH2O = 762 mmHg

PNH3 + 22.4 = 762
PNH3 = 762 – 22.4

= 739.6 mmHg

= 0.973 atm

PV = nRT

(0.973) (0.128) = n (0.08206) (297.15)

n = 5.11 x 10-3 mol

mass of NH3 = 5.11 x 10-3 x 17 = 0.0869 g

5. (a) Sulphur trioxide, SO3 gas decomposes to sulphur dioxide and oxygen gas
according to the following equation:

2SO3 (g) 2SO2 (g) + O2 (g)

When 1.0 mol of SO3 is placed into a 2 L vessel and heated to 344 K, the
system achieves equilibrium and 0.6 mol of SO3 gas is remained.

(i) Calculate the concentrations of each gas at equilibrium.

(ii) Calculate the equilibrium constant Kc at 344 K.

[7 marks]

(b) Phosphorus pentachloride, PCl5 is left in a sealed container to establish
equilibrium.

PCl5 (g) PCl3 (g) + Cl2 (g) ∆H = - ve

(i) Explain the effect of lowering temperature on the equilibrium constant,
Kp of the system.

(ii) Explain the effect of adding argon gas at constant volume on the
equilibrium position.

[4 marks]

5. (a) (i)

[ ]i / M 2SO3 (g) 2SO2 (g) + O2 (g)
[ ]c / M
[ ]eq / M 0.5 00
-2x +2x +x
0.5-2x 2x x

[SO3] eq = 0.5-2x = 0.3 M
x = 0.1 M

thus, at equilibrium:
[SO3] = 0.3 M

[SO2] = 0.2 M

[O2] = 0.1 M

(ii) Kc = [ ] [ ]
[ ]

= ( . ) ( . )
( . )

= 0.044

(b) (i) When temperature of the system is lowering, the equilibrium
position shift forward (or shift to the right) since it is an
exothermic process. Thus, more product is produce. Hence, Kp
value will increases.

(ii) Adding argon gas at constant volume to the system, the
equilibrium position will not change since partial pressure of each
gas does not change.

6. (a) Nitrous acid, HNO2 is used to distinguish between primary, secondary and
tertiary amines. At 25oC, an aqueous solution of 0.215 M HNO2 is 4.0%
dissociated.

(i) Write a balanced equation for the dissociation of HNO2 in water.

(ii) Determine the acid dissociation constant, Ka for HNO2.

(iii) Calculate the pH of the solution.

[6 marks]

(b) Calcium hydroxide, Ca(OH)2 is used to neutralize excess acidity in lakes and
soils. Calculate the pH of 0.30 M Ca(OH)2 solution.

[4 marks]

(c) In a weak acid-strong base titration, 25.00 ml of KOH solution, is titrated with
21.45 ml of 0.10 M HCN to reach the equivalence point.

(i) Write the balanced chemical reaction equation for the titration.

(ii) Calculate the concentration of KOH solution at the equivalence point.

(iii) Explain why potassium cyanide, KCN is not a neutral salt using
appropriate equations.

(iv) Suggest a suitable indicator for the titration.

[7 marks]

6. (a) (i) HNO2 (aq) + H2O (l) NO2+ (aq) + H3O+ (aq)

(ii)

[ ]i / M HNO2 (aq) H2O (l) NO2+ (aq) H3O+ (aq)
[ ]c / M 0.215 - 0 0
[ ]eq / M -x - +x +x
0.215-x - x x

HNO2 dissociated = x 100 = 4.0
.

x = 8.6 x 10-3

Thus, [HNO2] = 0.2064 M

[NO2+] = [H3O+] = 8.6 x 10-3 M

Ka = [ + ][ +]
[ ]

= ( . − )

.

= 3.58 x 10-4

(iii) p H = -log [H3O+]
= -log (8.6 x 10-3)

= 2.07
(b) Ca(OH)2 (aq) → Ca2+ (aq) + 2OH- (aq)

[ ]I /M 0.30 00

[ ]c /M - 0.30 + 0.30 + 2(0.30)

[ ]f /M 0 0.30 0.60

p OH = -log [OH- ]

= -log (0.60)

= 0.22
p H = 14 – p OH

= 14 – 0.22

(c) (i) = 13.78
KOH (aq) + HCN (aq) → KCN (aq) + H2O (l)

(ii) Concentration of KOH = 0

(iii) Dissociation of salt : KCN (aq) → K+ (aq) + CN- (aq)

Dissociation of KCN salt produce K+ and CN- ion. Since CN- ion is
strong conjugate base, thus CN- ion does hydrolyses.

Hydrolysis of CN- ion :

CN- (aq) + H2O (l) HCN (aq) + OH- (aq)

Therefore, KCN is not a neutral but basic salt since OH- is
produced.

(iv) Phenolphthalein @ thymol blue

SUGGESTED ANSWER PSPM I 2020/2021

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