ANSWERS for ENRICHMENT QUESTIONS Sem 1

ii. Explain whether OF2 is a polar or non-polar molecule.
 Polar molecule.
 F is more electronegative than O. Therefore, O-F bond is polar.

FO
F

 Since, molecule OF2 is unsymmetrical, thus, dipole moment can’t cancel
each other, μ≠0.
[8 marks]

c) Aluminium and sodium are metals.
i. Explain the formation of metallic bond in sodium using the electron sea
model.

Na+ Na+ Na+ Na+ Na+
e e e e ee

Na+ Na+ Na+ Na+ Na+
ee e eee

Na+ Na+ Na+ Na+ Na+
e ee

 Metallic bond form in Na metal is the electrostatic forces between
positively charge ion, Na+ and negatively charge free moving electrons.

ii. Why aluminium has higher boiling point than sodium?
 Aluminium ion, Al3+ is smaller than sodium ion, Na+. Thus, Al3+ has
stronger nucleus attraction towards the free moving electrons.
 Al has 3 valence electrons and sodium has only 1 valence electron.
 Strength of metallic bond is directly proportional to the number of
valence electron and inversely proportional to size of ion. Thus, Al has
stronger metallic bond than Na.

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4. a) A 10-L cylinder contains 4 g of hydrogen gas and 28 g of nitrogen gas. If the
temperature is 31oC,

i. Determine the total pressure of the gas mixture.

Mole of hydrogen gas = 2 mol
Mole of nitrogen gas = 1 mol
Total mol = 3 mol

PT = nT RT
V

 3 mol 0.08206 atm L mol-1 K-1 304.15 K

= 10 L

= 7.49 atm

ii. Calculate the partial pressure of hydrogen gas.

PH2  X H2 PT

  2  7.49 atm
 3 

= 4.99 atm

iii. What will happen if the gaseous mixture is heated to 550oC?

 At high temperature, the average kinetic energy of gas particles will
increase and cause the reaction between hydrogen and nitrogen to
occur producing gas ammonia.


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.

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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.

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 H 2O 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

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

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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.

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

Molarity = 2.74103 mol
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

Molarity of NaOH = 1.26103 mol
0.065L

= 0.0194 M

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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  HA (aq)  OH(aq)

 pH > 7.

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) 2AsO3(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

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Suggested Answer
Objective Question

ANSWER FOR OBJECTIVE QUESTIONS
SESSION 2018/2019

SET 1 B SET 2 C SET 3
1 A 1 B 1C
2 A 2 D 2B
3 C 3 D 3B
4 D 4 C 4D
5 A 5 A 5A
6 C 6 C 6A
7 D 7 B 7C
8 C 8 C 8C
9 A 9 A 9A
10 C 10 B 10 D
11 C 11 C 11 A
12 B 12 B 12 A
13 C 13 D 13 A
14 A 14 B 14 C
15 B 15 D 15 D
16 C 16 D 16 D
17 D 17 D 17 A
18 D 18 A 18 C
19 D 19 B 19 A
20 A 20 C 20 C
21 A 21 D 21 C
22 B 22 B 22 D
23 D 23 D 23 D
24 C 24 B 24 B
25 D 25 B 25 A
26 A 26 C 26 A
27 B 27 B 27 D
28 D 28 B 28 A
29 D 29 C 29 C
30 30 30 D

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