JAWAPAN 20162017

SUGGESTED ANSWER PSPM 1
SESI 2016/2017

1. (a) Sodium carbonate, Na2CO3 dissolves in ethanol to give 2.5 M solution.
Calculate the molarity of the solution if the density of the solution is 1.430 g mL-1.

Molarity = Mol of solute(mol)__ = 2.5 M @ mol L-1
Volume of solution(L)

Assume that Volume of solution = 1L = 1000 mL
Mol of solute (Na2CO3) = 2.5 mol
Mass of solute (Na2CO3) = 2.5 mol x 106 gmol-1
= 265 g

Density = Mass of solution (g)__ =1.430 g mL-1
Volume of solution(mL)

Mass of solution(Na2CO3)= 1430g

Mass of solution = mass of solute (Na2CO3) + mass of solvent(Ethanol)
mass of solvent(Ethanol) = 1430 g – 265 g
= 1165 g
= 1.165 g

Molality = Mol of solute (Na2CO3) (mol)
Mass of solvent (Ethanol)(Kg)

= 2.5 mol
1.165 Kg
= 2.146 m @ molal

(b) The following redox reaction occurs in acidic condition.

Zn + NO3-  Zn2+ + NH4+

i. Balance the redox equation using ion electron method.

4Zn → 4 Zn2+ + 8e

10H+ + NO3- + 8e → NH4+ + 3H2O______

4Zn +10H+ + NO3- → 4 Zn2+ + NH4+ + 3H2O

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ii. A 4.0 g impure sample of zinc reacted completely with 25.0 mL of 0.5 M
HNO3 solution. Calculate the percentage purity of the Zinc sample.

Mol of NO3- = 25.0mL x 0.5M
1000

= 0.0125 mol

1 mol of NO3- ≡ 4 mol of Zn
0.0125 mol of NO3- ≡ 0.0125 x 4

1

= 0.05 mol of Zn

Mass of Zn = 0.05 mol x 65.4 gmol-1
= 3.27 g

% purity of the Zinc sample = 3.27 g x 100%
4.0 g

= 81.75%

2. (a) An atom X combines with oxygen atoms to form two covalent molecules, XO2 and
XO3. XO2 is identified as a polar molecule while XO3 is a nonpolar molecule.
i. Draw a possible molecular geometry for each of the molecule.

XO2 XO3

X O
OO
X
OO

ii. Comment on the polarity of X-O bond. Give a possible reason.

X-O bond is polar.
O and X both are from group 16 but O is located at higher position than
O. Since electronegativity decreases going down the group, then O is more
electronegative than X.

iii. Explain why XO2 is polar while XO3 is nonpolar.

XO2 XO3

X O
OO
X
OO

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XO2 XO3
 Molecular geometry is bent.  Molecular geometry is symmetrical.
 The polarity of bond is identical and
 Vectors cannot cancel out.
in opposite direction, therefore can
 The combined vectors is not cancel one another out.
zero, µ≠0  The sum vectors is zero, µ = 0.

(b) Phosporyl chloride, POCl3, can be prepared by the reaction of phosphorus trichloride,
PCl3 with oxygen at 20-50oC.
Draw the most stable Lewis structure of PCl3 and POCl3, with P as the central atom.

PCl3 POCl3

Cl Cl
Cl P Cl Cl P Cl

O

i. What is the hybrid orbital of P atoms in both compounds?
sp3

ii. Show the overlapping of orbitals in PCl3.

P ground state 3p Cl: 3p
3s 3p 3s

P excited state
3s

P hybrid state

sp3

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iii. Compare the Cl-P-Cl bond angles in PCl3 and POCl3. Explain your answer.

PCl3 POCl3
 Has 4 pairs of e- around the  Has 4 pairs of e- around

central atom, e- arrangement is the central atom, e-
tetrahedral. arrangement is
tetrahedral.
 Has 3 bonding pair and 1 lone  All are bonding pair e-.
pair e-.
 According to VSEPR, the
 According to VSEPR, repulsion repulsion is equal.
between lone pair-bonding pair
is greater than bonding pair-  Bond angle is 109.5o
bonding pair.

 The bonding pair e- is pushed
closer towards each other.

 Bond angle is <109.5o

3. (a) A sample of potassium chlorate, KClO3, was decomposed upon heating, producing

potassium chloride, KCl, and oxygen gas, O2. The volume of gas collected by
displacement of water was 0.250 L at 26oC and a pressure of 765 mmHg. Given
water vapour pressure at 26oC is 25.2 mmHg, calculate

i. The partial pressure of O2.

PT = PO2 + PH2O
PH2O = 765 – 25.2 mmHg

= 739.8 mmHg

ii. The moles of O2 collected.

PV = nRT
n = (738/760 atm) (0.250 L)___
(0.8206 L mol-1 K-1) (26+273.15 K)
= 9.91 x 10-3 mol

iii. The amount of KClO3 (in g) decomposed.

2 KClO3 (s)  2 KCl (s) + 3 O2 (g)

3 mol O2 ≡ 2 mol KClO3
nKClO3 = ⅔ x 9.91 x 10-3

= 6.61 x 10-3 mol

mass KClO3 = 6.61 x 10-3 x [ 39.1 + 35.5 + (3x16)]
= 0.810 g

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(b) Nitrogen gas is a real gas that deviates from ideality. It is the most abundant gas
found in the atmosphere.
i. Why does nitrogen behave as a real gas at room temperature?

Nitrogen gas behave as a real gas at room temperature because nitrogen

gas has mass and volume. Its occupy space. So the intermolecular forces
can’t be ignored and it has volume.

At low temperature, nitrogen gas has low kinetic energy and moves at low
speed. Therefore the intermolecular forces can’t be ignored.

ii. State the conditions under which nitrogen gas can be liquefied.

At very high pressure and very low temperature.

iv. How can nitrogen gas be made to behave as an ideal gas?

At very low pressure, the nitrogen molecules a very far apart. Hence the
intermolecular force can be ignored. And at very low pressure the
volume of container is huge. Hence the volume of gas is not significant
and consider volume of gas equal to volume of container.

At very high temperature, the nitrogen gas molecules has very high
kinetic energy and move at very high speed. Therefore, the intermolecular
forces can be ignored.

4 (a) Methylamine, CH3NH2, is a weak base. This base has the degree of ionization of 4.7%

at 0.16 M concentration. Calculate
i. the concentrations of OH-, CH3NH3+ and H3O+,

() () () ()

[ ]i/ M 0.16 -- -
/M
-x +x +x
[ ]f/ M (0.16 – x) xx

M

Therefore, [ ] = [ ]=

[ ] [H3O+] = 1 x 10-14 M2
 [H3O+] = 1.33 x 10-12 M

ii. the pH and

pH = -log [H3O+]
= -log (1.33 x 10-12)
= 11.90

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iii. the base dissociation constant, Kb.
[]

() )
(
)
( )
(

(b) Would lead(II) iodide, PbI2, precipitate (Ksp PbI2 = 8.5 x 10-9) be produced when a
10.0 mL solution of 1.0 M lead(II) nitrate, Pb(NO3)2 and 40.0 mL solution of 2.0 x 10-3
M sodium iodide, NaI, is mixed?

Pb(NO3)2 (aq) + 2NaI (aq)  PbI2 (aq) + 2NaNO3 (aq)

[Pb(NO3)2] = 0.2 M
[NaI] = 1.6 x 10-3M

PbI2(s) Pb2+(aq) + 2I-(aq)

0.2 M 1.6 x 10-3M

Qsp= [Pb2+][I-]2
= 5.12 x 10-7

Qsp > Ksp
- Solution is supersaturated.
- Precipitate will form.

5 (a) State the difference between the ground state and excited state of an electron in an

atom. An electron of a hydrogen atom is excited to n = 6 and falls to a lower energy
level forming the Paschen series. Calculate, in kJ mol -1, the energy of electron at the

excited level and the energy emitted as the result of the transition. Determine the

shortest wavelength in nanometres (nm) in the Paschen series for the atom

hydrogen atom.

A ground state electron of an element is an outer orbital electron that is in the
lowest energy state possible for that electron while an excited state electron has
absorbed additional energy and exists at a higher energy level than a ground
state electron.

Energy of electron at the excited level

En = -RH/n2
= -2.18 x 10-18/62
=-6.055 x 10-20 J/1000
= -6.055 x 10-23kJ x (6.02x 1023)
=-36.45 kJ/mol

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Energy emitted

E = RH(1/ni2-1/nf2)
= 2.18 x 10-18(1/62 -1/32) x (6.02 x 1023)/1000
= -109.6 kJ/mol

Shortest wavelength

1/ = RH (1/n12 -1/n22) n1 < n2
1/ = 1.097 x 107 (1/32-1/2)
 = 8.2 x 10-7 x 109

 = 820.45 nm

(b) Define the first ionization energy.
Sketch a graph to show the energies involved in the removal of the first four
electrons of an aluminium atom. Explain your answer.

Ionization energy is minimum energy required to remove 1 mol electron from 1
mol of gaseous atom to form 1 mol of positively charge gaseous ion.

Energy

12 3 4 No. of electron

removed

Based on the graph above, the sharp increase occur between lE3 to lE4. 3
electrons in outermost shell and fourth electron in inner shell.

6 (a) CH3F and CF4 are tetrahedral molecules. Using the Lewis structure and dipole
moment of these molecules, deduce their polarity.

For CH3F molecule:

Element No. of valence electrons, e-
C 4e- x 1 = 4e-
H
F 1e- x 3 = 3e-
7e- x 1 = 7e-

TOTAL = 14e-

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Lewis structure:

Number of electron groups = 4
Electron pair arrangement = tetrahedral

Molecular shape = tetrahedral

Each C-H bond and C-F bond are polar. However, C-F bond has greater dipole
moment than C-H bond due to F atom is more electronegative than H atom.

The four dipole moments do not cancel each other.
The resultant dipole moment, µ is not equal to zero.
Therefore, the molecule of CH3F is a polar molecule.
For CF4 molecule:

Element No. of valence electrons, e-
C 4e- x 1 = 4e-
F 7e- x 4 = 28e-

TOTAL = 32e-

Lewis structure:

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Number of electron groups = 4
Electron pair arrangement = tetrahedral

Molecular shape = tetrahedral

Each C-F bond is polar and has the same dipole moment.
Since the molecule is symmetrical, each C-F bond dipole moment cancel each
other.

The resultant dipole moment, µ is equal to zero.
Therefore, the molecule of CF4 is a non-polar molecule.

(b) Iodine pentafluoride, IF5, is a colourless liquid with a pungent odour. By using VSEPR
theory, predict the geometry and bond angles of this compound.

Element No. of valence electrons, e-
I 7e- x 1 = 7e-
F 7e- x 5 = 35e-

TOTAL = 42e-

Lewis Structure :

Based on the Lewis structure, the electron pair arrangement is octahedral since
there are six electron pairs around the central atom I which is consist of five
bonding pairs and one lone pair.
According to the VSEPR theory, the electrons are located as far as possible to
minimise the repulsion between them. Hence the repulsion between lone pair

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electrons and bonding pair electrons is greater than the repulsion between the
bonding pair electrons and bonding pair electrons.

In order to minimise the repulsions, the six electron pairs are arranged as
follows:

Therefore, the molecular geometry of IF5 molecule is square pyramidal with all
of the bond angles are similar which is less than 90º.

Show how the orbitals of the central atom undergo hybridisation. Draw a labelled
diagram showing the overlapping of orbitals in the molecule.

Valence orbital diagram for iodine, I:

Ground ↑↓ ↑↓ ↑↓ ↑
state: 5p

5s 5d

Excited ↑↓ ↑↑↑ ↑↑
state: 5p 5d
5s

Hybrid ↑↓ ↑↑↑↑ ↑
state: sp3d2orbitals

5d

The six sp3d2 hybrid orbitals are produced by the hybridisation of one 5s, three
5p and two 5d orbitals of the valence shell of the iodine atom.

Valence electron in fluorine, F:

Ground ↑↓ ↑↓ ↑↓ ↑
state : 2p

2s

Each I-F bond is formed by the overlapping of one sp3d2 orbital of the iodine
atom with one 2p orbital of the fluorine atom.

The results of these overlapping formed five σ bonds.
One of the sp3d2 orbitals of the iodine atom is occupied by lone pair electrons.

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< 90°

Electron pair arrangement: Octahedral
Molecular geometry: Square pyramidal

7 (a) Ethane, C2H6, burns in the air according to the equation below:
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)

A mixture of C2H6 and O2 at 30 mmHg and 124 mmHg, respectively, was added into a
7.5L flask at a temperature of 27oC. The combustion reaction was carried out until
completion. Calculate the total pressure of the gas mixture obtained at the end of
the reaction, at the same temperature. Give your answer in mmHg.

PV = nRT
n of C2H6 =

=
= 0.012 mol (available)

PV = nRT
n of O2 =

=
= 0.0497 mol (available)

From balanced equation ;

2 mol C2H6 7 mol O2
0.012 mol C2H6 0.012 x 7/2

= 0.042 mol O2 (needed)

moles of O2 needed < moles of O2 available
O2 is excess reactant while Limiting reactant is C2H6

nO2 remained = 0.0497 mol – 0.042 mol
= 0.0077 mol

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From balanced equation;

2 mol C2H6 4 mol CO2
0.012 mol C2H6 = 0.024 mol CO2

2 mol C2H6 6 mol H2O
0.012 mol C2H6 = 0.036 mol H2O

nT = nO2 + nCO2 + nH2O
= 0.0077 + 0.024 + 0.036
= 0.0677 mol

PT = nTRT
V

= 0.0677 mol x 0.08206 Latmmol-1K-1 x 300.15K
7.5 L

= 0.2223 atm x 760mmHg
1 atm

= 168.948 mmHg

(b) For the following reaction, the value of Kp is 6.4 x 10-6 at 227oC.
2Cl2(g) + 2H2O(g)  4HCl(g) + O2(g)

In a 2.0L container, the initial quantity of Cl2 is 0.50 mol, H2O is 0.40 mol, HCl is 0.50
mol and O2 is 0.015 mol. Determine whether the reaction is at equilibrium at this
temperature. If not, predict to which direction will the reaction proceed?

Kp = Kc (RT)n
Kc = ( )

= 6.4 x 10-6______
(0.08206 x 500.15)5-4

= 1.5594 x 10-7

[ Cl2 ] = 0.50 mol
2.0 L

= 0.25 M

[H2O] = 0.40 mol
2.0 L

= 0.20 M

[ HCl] = 0.50 mol
2.0 L

= 0.25 M

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[O2] = 0.015 mol
2.0 L

= 0.0075M

Qc = [HCl]4[O2]
[Cl2]2[H2O]2

= (0.25)4(0.0075)
(0.25)2(0.2)2

= 0.0117

Qc > Kc
- The reaction is not at equilibrium.
- The equilibrium position will shift to left to reestablish the equilibrium.

8. Name the processes that occur when HNO2 solution is added separately into water, NaOH
solution and NaNO2 solution. Write the chemical equation for each process. In three
separate beakers, 20.0 mL of 0.08 M HNO2 was added to 40.0 mL of water, 40.0 mL of 0.03
M NaOH and 40.0 mL of 0.03 M NaNO2 , respectively. Determine the pH of the solution in
each beaker.
[Ka for HNO2 = 4.5 X 10 -4]

Dilution of weak acid
i) HNO2(aq) + H2O (l) → NO2- (aq) + H3O + (aq)

Neutralisation
ii) HNO2 (aq) + NaOH (aq) → NaNO2 (aq) + H2O (l)

Formation of buffer solution
iii) HNO2 (aq) + NaNO2 (aq) → acidic buffer solution

HNO2(aq) + H2O (l) → NO2-(aq) + H3O + (aq)

ni 2x10-2x0.08 - 00
= 0.0016
- +x +x
nΔ -x - x x
nf 0.0016 –x x/0.06 x/0.06
[ ]f (0.0016-x)/ 0.06

Ka = (x/0.06)(x/0.06)

( 0.0016-x)/0.06
4.5x 10-4 = (x/0.06)(x/0.06)

( 0.0016-x)/0.06
Since Ka is larger than 1 X 10-5,
x2 + 2.7 x 10-5 x – 4.32 x 10-8 = 0
x= 1.9478 x 10-4 mol
[H3O+] = 3.2463 x 10-3 M

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pH = -log [H3O+]
= -log (3.2463 x 10-3)
= 2.49

HNO2(aq) + NaOH(aq) → NaNO2(aq) + H2O (aq)

ni 0.0016 0.0012 00
nΔ -0.0012
-0.0012 +0.0012 +0.0012
nf 0.0004
0 0.0012 0.0012
[ ]f 0.0004/ 0.06
= 6.6667x10-3 - 0.0012/0.06 0.0012/0.06

= 0.02 =0.02

pH  -log Ka  log [NO - ]
2

[HNO 2 ]

 - log 4.5x10-4  log 0.02
6.6667 x 10 -3

 3.82

[ NO2-] = 1.2x10-3 M [HNO2]= 0.02667M

pH  -log Ka  log [NO - ]
2

[HNO 2 ]

 - log 4.5x10-4  log 1.2x10-3
0.02667

 2.0

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