PYQ SK015 2015/2016

SUGGESTED ANSWER PSPM 1
SESI 2015/2016

1. Compound Q contains carbon, hydrogen and nitrogen. Combustion of 0.250 g of Q produces
0.344 g of water, H2O, and 0.558 g of carbon dioxide, CO2. Determine
a) the empirical formula of Q.

Given:
Mass of Q = 0.250 g
H2O = 0.344 g
CO2 = 0.558 g

Empirical formula

Mass of C = 12g mol1  0.558g = 0.152 g
44g mol1

Mass of H = 2g mol1  0.344 g = 0.0382 g
18g mol1

Mass of N = 0.250 g – (0.152 + 0.0382)g = 0.0598 g

Element C H N
Mass(g) 0.152 0.0382
No. of mole 0.0127 0.0382 0.0598
Mole ratio 2.9 4.27×10─3
8.9
3 9 1

1

∴ Empirical formula = C3H9N

b) the molar mass of Q if the molecular formula is the same as the empirical formula
and

Molar mass of Q

Molecular formula = empirical formula = C3H9N
Molar mass = 3(12 g mol─1) + 9 (1 g mol─1) + 1 (14 g mol─1)

= 59 g mol─1

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c) the number of hydrogen atoms present in the above sample Q.

Number of hydrogen atom in Q

Mol of Q = 0.25g = 4.24×10─3 mol
59g mol1

1 mol Q ≡ 9 mol H
4.24×10─3 mol Q ≡ 0.0381 mol H

∴number of H atom = (0.0381 mol)(6.02×1023 atom/mol)
= 2.296×1022 atoms

2. An atom X has 5 valence electrons, X reacts with fluorine gas to form XF3 and XF5
compounds.
a) For each compound,
i. draw the Lewis structure,

XF3
Total valence electrons = 5 + 3(7) = 26

FXF

F

XF5
Total valence electrons = 5 + 5(7) = 40

FF

FXF

F

ii. predict the electron pair geometry and the molecular geometry and

XF3
 Number of electron pair around central atom = 4
 Electron pair arrangement = tetrahedral
 Based on VSEPR theory, electron pair will be located as far as

possible in order to minimize the repulsion between them.
 The strength of lone pair-bonding pair repulsion > bonding pair –

bonding pair repulsion.
 Molecular geometry = trigonal pyramidal

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XF5
 Number of electron pair around central atom = 5
 Electron pair arrangement = trigonal bipyramidal
 Based on VSEPR theory, electron pair will be located as far as

possible in order to minimize the repulsion between them.
 The strength bonding pair – bonding pair repulsions are equal.
 Molecular geometry = trigonal bipyramidal

iii. draw the molecular geometry and state the bond angle(s).

XF3

X
FF

F

Bond angle = <109.5o

XF5

FF

FXF

F

Bond angle = 120o and 90o

b) Predict the change in hybridization (if any) of the X atom in the following reaction:

XF3  F2  XF5
XF3 = sp3 to XF5 = sp3d

3. a) A gas mixture contains 82% (w/w) methane, CH4, and 18% (w/w) ethane, C2H6. If a
15.50 g sample of the gas mixture is placed in a 15-L container at 20.0oC, calculate
Given:

% w/w of methane = 82%, % w/w of ethane = 18%, Mass of gas mixture = 15.50 g
V = 15 L, T = 20.0oC = 293.15 K

i. the total moles of gases,

Mass of methane = 82 15.5g = 12.71 g

100

Mole of methane = 12.71g = 0.794 mol
16g mol1

Mass of ethane = 18 15.5g = 2.79 g

100

Mole of ethane = 2.79 g = 0.093 mol
30g mol1

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∴ nT = 0.794 mol + 0.093 mol = 0.887 mol
ii. the total pressure (atm) and

PT  nTRT
V

 0.887 mol 0.08206 atm L mol1 K1 293.15K

 15L

= 1.423 atm

iii. the partial pressure of each gas in the container.

PCH4   0.794 mol  1.423atm = 1.274 atm
0.887 mol

PC2H6  0.093mol 1.423atm = 0.149 atm
0.887 mol

b) Explain each of the following observations:
i. Methane, CH4 (16 g mol−1) has lower boiling point than propane, C3H8
(44 g mol−1) whereas water, H2O (18 g mol−1) has higher boiling point than
hydrogen sulphide, H2S (34 g mol−1).
Boiling point of methane < propane.
Reason:
 Molecular mass of methane < propane.
 Strength of Van der Waals forces of methane < propane.

Boiling point of H2S< H2O.
Reason:
 H2O can form hydrogen bond between molecules while H2S can

only form weak Van der Waals forces between molecules.
 Strength of Van der Waals forces < hydrogen bond.

ii. Carbon dioxide, CO2, molecule and diamond are both covalently bonded.
However, solid CO2, is softer and exhibits lower melting point as compared to
diamond.

CO2 solid is softer and has lower melting point than diamond.
Reason:
 CO2 molecules are held by weak Van der Waals forces.
 Each C atoms in diamond are held by strong covalent bond to

form tetrahedral arrangement in gigantic covalent network.

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4. a) FIGURE 1 shows a titration curve for 20.0 mL of an unknown concentration of NaOH
solution titrated against a standard solution of 1.00 M HCl.

pH

Z

22.5 25.0 Volume of HCl (mL)

FIGURE 1
i. Define standard solution, equivalent point and end point.

Standard solution
A solution of known concentration for use in volumetric analysis.

Oxford Dictionary

A solution of accurately known concentration. Raymond Chang

A solution of known concentration. Brown Le. May

Equivalent Point
The point in a titration at which reaction is complete. Oxford Dictionary

The point at which the acid has completely reacted with or been

neutralized by the base. Raymond

Chang

The point in a titration when the number of moles of the added species is

stoichiometrically equivalent to the other species. Silberberg

The point in a titration at which the added solute reacts completely with

the solute present in the solution. Brown Le. May

End Point

The point in a titration at which reaction is complete as shown by the

indicator. Oxford Dictionary

The pH at which the indicator changes colour. Raymond Chang

The point in a titration at which the indicator changes colour.
Silberberg

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ii. What is the pH at equivalent point, Z for this titration?

pH at point Z = 7

iii. Calculate the molarity of the NaOH solution used in the titration.
molarity of NaOH used

mol of HCl = (0.0225 L)(1.00 M) = 0.0225 mol

1 mol HCl ≡ 1 mol H+
0.0225 mol HCl ≡ 0.0225 mol H+

at equivalent point,
mol of H+ ion ≡ mol of OH─ ion
0.0225 mol of H+ ion ≡ 0.0225 mol of OH─ ion

1 mol NaOH ≡ 1 mol OH─
0.0225 mol NaOH ≡ 0.0225 mol OH─

Molarity of NaOH =  0.0225 mol  = 1.125 M
 0.02 L 

iv. Calculate the pH of the solution after the addition of 25.0 mL of HCl.
pH of solution after addition 25.0 mL of HCl.

Mol of HCl = (0.025 L)(1.00 M) = 0.025 mol

NaOH(aq)  HCl(aq)  NaCl(aq)  H2O(l)

ni 0.0225 0.025 - -
∆n -0.0225 -0.0225 +0.0225 -
2.5×10─3 0.0225 -
nf 0

[HCl] = 2.5103 mol = 0.056 M
0.045 L

1 mol HCl ≡ 1 mol H+
0.056 M HCl ≡ 0.056 M H+

pH = -log (0.056 M) = 1.26

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b) Pyridine, C5H5N, is a weak base. This base ionises in water to produce C5H5NH+ and
OH−. Calculate the pH of a 1.00 M pyridine solution if the dissociation constant, Kb, of
pyridine is 1.50×10−9.

C5H5N(aq) + H2O(l) → C5H5NH+(aq) + OH─(aq)

[ ]i 1.00 - - -

[ ]c -x - +x +x

[ ]e 1-x -x x

Kb = C5H5NH  OH 

C5H5N

1.50×10─9 = xx
1 x

Since α <<<< 0.5, therefore, 1-x≈1.0

1.50×10─9 = xx
1

x = [OH─] = 3.87×10─5 M

pOH = - log (3.87×10─5 M) = 4.41
pH = 9.59

5. a) Calculate the first three wavelengths of the possible transitions of an electron in the
Paschen series for a hydrogen atom. Show these wavelengths in a sketch of the line
spectrum for this emission series. Explain how these transitions occurred.

First transition, ni = 4 to nf = 3

1  RH  1  1   n2
  n12 , n1
 n 2 
2

  1 1 
1.097 107 m1  32  42 

  1.88×10─6 m = 1880 nm

Second transition, ni = 5 to nf = 3

1  RH  1  1   n2
  n12 , n1
 n 2 
2

  1 1 
1.097 107 m1  32  52 

  1.282×10─6 m = 1282 nm

58

Third transition, ni = 6 to nf = 3

1  RH  1  1   n2
  n12 , n1
 n 2 
2

  1 1 
1.097 107 m1  32  62 

  1.094×10─6 m = 1094 nm

Line spectrum

st 2nd line 3rd line

1 line

1.88 1.28 1.09

wavelength,  increase (x10 -6 m)

Explanation:
 When gas in discharge tube is heated, the electrons of the gaseous atom

will absorbed the heat supply and excited from lower energy level to
higher energy level.
 The electrons in excited state is unstable and will fall back to lower energy
level, n=3 by releasing specific amount of energy with specific wavelength
that produce line in the Paschen series.

b) The following data are given for atomic and ionic radii of halogens and halides

respectively.

TABLE 1

Atom Ion

Species F Cl Br I F─ Cl─ Br─ I─

Radii 0.72 0.99 1.14 1.35 1.36 1.81 1.95 2.16
(Å)

Discuss the trend of radii in terms of nuclear charge and valence electron of the
species.
Explain the ionization energy and electron affinity trends of these species going
down the group in the periodic table.

Size of I > Br > Cl > F
 All atoms are in the group 17.
 Down group 17 from F to I, the number of shell is increased.
 Thus, shielding effect in F < Cl < Br < I.
 Strength of nucleus attraction towards valence electrons in F > Cl > Br >

I.

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Size of atom < its respective ion
 When electron is added into the particular atom, the mutual repulsion

of the atom will increased.
 Expand the electron cloud of atom and enlarge the size.

Ionization energy of I < Br < Cl < F
 Size of atom I > Br > Cl > F.
 Outermost electron is located further from the nucleus.
 Strength of nucleus attraction towards valence electron in atom

I < Br < Cl < F.
 Energy needed to remove the outermost electron in I < Br < Cl < F.

Electron affinity of I < Br < Cl < F
 Size of atom I > Br > Cl > F.
 Outermost electron is located further from the nucleus.
 Strength of nucleus attraction towards valence electron in atom

I < Br < Cl < F.

6. Name the bonding theory that explains orbital hybridization.

Valence bond theory

Using orbital diagram, describe the hybridization of the central atom in sulphur
hexafluoride, SF6 and formaldehyde, H2CO, molecules.

SF6

Total valence electron = 6 + 6(7) = 48

0

F F0

0

F S0 F

0

0F F

0

 Number of electron pairs around central atom : 6

 Type of hybrid of S : sp3d2

Valence orbital diagram,
F ground state:

2s 2p Sexcited state: Shybrid:
Sground state:

3s 3p 3s 3p 3d sp 3d2

60

H2CO

Total valence electron = 2 + 4 + 6 = 12

0
0O

H C0 H

0

 Number of electron pairs around central atom : 3

 Type of hybrid of C : sp2

 Type of hybrid of O : sp2

Valence orbital diagram,
H ground state:

1s

Oground state: Oexcited state: Ohybrid:

2s 2p 2s 2p sp 2 2p
Cground state: Cexcited state: Chybrid:

2s 2p 2s 2p sp2 2p

Draw the orbitals overlaps of each molecule and state the expected bond angles.

2p

F 2p sH
F 

 s p2
2p
 
F
F  sp3d2 2p  sp2 C 
s p2
s p3d2 s p3d2 s
H
S s p2 s p2

 sp3d2 s p3d2 O

s p3d2 s p2

F 
2p F

2p Bond angle: 120o

Bond angle: 90o

61

7. a) Crystalline solid is classified based on the assembly of their respective basic particles.
State four types of crystalline solid and discuss the nature of the particle assembly in
each crystalline solid. Give one example for each type of the crystalline solid.

4 types of crystalline solids:
Ionic crystal

 Composed of positively charged particle (cation) and negatively charged
particle (anion).

 This particles are hold by electrostatic attraction/ionic bond.
 Example: NaCl.

Covalent crystal
 Atoms are held together by covalent bond to form giant covalent network.
 Example: diamond.

Simple covalent crystal
 Consist of molecules that are held together by van der Waals force or
hydrogen bond.
 Atoms in molecule are held by covalent bond.
 Example: solid I2.

Metallic crystal
 Metal atoms in a crystal can be imagined as an array of positive ions
immersed in a sea of delocalised electrons.
 Both positive ions and delocalised electrons are attracted by metallic
bond.
 Example: solid Na.

b) Consider the following reaction system;

2NO(g)  O2 (g)  2NO2(g)

This system is made up from 2.0×10─3 mol of each gaseous component in a 2-L closed
vessel. If the equilibrium constant, Kc, for the system is 7.7×107 at 600 K, prove that

the system is not in equilibrium at this temperature.

 [NO] = [O2] = [NO2] =
2.0103 mol = 1×10−3 M

2L

Qc =  NO2 2
NO2 O2 

 1103 M 2
   = 1103 M 2 1103 M

= 1000

∴Qc < Kc, thus the system is not in equilibrium.

62

When the system is cooled to 450 K and left to reach equilibrium, the final amount of
O2 detected in the vessel is 9.6×10─4 M. Calculate the equilibrium constant, Kc, of the

system at 450 K.

At 450 K, the equilibrium position will move to right because amount of O2 was
decreased after reach equilibrium at this temperature.

2NO(g) + O2(g) → 2NO2(g)
[ ]I 1×10−3 1×10−3 1×10−3

[ ]c -2x -x +2x
[ ]e 1×10−3 – 2x 1×10−3-x 1×10−3+2x

9.6×10−4 M = 1×10−3 M – x
x = 4×10−5M

[NO] = 9.2×10−4 M [O2] = 9.6×10−4 M [NO2] = 1.08×10−3 M

Kc at 450 K =  NO2 2
NO2 O2 

 1.08103 M 2
   = 9.2104 M 2 9.6104 M

= 1435.49

8. a) Milk of magnesia consists of saturated solution Mg(OH)2 and gelatinous Mg(OH)2.
Determine the molarity of Mg2+ in a saturated solution of Mg(OH)2 and in the
saturated solution of Mg(OH)2 containing 0.01 M NaOH.
[Given: Ksp Mg(OH)2 = 8.9×10─12]

In saturated solution of Mg(OH)2

Mg(OH)2 (s)  Mg 2 (aq)  2OH

xM xM 2x M

Ksp = [Mg2+][OH-]2
8.9×10−12 = (x)(2x)2

x = 1.305×10-4 M

∴ [Mg2+] = 1.305×10−4 M

In saturated solution containing 0.01 M NaOH

63

NaOH (aq)  Na (aq)  OH  (aq)
0.01 M 0.01 M 0.01 M

Mg(OH)2 (s)  Mg 2 (aq)  2OH

y M y M 0.01 + 2y M

Ksp = [Mg2+][OH-]2
8.9×10−12 = (y)(0.01+2y)2
Since k is too small, assume 2y + 0.01 M ≈ 0.01 M

8.9×10−12 = (y)(0.01)2
x = 8.9×10−8 M

∴ [Mg2+] = 8.9×10−8 M

b) A buffer solution is prepared by dissolving 0.125 mol of sodium nitrite, NaNO2, in 500
mL of 0.25 M nitrous acid, HNO2. What is the pH of the solution?
The above buffer solution can resist the change in pH when a small amount of strong

acid or base is added into it. Explain how this buffer solution maintain its pH. If an

amount of 0.001 mol of NaOH is added to the solution, determine the change in the

pH value of the buffer solution.
[Given: Ka HNO2 = 5.0×10─4]

HNO 3 (aq)  H2O(l)  H3O (aq)  NO3 (aq)

0.125mol 
[NaNO2] = 0.5L

= 0.25 M

pH = -log ka + log NaNO2 
HNO2 

= - log (5.1×10−4) + log 0.25
0.25

= 3.29

When small amount of strong acid is added into the solution. H+ ion of strong
acid will be neutralized by conjugate base, NO 2 - ion in the buffer solution and

produce HNO2.

NO2 (aq)  H (aq)  HNO 2 (aq)

When small amount of strong base added in the solution. OH- of strong base will
be neutralized by weak acid, HNO2 in the buffer solution and produce NO2- and

water.

64

HNO 2 (aq)  OH (aq)  NO2 (aq)  H2O (l)

If 0.001 mol NaOH added.

HNO 2 (aq)  OH (aq)  NO2 (aq)  H2O (l)

ni 0.125 0.001 0.125 -

nc -0.001 -0.001 +0.001 -

nf 0.124 0 0.126 -

pH = -log ka + log NaNO2 
HNO2 

 = - log (5.1×10−4) + log
0.126 45

 45
0.248

= 3.30

∴ change in the pH value of the buffer solution = 0.01

65