JAWAPAN 20132014

SUGGESTED ANSWER PSPM 1
SESI 2013/2014

1. (a) Vinyl chloride, C2H3Cl, is prepared by the reaction of ethyne gas, C2H2, with
hydrochloric acid, HCl. In preparation of vinyl chloride, 70.0 g of C2H2 is reacted with
102.0 g of HCl.
i. Write a balanced chemical equation for the reaction.

C2H2 (g)  HCl(g)  C2H3Cl(g)
ii. Identify the limiting reagent in the reaction.

n C2H2 = 2.692 mol
n HCl = 2.794 mol

1 mol C2H2 ≡ 1 mol HCl
2.692 mol C2H2 ≡ 2.692 mol HCl

Moles of HCl available (2.794 mol) > moles of HCl needed (2.692 mol)
∴HCl is excess reactant and C2H2 is limiting reactant

iii. Calculate the mass of the product.
1 mol C2H2 ≡ 1 mol C2H3Cl
2.692 mol C2H2 ≡ 2.692 mol C2H3Cl

n C2H3Cl = 2.692 mol
mass C2H3Cl = 2.692 x (12x2+ 3+35.5) = 168.25 g

iv. Calculate the mass of the excess reactant.

n HCl remained = 2.794- 2.692 = 0.102 mol

mass HCl remained =0.102 x 36.5 = 3.72 g

(b) Magnesium is an element with a proton number of 12. In a sample of magnesium, it
was found that magnesium atoms have three different nucleon numbers which are
24, 25 and 26.
i. write the symbol for one of these three atoms, showing its proton number
and nucleon number.

24 Mg
12

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ii. A sample of magnesium contains 78.6% of atoms with a nucleon number of
24, 10.1% of atoms with a nucleon number of 25 and 11.3% of atoms with a
nucleon number of 26. Calculate the relative atomic mass of magnesium.

Average atomic mass = Isotopic mass  % abundance

% abundance

  24  78.6  2510.1  2611.3 = 24.3u
78.6 10.111.3

relative atomic mass  24.3u
1
12 x 12.00 u

 24.3

iii. Sketch and label the expected mass spectrum of the magnesium.

% abundance

78.6

10.1 11.3

24 25 26 mass

2. (a) Ammonia, NH3, is a colourless gas with a characteristic pungent smell.
i. Using a Lewis dot symbols show the formation of ammonia from the
respective elements.

N + H+ H + H H NH

H

ii. Predict the hybridization of the central atom of ammonia and draw a diagram
of the molecule showing the overlapping of the orbitals.

H

HN

H

 NH3 has 4 electron pair which are 1 lone pair and 3 bonding pair
around central atom.

 Electron pair arrangement : Tetrahedral
 Molecular geometry: Trigonal pyramidal
 Type of hybridisation: sp3

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Valence orbital diagram: H: sp 3
N(ground state): H 1s
N
2s 2p sp 3

N(excited state):  3 3 
2s s
H sp sp s
H
2p

N(hybrid state):

sp3 s
H

iii. Describe the polarity of the ammonia molecule.

 N is more electronegative than H.

 Bond dipole cannot cancel each other.
 ≠ 0
 Polar molecule

(b) Use the valence shell electron repulsion (VSEPR) theory to explain the difference in

the bond angles between the following pairs of compound/ion.
i. CO2 and CO32-

2-
O

OC O OC

O

CO2 Has 2 electron pair around central atom and all are bonding pair .
 Electron pair arrangement: linear.
 Based on VSEPR theory, the electron pairs are located as far as
 possible in order to minimize the repulsion.
The strength of repulsion between bonding pair-bonding pair
 electrons are equals.

 Molecular geometry: Linear.
 Bond angle: 180o.

CO32-
 Has 3 electron pair around central atom and all are bonding pairs.
 Electron pair arrangement: trigonal planar.
 Based on VSEPR theory, the electron pairs are located as far as
possible in order to minimize the repulsion.

 The strength of repulsion between bonding pair-bonding pair
electrons are equals.

 Molecular geometry: trigonal planar.
 Bond angle: 120o.

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ii. PCl3 and H2O

Cl H
Cl P OH

Cl

PCl3
 Has 4 electron pair around central atom which are 3 bonding pairs

and 1 lone pair .

 Electron pair arrangement: tetrahedral.

 Based on VSEPR theory, the electron pairs are located as far as

possible in order to minimize the repulsion.

 The strength of repulsion between lone pair-bonding pair >

Bonding pair-bonding pair.

 Molecular geometry: trigonal pyramidal.
 Bond angle: < 109.5o.

H2O Has 4 electron pair around central atom which are 2 bonding pairs
 and 2 lone pairs .
Electron pair arrangement: tetrahedral.
 Based on VSEPR theory, the electron pairs are located as far as
 possible in order to minimize the repulsion.
The strength of repulsion between lone pair-lone pair>lone pair-
 bonding pair > Bonding pair-bonding pair.
Molecular geometry: bent.
 Bond angle: 104.5o.


3. (a) i. Define vapour pressure.

Pressure exerted by vapour molecule in equilibrium with its liquid state.

ii. Explain how temperature affects the vapour pressure of a liquid.

 Temperature increase.
 Average kinetic energy increase.
 More molecule can escape as vapour.
 Vapour pressure increase.

(b) i. State Dalton’s law of partial pressure.

Total pressure of a mixture of non-reacting gases is equal to the sum of
the pressures that each gas would exert if it were present alone.

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ii. FIGURE 2 shows two connecting vessels containing different gases. When the

valve between the vessels is opened, the gases are allowed to mix. Ignoring

the volume taken by the valve, calculate the partial pressure of each gas and
the final pressure of the mixture at 25 oC.

FIGURE 2

PH2  nRT   0.50  0.08206  298.15  3.1atm
V 4

PCO2  nRT   0.25  0.08206  298.15  1.5 atm
V 4

PT = 3.1 + 1.5 = 4.6 atm

(c) Differentiate between a crystalline solid and an amorphous solid in terms of their
particle arrangement. Draw the structures and state the types of crystal for diamond
and sodium chloride.

Crystalline Solid Amorphous Solid
Have regular 3-D arrangement. Does not have regular 3-D
arrangement.

C

C
C

C
C

Giant Covalent Crystal

Cl Cl- -+ +
Na
Na

Cl Cl- -
+ +
Na Na

Ionic Crystal

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4. (a) Define buffer solution.

A solution that maintain its pH when a small amount of a strong acid or a strong
base is added to it.

(b) Explain how the pH of a mixture of aqueous ammonium chloride and aqueous
ammonia remains constant when a small amount of strong base is added. Give
suitable equations in your explanation.

The mixture of aqueous ammonium chloride and ammonia form buffer solution.
When small amount of strong base is added, the OH- ion from the strong base is
neutralized by NH4+ of buffer to form NH3.

NH4 (aq)  OH  (aq)  NH3(aq)  H2O(l)

Thus the pH of the mixture is maintained.

(c) A buffer solution of pH 4.84 was prepared by dissolving a certain amount of sodium
ethanoate in 1.0 dm3 of 0.20 M ethanoic acid.

i. Calculate the mass of sodium ethanoate needed to prepare the above buffer

solution.

pH   log Ka  log CH3COO 

CH3COOH 

 4.84  log CH3COO 
1.8 105  log
0.2

[CH3COO-] = 0.252 M

n CH3COO- = n CH3COONa = 0.252 mol

mass CH3COONa = 0.252 x (24+32+23+3) = 20.66 g

ii. Determine the change in pH when 1.0 cm3 of 1.0 M hydrochloric acid is
added to 100 cm3 of the buffer solution.

[Ka for CH3COOH = 1.8 x 10-5M]

CH3COO (aq)  H  (aq)  CH3COOH (aq)

ni 0.252 1 x 10-3 0.2

nc -1 x 10-3 -1 x 10-3 +1 x 10-3

nf 0.251 0 0.201

[ ] 0.251 0.201

 pH  log 0.251
1.8 105  log 0.201 = 4.84

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5. (a) Bohr used the information from a line spectrum of a hydrogen atom to
explain the electronic structure of a one-electron system. A blue line in the spectrum
of hydrogen atom was observed as a result of a transition of electron from the fourth
to the second shells of an atom.
What is meant by a line spectrum?

Line spectrum is a spectrum that consist of discontinuous and discrete lines with
specific wavelength and frequency.

Calculate the wavelength and energy for this blue line.

ni = 4 nf = 2

1  RH  1  1   n2
  n12 n22  , n1
 

 1.097 107  1  1 
 22 42 

λ = 4.86 x 10-7 m = 486 nm

  E  hc  6.631034 3108 = 4.09 x 10-19 J
  4.86107

Energy emitted = 4.09 x 10-19 J

State two of Bohr’s postulates.

 Electrons in an atom move in certain circular orbits.
 An electron moving in an does not radiate or absorb energy.
 The moving electron has a specific amount of energy and its energy is

quantized.
 Energy is emitted or absorbed only when the electron moves to another

energy level.
 Electron at its excited states is unstable. It will fall back to lower energy

level and emit a specific amount of energy in the form of light.
(any two answer)

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(b) The first five ionization energy (kJ mol-1) of atoms X and Y in period 3 of the periodic

table are as follows:

TABLE 5

Atom/IE First Second Third Fourth Fifth

X 738 1450 7730 10500 13600

Y 578 1820 2750 11600 14800

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

Explain why the first ionization energy of X is higher than that of Y.
IE1 of X > Y

 No. of proton in X > Y.
 Effective nuclear charge of X > Y.
 Nucleus attraction toward valence electrons in X > Y.
 Size of X < Y.
 Energy needed to remove the valence electrons from X > Y.

Predict the number of electron valence for X and Y and write their respective
electronic configuration. Write the set of four quantum numbers (n,l,m,s) for the
outermost electron(s).

X: Drastic increase from IE1 to IE2.
 The 2nd electron is removed from inner shell.

 There are 1 valence electron.
 1s2 2s2 2p6 3s1

(n: 3, l: 0, m: 0, s:+1/2)

Y: Drastic increase from IE2 to IE3.
 The 3rd electron is removed from inner shell.

 There are 2 valence electron.
 1s2 2s2 2p6 3s2

(n: 3, l: 0, m: 0, s:+1/2) & (n: 3, l: 0, m: 0, s:-1/2)

6. (a) CO2 and BeH2 are triatomic covalent molecules. Describe in detail the
formation of the covalent bonds in these molecules and explain why CO2 obeys the
octet rule while BeH2 does not.

CO2
C has 4 valence electrons, O has 6 valence electrons.
C will shared its valence electrons with O to form covalent bonds.

C +O +O OCO

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BeH2
Be has 2 valence electrons, H has 1 valence electrons.
Be will shared its valence electrons with H to form covalent bonds.

Be + H + H H Be H

BeH2 has less than 8 electrons around central atom.
CO2 has 8 electrons around central atom.

(b) Formal charge is a useful guide in determining the best or preferred structure.
Explain this statement using [OCN]- ion as example.

00 0 0 -1 +1 0 -2
OCN OCN OCN

AB C

[OCN]- has 3 possible structures.

Among those 3 structures, structure A is the best or preferred structure because

negative charge on structure A lies on the more electronegative atom.

7. (a) State the relationships of the volume of an ideal gas with its pressure,
absolute temperature and the amount of gas respectively.

Based on ideal gas law,

 volume of fixed amount of gas is inversely proportional with its pressure
at constant temperature. (Boyle’s law)

 Volume of fixed amount of gas is directly proportional with its absolute
temperature at constant pressure. (Charles’ law)

 At fixed temperature and pressure, volume is directly proportional with
the amount of gas. (Avogadro’s law)

A student added an amount of hydrochloric acid to a rock sample and observed the
fizzing action indicating a gas being released. At 25 oC, the gas collected in a 0.220 L

gas bulb was 0.300 g and at a pressure of 0.757 atm. Calculate the molar mass and

the density of this gas.

Mr  mRT   0.300  0.08206  298.15  44.1 gmol1
PV  0.757   0.220

  m   0.300  1.36 gL1
V  0.220

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(b) Sulphur trioxide, SO3, in a 1 L closed container was left to dissociate according to the
equation below:

2SO3(g) 2SO2(g) + O2(g) H = -197.8 kJ

Derive the relationship between Kc and Kp for the above reaction.

SO2 2 O2      Kp PSO22PO2
SO3 2
Kc  2

PSO3

PSO2  SO2  RT PO2  O2  RT PSO3  SO3  RT

Kp  SO2  RT 2 O2  RT   SO2 2 O2    RT 2  RT   Kc  RT 
SO3  RT 2 SO3 2  RT 2

Predict the equilibrium position of the reaction under the following separate
conditions;
temperature is increased,

When T increase, the equilibrium position will move backward in order to lower
temperature. Hence, decrease the value of K.

pressure is decreased,

When pressure is decreased, the equilibrium position will move forward towards
higher number of mole in order to increase the pressure.

and SO2 is removed.

When SO2 is removed, the equilibrium position will move forward in order to
recover the loss of SO2.

8. In a titration experiment at 25 oC, a total of 30 mL of 0.1 M NaOH was added
dropwise into a conical flask containing 25 mL of 0.1 M HCl and a few drops of an indicator.
Show the variation of pH of the solution before the addition of NaOH, at half equivalence
point, at equivalence point and at the final volume of the titration. Sketch a graph of pH
against the volume of NaOH.
[given: Kw at 25 oC = 1 x 10-14]

Before addition of NaOH,
At this point the solution only contain of HCl that dissociate completely in water.

HCl(aq)  H  (aq)  Cl (aq)

pH = - log [H3O+] = - log (0.1 M) = 1

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At half equivalence point: 12.5 mL

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

ni 2.5 x 10-3 1.25 x 10-3 - -

nc -1.25 x 10-3 -1.25 x 10-3 +1.25 x 10-3

nf 1.25 x 10-3 0 1.25 x 10-3

[ ] 0.0333

Since NaCl is neutral salt and NaOH is completely neutralized by HCl, the [H3O+] is
mainly come from the dissociation of acid.

pH = - log (0.067) = 1.48

At equivalence point
The solution only contain NaCl salt.
Both Na+ and Cl- ions does not hydrolyzed water.
The salt form is neutral salt.
pH = 7

Final volume
The solution only contain NaOH and NaCl.
Since NaCl is neutral salt. The [OH-] only come from dissociation of NaOH.

pH = 14 – (-log (0.1)) = 13

pH < 13.

pH
13
11

7

3 V NaOH, mL
1

25

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What is acid-base indicator? By referring to TABLE 6 below, suggest the most suitable

indicator to be used in the titration above and give reasons.

TABLE 6

Indicator pH range

Phenophthalein 8.3-10.0

Cresol red 7.0-8.8

Bromothymol blue 6.0-7.6

Litmus 4.7-8.3

Acid-base indicator is a solution containing weak acid and weak base that can
change colour over range pH value.
Most suitable indicator: Bromothymol blue
 The pH range of indicator cover the pH of end point.
 The pH range start just before pH of end point.

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