ANSWER SCHEME SCORE A PSPM SK015 | 1

Chapter 1 : Matter

PSPM 2011 / 2012

Q. 1 (a) Saline solution is prepared by dissolving 9.0 gram of NaCl in deionized water in 3
500 mL volumetric flask. Calculate the molarity of the solution. marks

Answer (a) nsolute = 9.0g / 58.5 gmol-1

= 0.1538 mol

Molarity = nsolute / Vsolution ( L )

= 0.1538 mol / 0.5 L
= 0.3076 M or 0.31 M

Q. 1 (b) Nitric oxide, NO reacts with oxygen to produce nitrogen dioxide, NO2: 4
2NO (g) + O2 (g) → 2NO2 (g) marks

In an experiment, 0.53 mol of nitric oxide is mixed with 0.3 mol oxygen. Calculate
the volume of produced NO2 at STP.

Answer (b) From equation, 2 mol of NO react with 1 mol O2

0.53 mol NO react with 0.53 mol / 2 mol
= 0.265 mol O2 ( needed )

n O2 needed ( 0.265 mol ) < n O2 provided ( 0.3 mol )

So, O2 is excess reactant or NO is limiting reactant

From equation, 2 mol of NO produce 2 mol NO2

n NO2 = 0.53 mol

At STP , 1 mol gas ≡ 22.4 L

Volume of NO2 = 0.53 mol x 22.4 L
= 11.872 L

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Q. 1 (c) Freon-12, CCl2F2 is used in various manufacturing industries. It is produced from 8

the reaction of carbon tetrachloride, CCl4 with antimony trifluoride, SbF3. marks

3 CCl4 + 2SbF3 → 3CCl2F2 + 2SbCl3

If 130.0 g of CCl4 is mixed with 95.0 g of SbF3, calculate

(i) the amount (in grams) of CCl2F2 obtained.

Answer (c)(i) mol CCl4 = 130.0g / 154 gmol-1

= 0.8442 mol

mol SbF3 = 95.0g / 178.8 gmol-1

= 0.5313 mol

Ratio CCl4 = 0.844 / 3

= 0.281 or any method that show n SbF3 needed > n SbF3
provided
Ratio SbF3 = 0.531 / 2
= 0.266

SbF3 is limiting reactant

2 mol SbF3 ≡ 3 mol CCl2F2
0.531 mol SbF3 ≡ 0.7965 mol CCl2F2

Mass CCl2F2 = 0.797 x 121

= 96.437g

Q. 1 (c) (ii) the amount of excess reactant after the reaction is completed.

Answer (c)(ii) Excess mol of CCl4 = 0.8442 mol - 0.7965 mol

= 0.0477 mol

Excess mass of CCl4 = 0.0477 mol x 154 gmol-1

= 7.346 g

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PSPM 2012 / 2013

Q. 2 (a) TABLE 1 shows the atomic structure of four particles, represented by
the letter A to D. the particles are atoms and ions.

Particle Electrons Protons Neutrons

A 6 6 6
B 12 12 12
C 10 12 12
D 6 6 8
TABLE 1
Use the letter A to D to answer the following questions.

(i) Which two particles are atom and ion of the same element? Explain. 2
marks
Answer a(i) B and C
Same no of protons and neutrons but different number of electrons

Q. 2 (a) (ii) Which two particles are isotopes of the same element? Explain. 2
marks
Answer a(ii) A and D
Same no of electrons and protons but different number of neutrons

Q. 2 (b) Compound E is a hydrocarbon, CxHy. When 6.84 g of this compound is burnt 6

completely in pure oxygen, 21.5 g of CO2 and 8.87 g of H2O are obtained. marks
Determine the empirical formula of compound E. If the molar mass of the
compound is 128.2 g/mol, what is its molecular formula?
Answer (b)
CxHy → CO2 (g) + H2O (g)
E

Mole of CO2 = 21.5 = 0.489

44
Mole of H2O = 8.87 = 0.493
18
Mole H = 0.493 x 2 = 0.986; mol C = 0.489

@ mol H:C = 2:1

Empirical formula = CH2

n = 128.2 / 14 = 9.2

n≈9
Molecular formula = C9H18

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Q. 2 (c) The reaction between the solutions of permanganate ion, MnO4-, and oxalate 5
ion, C2O42-, in acidic medium is shown below: marks
- 2- heat
MnO 4 + C2O4 MnO 2 + CO2
Write;
(i) The half-reaction for the redox reaction.

Answer (c) (i) - +

MnO 4 + 4H + 3e MnO 2 + 2H 2O
2-
C2O4 2CO2 + 2e

Q. 2 (c) (ii) The balanced redox equation for the reaction.

Answer (c) (ii) - +

2x MnO 4 + 4H + 3e MnO 2 + 2H 2O

2-
3x C2O 4 2CO2 + 2e

- 2- +
2MnO 4 + 3C 2O 4 + 8H 2MnO 2 + 6CO 2 + 4H 2O

PSPM 2013 / 2014

Q. 3 (a) Vinyl chloride, C2H3Cl, is prepared by the reaction of ethyne gas, C2H2, with 10
hydrochloric acid, HCl. In a preparation of vinyl chloride, 70.0 g of C2H2 is reacted marks
with 102.0 g of HCl.
(i) Write a balanced chemical equation for the reaction.
(ii) Identify the limiting reagent in the reaction.

Answer (a)(i) C2H2 + HCl C2H3Cl

(a)(ii) Mol of C2H2 = 70.0/26 @

= 2.692 mol C2H2

mol of HCl = 102.0/36.5 @

= 2.795 mol HCl

1 mol C2H2 ≡ 1 mol HCl

2.692 mol HCl (needed) < 2.795 mol HCl (available)

Limiting reagent is C2H2

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Q. 3 (a) (iii) Calculate the mass of the product.

(iv) Calculate the mass of the excess reactant.
1 mol C2H2 ≡ 1 mol C2H3Cl
Answer (a)(iii) 2.692 mol C2H2 ≡ 2.692 mol C2H3Cl

mass C2H3Cl = 2.692 mol x 62.5 g mol-1

= 168.3 g (@ 168 SF)
(a)(iv) Mol of excess reactant, HCl = 2.795 - 2.692 @
= 0.103 mol

Mass of excess HCl = 0.103 mol x 36.5 g mol-1

= 3.8 g (@ 3.76 SF)

Q. 3 (b) Magnesium is an element with a proton number of 12. In a sample of 5

magnesium, it was found that magnesium atoms have three different nucleon marks
numbers which are 24, 25 and 26.
(i) Write the symbol for one of these three atoms, showing its proton number
of nucleon number.
24 25 26
Answer (b)(i) Mg @ Mg @ Mg
12 12 12

Q. 3 (b) (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.

Answer (b)(ii) Relative atomic mass of Mg

= (24 x 78.6) + (25 x 10.1) + (26 x 11.3)
78.6 + 10.1 + 11.3
= 24.3

Q. 3 (b) (iii) Sketch and label the expected mass spectrum of the magnesium.

Answer (b)(iii)
Relative abundance (%)

78.6%
75.0

50.0

25.0 11.3%
10.1%

24 25 26 m/e

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PSPM 2014 / 2015

Q. 4 (a) Consider the following reaction: 5
2NO(g) + 2CO(g) N2(g) + 2CO2(g) marks
If 300.00 g of CO has reacted, calculate the mass percentage, % w/w
of CO2 in the mixture of product.

Answer (a)
2NO (g) + 2CO (g) N2 (g) + 2CO2 (g)

2 mol NO ≡ 2 mol CO ≡ 1 mol N2 ≡ 2 mol CO2

Mol CO = 300 g
28 g mol-1
= 10.71 mol

Mass N2 = 10.71 x 28 g mol-1

2
= 150.08 g

Mass CO2 = 10.71 x 44 g mol-1

= 471.24 g

% CO2 = 471.24 x 100

(471.24 + 150.08)
= 75.8%

Q. 4 (b) A 30.00 mL of 0.025 M sodium dichromate, Na2Cr2O7, solution is

Titrated with iron(II) sulphate, Fe2SO4 solution in acidic condition,
according to the following reaction:

Cr2O72- + Fe2+ Cr3+ + Fe3+

The titration requires 40.00 mL Fe2SO4 of solution to reach the end point.

(i) Balance the redox equation. 3

marks
Answer (b)(i)
Cr2O72- + 14H+ + 6e 2Cr3+ + 7H2O
2+
(Fe Fe3+ + e) x 6
________________________________________________
Cr2O72- + 6Fe2+ + 14H+ 2Cr3+ + 6Fe3+ + 7H2O

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Q. 4 (b) (ii) Calculate the mass of Na2Cr2O7 needed to prepare a 0.025 M solution in a 3
50 mL volumetric flask. marks
Answer (b)(ii)
Mol Na2Cr2O7 = 50.0 x 10-3 L x 0.025 mol L-1
= 1.25 x 10-3 mol

Mass Na2Cr2O7 = 1.25 x 10-3 mol x 262.0 g

= 0.33 g
Q. 4 (b) (iii) Determine the molarity of the Fe2+ solution. 4
marks
Answer (b)(iii)
Mol Cr2O72- = 30.00 x 10-3 L x 0.025 mol L-1
= 7.5 x 10-4 mol

1 mol Cr2O72- ≡ 6 mol Fe2+

7.5 x 10-4 mol Cr2O72- ≡ 6 x 7.5 x 10-4 mol
= 4.5 x 10-3 mol Fe2+
Molarity = mol
volume
= 4.5 x 10-3
40 x 10-3
= 0.112 M
PSPM 2015 / 2016
Q. 5 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, 11
marks
Answer (a) mass of H = 0.344 x 2
mass of C = 0.558 x 12
18
44
= 0.03822 g
= 0.15218 g
mass of N = 0.250 – (0.15218 + 0.03822)
= 0.0596 g
Element Carbon Hydrogen Nitrogen
Mass 0.15218 0.03822 0.0596
0.15218 0.03822 0.0596
Number of moles 12 1 14
= 0.01268 = 0.03822 = 0.00426
0.01268 0.03822 0.00426
Smallest ratio 0.00426 0.00426 0.00426
= 2.97 = 8.97 =1
3 9 1

Empirical formula of Q is C3H9N

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Q. 5 (b) the molar mass of Q if the molecular formula is the same as the empirical 1
formula and mark
Answer (b)
Molar mass of Q = 3(12) + 9(1) + 1(14)

= 59.0 g mol-1
Q. 5 (c) the number of hydrogen atoms present in the above sample Q. 3
marks
Answer (c)
1 mol C3H9N ≡ 9 mol of H

Mol of Q = 0.25
59.0
= 4.24 x 10-3

No of H atom = 9 x 4.24 x 10-3 x 6.02 x 1023

= 2.295 x 1022
= 2.30 x 1022

PSPM 2016 / 2017

7
Q. 6 (a) Sodium carbonate, Na2CO3, dissolves in ethanol to give 2.5 M marks
solution. Calculate the molality of the solution if the density
of the solution is 1.430 g mL-1.

Answer (a)
mass of solvent = mass of solute Na2CO3 + mass of solution
mass of solution = density x volume
= 1.430 g mL─1 x 1000 mL
= 1430 g
Mol of solute, Na2CO3 = molarity x volume
= 2.5mol L─1 x 1 L
= 2.5 mol

Mass of solute, Na2CO3 = 2.5 mol x 106g mol─1

= 265 g

Mass of solution = mass of solute + mass of solvent

Mass of solvent = 1430 ─ 265
= 1165 g
= 1.165 kg

mol of solute
Molality 
mass of solvent (kg)
2.5

1.165
= 2.15 m

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Q. 6 (b) The following redox reaction occurs in acidic condition. 8

marks
Zn + NO3– → Zn2+ + NH4+

(i) Balance the redox equation using ion-electron method.

Answer (b) (i) NO3─ + 10H+ + 8e─ → NH4+ + 3H2O

( Zn → Zn2+ + 2e─ ) x4
NO3─ + 4Zn + 10H+ → NH4+ + 4Zn2+ + 3H2O

Q. 6 (b) (ii) A 0.4 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.
Answer (b)(ii)
mol HNO3 = 25.0 x 10─4 x 0.5
= 0.0125 mol

1 mol NO3─ ≡ 4 mol Zn

─
0.0125 mol NO3 ≡ 4 x 0.0125 mol Zn
= 0.05 mol Zn

Mass of Zn = 0.05 x 65.4

= 3.27 g

% purity of Zn =
= 81.8%

PSPM 2017 / 2018

Q. 7 (a) A 25.0 mL of 0.050 M silver nitrate solution is mixed with 25.0 mL of 0.050 M of 6
calcium bromide solution to give 0.105 g of solid silver bromide. marks
(i) Write the balanced chemical equation for this reaction.

Answer (a)(i) 2AgNO3 + CaBr2 → 2AgBr + Ca(NO3)2

Q. 7 (a) (ii) Determine the limiting reactant.

Answer (a)(ii) n AgNO3 = 25.00 x 0.050

1000
= 1.25 x 10-3 mol

n CaBr2 = 25.00 x 0.050

1000
= 1.25 x 10-3 mol

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2 mol AgNO3 ≡ 1 mol CaBr2

1.25 x 10-3 mol AgNO3 ≡ 1.25 x 10-3 x 1
2
= 6.25 x 10-4 CaBr2 (needed)
AgNO3 is the limiting reactant.

Q. 7 (a) (iii) Calculate the percentage yield of silver bromide.

Answer (a)(iii) Mass of AgBr = 1.25 x 10-3 x 187.8 gmol-1
= 0.235g AgBr
% yield = actual yield x 100
Theoretical yield
= 0.105 x 100
0.235
= 44.7%

Q. 7 (b) An organic compound contains only carbon, hydrogen and oxygen with a mass 9
composition of 41.40% C, 3.47% H and 55.13% O. If 0.05 mol of this compound marks
weighs 5.80 g, determine its

(i) empirical formula.

Answer (b)(i)
Element C H O
Mass 41.4 3.47 55.13
n 41.4 3.47 55.13
12 1 16
= 3.45 =3.47 = 3.44
Smallest ratio 3.45 3.47 3.44
3.44 3.44 3.44
= 1 = 1 = 1

Empirical formula is CHO

Q. 7 (b) (ii) molecular formula

Answer (b)(ii) Molar mass empirical formula = 12.0 + 1.0 + 16.0 = 29.0 gmol-1
Molar mass molecular formula = 5.80 = 116 gmol-1
0.05
Molecular formula is C4H4O4

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PSPM 2018 / 2019

Q. 8 (a) Bromine has proton number of 3.5. FIGURE 1 shows a mass spectrum of 4m
bromine.

%
abundan
ce
51

49

m/e

FIGURE 1

i) Write the notations for all isotopes of bromine.

ii) Calculate the relative atomic mass of bromine.

Answer (a)(i) 79
35Br

81
35Br

(a) (ii)

Q. 8 (b) A reagent bottle contains a stock solution of 0.90% by mass of sodium chloride,
NaCl. The density of the solution is 1.00g cm-3. Calculate

i) The mole fraction of NaCl.

ii) The molality of NaCl solution.
iii) The volume of the stock solution required to prepare 100 mL of 0.01 M
NaCl solution.

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Answer (b) (i) Assume mass of solution = 100 g @

Mass of NaCl = 0.9 g @
Mass of water = 99.1 g

(b) (ii)

(b) (iii)
V= 0.1 L (ρ = 1.00 g cm-3)
Molarity =

=
= 0.154 M

M1V1 = M2V2

V1 =

= 6.50 mL
Percentage yield =

Theoretical yield = 941.176

Number of mol SiCl4 =

= 5.533 mol

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Q. 8 (c) Silicon tetrachloride, SiCl4 can be prepared by heating silicon in excess chlorine
gas.

Si (s) + 2Cl2 (g) → SiCl4 (l)

(i) Calculate the mass of silicon needed to produce 400g of SiCl4, if the percentage
yield is 42.5%.

(ii) If 15 mol of chlorine is used, determine the amount (mole) of unreacted

chlorine.

Answer (c) (i) From balanced equation,

(i) 1 mol Si ≡ 1 mol SiCl4
5.533 mol Si ≡ 5.533 mol SiCl4

Mass of Si = 5.533 mol X 28.1 g mol-1

= 155.48 g

(c) (ii) 1 mol Si ≡ 1 mol Cl2

5.533 mol Si ≡ 5.533 mol X 2 mol Cl2
= 11.066 mol

Therefore, unreacted Cl2 = ( 15 – 11.066 )

= 3.93 mol

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