Here are a few common misconceptions and challenges students face in their two-year pursuit of O levels Chemistry.

1. Students are often confused when negative melting and boiling points are given.

Example: 

Substance X melts at -80oC and boils at -50oC. What is the physical state of X at -100oC?

Did you get this right? Substance X is a solid at -100oC!

-100oC is a lower temperature than -80oC. Draw a number line if you are confused.

2. Students are confused about diffusion using the porous pot.

How will the water level at X change over a period of time?

Did you get this right? Level X drops first then rises back to its original position.

Hydrogen, H₂ has an Mr of 2. Carbon dioxide, CO₂ has an Mr of 44. 

  H₂ will diffuse into the porous pot faster than CO₂ diffusing out. 

  Hence, pressure inside the porous pot increases and level X drops. 

 Level X will rise to the original position after some time, when the concentration of gases inside and outside the porous pot becomes the same. 

3. Some students are confused with the terms neutron and nucleon.

Neutron is a subatomic particle found in the nucleus of an atom.

Nucleon number refers to the total number of protons and neutrons in the nucleus of an atom. 

Nucleon number is also known as mass number.

4. Proton number is also known as atomic number. These terms refer to the number of protons found in the nucleus of an atom. 

It is incorrect to write ‘neutron number’ or ‘electron number’ though. Do phrase it as ‘number of neutrons’ and ‘number of electrons’ instead.

5. Students are also often confused between the term ‘mass number’ and ‘relative atomic mass’. Why is relative atomic mass not a whole number?

Firstly, we have to understand that naturally occurring elements have isotopes that exist in different proportions. 

Relative atomic mass is calculated by taking the weighted average mass of all the isotopes of an element. 

Example: Chlorine has two isotopes, chlorine-35 and chlorine-37. Mass number of these 2 chlorine atoms are 35 and 37 respectively. However, relative atomic mass of chlorine is 35.5 (as seen in your Periodic Table!).

6. Hardness of a substance has nothing to do with the energy needed to overcome the forces of attraction in the structure of the substance. A substance is hard due to the numerous strong bonds throughout a giant structure.

Example: Diamond is hard and commonly use as tip of drilling bits. There are numerous strong covalent bonds throughout the giant molecular structure of diamond.

Diamond’s high melting point is due to a lot of energy needed to overcome all the strong covalent bonds in the giant molecular structure.

7. Some students think that a base is insoluble while alkali is soluble.

This is incorrect! A base refers to any metal oxide or metal hydroxide, regardless of solubility. An alkali is a soluble base. 

Common alkali that you should memorise are sodium hydroxide, potassium hydroxide, calcium hydroxide and aqueous ammonia.

8. Do not confuse a ‘chemical test’ with a ‘simple test’. A chemical test suggests that there must be a reaction involved.

For instance, a simple test for an acid is to use blue litmus paper or add universal indicator. The blue litmus paper turns red and the green universal indicator turns red / yellow / orange.

However, to describe a chemical test for an acid, you have to suggest a reaction involving an acid. A possible answer looks like this: “Add magnesium metal to the acid. Effervescence is observed. Gas produced extinguishes a lighted splint at the mouth of the test tube with a ‘pop’ sound.” 

9. The general formula for carboxylic acids is C_nH_2n+1COOH. Students often write C₂H₅COOH for the formula of ethanoic acid, as they think that n=2. This is incorrect. For carboxylic acids, remember to count the carbon atom in the functional group, -COOH. Hence, for ethanoic acid, n=1 and formula is CH3COOH.

10. Students are often lost on how to start when they have to draw polymers from a given monomer. Remember the “magic CAPITAL H”. Always rearrange the atoms into a “H” shape, with the C=C bond in the middle first.

The correct polymer is

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O Levels Chemistry (6092) 2022 Answers

A1(a) M [1]

(b) R [1]

(c) MQ₃ [1]

(d) T [1]

(e) Q and Z [1]

(f) RQ₄ [1]

A2(a) Particles are always in constant random motion. The perfume particles will diffuse from a region of higher concentration to a region of lower concentration. [1]

(b) More particles should be added to the diagrams / particles should be more closely packed together. [1] 

The particles of both liquids are not spread out evenly after a few seconds. [1]

(c) Difference in A and B: 

Ions moved a greater distance in beaker B than A. [1] 

Temperature in beaker B is higher, hence the ions gain more energy and move faster and further. [1]

Difference in A and C: 

Ions moved a greater distance in beaker C than A. [1] 

V⁺² has a smaller relative atomic mass / Ar than Cu+2, hence move faster and further. [1] 

A3(a) B: carboxylic acid

C: alkene

D: alcohol [1]

(b)(i) test: add aqueous bromine [1]

result: changes from reddish brown to colourless [1]

(ii) B is a weak acid which ionizes partially in water to form a low concentration of H+ ions. [1] 

(iii) D is oxidised [1] to form a carboxylic acid/ethanoic acid [1].  

(c)(i) 

empirical formula = CH₂O [1]

(ii) compound: B

reasoning: B has a molecular formula of C₂H₄O₂, hence the simplest ratio of B is CH₂O. [1]

A4(a) chlorine: yellowish-green gas

bromine: reddish-brown liquid

iodine: black solid

[1] for colour [1] for state

(b)(i) Reactivity decreases down Group VII. [1]

Table 4.2 shows that only a more reactive halogen is able to displace a less reactive halogen from its solution. For instance, chlorine displaces bromine and iodine from bromide and iodide. Iodine is unable to displace chlorine and bromine from chloride and bromide. [1]

(ii) In A, bromine displaces iodine from iodide to form brown aqueous iodine. 

In B, aqueous iodine remains brown and unreacted as it is unable to displace bromide ions. 

(c) Fluorine gains electrons to form fluoride.

(d)(i) Chlorine atoms are regenerated / reformed / remains chemically unchanged at the end of the second reaction, which can be reused in the first reaction. [1]

(ii) CFC molecules break down to form chlorine atoms which are not used up. The chlorine atoms remain in the atmosphere, hencing reacting with thousands of ozone molecules and deplete the ozone layer. [1] Excessive harmful UV rays enter the earth and cause skin cancer/eye cataract. [1] 

A5(a) 2H₂ + O₂ 🡪 2H₂O [1]

(b) 2 vol H₂: 1 vol O₂

500 cm³ : 250 cm³ [1]

Volume of air = 100/21 ×250 = 1190 cm³ (3 sf) [1]

(c) Aqueous KOH contains free-moving / mobile ions which conduct electricity. Pure water does not conduct electricity as it only have simple covalent molecules.

(d) [1] “water” as product

[1] Ea(2) lower than Ea(1) 

[1] Ea with upward arrows, starting from reactants’ energy level

[1] product lower energy level than reactant and enthalpy change using downward arrow

A6(a) Heat [1] the dilute aqueous ammonium sulfate gently [1] to obtain concentrated ammonium sulfate. Do not overheat to dryness. 

(b)(i) (NH₄)2SO₄ + O₂ 🡪 4H₂O + N₂ + SO₂

[1] for correct formulas of all substances [1] for balancing

(b)(ii) nitrogen: increases from -3 to 0 [1]

sulfur: decreases from +6 to +4 [1]

(c) manufacture detergents / manufacture fertilisers / electrolyte in car batteries [1]

(d) Process 1 does not follow the principle of using minimum amount of energy, as energy is needed to heat the dilute solution and to heat the furnace to 1000^oC in process 2.

Both processes use raw materials which are either renewable (O₂ from air) or waste material such as ammonium sulfate, hence they follow the principle.

Both processes only produce harmless substances such as water and nitrogen, hence follow the principle.

B7(a) similarities: both have side chains / both are branched [1]

differences: LDPE has side chain that is further branched while  LLDPE has straight side chains

(b) 1 molecule contains 4000 to 40000 carbon atoms.

1 LDPE unit shown has 16 carbon atoms and 32 hydrogen atoms.

Mr = 224

Minimum number of LDPE unit = 4000/16 = 125 units

Minimum mass of 1 mol LDPE molecule = 125 x 224 = 28000 g

Minimum mass of 1 LDPE molecule = 28000/6 ×10²³ = 4.67 x 10^-20 g

Maximum number of LDPE unit = 40000/16 = 1250 units

Maximum mass of 1 mol LDPE molecule = 1250 x 224 = 280000 g

Maximum mass of 1 LDPE molecule = 280000/6 ×10²³ = 4.67 x 10^-19 g

(c)(i) LDPE and LLDPE have side chains but HDPE does not. [1]

Hence HDPE molecules are arranged closer together and HDPE has higher density. [1]

(d)(i) 

(d)(ii) 

B8(a) Li and Me metals tend to lose electrons to form cations [1] and have fully filled valence electron shells.

Non-metals like B, C, N, O and F tend to gain electrons to form anions, or share electrons, [1] and have fully filled valence electron shells [1]. 

(b) Both lithium and graphite have mobile electrons to conduct electricity. [1]

Oxygen only has simple covalent molecules and does not have mobile electrons, hence cannot conduct electricity. [1]

A lot of energy is needed to overcome the strong covalent bonds between carbon atoms throughout graphite’s giant covalent structure. Hence, graphite has a high melting point. [1]

Little energy is needed to overcome the weak intermolecular forces between the simple molecules, hence oxygen has low melting point. [1]

Lithium is a group I metal, hence it has a relatively low melting point. More energy is needed to overcome the stronger metallic bonds between the cations and sea of electrons in the giant metallic structure, compared to that for oxygen. [1]

B9 (Either)

(a) In both experiments, H⁺ ions are preferentially discharged over Na⁺ ions at the cathode, as H⁺ ions are lower in the electrochemical series. [1]

2H⁺ + 2e^– 🡪 H₂ [1]

In both experiments, OH^– ions are discharged at the anode. OH^– ions are the only anions in experiment 1, OH^– ions are lower than Cl^– ions in the electrochemical series for experiment 2. [1]

4OH^– – 4e^–🡪 2H₂O + O₂ [1]

The same ions are discharged at both cathode and anode in both experiments, hence the ratio of gases collected is the same. [1]

For the same number of mole of electrons flowing through the circuit, twice the volume of H₂ is produced compared to O₂. [1]

(b) Experiment 3 produces chlorine gas at the anode while experiment 2 produces oxygen.

Experiment 3 has NaOH left as the electrolyte while experiment 2 has NaCl left. 

(c) colour in experiment 2: green

colour in experiment 3: purple / violet

In experiment 2, H⁺ ions and OH^– ions are discharged at the cathode and anode respectively, hence concentration of the two ions are equal in the electrolyte.

In experiment 3, H⁺ ions and Cl^– ions are discharged at the cathode and anode respectively, hence concentration of OH^– ions is higher than H⁺ ions in the electrolyte.

B9 (Or)

(a) Experiment 1: 

Pinkish-brown solid forms on the zinc rod. Blue aqueous CuSO₄ turns colourless. 

More reactive zinc displaces less reactive copper from copper (II) sulfate.

Zn 🡪 Zn²⁺ + 2e^–

Cu²⁺ + 2e^– 🡪 Cu 

Experiment 2: 

Pinkish-brown solid formed on cathode. Blue aqueous CuSO₄ turns colourless. Effervescence at anode.

Cu²⁺ is preferentially discharged to form Cu metal as it is lower than H⁺ in electrochemical series.

OH^– is hence preferentially discharged to form oxygen gas as it is lower than SO₄^2- in electrochemical series

Cu²⁺ + 2e^– 🡪 Cu

4OH^– 🡪 2H₂O + O₂ + 4e^–

(b) Experiment 1: 

Mass of solution increases. Zn displaces Cu from CuSO₄ to form ZnSO₄. Zn has a higher Ar than Cu, hence ZnSO₄ has a lower mass than CuSO₄.

Experiment 2: 

Mass of solution decreases.

Cu²⁺ and OH^– ions are discharged. O₂ gas escapes the solution.

(c) Blue solution turns colourless for experiment 2 but remains blue for experiment 3. 

Anode remains same size for experiment 2 but becomes smaller for experiment 3.

For every Cu²⁺ discharged at cathode, one Cu is oxidised to form one Cu²⁺ at the anode in experiment 3. Hence, concentration of Cu²⁺ ions remain unchanged.