Consider the following reaction equation: $2SO_3(g) → 2SO_2(g) + O_2$(g) ∆H = +198 kJ/mol Which of the following statements about the reaction is correct?

D. 198 kJ of energy is absorbed

A. 198 kJ of energy is given off

C. The reaction container would feel warm.

D. 198 kJ of energy is absorbed

The following statements are correct except

C. there is large decrease in the volume of a solid metal when pressure is applied to it.

A. energy is released when liquids change to solids.

B. carbon atoms in gaseous methane are further apart than those in solid diamond.

C. there is large decrease in the volume of a solid metal when pressure is applied to it.

D. particles move faster in the gaseous state than in the liquid state.

Which of the statements about gases is not correct?

D. Gases are highly soluble in water at high temperatures

A. Gases have low densities compared to solids and liquids of equal mass.

B. The total kinetic energy of the gas is not affected by collision

C. Molecules of the gas are in constant motion

D. Gases are highly soluble in water at high temperatures

Which of the following reactions represents the hydrolysis of an alkanoate?

D. $CH_3COOCH_2CH_3 + H_2O ⇌H^+⇌ CH_3COOH + CH_3CH_2OH$

A. $CH_3COOH + OH^- ⇌H^+⇌ CH_3COO^- + H_2O$

B. $CH_3COOH + C_2H_5OH ⇌H^+⇌ CH_3COOC_2H_5 + H_2O$

C. $CH_3COOC_2H_5 + H_2O ⇌H^+⇌ CH_3COO^- + CH_3CH_2OH$

D. $CH_3COOCH_2CH_3 + H_2O ⇌H^+⇌ CH_3COOH + CH_3CH_2OH$

An example of a crystalline substance that does not possess water of crystallization is

B. sodium tetraoxosulphate (VI).

C. iron (II) tetraoxosulphate (VI)

D. sodium trioxocarbonate (IV).

Carbon is deposited in the exhaust pipes of cars because of

B. incomplete combustion of petrol

A. presence of additives in petrol

B. incomplete combustion of petrol

C. presence of carbon in petrol

D. contamination of petrol with diesel

Alkenes can be manufactured by

B. the cracking of hydrocarbons.

D. addition of hydrogen to unsaturated vegetable oils.

Arrange the following compounds in decreasing order of their boiling points: $NH_3$, HF, $SiH_4$, $CH_4$

B. $CH_4$, $SiH_4$, HF, $NH_3$,

Which of the following statements is correct?

D. An alkane with 49 carbon atoms contains 100 hydrogen atoms

A. Addition reactions occur between alkanes and chlorine.

B. Butane has a lower boiling point than propane.

D. An alkane with 49 carbon atoms contains 100 hydrogen atoms

Which of the following statements about the collision theory is correct?

C. The rate of reaction is proportional to the number of effective collisions

A. Ineffective collision brings about chemical reaction

B. All collisions bring about reactions

C. The rate of reaction is proportional to the number of effective collisions

D. Collision of molecules will split the molecules to react

Dilute trioxonitrate (V) acid does not produce hydrogen when it reacts with metals because

D. it is a strong oxidizing agent.

C. there is no visible reaction.

D. it is a strong oxidizing agent.

Which of the following equations does not illustrate correctly one of the reactions of chlorine?

B. $Cl_2 + 2NaF → 2NaCl + F_2$

A. $Ca(OH)_2 + Cl_2 → CaOCl_2 + H_2O$

B. $Cl_2 + 2NaF → 2NaCl + F_2$

C. $Cl_2 + 2NaOH → NaCl + NaClO + H_2O$

The reduction half equation of the following reaction is: $Zn_{(s)} + CuSO_{4(aq)} → ZnSO_{4(aq)} + Cu_{(s)}$

C. $Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}$

A. $Zn_{(s)} → Zn^{2+}_{(aq)} + 2e^-$

B. $CuSO_{4(s)} + H_2O_{(l)}$ → $Cu^{2+}_{(aq)}$ + $SO_{4(aq)}^{2-}$

C. $Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}$

D. $Cu^{2+}_{(aq)} + e^- → Cu_{(s)}$

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WAEC (WASSCE) 2023

2023 WAEC (WASSCE) Chemistry Past Questions and Answers

The 2023 WAEC (WASSCE) Chemistry past questions with answers: 50 objective questions from the 2023 paper in the DeeepSolve question bank, each with its answer behind "Show answer". Try each question before you check, then practise the paper as a timed WAEC-style CBT.

Practise this paper as a timed WAEC-style CBT All WAEC (WASSCE) Chemistry years

Questions

1.
Consider the following reaction equation: $2SO_3(g) → 2SO_2(g) + O_2$ (g) ∆H = +198 kJ/mol Which of the following statements about the reaction is correct?
1. A.198 kJ of energy is given off
2. B.The reaction is exothermic
3. C.The reaction container would feel warm.
4. D.198 kJ of energy is absorbed
Show answer
Answer: D. 198 kJ of energy is absorbed
AI explanation
Step 1: Analyze the question We are given a thermochemical equation with ΔH = +198 kJ/mol and asked which statement about it is correct.
Step 2: The concept The sign of ΔH tells you whether a reaction is endothermic or exothermic:
ΔH positive (+): heat is absorbed from the surroundings → endothermic. The container feels cold.
ΔH negative (−): heat is released to the surroundings → exothermic. The container feels warm.
Step 3: Apply it to the reaction 2SO₃(g) → 2SO₂(g) + O₂(g) ΔH = +198 kJ/mol
Since ΔH is positive:
The reaction is endothermic.
Energy is absorbed (198 kJ for the equation as written, i.e., for 2 moles of SO₃ decomposing).
The surroundings lose heat, so the container would feel cool, not warm.
Step 4: Check each option
A. 198 kJ of energy is given off: Wrong. Energy is absorbed, not released.
B. The reaction is exothermic: Wrong. It is endothermic.
C. The reaction container would feel warm: Wrong. It would feel cold.
D. 198 kJ of energy is absorbed: Correct.
Answer: D
Exam tip: Remember "Positive = Put in" (heat is taken in, endothermic) and "Negative = Net out" (heat is given out, exothermic). Also, endothermic reactions feel cold because they draw heat from their surroundings.
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2.
What takes place at the cathode during electrolysis?
1. A.Anions lose electrons
2. B.Anions are oxidized
3. C.Cations are oxidized
4. D.Cations are discharged.
Show answer
Answer: D. Cations are discharged.
AI explanation
Analyze the question: It asks what happens at the cathode, the negative electrode, during electrolysis.
Concept:
The cathode is the negative electrode. It attracts cations (positive ions).
The anode is the positive electrode. It attracts anions (negative ions).
Reduction happens at the cathode, where ions gain electrons. Oxidation happens at the anode, where ions lose electrons. Remember OIL RIG: Oxidation Is Loss, Reduction Is Gain.
When an ion gains or loses electrons to become a neutral atom or molecule, we say it is discharged.
Step-by-step check of the options:
A. Anions lose electrons: Anions do lose electrons, but they do so at the anode, not the cathode. ✗
B. Anions are oxidized: Again, this happens at the anode. ✗
C. Cations are oxidized: Cations gain electrons at the cathode, so they are reduced, not oxidized. ✗
D. Cations are discharged: Cations migrate to the cathode, gain electrons and become neutral atoms. For example, Cu²⁺ + 2e⁻ → Cu, or 2H⁺ + 2e⁻ → H₂. ✓
Tip: Remember "CAT has Cations" (the cathode attracts cations) and "AN OX" (anode is where oxidation occurs). Also, "RED CAT": reduction occurs at the cathode.
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3.
How many grammes of NaOH(s) would be needed to produce 100.0 cm3 of a 0.20 mol $dm^{-3}$ NaOH, (aq)? [NaOH = 40.0]
1. A.800.0g
2. B.0.02 g
3. C.20.0 g
4. D.0.80 g
Show answer
Answer: D. 0.80 g
AI explanation
Step 1: Analyze the question We need the mass of NaOH required to make 100.0 cm³ of solution with a concentration of 0.20 mol/dm³.
Step 2: Concept
Moles = concentration × volume (in dm³)
Mass = moles × molar mass
Step 3: Solution
Convert the volume to dm³: 100.0 cm³ = 100.0 ÷ 1000 = 0.100 dm³
Moles of NaOH: n = C × V = 0.20 × 0.100 = 0.020 mol
Mass of NaOH: mass = n × M = 0.020 × 40.0 = 0.80 g
Step 4: Answer The mass needed is 0.80 g, which is option D.
Why the other options are wrong:
A (800 g): the volume was not converted from cm³ to dm³.
B (0.02 g): this is the number of moles, and the student forgot to multiply by the molar mass.
C (20.0 g): this comes from a wrong conversion, such as using 100 dm³ in place of 0.1 dm³.
Tip: Always convert cm³ to dm³ first by dividing by 1000. Then follow the chain: C × V → moles → × M → mass. A quick shortcut is mass = C × M × V, so 0.20 × 40 × 0.1 = 0.80 g.
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4.
The following statements are correct except
1. A.energy is released when liquids change to solids.
2. B.carbon atoms in gaseous methane are further apart than those in solid diamond.
3. C.there is large decrease in the volume of a solid metal when pressure is applied to it.
4. D.particles move faster in the gaseous state than in the liquid state.
Show answer
Answer: C. there is large decrease in the volume of a solid metal when pressure is applied to it.
AI explanation
Analyze the question: This is an "except" question. Three statements are true and one is false. We need to find the false one.
Concept: This tests the kinetic theory of matter. In solids, particles are closely packed in fixed positions with strong forces between them. In liquids, particles are close but can move past one another. In gases, particles are far apart and move rapidly and randomly.
Checking each option:
A. Energy is released when liquids change to solids. TRUE. Freezing (solidification) is exothermic. Particles lose kinetic energy and form stronger attractions, so heat is given out (the heat of fusion is released).
B. Carbon atoms in gaseous methane are further apart than those in solid diamond. TRUE. Gases have particles widely separated. Diamond is a giant covalent network in which the carbon atoms are tightly bonded. So the carbon atoms in methane gas are farther apart.
C. There is a large decrease in the volume of a solid metal when pressure is applied to it. FALSE. Solids are almost incompressible because their particles are already tightly packed with very little space between them. Applying pressure causes only a negligible change in volume.
D. Particles move faster in the gaseous state than in the liquid state. TRUE. Gas particles have higher kinetic energy and move faster than liquid particles at their respective temperatures.
Answer: The statement that is NOT correct is C.
Tip: For "except" questions, test each option and look for the odd one out. Remember: solids are practically incompressible, liquids are slightly compressible, and gases are highly compressible.
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5.
Why are $H_2SO_4$ and $CaCl_2$ , not suitable for drying ammonia gas? They
1. A.are poisonous.
2. B.pollute the gas
3. C.are corrosive
4. D.react with the gas
Show answer
Answer: D. react with the gas
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6.
The molecule that has a non-polar covalent bond is
1. A.HCl
2. B. $NH_3$
3. C. $Cl_2$
4. D. $H_2O$
Show answer
Answer: C. $Cl_2$
AI explanation
Step 1: Analyze the question We need the molecule whose covalent bond is non-polar, meaning the electrons are shared equally between the bonded atoms.
Step 2: The concept Bond polarity depends on the difference in electronegativity between the bonded atoms:
Different atoms (different electronegativities) give unequal sharing, so the bond is polar covalent. The more electronegative atom gets a partial negative charge (δ−).
Identical atoms (same electronegativity) give equal sharing, so the bond is non-polar covalent.
Step 3: Check each option
| Option | Bond | Electronegativity difference | Type | |---|---|---|---| | A. HCl | H–Cl | Cl (3.0) vs H (2.1), about 0.9 | Polar | | B. NH₃ | N–H | N (3.0) vs H (2.1), about 0.9 | Polar | | C. Cl₂ | Cl–Cl | 0 (same atom) | Non-polar | | D. H₂O | O–H | O (3.5) vs H (2.1), about 1.4 | Polar |
In Cl₂, both chlorine atoms pull on the shared pair of electrons with exactly the same strength. The electrons are shared equally, so there is no partial charge on either atom.
Step 4: Answer The molecule with a non-polar covalent bond is Cl₂, which is option C.
Exam tip: Look for diatomic molecules made of one element (H₂, O₂, N₂, Cl₂, F₂, Br₂, I₂). They always have non-polar covalent bonds because the two atoms are identical. If the atoms in the bond are different, the bond is almost always polar.
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7.
The relative molar mass of a gaseous hydrocarbon is 30. Determine its vapour density.
1. A.45
2. B.30
3. C.15
4. D.60
Show answer
Answer: C. 15
AI explanation
Analyze the question: You're given the relative molar mass of a gaseous hydrocarbon (30) and asked for its vapour density. Concept: Vapour density (V.D.) compares the mass of a gas to the mass of an equal volume of hydrogen under the same conditions: V.D. = Molar mass of gas ÷ Molar mass of H₂ Since H₂ has a molar mass of 2 (2 × 1), this gives the key relationship: Relative molar mass = 2 × Vapour density so V.D. = Relative molar mass ÷ 2 Solution: V.D. = 30 ÷ 2 = 15 Check: The hydrocarbon with a molar mass of 30 is ethane (C₂H₆): (2 × 12) + (6 × 1) = 30. Its vapour density is 30/2 = 15. ✔ Answer: C. 15 Exam tip: Remember "Molar mass is double the vapour density" (M = 2 × V.D.). Students often pick 30 by confusing the two, or choose a larger number by multiplying instead of dividing. Vapour density has no units because it is a ratio. If the question compares to air instead, use the molar mass of air (about 29).
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8.
The periodic property that is used to determine whether a covalent molecule is polar or not is
1. A.atomic radius
2. B.ionization energy
3. C.electron affinity
4. D.electronegativity
Show answer
Answer: D. electronegativity
AI explanation
Step 1: Analyze the question We need the periodic property that tells us whether a covalent bond or molecule is polar (has partial positive and negative ends).
Step 2: The concept In a covalent bond, two atoms share electrons. If both atoms attract the shared electrons equally, the bond is non-polar (e.g., Cl-Cl, H-H). If one atom attracts the shared electrons more strongly, the electrons are pulled toward it. That atom becomes partially negative (δ-) and the other partially positive (δ+). This makes the bond polar (e.g., H-Cl).
The ability of an atom to attract shared electrons in a bond is called electronegativity.
Step 3: Apply it
The difference in electronegativity between the bonded atoms shows how polar the bond is. • Difference of 0 (or very small): non-polar covalent • Moderate difference: polar covalent • Very large difference: ionic
For a whole molecule, you also need to consider the shape. CO₂ has polar bonds, but the molecule is non-polar because the bond dipoles cancel by symmetry. Electronegativity is still the property that decides the bond polarity in the first place.
Step 4: Check the other options
A. Atomic radius: the size of an atom. It does not measure how strongly shared electrons are attracted.
B. Ionization energy: the energy needed to remove an electron from a gaseous atom. It relates to forming positive ions, not to sharing electrons.
C. Electron affinity: the energy change when an isolated gaseous atom gains an electron. It is related to electronegativity but applies to free atoms, not to atoms in a bond.
D. Electronegativity: the tendency of an atom to attract shared electrons in a covalent bond. This is the one used to judge polarity.
Answer: D. electronegativity
Exam tip: Remember "Polar = Pull". Electronegativity is the pulling power on shared electrons. If a question mentions bond polarity or the type of bond (ionic, polar covalent, non-polar covalent), think electronegativity difference.
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9.
Which of the statements about gases is not correct?
1. A.Gases have low densities compared to solids and liquids of equal mass.
2. B.The total kinetic energy of the gas is not affected by collision
3. C.Molecules of the gas are in constant motion
4. D.Gases are highly soluble in water at high temperatures
Show answer
Answer: D. Gases are highly soluble in water at high temperatures
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10.
Which of the following elements has its valence electrons in the s-orbital?
1. A.Aluminum
2. B.Phosphorus
3. C.Sodium
4. D.Carbon
Show answer
Answer: C. Sodium
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11.
Which of the following gases has the lowest rate of diffusion? [H=1.0, C=12.0, N = 14.0, O= 16.0]
1. A.Nitrogen
2. B.methane
3. C.Ammonia
4. D.Oxygen
Show answer
Answer: D. Oxygen
AI explanation
Analyze the question: We need the gas that diffuses most slowly. This is a Graham's Law question.
Concept: Graham's Law of Diffusion says the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (or density):
Rate ∝ 1/√M
So the heavier the gas, the slower it diffuses. The lightest gas diffuses fastest.
Calculate the molar masses:
A. Nitrogen (N₂) = 2 × 14 = 28 g/mol
B. Methane (CH₄) = 12 + (4 × 1) = 16 g/mol
C. Ammonia (NH₃) = 14 + (3 × 1) = 17 g/mol
D. Oxygen (O₂) = 2 × 16 = 32 g/mol
Compare: Oxygen has the highest molar mass (32 g/mol), so it has the lowest rate of diffusion. The order of rates from fastest to slowest is:
CH₄ (16) > NH₃ (17) > N₂ (28) > O₂ (32)
Tip: In "lowest rate of diffusion" questions, just compute the molar masses. Highest molar mass means slowest, and lowest molar mass means fastest. Remember that nitrogen and oxygen are diatomic (N₂, O₂), so don't forget to double their atomic masses. That is a common trap in this question!
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12.
Which of the following properties does not give evidence of the kinetic theory of matter?
1. A.Diffusion
2. B.Melting
3. C.Polymerization
4. D.Evaporation
Show answer
Answer: C. Polymerization
AI explanation
Step 1: Analyze the question We need the property that does NOT support the kinetic theory of matter.
Step 2: Concept The kinetic theory says matter is made of tiny particles that are in constant random motion. The particles' energy and the forces between them determine whether the matter is a solid, liquid or gas.
Evidence for the theory includes:
Diffusion: particles move randomly and spread from high to low concentration (e.g. the smell of cooking rice spreading through a house).
Melting: heating gives particles more kinetic energy, so they overcome the forces holding them in fixed positions.
Evaporation: faster-moving surface particles escape from the liquid.
Others include Brownian motion, gas pressure and the expansion of gases.
Step 3: Examine each option
A. Diffusion: direct evidence of particle motion, so it supports the theory.
B. Melting: shows particles gaining energy and changing arrangement, so it supports the theory.
C. Polymerization: this is a chemical reaction in which small monomer molecules join by covalent bonds to form long chains. It is about bonding and chemical change, not the physical motion and energy of particles, so it does not give evidence of the kinetic theory.
D. Evaporation: particles escape from the liquid surface, so it supports the theory.
Step 4: Answer The correct option is C. Polymerization.
Tip: Kinetic theory questions are about physical changes and particle movement (diffusion, Brownian motion, melting, boiling, evaporation). If an option describes a chemical reaction that makes new substances, such as polymerization, it is usually the odd one out.
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13.
Which of the following reactions represents the hydrolysis of an alkanoate?
1. A. $CH_3COOH + OH^- ⇌H^+⇌ CH_3COO^- + H_2O$
2. B. $CH_3COOH + C_2H_5OH ⇌H^+⇌ CH_3COOC_2H_5 + H_2O$
3. C. $CH_3COOC_2H_5 + H_2O ⇌H^+⇌ CH_3COO^- + CH_3CH_2OH$
4. D. $CH_3COOCH_2CH_3 + H_2O ⇌H^+⇌ CH_3COOH + CH_3CH_2OH$
Show answer
Answer: D. $CH_3COOCH_2CH_3 + H_2O ⇌H^+⇌ CH_3COOH + CH_3CH_2OH$
AI explanation
Analyze the question: Hydrolysis of an alkanoate (an ester) means the ester reacts with water to break down into its parent carboxylic acid and alcohol. With an acid catalyst (H⁺), the reaction is reversible.
Concept: RCOOR' + H₂O ⇌ (H⁺) RCOOH + R'OH
This is the reverse of esterification (Fischer esterification).
Check each option:
A. CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O is a neutralisation (acid-base) reaction, not hydrolysis.
B. CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O is esterification, the formation of an ester. This is the reverse of hydrolysis.
C. CH₃COOC₂H₅ + H₂O → CH₃COO⁻ + CH₃CH₂OH starts with an ester and water, but the product is the carboxylate ion. That is characteristic of alkaline hydrolysis (saponification, with OH⁻), not the acid-catalysed pathway shown with H⁺. It is also not balanced for charge, since the H⁺ is not accounted for. So it is not the best representation.
D. CH₃COOCH₂CH₃ + H₂O ⇌ (H⁺) CH₃COOH + CH₃CH₂OH takes ethyl ethanoate plus water and gives ethanoic acid and ethanol. This is a correct, balanced acid-catalysed hydrolysis of an ester.
Tip: Remember "ester + water → acid + alcohol" for hydrolysis, and "acid + alcohol → ester + water" for esterification. If the ester is hydrolysed with a base (NaOH), you get the carboxylate salt and alcohol instead, and this is called saponification.
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14.
An example of a crystalline substance that does not possess water of crystallization is
1. A.potassium trioxonitrate (V)
2. B.sodium tetraoxosulphate (VI).
3. C.iron (II) tetraoxosulphate (VI)
4. D.sodium trioxocarbonate (IV).
Show answer
Answer: A. potassium trioxonitrate (V)
AI explanation
Analyze the question: We need a crystalline salt that is anhydrous in its normal crystalline form, meaning it has no water of crystallization.
Concept: Water of crystallization is the fixed number of water molecules that some salts hold within their crystal structure. Examples are:
Na₂SO₄·10H₂O (Glauber's salt)
FeSO₄·7H₂O (green vitriol)
Na₂CO₃·10H₂O (washing soda)
Salts like KNO₃, NaCl, and KCl crystallize without water.
Going through the options:
A. Potassium trioxonitrate (V), KNO₃: It crystallizes in anhydrous form. No water of crystallization. ✔
B. Sodium tetraoxosulphate (VI), Na₂SO₄: The common crystalline form is the decahydrate, Na₂SO₄·10H₂O. It does have water of crystallization. ✘ (The anhydrous form exists, but it is not the typical crystalline form, and it is not what the question is after.)
C. Iron (II) tetraoxosulphate (VI), FeSO₄: Crystallizes as FeSO₄·7H₂O. ✘
D. Sodium trioxocarbonate (IV), Na₂CO₃: Crystallizes as Na₂CO₃·10H₂O (washing soda). ✘
Answer: The salt that has no water of crystallization is potassium trioxonitrate (V), KNO₃, which is option A.
Exam tip: Remember that nitrates and chlorides of Na and K (KNO₃, NaNO₃, NaCl, KCl) are usually anhydrous. The hydrated ones that WAEC commonly tests are:
Na₂CO₃·10H₂O
FeSO₄·7H₂O
CuSO₄·5H₂O
MgSO₄·7H₂O
Na₂SO₄·10H₂O
Keep this list handy, and any salt not on it is likely your anhydrous one.
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15.
Which of the following laws or theory cannot be explained by the application of the kinetic theory of gases?
1. A.Dalton's atomic theory
2. B.Charle's law
3. C.Boyle's law
4. D.Gay- Lussac's law
Show answer
Answer: A. Dalton's atomic theory
AI explanation
Step 1: Analyze the question We need to find the option that the kinetic theory of gases does not explain. The kinetic theory deals with the behaviour of gas particles: their motion, collisions, pressure, temperature and volume.
Step 2: Recall the concept The kinetic theory of gases says that:
Gases consist of tiny particles in constant, random motion.
The particles collide elastically with each other and with the walls of the container, and these collisions produce pressure.
The average kinetic energy of the particles is proportional to the absolute temperature (K).
The particles have negligible volume, and there are negligible forces between them.
Step 3: Test each option
B. Charles' law (V ∝ T at constant P): Explained. Raising the temperature increases the particles' kinetic energy, so they hit the walls harder and more often. The volume must expand to keep the pressure constant.
C. Boyle's law (P ∝ 1/V at constant T): Explained. A smaller volume means more frequent collisions with the walls, so the pressure is higher.
D. Gay-Lussac's law (P ∝ T at constant V): Explained. Higher temperature means faster particles and more forceful collisions, so the pressure rises.
A. Dalton's atomic theory: This theory says matter is made of indivisible atoms, that atoms of an element are identical, and that atoms combine in simple whole-number ratios. It concerns the nature and combination of atoms, not the motion, pressure or volume of gases. The kinetic theory does not explain it.
(Dalton's law of partial pressures can be explained by kinetic theory, but Dalton's atomic theory is a different thing.)
Step 4: Answer The correct answer is A. Dalton's atomic theory.
Tip: Boyle, Charles and Gay-Lussac all link pressure, volume and temperature, which are the variables kinetic theory describes. If an option is about atoms combining or the structure of matter instead, it is probably the odd one out.
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16.
The vapour density of an organic compound with the molecular formula $C_2H_4O_2$ , is [H=1.0, C=12.0, O= 16.0]
1. A.40
2. B.120
3. C.65
4. D.30
Show answer
Answer: D. 30
AI explanation
Analyze the question: You're asked for the vapour density of C₂H₄O₂. Vapour density (V.D.) compares the density of a gas or vapour to that of hydrogen.
Concept:
Vapour density = Molar mass ÷ 2
(This is because Molar mass = 2 × Vapour density, since a hydrogen molecule H₂ has a mass of 2.)
Solution:
Calculate the molar mass of C₂H₄O₂: • C: 2 × 12 = 24 • H: 4 × 1 = 4 • O: 2 × 16 = 32 • Total = 24 + 4 + 32 = 60 g/mol
Calculate the vapour density: • V.D. = 60 ÷ 2 = 30
Answer: Option D (30).
Tip: Whenever you see "vapour density" in an exam, just find the molar mass and halve it. Also, the reverse works: if the question gives V.D. and asks for molar mass, double it. Watch out for option B (120), which is what you'd get if you multiplied by 2 instead of dividing.
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17.
Petrochemistry is an example of
1. A.applied chemistry
2. B.Biochemistry
3. C.Pure chemistry
4. D.Environmental chemistry
Show answer
Answer: A. applied chemistry
AI explanation
Step 1: Analyze the question The question asks which branch of chemistry petrochemistry belongs to.
Step 2: Explain the concept
Pure chemistry studies chemical principles and theory for the sake of knowledge (e.g., atomic structure, bonding).
Applied chemistry uses chemical principles to solve practical, industrial, or commercial problems (e.g., petrochemistry, pharmaceuticals, metallurgy, food technology).
Biochemistry studies chemical processes in living organisms (e.g., enzymes, proteins, metabolism).
Environmental chemistry studies chemical processes in the environment (e.g., pollution, water and soil chemistry).
Step 3: Apply to the question Petrochemistry deals with petroleum and natural gas and how they are processed into useful products such as petrol, plastics, fertilizers, synthetic fibres, detergents, and paints. This is a practical, industrial use of chemistry, not a study of theory alone. It is not about living organisms, and it is not mainly about studying the environment.
Step 4: Answer Petrochemistry is an example of applied chemistry, which is option A.
Tip: Ask yourself, "Is this chemistry being used to make or do something practical?" If yes, it is usually applied chemistry. Think of Nigeria's refineries and the Dangote petrochemical industry, which are real-life examples of applied chemistry.
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18.
How many unpaired electrons are present in $_{26}Fe^{3+}$ ?
1. A.4
2. B.2
3. C.3
4. D.5
Show answer
Answer: D. 5
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19.
Isoelectronic species have the same number of
1. A.ions
2. B.electrons
3. C.protons
4. D.neutrons
Show answer
Answer: B. electrons
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20.
The gas that is less dense than air is
1. A.nitrogen
2. B.chlorine
3. C.Carbon(IV) oxide
4. D.Oxygen
Show answer
Answer: A. nitrogen
AI explanation
Step 1: Analyze the question We need the gas that is less dense than air. Density of a gas is proportional to its molar mass (at the same temperature and pressure), so we compare molar masses with the average molar mass of air. Step 2: Concept Air is about 78% nitrogen, 21% oxygen, and 1% other gases. Its average molar mass is about 28.8 g/mol (often rounded to 29). A gas with a molar mass lower than about 29 is less dense than air. A gas with a higher molar mass is denser. Step 3: Compare the options | Gas | Formula | Molar mass (g/mol) | Compared with air (≈29) | |---|---|---|---| | Nitrogen | N₂ | 28 | Slightly lighter | | Chlorine | Cl₂ | 71 | Denser | | Carbon(IV) oxide | CO₂ | 44 | Denser | | Oxygen | O₂ | 32 | Denser | Step 4: Answer Only nitrogen (M = 28) has a molar mass below that of air (about 29), so it is slightly less dense than air. Tip: Memorize the "reference point": air ≈ 29. Any common gas with a molar mass below 29 (H₂, He, NH₃, CH₄, N₂, CO, C₂H₂) is lighter than air, and gases above 29 (O₂, CO₂, Cl₂, SO₂, HCl) are heavier. This also tells you how to collect gases: lighter-than-air gases by downward displacement of air, and heavier ones by upward displacement.
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21.
Carbon is deposited in the exhaust pipes of cars because of
1. A.presence of additives in petrol
2. B.incomplete combustion of petrol
3. C.presence of carbon in petrol
4. D.contamination of petrol with diesel
Show answer
Answer: B. incomplete combustion of petrol
AI explanation
Analyze the question: Why does soot (carbon) collect in a car's exhaust pipe?
Concept: Petrol is a mixture of hydrocarbons (mainly octane, C₈H₁₈). When hydrocarbons burn, the products depend on how much oxygen is available:
Complete combustion (plenty of oxygen): 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O Only carbon dioxide and water form, with no soot.
Incomplete combustion (limited oxygen): C₈H₁₈ + O₂ → CO + H₂O (carbon monoxide) or → C (soot) + H₂O Carbon monoxide and free carbon (soot) form.
Solution:
Inside the engine cylinder, the air supply is often not enough to burn all the fuel completely. Some carbon is therefore released as fine black soot and deposited in the exhaust pipe.
A. Additives: these are meant to improve burning (e.g., anti-knock agents). They are not the main cause of carbon deposits.
B. Incomplete combustion of petrol: this is correct, because insufficient oxygen leaves unburnt carbon.
C. Presence of carbon in petrol: petrol contains hydrocarbons (compounds of carbon and hydrogen), not free carbon. This option is misleading.
D. Contamination with diesel: this is not the general cause.
Answer: B. Incomplete combustion of petrol.
Exam tip: Whenever you see "soot," "black smoke," or "carbon monoxide," think incomplete combustion = limited oxygen. Remember: Lots of O₂ gives CO₂ + H₂O; little O₂ gives CO + C (soot).
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22.
How many moles of copper would be deposited by passing 1 Faraday of electricity through a $CuCl_2$ solution?
1. A.2 moles
2. B.0.5moles
3. C.0.025moles
4. D.1 mole
Show answer
Answer: B. 0.5moles
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23.
Alkenes can be manufactured by
1. A.the combustion of alkanes.
2. B.the cracking of hydrocarbons.
3. C.polymerization reactions.
4. D.addition of hydrogen to unsaturated vegetable oils.
Show answer
Answer: B. the cracking of hydrocarbons.
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24.
Determine the volume of 0.100 mol of HCl in $0.250 mol dm^3$ of solution.
1. A. $200cm^3$
2. B. $100cm^3$
3. C. $20cm^3$
4. D. $400cm^3$
Show answer
Answer: D. $400cm^3$
AI explanation
Analyze the question: We have 0.100 mol of HCl, and the concentration of the solution is 0.250 mol/dm³. We need the volume of solution that contains this amount of HCl. Concept: Concentration (molarity) is moles of solute per unit volume of solution: C = n / V, so V = n / C Solution: V = 0.100 mol ÷ 0.250 mol/dm³ V = 0.4 dm³ Convert to cm³ (1 dm³ = 1000 cm³): V = 0.4 × 1000 = 400 cm³ Answer: Option D, 400 cm³. Tip: Always check the units. Mole ÷ (mol/dm³) gives dm³, and you must multiply by 1000 to get cm³. A quick check: 0.250 mol/dm³ is a quarter of a mole per dm³, so you need 4 dm³ for 1 mol, and therefore 0.4 dm³ for 0.1 mol.
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25.
Going down group II in the periodic table normally
1. A.ionization energy increases
2. B.melting point increases
3. C.shielding effect decreases
4. D.electronegativity increases
Show answer
Answer: B. melting point increases
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26.
An element X has two isotopes, $^{65}X_{30}$ and $^{66}X_{30}$ with relative abundance of 60 % and 40% respectively. The relative atomic mass of X is
1. A.66.00
2. B.65.50
3. C.65.00
4. D.65.40
Show answer
Answer: D. 65.40
AI explanation
Analyze the question: An element has two isotopes with different mass numbers (65 and 66) and different abundances. We need the average atomic mass, weighted by abundance.
Concept: Relative atomic mass is the weighted average of the isotopic masses:
$A_r = \frac{(\text{mass}_1 \times \%_1) + (\text{mass}_2 \times \%_2)}{100}$
Solution:
Isotope 1: 65 × 60 = 3900
Isotope 2: 66 × 40 = 2640
$A_r = \frac{3900 + 2640}{100} = \frac{6540}{100} = 65.40$
Check: The answer must lie between 65 and 66. Since the lighter isotope (65) is more abundant (60%), the average should be closer to 65 than to 66. 65.40 fits this.
Answer: The relative atomic mass of X is 65.40, which is option D.
Tip: A shortcut is to start from the lighter mass and add only the extra contribution: 65 + (1 × 0.40) = 65.40. Since the isotopes differ by 1 unit, you just add the fractional abundance of the heavier isotope to 65. Always check that your answer lies between the lightest and heaviest isotope masses.
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27.
Arrange the following compounds in decreasing order of their boiling points: $NH_3$ , HF, $SiH_4$ , $CH_4$
1. A.HF, $NH_3$ , $SiH_4$ , $CH_4$
2. B. $CH_4$ , $SiH_4$ , HF, $NH_3$ ,
3. C. $NH_3$ , $CH_4$ , $SiH_4$ , HF
4. D. $CH_4$ , $SiH_4$ , $NH_3$ , HF
Show answer
Answer: A. HF, $NH_3$ , $SiH_4$ , $CH_4$
AI explanation
Analyze the question: We need to rank four hydrides from highest to lowest boiling point. Boiling point depends on the strength of the intermolecular forces.
Concept:
Hydrogen bonding is the strongest of these forces. It occurs when H is bonded to a very electronegative atom (F, O, N).
Without hydrogen bonding, only weak van der Waals (London dispersion) forces act. These get stronger as molecular mass increases.
Solution:
HF and NH₃ both form hydrogen bonds, so they boil at much higher temperatures than SiH₄ and CH₄. • HF boiling point is about +19.5 °C. F is the most electronegative element, so the H–F bond is very polar and the hydrogen bonds are very strong. • NH₃ boiling point is about -33 °C. N is less electronegative than F, and each NH₃ has only one lone pair to accept hydrogen bonds, so the bonding is weaker than in HF. • So HF > NH₃.
SiH₄ and CH₄ are non-polar and have only London dispersion forces. • SiH₄ has the larger molar mass (32 g/mol) and a bigger electron cloud. Its boiling point is about -112 °C. • CH₄ has a molar mass of 16 g/mol. Its boiling point is about -161 °C. • So SiH₄ > CH₄.
Decreasing order: HF > NH₃ > SiH₄ > CH₄
This matches option A.
Why the others are wrong:
B, C and D place CH₄ or SiH₄ above the hydrogen-bonded compounds, or put NH₃ above HF. Both orderings contradict the intermolecular force strengths.
Tip: Remember the order of strength: hydrogen bonding > dipole-dipole > London dispersion. Within a group of non-hydrogen-bonded molecules, the heavier one boils higher. A quick way to remember the hydrogen-bonded atoms is "FON" (F, O, N).
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28.
How many moles are there in 3.0 g of $O_2$ ? [O = 16.0]
1. A.0.0093 moles.
2. B.0.0930 moles.
3. C.0.0625 moles.
4. D.0.6250 moles.
Show answer
Answer: B. 0.0930 moles.
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29.
The formation of a bond between hydrogen and a highly electronegative atom results in
1. A.polarity
2. B.electrovalent bond.
3. C.dipole
4. D.metallic bond
Show answer
Answer: A. polarity
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30.
Which of the following statements is correct?
1. A.Addition reactions occur between alkanes and chlorine.
2. B.Butane has a lower boiling point than propane.
3. C.Pentane has five isomers.
4. D.An alkane with 49 carbon atoms contains 100 hydrogen atoms
Show answer
Answer: D. An alkane with 49 carbon atoms contains 100 hydrogen atoms
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31.
Thermal cracking of alkanes usually
1. A.is an exothermic process
2. B.produces only small alkanes
3. C.involves decomposition
4. D.requires hydrogen
Show answer
Answer: C. involves decomposition
AI explanation
Step 1: Analyze the question We need the statement that correctly describes thermal cracking of alkanes.
Step 2: The concept Cracking is the breakdown of large hydrocarbon molecules into smaller, more useful ones. In thermal cracking, high temperature (about 450-750°C) and high pressure are used, often without a catalyst.
Example: C₁₀H₂₂ → C₈H₁₈ + C₂H₄ (decane → octane + ethene)
Step 3: Check each option
A. Exothermic process: Wrong. Cracking breaks strong C-C bonds, so it absorbs a lot of heat. It is endothermic.
B. Produces only small alkanes: Wrong. It produces a mixture of smaller alkanes and alkenes (such as ethene and propene), and sometimes hydrogen.
C. Involves decomposition: Correct. A large molecule is broken down by heat into smaller molecules. This is thermal decomposition.
D. Requires hydrogen: Wrong. Hydrogen is needed in hydrocracking, not in ordinary thermal cracking. Thermal cracking needs only heat.
Step 4: Answer
Tip: Remember "cracking = breaking by heat". If you see "decomposition" or "breakdown" with cracking, it is usually the right choice. Also remember that cracking gives alkenes too, so any option saying "only alkanes" is a trap.
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32.
Which of the following statements about the collision theory is correct?
1. A.Ineffective collision brings about chemical reaction
2. B.All collisions bring about reactions
3. C.The rate of reaction is proportional to the number of effective collisions
4. D.Collision of molecules will split the molecules to react
Show answer
Answer: C. The rate of reaction is proportional to the number of effective collisions
AI explanation
Step 1: Analyze the question We are asked to pick the correct statement about collision theory, which explains how and why chemical reactions occur and how fast they happen.
Step 2: The concept Collision theory says that for a reaction to occur:
Reactant particles must collide.
They must collide with enough energy, at least the activation energy.
They must collide with the proper orientation.
A collision that meets these conditions is an effective collision. Only effective collisions lead to products. The more effective collisions per unit time, the faster the reaction.
Step 3: Check each option
A. Ineffective collision brings about chemical reaction. Wrong. Ineffective collisions lack sufficient energy or proper orientation, so the particles just bounce off each other.
B. All collisions bring about reactions. Wrong. Most collisions are ineffective. If all were effective, nearly every reaction would be instantaneous.
C. The rate of reaction is proportional to the number of effective collisions. Correct. Increasing concentration, temperature, surface area, or adding a catalyst increases the frequency of effective collisions, and so increases the rate.
D. Collision of molecules will split the molecules to react. Wrong. Collisions do not automatically split molecules. Bonds break only if the colliding particles have enough energy.
Step 4: Answer The correct option is C.
Tip: In collision theory questions, look for the words "effective" or "sufficient energy" and "proper orientation". Statements saying "all collisions" or "ineffective collisions" cause reaction are almost always wrong.
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33.
Copper (II) ions are able to participate in co-ordinate covalent bonding because they
1. A.are coloured
2. B.are positively charged
3. C.have vacant d-orbital
4. D.have unpaired electrons
Show answer
Answer: C. have vacant d-orbital
AI explanation
Analyze the question: It asks why Cu²⁺ ions can form coordinate (dative) covalent bonds, for example in [Cu(NH₃)₄]²⁺ or [Cu(H₂O)₄]²⁺.
Concept: A coordinate covalent bond forms when one species (the ligand) donates a lone pair of electrons into an empty orbital of another species (the acceptor). The acceptor must therefore have vacant, low-energy orbitals available to receive the electron pairs. Transition metal ions are good acceptors because they have vacant d (and also s and p) orbitals.
Step-by-step solution:
Electronic configuration of Cu: [Ar] 3d¹⁰ 4s¹
Cu²⁺ loses 2 electrons (one 4s and one 3d): [Ar] 3d⁹
The 3d⁹ configuration leaves one d-orbital with a vacancy. The ion also has empty 4s and 4p orbitals available.
Ligands such as NH₃ and H₂O have lone pairs. They donate these pairs into the vacant orbitals of Cu²⁺, forming coordinate bonds.
Checking the options:
A. are coloured: Colour is a result of d-d electronic transitions. It is not the reason for bonding.
B. are positively charged: Positive charge attracts ligands, but many positive ions (like Na⁺) do not form coordinate bonds readily. Charge alone is not the key.
C. have vacant d-orbital: The ion can accept lone pairs into vacant orbitals. This is the standard WASSCE explanation.
D. have unpaired electrons: Cu²⁺ does have one unpaired electron, but that explains paramagnetism, not coordinate bonding.
Answer: C
Exam tip: For "why can a metal ion form complexes or coordinate bonds?", think "vacant orbitals to accept lone pairs". For "why is it coloured or paramagnetic?", think "d-d transitions" and "unpaired electrons".
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34.
A compound that could be dried by using conc. tetraoxosulphate (VI) acid and not by calcium oxide is likely to be
1. A.a deliquescent salt
2. B.a neutral salt
3. C.an alkaline gas
4. D.an acid anhydride
Show answer
Answer: D. an acid anhydride
AI explanation
Analyze the question: We need a compound (usually a gas) that can be dried with concentrated H₂SO₄ (an acidic drying agent) but not with CaO (a basic drying agent). So the compound must react with CaO, which means it is acidic, and it must not react with H₂SO₄.
Concept:
Conc. H₂SO₄ is an acidic drying agent. It dries neutral and acidic gases (e.g. CO₂, SO₂, HCl, Cl₂) but not basic gases like NH₃, which it would react with.
CaO is a basic drying agent. It dries neutral and basic gases (e.g. NH₃) but not acidic gases, because it reacts with them: CaO + CO₂ → CaCO₃.
Going through the options:
A. Deliquescent salt: This absorbs water from the air and is not a gas to be dried in this sense. It does not fit the pattern.
B. Neutral salt: It would react with neither agent, so both could dry it. It does not fit.
C. Alkaline gas (e.g. NH₃): It reacts with conc. H₂SO₄, so it is dried by CaO and not by H₂SO₄. This is the opposite of what the question asks.
D. Acid anhydride: Acid anhydrides are acidic oxides such as CO₂, SO₂ and SO₃ (with SO₃ being a special case, since it dissolves in conc. H₂SO₄). They react with basic CaO, so CaO cannot be used. Most, like CO₂ and SO₂, do not react with conc. H₂SO₄, so it can dry them. This fits.
Tip: Match opposites. Acidic gases need acidic drying agents (conc. H₂SO₄), and basic gases need basic drying agents (CaO). Remember that NH₃ is the classic gas dried by CaO only.
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35.
A consequence of global warming is
1. A.air pollution
2. B.High Humidity
3. C.flooding
4. D.Water pollution
Show answer
Answer: C. flooding
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36.
Dilute trioxonitrate (V) acid does not produce hydrogen when it reacts with metals because
1. A.it reacts with the product.
2. B.it is highly corrosive.
3. C.there is no visible reaction.
4. D.it is a strong oxidizing agent.
Show answer
Answer: D. it is a strong oxidizing agent.
AI explanation
Analyze the question: Trioxonitrate(V) acid is HNO₃ (nitric acid). Normally, acids react with reactive metals to give a salt and hydrogen gas. The question asks why HNO₃ does not give hydrogen.
Concept: Nitric acid is a strong oxidizing agent. The nitrate ion (NO₃⁻) in acidic solution is reduced more easily than H⁺ ions. Also, any hydrogen that forms is quickly oxidized to water by the acid.
Solution:
A normal acid-metal reaction is: Metal + acid → salt + H₂
With HNO₃, the nitrate ion takes the electrons that would have reduced H⁺ to H₂.
The nitrogen is reduced to nitrogen oxides instead: • Dilute HNO₃ with Cu: 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O • Concentrated HNO₃ with Cu: Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
So the products are water and nitrogen oxides (NO, NO₂, sometimes N₂O or NH₄⁺), not hydrogen gas. • (Only very dilute, cold HNO₃ reacting with magnesium or manganese gives some hydrogen.)
Checking the options:
A. "It reacts with the product" is vague. The real reason is the oxidizing power of the acid, so this is not the best answer.
B. Being corrosive does not affect whether H₂ forms.
C. There is a visible reaction (brown fumes of NO₂, for example), so this is wrong.
D. It is a strong oxidizing agent. This is correct.
Answer: D
Tip: Remember "HNO₃ = Oxidizer, no H₂." Whenever you see nitric acid with a metal, think of nitrogen oxides and water, not hydrogen. Only HCl and dilute H₂SO₄ give the classic metal + acid → salt + H₂ reaction.
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37.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.
38.
Which of the following equations does not illustrate correctly one of the reactions of chlorine?
1. A. $Ca(OH)_2 + Cl_2 → CaOCl_2 + H_2O$
2. B. $Cl_2 + 2NaF → 2NaCl + F_2$
3. C. $Cl_2 + 2NaOH → NaCl + NaClO + H_2O$
4. D. $H_2S + Cl_2 → 2HCl + S$
Show answer
Answer: B. $Cl_2 + 2NaF → 2NaCl + F_2$
AI explanation
Analyze the question: We need the equation that does NOT correctly show a reaction of chlorine. We check each option against known chlorine chemistry.
Concept: Chlorine is a strong oxidizing agent. It can displace less reactive halogens (Br⁻, I⁻) from their salts, but not more reactive ones. The reactivity of the halogens decreases down the group: F₂ > Cl₂ > Br₂ > I₂.
Checking each option:
A. Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O This is the production of bleaching powder (calcium chlorate(I)/calcium oxychloride) from chlorine and slaked lime. It is correct and balanced.
B. Cl₂ + 2NaF → 2NaCl + F₂ This would need chlorine to displace fluorine. Fluorine is more reactive (more electronegative) than chlorine, so chlorine cannot displace it. This reaction does not occur, so the equation is incorrect.
C. Cl₂ + 2NaOH → NaCl + NaClO + H₂O Cold, dilute NaOH reacts with chlorine to give sodium chloride and sodium hypochlorite (bleach). It is correct and balanced.
D. H₂S + Cl₂ → 2HCl + S Chlorine oxidizes hydrogen sulfide to sulfur and is reduced to HCl. It is correct and balanced.
Answer: Option B is the equation that does not illustrate a real reaction of chlorine.
Tip: For halogen displacement questions, remember that a halogen can only displace another halogen below it in the group. So Cl₂ can displace Br⁻ and I⁻, but never F⁻. A quick memory aid is "Fluorine is the Fiercest."
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39.
If $100^3$ of a saturated solution of sodium tetraoxosulphate (VI) at 30° C contains 10.5 g of the salt, what would be its solubility at this temperature? [( $Na_2SO_4$ = 142]
1. A.2.15 mol $dm^{-3}$
2. B.0.74 mol $dm^{-3}$
3. C.0.60 mol $dm^{-3}$
4. D.0.57 mol $dm^{-3}$
Show answer
Answer: B. 0.74 mol $dm^{-3}$
AI explanation
Analyze the question: The question says "100³" but this is almost certainly a typo for 100 cm³. A saturated solution of 100 cm³ contains 10.5 g of Na₂SO₄. We need the solubility in mol dm⁻³.
Concept: Solubility in mol dm⁻³ = (mass dissolved in g per dm³) ÷ molar mass
Solution:
Convert the mass to g per dm³. 100 cm³ holds 10.5 g, so 1000 cm³ holds: 10.5 × 10 = 105 g/dm³
Convert to moles using the molar mass (Na₂SO₄ = 142 g/mol): Solubility = 105 ÷ 142 = 0.739 mol/dm³ ≈ 0.74 mol dm⁻³
Check against the options: 0.
Answer: B
Tip: For solubility questions, always follow two steps: (1) scale the mass up to 1 dm³ (1000 cm³), then (2) divide by the molar mass. Watch the units. If you see "100 cm³", multiply by 10 to get per dm³.
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40.
The separation technique that is suitable for separating iodine from tetrachloromethane is
1. A.solvent extraction
2. B.decantation
3. C.fractional distillation
4. D.filtration
Show answer
Answer: C. fractional distillation
AI explanation
Analyze the question: Iodine dissolves in tetrachloromethane (CCl₄), giving a purple solution. So we need to separate a dissolved solute (iodine) from its solvent (CCl₄).
Concept:
Solvent extraction separates a substance based on its different solubility in two immiscible solvents. Iodine could be pulled out of CCl₄ into another solvent, but that would need a suitable second solvent (e.g., aqueous KI) and is not the usual way to recover iodine from CCl₄.
Decantation separates an insoluble solid from a liquid, or two immiscible liquids. It does not work here because the iodine is dissolved.
Filtration separates an insoluble solid from a liquid. It does not work on a solution.
Fractional distillation separates miscible liquids or a solution by differences in boiling points. CCl₄ boils at about 77 °C. Iodine is a solid with a much higher boiling point (about 184 °C) and also sublimes. Heating the mixture drives off the CCl₄ first, leaving the iodine behind.
Solution:
Iodine and CCl₄ form a homogeneous solution, so decantation and filtration are ruled out.
The components have very different boiling points (CCl₄ ≈ 77 °C, I₂ ≈ 184 °C).
Heating and distilling collects the CCl₄ as distillate, and the iodine remains in the flask.
Strictly speaking, simple distillation is the more exact term, since only one component is volatile. Among the options given, fractional distillation is the one that fits, as it is the distillation-based method.
Answer: C. fractional distillation
Tip: First ask, "Is the solid dissolved or not?" If it is dissolved, filtration and decantation are out. Then compare boiling points: a large difference points to distillation.
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41.
A mixture containing two salts of different solubility can be separated by
1. A.chromatography.
2. B.crystallization.
3. C.distillation.
4. D.evaporation.
Show answer
Answer: B. crystallization.
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42.
The reduction half equation of the following reaction is: $Zn_{(s)} + CuSO_{4(aq)} → ZnSO_{4(aq)} + Cu_{(s)}$
1. A. $Zn_{(s)} → Zn^{2+}_{(aq)} + 2e^-$
2. B. $CuSO_{4(s)} + H_2O_{(l)}$ → $Cu^{2+}_{(aq)}$ + $SO_{4(aq)}^{2-}$
3. C. $Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}$
4. D. $Cu^{2+}_{(aq)} + e^- → Cu_{(s)}$
Show answer
Answer: C. $Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}$
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43.
Species that occur in a reaction pathway but not in the overall reaction are known as
1. A.Reactants
2. B.Inhibitors
3. C.Products
4. D.Intermediates
Show answer
Answer: D. Intermediates
AI explanation
Step 1: Analyze the question The question asks for the name of species that appear in the steps of a reaction mechanism (the reaction pathway) but do not appear in the overall balanced equation.
Step 2: Explain the concept Many reactions happen in several steps, called elementary steps. In these steps:
Reactants are the starting materials. They appear on the left of the overall equation.
Products are the final substances formed. They appear on the right of the overall equation.
Intermediates are formed in one step and used up in a later step. They never show up in the overall equation because they cancel out.
Inhibitors are substances that slow down or stop a reaction. They are not defined by the pathway-versus-overall idea.
Step 3: Example Consider a two-step mechanism:
Step 1: A + B → C
Step 2: C + D → E
Adding the steps gives A + B + C + D → C + E. Cancelling C from both sides gives the overall reaction A + B + D → E.
Species C is produced in step 1 and consumed in step 2. It is present in the pathway but absent from the overall reaction, so C is an intermediate.
Step 4: Check the options
A. Reactants: these appear in the overall reaction. ✗
B. Inhibitors: these slow reactions and have nothing to do with the cancellation of species. ✗
C. Products: these appear in the overall reaction. ✗
D. Intermediates: formed and then consumed, so absent from the overall equation. ✓
Answer: D. Intermediates
Exam tip: Think "in-between". An intermediate exists only between steps, so it is made and then used up. If you can cancel a species out when adding the steps, it is an intermediate. Don't confuse it with a catalyst, which is used first and then regenerated. Both vanish from the overall equation, but an intermediate is made first and then consumed, while a catalyst is consumed first and then remade.
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44.
Positive ions in a sea of electrons are found in
1. A.covalent bonds
2. B.metallic bonds
3. C.dative bonds
4. D.ionic bonds
Show answer
Answer: B. metallic bonds
AI explanation
Step 1: Analyze the question The question describes a structure with "positive ions in a sea of electrons." We need to identify which type of bonding matches this description.
Step 2: Explain the concept Each type of bond has its own features:
Covalent bond: atoms share pairs of electrons (e.g., H₂, CH₄).
Metallic bond: metal atoms lose their valence electrons, which become delocalised. The result is a lattice of positive metal ions surrounded by a "sea" of free-moving electrons. The electrostatic attraction between the ions and the electrons holds the metal together.
Dative (coordinate) bond: both electrons in the shared pair come from one atom (e.g., NH₄⁺, H₃O⁺).
Ionic bond: electrons are transferred from a metal to a non-metal, forming oppositely charged ions held together by electrostatic attraction (e.g., NaCl). There is no "sea of electrons."
Step 3: Apply to the question The phrase "positive ions in a sea of electrons" is the classic description of metallic bonding. It also explains why metals conduct electricity (the free electrons move), and why they are malleable and ductile (layers of ions can slide without breaking the bonding).
Step 4: Answer The correct option is B. metallic bonds.
Tip: Link the key words to the bond type:
"Sea of electrons" or "delocalised electrons" → metallic
"Sharing" → covalent
"Transfer" → ionic
"One atom donates both electrons" → dative
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45.
The primary component of natural gas
1. A.ethane
2. B.methane
3. C.butane
4. D.propane
Show answer
Answer: B. methane
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46.
The following steps are scientific methods except
1. A.experimentation
2. B.problem identification
3. C.Analysis
4. D.Open-mindedness
Show answer
Answer: D. Open-mindedness
AI explanation
Analyze the question: The question asks which option is not a step in the scientific method.
Concept: The scientific method is a systematic procedure for investigating problems. The usual steps are:
Identifying or stating the problem
Observation and gathering information
Forming a hypothesis
Experimentation (testing the hypothesis)
Analysis and interpretation of results (data analysis)
Drawing a conclusion
Reporting or communicating the results
Checking each option:
A. Experimentation: This is a step, because the hypothesis is tested by experiment.
B. Problem identification: This is a step, because it is usually the very first stage.
C. Analysis: This is a step, because results are analyzed and interpreted after the experiment.
D. Open-mindedness: This is not a step. It is a scientific attitude (a quality or disposition of a scientist), like honesty, curiosity, objectivity, and patience. It describes how a scientist should behave, not a stage in the procedure.
Answer: D
Exam tip: Keep scientific methods (steps/procedures) separate from scientific attitudes (traits). Steps are actions, such as identifying, observing, hypothesizing, experimenting, analyzing, and concluding. Attitudes are qualities, such as open-mindedness, honesty, curiosity, and objectivity. If an option describes a personal quality rather than an action, it is probably the odd one out.
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47.
The pair of compounds that belongs to the same homologous series is
1. A. $C_2H_6$ and $C_4H_{10}$
2. B. $C_3H_8$ and $C_3H_6$
3. C. $C_2H_4$ and $C_4H_{10}$
4. D. $C_4H_{10}$ and $C_5H_{10}$
Show answer
Answer: A. $C_2H_6$ and $C_4H_{10}$
AI explanation
Analyze the question: A homologous series is a family of organic compounds with the same general formula and the same functional group. Successive members differ by a CH₂ unit.
Concept: general formulas
Alkanes: CₙH₂ₙ₊₂
Alkenes: CₙH₂ₙ
Alkynes: CₙH₂ₙ₋₂
Check each option:
A. C₂H₆ and C₄H₁₀ • C₂H₆: n = 2, so 2n+2 = 6 ✔ (alkane, ethane) • C₄H₁₀: n = 4, so 2n+2 = 10 ✔ (alkane, butane) • Both are alkanes and differ by 2 CH₂ units. They are in the same homologous series.
B. C₃H₈ and C₃H₆ • C₃H₈ is an alkane (propane). C₃H₆ is an alkene (propene). Different series. ✘
C. C₂H₄ and C₄H₁₀ • C₂H₄ is an alkene (ethene). C₄H₁₀ is an alkane (butane). ✘
D. C₄H₁₀ and C₅H₁₀ • C₄H₁₀ is an alkane (butane). C₅H₁₀ is an alkene (pentene) or a cycloalkane. ✘
Answer: Option A.
Tip: Quickly test each formula against CₙH₂ₙ₊₂ (alkane), CₙH₂ₙ (alkene) and CₙH₂ₙ₋₂ (alkyne). If two compounds in a pair fit the same formula, they belong to the same homologous series. If the carbon count is the same but the hydrogen count differs (as in B), they are in different series.
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48.
Which of the following equimolar solutions has the highest conductivity?
1. A. $H_2SO{_4(aq)}$
2. B. $NaOH{_(aq)}$
3. C. $H_2CO{_3(aq)}$
4. D. $CH_3COOH{_4(aq)}$
Show answer
Answer: A. $H_2SO{_4(aq)}$
AI explanation
Analyze the question: Conductivity of a solution depends on the concentration of mobile ions. For equimolar solutions (same concentration), the one that produces the most ions has the highest conductivity.
Concept:
Strong electrolytes ionize completely and give many ions.
Weak electrolytes ionize only slightly and give few ions.
Also, H⁺ ions are very mobile, which increases conductivity further.
Analyze each option:
A. H₂SO₄: a strong acid that ionizes almost completely. H₂SO₄ → 2H⁺ + SO₄²⁻ Each mole gives about 3 moles of ions, including highly mobile H⁺.
B. NaOH: a strong base that ionizes completely. NaOH → Na⁺ + OH⁻ Each mole gives 2 moles of ions.
C. H₂CO₃: a weak acid that ionizes only slightly, so few ions.
D. CH₃COOH (written as CH₃COOH₄, a typo): a weak acid that ionizes only slightly, so few ions.
Compare: Strong electrolytes (A and B) beat weak ones (C and D). Between A and B, H₂SO₄ releases more ions per mole (about 3 versus 2), and H⁺ has a very high ionic mobility. So H₂SO₄ has the highest conductivity.
Answer: A. H₂SO₄(aq)
Tip: For "equimolar" conductivity questions, first pick out the strong electrolytes, then choose the one that gives the most ions per formula unit. Remember the strong acids: H₂SO₄, HCl and HNO₃. Weak acids such as ethanoic acid and carbonic acid always rank low.
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49.
The best indicator to use for the titration of ethanoic acid with sodium hydroxide is
1. A.methyl orange
2. B.phenolphthalein
3. C.screened methyl orange
4. D.methyl red
Show answer
Answer: B. phenolphthalein
AI explanation
Step 1: Analyze the question Ethanoic acid (CH₃COOH) is a weak acid, and sodium hydroxide (NaOH) is a strong base. We need the indicator that changes colour at the right pH for this pairing.
Step 2: The concept The indicator must change colour within the steep part of the titration curve, around the equivalence point.
At the equivalence point, the product is sodium ethanoate (CH₃COONa), a salt of a weak acid and a strong base. It hydrolyses in water:
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
This produces excess OH⁻ ions, so the solution is slightly basic, with a pH of about 8.5 to 9.
Step 3: Match with the indicators
| Indicator | pH range | Suitable? | |---|---|---| | Methyl orange | 3.1 – 4.4 | No (acidic range) | | Screened methyl orange | about 3.2 – 4.2 | No (acidic range) | | Methyl red | 4.4 – 6.2 | No (too acidic) | | Phenolphthalein | 8.3 – 10.0 | Yes, matches the equivalence point |
Step 4: Answer Phenolphthalein changes colour (colourless in acid to pink in base) in the pH range of 8.3 to 10, which covers the equivalence point of a weak acid and strong base titration. The correct answer is B.
Exam tip: Remember the pairings:
Strong acid + strong base: almost any indicator works
Weak acid + strong base: phenolphthalein
Strong acid + weak base: methyl orange
Weak acid + weak base: no suitable indicator
A quick memory trick: "Phenolphthalein is for Poor (weak) acids."
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50.
The salt solution formed firm a reaction between ethanoic acid and sodium hydroxide solution would be
1. A.amphoteric.
2. B.neutral
3. C.basic
4. D.acidic
Show answer
Answer: C. basic
AI explanation
Analyze the question: Ethanoic acid (CH₃COOH) is a weak acid. Sodium hydroxide (NaOH) is a strong base. The question asks whether the salt solution formed (sodium ethanoate) is acidic, basic, neutral or amphoteric.
Concept: Salt hydrolysis The nature of a salt solution depends on the strength of the acid and base that formed it:
Strong acid + strong base → neutral salt (e.g. NaCl)
Strong acid + weak base → acidic salt solution (e.g. NH₄Cl)
Weak acid + strong base → basic (alkaline) salt solution (e.g. CH₃COONa)
Weak acid + weak base → depends on the relative strengths
Solution:
Reaction: CH₃COOH + NaOH → CH₃COONa + H₂O
Sodium ethanoate dissolves and ionises completely: CH₃COONa → CH₃COO⁻ + Na⁺
Na⁺ comes from a strong base, so it does not react with water.
The ethanoate ion CH₃COO⁻ is the conjugate base of a weak acid, so it picks up H⁺ from water (hydrolysis): CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
This produces excess OH⁻ ions, so the solution has pH > 7, which means it is basic.
Answer: The salt solution is basic, which is option C.
Tip: Remember "the strong one wins." The salt solution takes on the character of the stronger parent. Weak acid + strong base gives a basic solution. Strong acid + weak base gives an acidic solution. Equal strengths give a neutral solution.
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2023 WAEC (WASSCE) Chemistry Past Questions and Answers

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