The IUPAC name for CH$_{3}$CHCH$_{3}$CHCLCHCH$_{3}$CH$_{2}$CH$_{3}$ is

D. 3 - chloro - 2, 4 - dimethylhexane

A. 2, 4 - dimethyl - 3 - chlorohexane

B. 3, 5 - dimethyl - 4 - chlorohexane

C. 4 - chloro - 3, 5 - dirnethylhexane

D. 3 - chloro - 2, 4 - dimethylhexane

Alkanols can be manufactured from alkenes by the initial reaction of alkenes with

B. concentrated tetraoxosulphate(VI) acid

A. bromine in tetrachloromethane

B. concentrated tetraoxosulphate(VI) acid

C. aqueous potassium tetraoxomanganate(VII).

Before a reaction could take place, there should be

C. breakage of bonds of reactants

A. ionization of reactant particles

B. breakage of bonds of products

C. breakage of bonds of reactants

D. ionization of product particles

The cleansing effect of soap is low in acidic water because of

A. the formation of unionized fatty acid.

B. increase in the pH of the soap molecules.

C. anti-biodegradable effect of hydrogen ions

D. the hardness of the acidic water.

Which of the following statements about atoms of a metal is correct? They

D. are held together by a sea of electron cloud

C. are held together by covalent bond

D. are held together by a sea of electron cloud

When air in a syringe is compressed such that there is no change in temperature, the

C. intermolecular space increases

Consider the following reaction equation N$_{2}$(g) + 3H$_{2(g)}$ ⇌ 2NH3(g) ∆H = -92kJ Increasing the temperature of the reaction would

D. decrease the yield of ammonia

A. shift the equilibrium to the right

B. increase the yield of ammonia

C. decrease the amount of hydrogen

D. decrease the yield of ammonia

Electropositivity of elements across the periodic table normally

A. remains constant down the group

If 5.0 g of marble reacts with 25.0 cm3 hydrochloric acid, which of the following combinations has the fastest reaction rate?

C. Powdered marble and 2.5 mol dm$^{-3}$ HCl$_{(aq)}$

A. Marble chips and 2.0 mol dm$^{-3}$ HCl$_{(aq)}$

B. Marble chips and 2.5 mol dm$^{-3}$ HCl$_{(aq)}$

C. Powdered marble and 2.5 mol dm$^{-3}$ HCl$_{(aq)}$

D. Powdered marble and 2.0 mol dm$^{-3}$ HCl$_{(aq)}$

Which of the following statements about an equilibrium system is correct?

A. Forward and backward reactions occur at the same rate

B. The concentrations of reactants must equal that of the products

C. The concentrations of, reactants and products can be changed by adding a catalyst,

D. The concentrations of reactants and products are not affected by a change in temperature

Which of the following processes occur during fractional distillation of petroleum?

D. Evaporation and condensation

The electron configuration of carbon atom in its excited state is

C. 1s$^{2}$2s$^{1}$2px$^{1}$2py$^{1}$2pz$^{1}$

A. 1s$^{2}$2s$^{2}$2px$^{2}$2py$^{0}$

B. 1s$^{2}$2s$^{2}$2px$^{1}$2py$^{1}$

C. 1s$^{2}$2s$^{1}$2px$^{1}$2py$^{1}$2pz$^{1}$

D. 1s$^{1}$2s$^{2}$2px$^{1}$2py$^{1}$

Increasing the temperature generally

B. increases the solubility of a solid in a liquid but decreases the solubility of a gas in a liquid.

A. decreases the solubility of a solid in a liquid but increases the solubility of a gas in a liquid.

B. increases the solubility of a solid in a liquid but decreases the solubility of a gas in a liquid.

C. increases the solubility of both a solid and a gas in a liquid

D. decreases the solubility of both a solid and a gas in a liquid

A colourless gas with a pungent smell is evolved when dilute hydrochloric acid is added to a sample of a salt. The gas evolved could turn

B. acidified KMnO$_{4}$ solution colourless

A. acidified K$_{2}$Cr$_{2}$O$_{7}$ solution colourless

B. acidified KMnO$_{4}$ solution colourless

C. Fe(NO$_{3}$)$_{2}$ solution green

D. Pb(NO$_{3}$)$_{2}$ paper black

Ionization energy increases across the period in the periodic table because

B. effective nuclear charge increases

Which of the following statements about the standard hydrogen electrode is not correct?

D. The temperature is kept at 20 °C

A. The hydrogen gas is at a pressure of 1 atmosphere

B. A solution containing 1 mol $dm^{-3}$ of H+ (aq) ions is used

C. A platinum electrode is used

D. The temperature is kept at 20 °C

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WAEC (WASSCE) 2022

2022 WAEC (WASSCE) Chemistry Past Questions and Answers

The 2022 WAEC (WASSCE) Chemistry past questions with answers: 50 objective questions from the 2022 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.
What amount of electricity is required to deposit one mole of aluminium from a solution of AlCl3?
1. A.One Faraday
2. B.One ampere
3. C.Three Faradays
4. D.Three amperes
Show answer
Answer: C. Three Faradays
AI explanation
Analyze the question: We need the quantity of electricity (charge) needed to deposit 1 mole of aluminium from AlCl₃ solution.
Concept: The charge needed depends on the number of electrons gained per ion. One mole of electrons carries 1 Faraday (about 96,500 C). Faraday is a unit of charge (quantity of electricity), while the ampere is a unit of current, not an amount of electricity.
Solution:
In solution, AlCl₃ gives Al³⁺ ions.
At the cathode: Al³⁺ + 3e⁻ → Al
To deposit 1 mole of Al, 3 moles of electrons are needed.
Charge = 3 × 96,500 C = 289,500 C, which is 3 Faradays.
Check the options:
A. One Faraday deposits only 1/3 mole of Al, so this is wrong.
B and D. The ampere is a unit of current, not an amount of electricity, so both are wrong.
C. Three Faradays matches our working.
Tip: The number of Faradays needed to deposit 1 mole of a metal equals the charge on its ion. For example, Na⁺ needs 1 F, Cu²⁺ needs 2 F, and Al³⁺ needs 3 F. Also remember that amperes measure current, while Faradays and coulombs measure charge.
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2.
Consider the following reaction equation: CaO + SiO $_{2}$ ------> CaSiO $_{3}$ Silicon(IV) oxide is acting as
1. A.a basic oxide
2. B.a reducing agent
3. C.an acidic oxide
4. D.an oxidizing agent
Show answer
Answer: C. an acidic oxide
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3.
An oxide has the following properties. It I. is a white powder II. reacts with HCI. III. reacts with NaOH IV. is insoluble in water The oxide is
1. A.alkaline
2. B.amphoteric
3. C.acidic
4. D.neutral
Show answer
Answer: B. amphoteric
AI explanation
Analyze the question: We're given four properties of an oxide and asked to classify it. The key clues are properties II and III: it reacts with both an acid (HCl) and a base (NaOH).
Concept: Types of oxides
| Type | Behavior | Examples | |---|---|---| | Basic/alkaline | Reacts with acids only (alkaline if it dissolves in water) | Na₂O, CaO, MgO | | Acidic | Reacts with bases only | CO₂, SO₂, P₄O₁₀ | | Neutral | Reacts with neither | CO, H₂O, N₂O | | Amphoteric | Reacts with both acids and bases | ZnO, Al₂O₃, PbO |
Step-by-step solution
White powder: this is true of ZnO and Al₂O₃, so it doesn't rule anything out.
Reacts with HCl: this shows basic character, which rules out acidic and neutral oxides. • ZnO + 2HCl → ZnCl₂ + H₂O
Reacts with NaOH: this shows acidic character, which rules out purely basic/alkaline oxides. • ZnO + 2NaOH + H₂O → Na₂[Zn(OH)₄]
Insoluble in water: this rules out an alkaline oxide, since alkaline oxides dissolve in water to form hydroxides (e.g. Na₂O + H₂O → 2NaOH).
Reacting with both an acid and a base is the definition of an amphoteric oxide. An example is zinc oxide or aluminium oxide.
Answer: B (amphoteric)
Exam tip: Remember "amphoteric = both." When a question says an oxide reacts with both an acid and an alkali, think of ZnO, Al₂O₃, PbO (a mnemonic: "Zebras, Ants and Pigs are amphoteric"). Also, if it reacts with both, you can ignore details like color and solubility.
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4.
What number of moles of oxygen would exert a pressure of 10 atm at 320 K in an 8.2 dm $^{3}$ cylinder? [R=0.082 atm dm $^{-3}$ mol $^{-1}$ K $^{-1}$ ]
1. A.0.32
2. B.1.52
3. C.3.13
4. D.31.25
Show answer
Answer: C. 3.13
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5.
Which of the following sugars is a product of the condensation of monosaccharides?
1. A.galactose
2. B.maltose
3. C.glucose
4. D.fructose
Show answer
Answer: B. maltose
AI explanation
Analyze the question: It asks which sugar is formed when monosaccharides (simple sugars) join together by condensation. That means we are looking for a disaccharide, not a monosaccharide.
Explain the concept:
Monosaccharides are single sugar units that cannot be hydrolysed into simpler sugars. Examples are glucose, fructose and galactose (all C₆H₁₂O₆).
Disaccharides are formed when two monosaccharides join by a condensation reaction. A molecule of water is lost and a glycosidic bond forms between them.
Common disaccharides: • Maltose = glucose + glucose • Sucrose = glucose + fructose • Lactose = glucose + galactose
Solution:
Condensation of two glucose units:
C₆H₁₂O₆ + C₆H₁₂O₆ → C₁₂H₂₂O₁₁ + H₂O (glucose + glucose → maltose + water)
Check each option:
A. Galactose is a monosaccharide, so it is not a product of condensation.
B. Maltose is a disaccharide formed from two glucose units, so it is a product of condensation.
C. Glucose is a monosaccharide.
D. Fructose is a monosaccharide.
Answer: B. maltose.
Exam tip: Sugar names ending in -ose can be monosaccharides or disaccharides, so memorize the three disaccharides: MaSLo, meaning Maltose, Sucrose, Loctose (lactose). Each contains glucose. Anything else in the list is usually a monosaccharide.
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6.
Which of the following acids would form normal salt only?
1. A.Tetraoxosulphate(VI) acid
2. B.Trioxonitrate(V) acid
3. C.Tetraoxophosphate(V) acid
4. D.Trioxosulphate(IV) acid
Show answer
Answer: B. Trioxonitrate(V) acid
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7.
The atom with the electron configuration 1s $^{2}$ 2s $^{2}$ 2p63s $^{2}$ 3p $^{6}$ 3d $^{10}$ 4s $^{2}$ 4p $^{4}$ is in
1. A.period 4, p- block
2. B.period 3, p- block
3. C.period 4, d- block
4. D.period 3, d- block
Show answer
Answer: A. period 4, p- block
AI explanation
Step 1: Analyze the question We need to find the period and block of an element from its electron configuration.
Step 2: Concept
Period = the highest principal quantum number (n) occupied. Here the highest is n = 4, so the element is in period 4.
Block = the type of subshell that holds the last electron added. The last electrons go into 4p, so it is in the p-block.
Step 3: Solution Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁴
Total electrons = 2 + 2 + 6 + 2 + 6 + 10 + 2 + 4 = 34, so the element is selenium (Se).
The outermost shell is n = 4, so it is in period 4.
The valence electrons are 4s² 4p⁴, which is 6 valence electrons, so it is in Group 16 (VIA). The last subshell filling is 4p, so it is in the p-block.
The 3d subshell is full and lies inside the 4th shell, so it does not make the element a d-block element. d-block elements have their last electron in a d subshell (such as 3d¹ to 3d⁹).
Step 4: Answer Period 4, p-block, which is option A.
Tip: Look at the last subshell in the filling order, and use the highest shell number for the period. Do not be misled by 3d¹⁰ appearing in the middle. If the configuration ends in p, it is a p-block element.
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8.
The IUPAC name for CH $_{3}$ CHCH $_{3}$ CHCLCHCH $_{3}$ CH $_{2}$ CH $_{3}$ is
1. A.2, 4 - dimethyl - 3 - chlorohexane
2. B.3, 5 - dimethyl - 4 - chlorohexane
3. C.4 - chloro - 3, 5 - dirnethylhexane
4. D.3 - chloro - 2, 4 - dimethylhexane
Show answer
Answer: D. 3 - chloro - 2, 4 - dimethylhexane
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9.
Oxochlorate(I) acid is used as a bleaching agent because it is
1. A.a weak acid
2. B.a reducing agent.
3. C.an oxidizing agent
4. D.a strong acid.
Show answer
Answer: C. an oxidizing agent
AI explanation
Analyze the question Oxochlorate(I) acid is HOCl (hypochlorous acid). The question asks why it works as a bleaching agent.
Explain the concept Bleaching by chlorine-based compounds is an oxidation process. Coloured substances (dyes and stains) contain chromophores, which are groups that absorb visible light. An oxidizing agent breaks down or changes these chromophores, so the substance loses its colour.
Step-by-step solution
HOCl contains chlorine in the +1 oxidation state, which is the meaning of "(I)" in oxochlorate(I).
Chlorine is very stable as chloride (Cl⁻, oxidation state −1), so HOCl readily gains electrons: HOCl + H⁺ + 2e⁻ → Cl⁻ + H₂O
Because it gains electrons, HOCl is reduced, and so it is an oxidizing agent.
In bleaching, HOCl releases nascent (atomic) oxygen: HOCl → HCl + [O] The [O] oxidizes the coloured dye to a colourless compound: Dye (coloured) + [O] → Oxidized dye (colourless)
Checking the options
A. Weak acid: HOCl is a weak acid, but acid strength is not what causes bleaching.
B. Reducing agent: wrong, because Cl(+1) is reduced, not oxidized.
C. Oxidizing agent: correct, because bleaching works by oxidation.
D. Strong acid: wrong, because HOCl is weak, and acid strength is irrelevant to bleaching anyway.
Exam tip Bleaching agents such as chlorine, HOCl, NaOCl (household bleach) and H₂O₂ are generally oxidizing agents. The exception is SO₂, which bleaches by reduction. So "bleach = oxidize" works for most WASSCE questions.
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10.
Which of the following properties indicate that an element is a metal. It I. reacts with oxygen to form an acidic oxide II. forms ionic chlorides III. has variable oxidation states IV. displaces hydrogen from dilute HCl
1. A.I and III only
2. B.I and II only
3. C.II and IV only
4. D.I, II, III and IV
Show answer
Answer: C. II and IV only
AI explanation
Analyze the question: We need to find which properties are characteristic of metals.
Check each statement:
I. Reacts with oxygen to form an acidic oxide: False. Metals form basic oxides (or amphoteric ones). Non-metals form acidic oxides, such as CO₂ and SO₂.
II. Forms ionic chlorides: True. Metals lose electrons to form cations, so their chlorides are ionic (NaCl, MgCl₂). Some metal chlorides like AlCl₃ are partly covalent, but this is the general metallic property.
III. Has variable oxidation states: This is typical of transition metals (Fe²⁺/Fe³⁺, Cu⁺/Cu²⁺), but not of all metals. Na, K, Mg and Ca have only one oxidation state. Non-metals such as sulfur and nitrogen also show variable oxidation states, so it does not reliably indicate a metal.
IV. Displaces hydrogen from dilute HCl: True for metals above hydrogen in the reactivity series (Mg, Zn, Fe). This is a classic test of metallic character, although unreactive metals like Cu and Ag do not do it.
Conclusion: The properties that indicate a metal are II and IV only.
Tip: Metals form basic oxides, ionic chlorides, and release H₂ from acids. Non-metals form acidic oxides and covalent chlorides. Don't be misled by "variable oxidation states", as it is not exclusive to metals.
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11.
If 60 g of M combines with 24 g of oxygen, what would the empirical formula of the oxide be? ( molar mass of M = 120, O = 16)
1. A.MO
2. B.MO $_{2}$
3. C.MO $_{3}$
4. D.MO $_{4}$
Show answer
Answer: C. MO $_{3}$
AI explanation
Step 1: Analyze the question We are given the masses of M and oxygen in an oxide, and we need the simplest whole-number ratio of atoms (the empirical formula).
Step 2: Concept Empirical formula steps:
Find moles of each element (mass ÷ molar mass)
Divide by the smallest number of moles
Convert to whole numbers
Step 3: Solution
Moles of M = 60 ÷ 120 = 0.5 mol
Moles of O = 24 ÷ 16 = 1.5 mol
Divide by the smallest (0.5):
M = 0.5 ÷ 0.5 = 1
O = 1.5 ÷ 0.5 = 3
The ratio M : O = 1 : 3, so the empirical formula is MO₃.
Step 4: Answer
Tip: Remember "Mass → Moles → Divide → Whole number". If the ratio gives a decimal like 1.5, multiply everything by 2. Here the numbers divide out cleanly, so no extra step is needed.
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12.
Charcoal is used in the decolourization of sugar because of its
1. A.absorption property.
2. B.amorphous property.
3. C.oxidizing property.
4. D.adsorption property.
Show answer
Answer: D. adsorption property.
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13.
Graphite is used as a dry lubricant due to the presence of
1. A.mobile electrons
2. B.free electrons
3. C.octahedral structures
4. D.layered structures
Show answer
Answer: D. layered structures
AI explanation
Step 1: Analyze the question The question asks why graphite can act as a dry lubricant, meaning why it feels slippery and lets surfaces slide over each other.
Step 2: Explain the concept Graphite is a giant covalent structure of carbon with these features:
Each carbon atom is covalently bonded to three other carbon atoms in flat hexagonal rings. This forms a flat sheet, or layer.
The fourth valence electron of each carbon is delocalised (mobile) within the layer. This is why graphite conducts electricity.
The layers are held together only by weak van der Waals forces.
Step 3: Apply it Because the forces between the layers are weak, the layers slide over one another easily when pressure is applied. This sliding is what makes graphite slippery and useful as a dry lubricant, for example in machine parts and pencil "lead".
Step 4: Check each option
A. Mobile electrons: These explain graphite's electrical conductivity, not its lubricating property.
B. Free electrons: This is the same idea as A, so it also relates to conductivity.
C. Octahedral structures: Graphite has no such structure. Its structure is planar hexagonal.
D. Layered structures: Correct. Layers held by weak forces slide over each other.
Answer: D. Layered structures
Exam tip: Link each property of graphite to its cause:
Conducts electricity → delocalised (mobile) electrons
Lubricant and soft → layers with weak van der Waals forces
Diamond is hard → a rigid 3D network of covalent bonds
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14.
Alkanols can be manufactured from alkenes by the initial reaction of alkenes with
1. A.bromine in tetrachloromethane
2. B.concentrated tetraoxosulphate(VI) acid
3. C.aqueous potassium tetraoxomanganate(VII).
4. D.sodium hydroxide solution.
Show answer
Answer: B. concentrated tetraoxosulphate(VI) acid
AI explanation
Analyze the question: We want to make an alkanol (alcohol) from an alkene. The question asks which reagent is used in the first (initial) step.
Concept: Industrially, alcohols such as ethanol are made from alkenes by hydration. In the laboratory and in some industrial processes, this happens in two stages:
Addition of concentrated H₂SO₄ (tetraoxosulphate(VI) acid) across the C=C double bond, forming an alkyl hydrogen sulphate: CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₃H (ethyl hydrogen sulphate)
Hydrolysis by adding water and warming: CH₃CH₂OSO₃H + H₂O → CH₃CH₂OH + H₂SO₄
Check the other options:
A. Br₂ in CCl₄: This is the test for unsaturation. It gives a dibromoalkane (e.g. 1,2-dibromoethane), not an alcohol.
C. Aqueous KMnO₄: This oxidises the alkene to a diol (e.g. ethane-1,2-diol), not a simple alkanol. It is also the Baeyer test for unsaturation.
D. NaOH solution: Alkenes do not react with it under normal conditions.
Answer: The initial reaction is with concentrated tetraoxosulphate(VI) acid, which is option B.
Tip: Remember "alkene → alcohol = acid first, then water." Br₂ means test for unsaturation, and KMnO₄ means diol or oxidation.
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15.
Which of the following elements would displace copper from a solution of copper ions?
1. A.silver
2. B.gold
3. C.lead
4. D.mercury
Show answer
Answer: C. lead
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16.
Which of the following oxides is amphoteric?
1. A.Carbon(II) oxide
2. B.Nitrogen(IV) oxide
3. C.Lead(II) oxide
4. D.Calcium oxide
Show answer
Answer: C. Lead(II) oxide
AI explanation
Step 1: Analyze the question We need the oxide that is amphoteric, meaning it reacts with both acids and bases to form salts and water.
Step 2: The concept
Acidic oxides (non-metal oxides such as CO₂, SO₂, NO₂) react with bases.
Basic oxides (metal oxides such as CaO, Na₂O, MgO) react with acids.
Neutral oxides (CO, H₂O, N₂O) react with neither.
Amphoteric oxides react with both acids and bases. The common ones are ZnO, Al₂O₃, PbO, SnO (remember "Zebra And Pig Stay"... or simply Zn, Al, Pb, Sn).
Step 3: Check each option
A. Carbon(II) oxide (CO): Neutral oxide. It does not react with acids or bases under normal conditions.
B. Nitrogen(IV) oxide (NO₂): Acidic oxide. It dissolves in water to give HNO₃ and HNO₂, and it reacts with bases.
C. Lead(II) oxide (PbO): Amphoteric. • With acid: PbO + 2HNO₃ → Pb(NO₃)₂ + H₂O • With base: PbO + 2NaOH + H₂O → Na₂[Pb(OH)₄] (sodium plumbite)
D. Calcium oxide (CaO): Basic oxide. It reacts with acids only: CaO + 2HCl → CaCl₂ + H₂O.
Step 4: Answer Lead(II) oxide is the amphoteric oxide, which is option C.
Tip: Metals that give amphoteric oxides sit near the "staircase" between metals and non-metals in the periodic table or are post-transition metals. Memorize Zn, Al, Pb, Sn. Group 1 and 2 metal oxides (like CaO) are basic, and non-metal oxides are acidic or neutral.
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17.
Before a reaction could take place, there should be
1. A.ionization of reactant particles
2. B.breakage of bonds of products
3. C.breakage of bonds of reactants
4. D.ionization of product particles
Show answer
Answer: C. breakage of bonds of reactants
AI explanation
Analyze the question: It asks what must happen before a chemical reaction can take place. This is about how reactions proceed, which is collision theory and bond energy.
Explain the concept: For a chemical reaction to occur:
Reactant particles must collide with enough energy (the activation energy).
The bonds holding the reactant particles together must be broken. This requires energy, so it is endothermic.
New bonds then form to make the products. This releases energy, so it is exothermic.
Bond breaking in the reactants always comes first. The products do not exist until the reaction has happened, so their bonds cannot be broken beforehand.
Evaluate the options:
A. Ionization of reactant particles: Not necessary. Many reactions involve covalent molecules and never form ions, for example CH₄ + 2O₂ → CO₂ + 2H₂O.
B. Breakage of bonds of products: Wrong. The products have not formed yet.
C. Breakage of bonds of reactants: Correct. Existing bonds must be broken (or at least weakened) so that atoms can rearrange into products.
D. Ionization of product particles: Wrong, for the same reason as B.
Example: In H₂ + Cl₂ → 2HCl, the H–H and Cl–Cl bonds break first. Then the H–Cl bonds form.
**Answer: C. Breakage of bonds of reactants.
Exam tip: Remember "Break first, make later." Reactant bonds break (energy absorbed), then product bonds form (energy released). Any option that mentions products being broken before the reaction can be eliminated straight away.
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18.
The following compounds are condensation polymers except
1. A.nylon
2. B.protein
3. C.starch
4. D.polyethene
Show answer
Answer: D. polyethene
AI explanation
Step 1: Analyze the question We are asked which compound is NOT a condensation polymer. So three options are condensation polymers, and one is not (probably an addition polymer).
Step 2: The concept
Condensation polymerisation: monomers join together and lose a small molecule (usually water, or HCl) at each linkage.
Addition polymerisation: monomers with a C=C double bond join together with no loss of any molecule. Nothing is lost, so the polymer has the same empirical formula as the monomer.
Step 3: Check each option
A. Nylon: It is made from a diamine and a dicarboxylic acid (or diacid chloride). Water (or HCl) is lost as the amide links form. This is a condensation polymer (a polyamide).
B. Protein: Amino acids join through peptide bonds (-CONH-), and water is lost at each link. This is a condensation polymer.
C. Starch: Glucose units join through glycosidic bonds, and water is lost at each link. This is a condensation polymer.
D. Polyethene: It is made from ethene, CH₂=CH₂, by opening the double bond:
n CH₂=CH₂ → -(CH₂-CH₂)ₙ-
Nothing is eliminated, so this is an addition polymer.
Step 4: Answer Polyethene is the only one that is not a condensation polymer.
Exam tip: Remember that condensation polymers lose water, and the natural ones are built from sugars (starch, cellulose), amino acids (proteins). Synthetic ones to remember are nylon and terylene. Addition polymers come from alkenes: polyethene, PVC, polystyrene, and perspex. If the monomer has C=C, think addition.
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19.
A hydrogen chloride gas reacted with oxygen gas to yield water and chlorine gas. The mole ratio of the hydrogen chloride gas to water is
1. A.1:3
2. B.2:1
3. C.3:1
4. D.4:1
Show answer
Answer: B. 2:1
AI explanation
Analyze the question: We need the balanced equation for HCl reacting with O₂ to give H₂O and Cl₂, then compare the coefficients of HCl and H₂O.
Concept: The mole ratio of reactants and products comes from the coefficients of the balanced equation.
Solution:
Unbalanced equation: HCl + O₂ → H₂O + Cl₂
Balance step by step:
Oxygen: 2 O atoms on the left (in O₂), so we need 2 H₂O on the right.
Hydrogen: 2 H₂O contains 4 H, so we need 4 HCl on the left.
Chlorine: 4 HCl contains 4 Cl, so we need 2 Cl₂ on the right.
Balanced equation: 4HCl + O₂ → 2H₂O + 2Cl₂
Mole ratio of HCl : H₂O = 4 : 2 = 2 : 1
Checking the options: The ratio 4:2 simplifies to 2:1, which is option B. Option D (4:1) is a common trap. It comes from reading the coefficient of HCl (4) and forgetting that H₂O has a coefficient of 2, not 1.
Answer: B.
Tip: Always balance the equation first, then simplify the ratio of the coefficients. Don't just take the first numbers you see. Balance O first, then H, then Cl, since oxygen appears in only one place on each side.
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20.
The cleansing effect of soap is low in acidic water because of
1. A.the formation of unionized fatty acid.
2. B.increase in the pH of the soap molecules.
3. C.anti-biodegradable effect of hydrogen ions
4. D.the hardness of the acidic water.
Show answer
Answer: A. the formation of unionized fatty acid.
AI explanation
Step 1: Analyze the question We need to explain why soap works poorly in acidic water.
Step 2: The concept Soap is the sodium (or potassium) salt of a long-chain fatty acid, e.g. sodium stearate, C₁₇H₃₅COO⁻Na⁺. Its cleansing action depends on the ionized carboxylate head (C₁₇H₃₅COO⁻), which is hydrophilic, and on the long hydrocarbon tail, which is hydrophobic. Together they let soap form micelles that trap grease and dirt.
Step 3: What happens in acid Acidic water contains a high concentration of H⁺ ions. These ions react with the soap anion:
C₁₇H₃₅COO⁻ + H⁺ → C₁₇H₃₅COOH
The product, stearic acid, is a weakly ionized (essentially unionized) fatty acid. It is insoluble in water and forms a scum-like precipitate. This means:
there is less soap anion available to form micelles and emulsify grease, so the cleansing effect falls.
Step 4: Check the options
A. Formation of unionized fatty acid: Correct, as shown above.
B. Increase in the pH of the soap molecules: Acidic water has a low pH, and pH is not a property of soap molecules, so this is wrong.
C. Anti-biodegradable effect of hydrogen ions: This is not a real concept here.
D. Hardness of the acidic water: Hardness is caused by Ca²⁺ and Mg²⁺ ions, not by acidity. That is a different problem, where insoluble calcium or magnesium soap scum forms.
Answer: A
Tip: Remember the pair: acid water → H⁺ converts soap to unionized fatty acid; hard water → Ca²⁺/Mg²⁺ form insoluble scum. Both reduce lathering, but the causes are different, and WASSCE likes to test this difference.
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21.
If the molar mass of X(HCO3) $_{2}$ is 162 g mol $^{-1}$ , determine the relative atomic mass of X. [H = 1.0, C = 12.0, 0 = 16.0]
1. A.40
2. B.48
3. C.61
4. D.101
Show answer
Answer: A. 40
AI explanation
Analyze the question: We're given the molar mass of a compound, X(HCO₃)₂, and asked to find the atomic mass of X by subtracting the mass of everything else.
Concept: Molar mass of a compound = sum of the atomic masses of all atoms in its formula. The subscript 2 outside the bracket means everything inside the bracket is doubled.
Solution:
Mass of one HCO₃ group:
H = 1
C = 12
O₃ = 3 × 16 = 48
Total = 1 + 12 + 48 = 61
Mass of two HCO₃ groups = 2 × 61 = 122
Now: X + 122 = 162 X = 162 − 122 = 40
This matches calcium, Ca = 40, so the compound is calcium hydrogencarbonate, Ca(HCO₃)₂, which is a familiar compound (it causes temporary hardness of water).
Answer: A. 40
Tip: Do not pick option C (61) just because it is the mass of one HCO₃ group. Always multiply the bracketed group by the subscript outside it. Also, when the answer matches a known element's mass (Ca = 40), that is a good confirmation.
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22.
Which of the following statements about atoms of a metal is correct? They
1. A.readily accept electrons
2. B.are soft
3. C.are held together by covalent bond
4. D.are held together by a sea of electron cloud
Show answer
Answer: D. are held together by a sea of electron cloud
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23.
Which of the following bond types is intermolecular?
1. A.Covalent bond
2. B.Hydrogen bond
3. C.Ionic bond
4. D.Metallic bond
Show answer
Answer: B. Hydrogen bond
AI explanation
Step 1: Analyze the question The question asks which bond type acts between molecules (intermolecular), not within them (intramolecular).
Step 2: Key concept
Intramolecular (strong) bonds hold atoms together within a substance: • Covalent bond: sharing of electrons between atoms • Ionic bond: transfer of electrons, giving electrostatic attraction between ions • Metallic bond: attraction between positive metal ions and a "sea" of delocalised electrons
Intermolecular (weak) forces act between separate molecules: • Van der Waals (London dispersion) forces • Dipole-dipole attractions • Hydrogen bonds
Step 3: Check each option
A. Covalent bond: It holds atoms together inside a molecule (e.g., the O-H bonds inside a water molecule). It is intramolecular.
B. Hydrogen bond: It forms between a hydrogen atom bonded to a highly electronegative atom (N, O, or F) in one molecule and a lone pair on N, O, or F in a neighbouring molecule. For example, it links one water molecule to another. It is intermolecular.
C. Ionic bond: It holds ions together in a crystal lattice (e.g., NaCl). It is not intermolecular.
D. Metallic bond: It holds atoms together in a metal. It is not intermolecular.
Step 4: Answer The hydrogen bond is the intermolecular one, which is option B.
Exam tip: Remember "N-O-F" for hydrogen bonding (H attached to N, O, or F). This is why water has a high boiling point compared with H₂S. Also, if a question says "between molecules", think of hydrogen bonds or van der Waals forces. Covalent, ionic, and metallic bonds are the "within" bonds.
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24.
Which of the following statements about liquids is/are true? I. Liquids maintain their volume at constant temperature. II. Liquids have fixed shape III. Liquids do not diffuse IV. Change in pressure affects volume of liquids
1. A.I only
2. B.IV only
3. C.I and IV only
4. D.II and III only
Show answer
Answer: A. I only
AI explanation
Analyze the question: We must decide which statements about liquids are true.
Concept: Liquids have particles that are close together but free to slide past one another. This gives them a definite volume but no definite shape. They are nearly incompressible.
Evaluate each statement:
I. Liquids maintain their volume at constant temperature. True. A liquid has a fixed volume, whatever container it is in, as long as temperature (and pressure) stay constant.
II. Liquids have fixed shape. False. A liquid takes the shape of its container.
III. Liquids do not diffuse. False. Liquids do diffuse. For example, ink or potassium permanganate spreads through water, though more slowly than gases do.
IV. Change in pressure affects the volume of liquids. Mostly false at this level. Liquids are almost incompressible, so pressure changes cause negligible change in volume. Strictly, the change is not exactly zero, but in WASSCE-level chemistry the accepted view is that liquids do not noticeably compress. Gases are the ones whose volume depends strongly on pressure.
So only statement I is true.
Answer: A. I only.
Tip: Remember the pattern for the states of matter:
Solid: fixed shape and fixed volume
Liquid: fixed volume, no fixed shape
Gas: no fixed shape and no fixed volume (compressible)
Option C is the usual trap. Remember that "almost incompressible" means pressure is treated as having no effect on a liquid's volume.
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25.
What is the percentage composition of carbon in Ca(HCO3) $_{2}$ ? [ Ca = 40.0, 0= 16.0, C = 12.0, H= 1.0]
1. A.22.2%
2. B.14.8%
3. C.7.4%
4. D.3.7%
Show answer
Answer: B. 14.8%
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26.
When air in a syringe is compressed such that there is no change in temperature, the
1. A.air liquefies
2. B.pressure increases
3. C.intermolecular space increases
4. D.density decreases
Show answer
Answer: B. pressure increases
AI explanation
Analyze the question: Air in a syringe is compressed (the volume is reduced) while the temperature stays constant. We need to say what happens.
Concept: Boyle's Law At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume:
P₁V₁ = P₂V₂
Step-by-step solution:
Pushing the plunger reduces the volume of the air (V decreases).
Temperature is constant, so Boyle's Law applies: if V decreases, P must increase.
Check each option: • A. Air liquefies: Wrong. Liquefying a gas needs very low temperature and very high pressure. Simple compression at room temperature in a syringe will not do this. • B. Pressure increases: Correct. Smaller volume means the molecules hit the walls more often, so the pressure rises. • C. Intermolecular space increases: Wrong. Compression pushes the molecules closer together, so the space between them decreases. • D. Density decreases: Wrong. The same mass is squeezed into a smaller volume, so density (mass/volume) increases.
Answer: B. Pressure increases
Exam tip: For any "compression at constant temperature" question, remember: Volume ↓, Pressure ↑, Density ↑, Spacing ↓. Boyle's Law is about pressure and volume moving in opposite directions.
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27.
If 50 cm $^{3}$ of a saturated solution of KNO $_{3}$ at 40 °C contained 5.05 g of the salt, its solubility at the same temperature would be [KNO $_{3}$ = 101]
1. A.1.0 mol dm $^{-3}$
2. B.1.5 mol dm $^{-3}$
3. C.2.0 mol dm $^{-3}$
4. D.5.0 mol dm $^{-3}$
Show answer
Answer: A. 1.0 mol dm $^{-3}$
AI explanation
Analyze the question: Solubility here is expressed in mol dm⁻³, so we need the concentration of KNO₃ in the saturated solution.
Concept:
Moles = mass ÷ molar mass
Concentration (mol dm⁻³) = moles ÷ volume in dm³
Solution:
Moles of KNO₃ = 5.05 g ÷ 101 g/mol = 0.05 mol
Volume = 50 cm³ = 50 ÷ 1000 = 0.05 dm³
Concentration = 0.05 mol ÷ 0.05 dm³ = 1.0 mol dm⁻³
Tip: Always convert cm³ to dm³ by dividing by 1000 before calculating concentration. A quick shortcut is that concentration = (mass ÷ molar mass) × (1000 ÷ volume in cm³). Here that is 0.05 × 20 = 1.0. Also notice the numbers were chosen to be neat (5.05 is exactly 101 ÷ 20), which is a good sign you're on the right track.
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28.
Which of the following scientists formulated the law of conservation of mass?
1. A.A. Lavoisier
2. B.J. Dalton
3. C.R. Boyle
4. D.J. Proust
Show answer
Answer: A. A. Lavoisier
AI explanation
Analyze the question: The question asks which scientist is credited with the law of conservation of mass. This law says that matter is neither created nor destroyed in a chemical reaction, so the total mass of the reactants equals the total mass of the products.
Explain the concept: Several scientists are linked to the early laws of chemistry, and the options mix them up on purpose:
Antoine Lavoisier (1789): Law of Conservation of Mass. He carefully weighed substances before and after reactions in sealed vessels, and the mass stayed the same.
John Dalton: Atomic Theory and the Law of Multiple Proportions. His theory explained why mass is conserved, because atoms are only rearranged and never created or destroyed. But he did not formulate the law itself.
Robert Boyle: Boyle's Law (pressure and volume of a gas at constant temperature, P ∝ 1/V).
Joseph Proust: Law of Definite (Constant) Proportions. A pure compound always contains the same elements in the same proportion by mass.
Solution, step by step:
The law in question is about mass being conserved in a reaction.
Dalton explained it with atomic theory but did not formulate it. Boyle is the gas law man. Proust is definite proportions.
That leaves Lavoisier, the "father of modern chemistry," who established the law through quantitative experiments.
Answer: A. A. Lavoisier
Exam tip: Match each scientist to one law:
Lavoisier → Law of conservation of mass
Proust → Proportions (definite)
Dalton → Different (multiple) proportions and atoms
Boyle → Balloon (gas pressure and volume)
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29.
Consider the following reaction equation N $_{2}$ (g) + 3H $_{2(g)}$ ⇌ 2NH3(g) ∆H = -92kJ Increasing the temperature of the reaction would
1. A.shift the equilibrium to the right
2. B.increase the yield of ammonia
3. C.decrease the amount of hydrogen
4. D.decrease the yield of ammonia
Show answer
Answer: D. decrease the yield of ammonia
AI explanation
Analyze the question: The Haber process reaction is exothermic (ΔH = -92 kJ). We are asked what happens to the equilibrium when the temperature is increased.
Concept: Le Chatelier's Principle When a system at equilibrium is disturbed, it shifts in the direction that opposes the change.
Increasing temperature means adding heat. The system responds by absorbing that heat, so it favors the endothermic direction.
Decreasing temperature means removing heat. The system favors the exothermic direction.
Solution:
Forward reaction: N₂ + 3H₂ → 2NH₃, ΔH = -92 kJ, which is exothermic (releases heat).
Backward reaction: 2NH₃ → N₂ + 3H₂, which is endothermic (absorbs heat).
Raising the temperature makes the system absorb heat, so the equilibrium shifts to the left (backward, endothermic direction).
Results of shifting left: • NH₃ is decomposed, so the yield of ammonia decreases. • N₂ and H₂ increase, so the amount of hydrogen increases.
Checking the options:
A. Shift to the right: wrong, it shifts left.
B. Increase the yield of ammonia: wrong.
C. Decrease the amount of hydrogen: wrong, hydrogen increases.
D. Decrease the yield of ammonia: correct.
Tip: For exothermic reactions, think of heat as a product: "Heat + reactants ⇌ products + heat". Adding heat is like adding a product, so it pushes the reaction backward. In industry, the Haber process uses a compromise temperature of about 450°C, which is high enough for a reasonable rate but not so high that it ruins the yield.
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30.
The first definition of an element was made by
1. A.J. Dalton.
2. B.A. Lavoisier.
3. C.R. Boyle.
4. D.J. J. Thompson.
Show answer
Answer: C. R. Boyle.
AI explanation
Analyze the question: It asks who first defined an element. This is a history-of-chemistry question.
Concept: The idea of an "element" developed over time:
Ancient Greeks believed in four "elements": earth, air, fire and water.
Robert Boyle (1661), in his book The Sceptical Chymist, rejected the Greek idea. He defined an element as a substance that cannot be broken down into simpler substances by chemical means. This is regarded as the first modern definition of an element.
Antoine Lavoisier (1789) later compiled a list of elements, treating them as substances that could not be decomposed further. His work built on Boyle's idea.
John Dalton (1803-1808) proposed the atomic theory, in which elements are made of identical atoms.
J. J. Thomson (1897) discovered the electron.
Solution:
Dalton: gave the atomic theory, not the first definition. ✗
Lavoisier: gave the first list of elements and refined the definition, but came after Boyle. ✗
Boyle: gave the first definition of an element. ✓
Thomson: discovered the electron. ✗
The correct answer is **C. R.
Tip: Link each scientist to one key word:
Boyle = first definition of an element (and gas pressure-volume law)
Lavoisier = oxygen and conservation of mass
Dalton = atomic theory
Thomson = electron
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31.
Electropositivity of elements across the periodic table normally
1. A.remains constant down the group
2. B.increases across the period
3. C.decreases across the period
4. D.decreases down the group
Show answer
Answer: C. decreases across the period
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32.
If 5.0 g of marble reacts with 25.0 cm3 hydrochloric acid, which of the following combinations has the fastest reaction rate?
1. A.Marble chips and 2.0 mol dm $^{-3}$ HCl $_{(aq)}$
2. B.Marble chips and 2.5 mol dm $^{-3}$ HCl $_{(aq)}$
3. C.Powdered marble and 2.5 mol dm $^{-3}$ HCl $_{(aq)}$
4. D.Powdered marble and 2.0 mol dm $^{-3}$ HCl $_{(aq)}$
Show answer
Answer: C. Powdered marble and 2.5 mol dm $^{-3}$ HCl $_{(aq)}$
AI explanation
Analyze the question: We need the combination that gives the fastest rate for CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. The mass of marble (5.0 g) and the volume of acid (25.0 cm³) are the same in every option. Only the form of the marble (chips or powder) and the acid concentration change.
Concept: Two factors from collision theory matter here.
Surface area: Powder has a much larger exposed surface than chips, so more particles can collide with the acid at once. This increases the rate.
Concentration: A higher concentration of HCl means more acid particles per unit volume. Collisions are then more frequent, so the rate increases.
Solution:
Powdered marble beats marble chips, which rules out A and B.
Between C and D, both use powder, so concentration decides. 2.5 mol dm⁻³ is higher than 2.0 mol dm⁻³.
Option C therefore combines the larger surface area with the higher concentration.
Answer: C. Powdered marble and 2.5 mol dm⁻³ HCl
Exam tip: For "fastest rate" questions, pick the option that makes every factor favourable: powder (not lumps), higher concentration, and higher temperature if it is listed. Remember "Powder + Power (high concentration) = Faster."
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33.
Which of the following statements about an equilibrium system is correct?
1. A.Forward and backward reactions occur at the same rate
2. B.The concentrations of reactants must equal that of the products
3. C.The concentrations of, reactants and products can be changed by adding a catalyst,
4. D.The concentrations of reactants and products are not affected by a change in temperature
Show answer
Answer: A. Forward and backward reactions occur at the same rate
AI explanation
Step 1: Analyze the question We need the one true statement about a system in chemical equilibrium.
Step 2: The concept A chemical equilibrium is dynamic. The reactions don't stop. The forward reaction and the backward reaction continue, but at equal rates. As a result, the concentrations of reactants and products stay constant (not necessarily equal).
Step 3: Check each option
A. Forward and backward reactions occur at the same rate. ✅ True. This is the defining condition of equilibrium.
B. Concentrations of reactants must equal those of products. ❌ False. They only stay constant. Depending on the equilibrium constant (K), one side may be much larger than the other.
C. Concentrations can be changed by adding a catalyst. ❌ False. A catalyst speeds up both the forward and backward reactions equally. It helps equilibrium be reached faster but does not change the equilibrium concentrations or the position of equilibrium.
D. Concentrations are not affected by a change in temperature. ❌ False. Changing temperature shifts the equilibrium position (Le Chatelier's principle) and changes the value of K, so concentrations change.
Step 4: Answer The correct option is A.
Exam tip: Remember "Equal Rates, Constant concentrations" at equilibrium. Rates are equal, concentrations are constant (not equal). A catalyst affects only the speed, never the position. Only changes in temperature, pressure (for gases), or concentration shift the position.
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34.
Which of the following gases is alkaline?
1. A.NO $_{2}$
2. B.CO $_{2}$
3. C.NH $_{3}$
4. D.N $_{2}$ O
Show answer
Answer: C. NH $_{3}$
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35.
Which of the following processes occur during fractional distillation of petroleum?
1. A.Condensation and diffusion
2. B.Diffusion and evaporation
3. C.Diffusion and sublimation
4. D.Evaporation and condensation
Show answer
Answer: D. Evaporation and condensation
AI explanation
Step 1: Analyze the question We are asked which physical processes take place during fractional distillation of petroleum (crude oil).
Step 2: The concept Fractional distillation separates a mixture of liquids according to their different boiling points. It involves two main changes of state:
Evaporation (vaporisation): Crude oil is heated in a furnace until most of it turns into vapour.
Condensation: The vapour rises up the fractionating column, where the temperature falls with height. Each fraction condenses back into a liquid at the level where the temperature is just below its boiling point.
Step 3: Step-by-step solution
Crude oil is heated to about 400 °C, so it evaporates.
The hot vapours enter the fractionating column near the bottom.
As they rise, they cool. Fractions with high boiling points (bitumen, lubricating oil, fuel oil) condense low down. Those with low boiling points (petrol, refinery gases) condense higher up or remain as gas.
The liquids are collected at different levels.
Step 4: Eliminate the wrong options
Sublimation (solid → gas directly) does not occur here, so C is out.
Diffusion is not the main process that separates the fractions, so A, B and C are out.
Only evaporation and condensation describe the process.
Answer: The correct option is D. Evaporation and condensation.
Exam tip: Any distillation, whether simple or fractional, is "evaporate, then condense". Remember it as "Boil up, cool down". If an option includes sublimation, it is usually wrong, since sublimation is for substances like iodine, ammonium chloride and dry ice.
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36.
The electron configuration of carbon atom in its excited state is
1. A.1s $^{2}$ 2s $^{2}$ 2px $^{2}$ 2py $^{0}$
2. B.1s $^{2}$ 2s $^{2}$ 2px $^{1}$ 2py $^{1}$
3. C.1s $^{2}$ 2s $^{1}$ 2px $^{1}$ 2py $^{1}$ 2pz $^{1}$
4. D.1s $^{1}$ 2s $^{2}$ 2px $^{1}$ 2py $^{1}$
Show answer
Answer: C. 1s $^{2}$ 2s $^{1}$ 2px $^{1}$ 2py $^{1}$ 2pz $^{1}$
AI explanation
Analyze the question: You need the excited-state electron configuration of carbon (atomic number 6, so 6 electrons).
Concept:
Ground state: electrons fill the lowest energy orbitals first (Aufbau principle), and Hund's rule applies to the p orbitals. Carbon ground state: 1s² 2s² 2px¹ 2py¹ (2pz⁰)
Excited state: an electron is promoted to a higher-energy orbital, but the total number of electrons stays at 6. For carbon, one 2s electron moves into the empty 2pz orbital. This gives four unpaired electrons, which is why carbon forms four bonds (as in CH₄).
Check each option:
A. 1s² 2s² 2px² 2py⁰ has 6 electrons, but it is just a different arrangement with paired electrons in 2px. It breaks Hund's rule, so it is a higher-energy, unusual state and not the standard excited state for bonding.
B. 1s² 2s² 2px¹ 2py¹ is the ground state of carbon, with 6 electrons.
C. 1s² 2s¹ 2px¹ 2py¹ 2pz¹ has 2 + 1 + 3 = 6 electrons. One 2s electron has been promoted to 2pz, giving four unpaired electrons. This is the classic excited state.
D. 1s¹ 2s² 2px¹ 2py¹ has 1 + 2 + 2 = 5 electrons, so it is wrong for carbon.
Tip: For carbon's excited state, remember "2s¹ 2p³". Promoting one electron from 2s to 2p gives four unpaired electrons, which explains carbon's tetravalency and sp³ hybridisation. Always count the electrons first, since this quickly eliminates options like D.
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37.
Which of the following sketches is a graphical representation of Boyle's law?
1. A.A
2. B.B
3. C.C
4. D.D
Show answer
Answer: C. C
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38.
Which of the following compounds would react rapidly with bromine?
1. A.Benzene
2. B.Hexane
3. C.Hexene
4. D.Hexanol
Show answer
Answer: C. Hexene
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39.
The IUPAC name of the compound CH3CH(CH3)CHCH2 is
1. A.2 - methyl but - 1 - ene
2. B.2 - methyl but - 2 - ene
3. C.3 - methyl but - 1 - ene
4. D.3 - methyl but - 2 - ene
Show answer
Answer: C. 3 - methyl but - 1 - ene
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40.
The products of the electrolysis of dilute sodium chloride using carbon electrodes are Anode Cathode
1. A.chlorine and sodium.
2. B.oxygen and hydrogen.
3. C.chlorine and hydrogen.
4. D.hydrogen and oxygen.
Show answer
Answer: B. oxygen and hydrogen.
AI explanation
Analyze the question: We need the products at the anode and the cathode when dilute sodium chloride solution is electrolysed using inert carbon electrodes. The order matters: anode first, then cathode.
Concept: Selective (preferential) discharge
The ions in the solution are:
From NaCl: Na⁺ and Cl⁻
From water: H⁺ and OH⁻
At the cathode (negative electrode), cations compete. H⁺ is lower than Na⁺ in the electrochemical series, so H⁺ is discharged: 2H⁺ + 2e⁻ → H₂
At the anode (positive electrode), anions compete. Three factors matter:
Position in the electrochemical series: OH⁻ is preferred over Cl⁻.
Concentration: in a concentrated NaCl solution, the high concentration of Cl⁻ overrides this and chlorine is discharged. In a dilute solution, there is little Cl⁻, so OH⁻ is discharged.
Electrode: carbon is inert, so it does not react.
Anode reaction in dilute solution: 4OH⁻ → 2H₂O + O₂ + 4e⁻
Products:
Anode: oxygen
Cathode: hydrogen
Checking the options:
A. Chlorine and sodium: sodium is never discharged from aqueous solution. Wrong.
B. Oxygen (anode) and hydrogen (cathode): matches. Correct.
C. Chlorine and hydrogen: this is for concentrated brine. Wrong.
D. Hydrogen and oxygen: the electrodes are reversed. Wrong.
Answer: B
Exam tip: Always check the word "dilute" or "concentrated" in electrolysis questions.
Concentrated NaCl gives Cl₂ at the anode and H₂ at the cathode.
Dilute NaCl gives O₂ at the anode and H₂ at the cathode.
Also remember "Anode = Anions (negative ions) go there, and Oxidation happens there" (AN OX), and read the order "Anode, Cathode" carefully in the options.
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41.
The by-product of fermentation of sugar is
1. A.carbon(IV) oxide.
2. B.ethanoic acid.
3. C.propanol
4. D.propan-1, 2, 3- triol.
Show answer
Answer: A. carbon(IV) oxide.
AI explanation
Analyze the question: Fermentation of sugar is a reaction in which enzymes break sugar down. The question asks for the by-product, not the main product.
Explain the concept: Fermentation is the breakdown of sugars (such as glucose) by enzymes in yeast. The enzyme zymase converts glucose into ethanol (the main product) and carbon(IV) oxide (the by-product).
Step-by-step solution:
Sugar (sucrose) is first hydrolysed by the enzyme invertase into glucose and fructose: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆
Zymase in yeast then converts glucose to ethanol and carbon(IV) oxide: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
Ethanol (C₂H₅OH) is the main product, and CO₂ is the by-product, which is the gas that bubbles out. This is why dough rises in baking and why palm wine froths.
Checking the options:
A. Carbon(IV) oxide is correct, as it is released alongside ethanol.
B. Ethanoic acid forms only if ethanol is further oxidised, as when wine turns sour. It is not the by-product of the fermentation itself.
C. Propanol is not formed.
D. Propan-1,2,3-triol (glycerol) is a by-product of saponification or fat hydrolysis, not sugar fermentation.
**Answer: A. carbon(IV) oxide.
Exam tip: Remember "Fermentation = Ethanol + CO₂". Ethanol is the main product, CO₂ is the by-product. You can confirm the CO₂ in the lab because it turns lime water milky.
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42.
A white precipitate was formed when BaCl $_{2(aq)}$ was added to an aqueous solution of a salt X. The precipitate dissolved in dilute HCl with rapid effervescence. Salt X is likely to contain
1. A.SO $_{4}$ $^{2-}$ ions
2. B.SO $_{3}$ $^{2-}$ ions
3. C.NO $_{3}$ - ions
4. D.CO $_{3}$ $^{2-}$ - ions
Show answer
Answer: D. CO $_{3}$ $^{2-}$ - ions
AI explanation
Step 1: Analyze the question Adding BaCl₂ to a solution of salt X gives a white precipitate. That precipitate dissolves in dilute HCl with rapid effervescence (fizzing). We need to find the anion in X.
Step 2: Concept Ba²⁺ forms white precipitates with several anions:
SO₄²⁻ gives BaSO₄, which is insoluble in dilute HCl. This is the standard test for sulphate.
SO₃²⁻ gives BaSO₃, which dissolves in dilute HCl, releasing SO₂ gas. SO₂ has a choking smell and turns acidified K₂Cr₂O₇ from orange to green.
CO₃²⁻ gives BaCO₃, which dissolves in dilute HCl, releasing CO₂ gas. CO₂ is odourless and turns limewater milky.
NO₃⁻ gives no precipitate, because all nitrates are soluble.
Step 3: Work through the options
A. SO₄²⁻: BaSO₄ does not dissolve in HCl, so this is ruled out.
B. SO₃²⁻: BaSO₃ does dissolve in HCl, but it gives SO₂, which is a gas with a pungent smell. Sulphite is a plausible candidate, and the question gives no test to tell the gas apart.
C. NO₃⁻: There is no precipitate, so this is ruled out.
D. CO₃²⁻: BaCO₃ is a white precipitate. It dissolves in dilute HCl with rapid effervescence of CO₂:
BaCO₃(s) + 2HCl(aq) → BaCl₂(aq) + H₂O(l) + CO₂(g)
Step 4: Choose between B and D Both carbonate and sulphite fit the observations. The question says "rapid effervescence" and mentions no pungent smell. WASSCE treats this as the classic carbonate test: a white precipitate that dissolves in dilute acid with effervescence.
Answer: D, CO₃²⁻ ions
Exam tip: Remember "BaSO₄ stays, BaCO₃ goes." A white precipitate that persists in acid means sulphate. One that dissolves with fizzing means carbonate (or sulphite, if the gas smells sharp). If the question mentions a choking smell, think sulphite.
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43.
Increasing the temperature generally
1. A.decreases the solubility of a solid in a liquid but increases the solubility of a gas in a liquid.
2. B.increases the solubility of a solid in a liquid but decreases the solubility of a gas in a liquid.
3. C.increases the solubility of both a solid and a gas in a liquid
4. D.decreases the solubility of both a solid and a gas in a liquid
Show answer
Answer: B. increases the solubility of a solid in a liquid but decreases the solubility of a gas in a liquid.
AI explanation
Step 1: Analyze the question The question asks how a rise in temperature generally affects solubility, for both solids and gases dissolving in liquids.
Step 2: The concepts
Solids in liquids: Dissolving most solids is endothermic, meaning it absorbs heat. By Le Chatelier's principle, adding heat (raising the temperature) shifts the equilibrium toward more dissolving. So solubility increases with temperature. Examples are sugar, KNO₃ and NaNO₃ in water. (There are exceptions, such as calcium hydroxide, but the question says "generally".)
Gases in liquids: Dissolving a gas is usually exothermic, meaning it releases heat. Raising the temperature shifts the equilibrium the other way, so gas escapes from the solution. Solubility decreases with temperature. Also, gas molecules gain kinetic energy at higher temperatures and escape more easily. This is why a warm soft drink goes flat faster than a cold one, and why fish struggle in very warm water, which holds less dissolved oxygen.
Step 3: Check the options
A: Reversed. The solid and gas effects are swapped, so it is wrong.
B: Solids increase, gases decrease. This is correct.
C: Wrong, because gas solubility does not increase.
D: Wrong, because solid solubility does not decrease in general.
Step 4: Answer
Tip: Remember "Solids like it hot, gases like it cold." Solids dissolve better when heated, and gases stay dissolved better when cooled. Think of a cold bottle of Coke keeping its fizz.
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44.
A colourless gas with a pungent smell is evolved when dilute hydrochloric acid is added to a sample of a salt. The gas evolved could turn
1. A.acidified K $_{2}$ Cr $_{2}$ O $_{7}$ solution colourless
2. B.acidified KMnO $_{4}$ solution colourless
3. C.Fe(NO $_{3}$ ) $_{2}$ solution green
4. D.Pb(NO $_{3}$ ) $_{2}$ paper black
Show answer
Answer: B. acidified KMnO $_{4}$ solution colourless
AI explanation
Analyze the question: A salt reacts with dilute HCl to give a colourless gas with a pungent smell. The most likely gas is SO₂, which comes from a sulphite (SO₃²⁻ + 2H⁺ → SO₂ + H₂O). The other candidates are H₂S (rotten-egg smell, so not "pungent" in the usual textbook sense) and CO₂ (odourless).
Concept: SO₂ is a reducing agent. Its reactions with oxidising agents are:
Acidified K₂Cr₂O₇: orange → green (Cr³⁺). It does not go colourless.
Acidified KMnO₄: purple → colourless (Mn²⁺).
Pb(NO₃)₂ paper turns black only with H₂S (forming PbS). SO₂ does not do this.
Fe(NO₃)₂ is already pale green, and SO₂ would not turn it green.
Check each option:
A. Wrong. Dichromate turns green with SO₂, not colourless.
B. Correct. SO₂ decolourises acidified permanganate: 5SO₂ + 2MnO₄⁻ + 2H₂O → 5SO₄²⁻ + 2Mn²⁺ + 4H⁺
C. Wrong. Fe²⁺ is already green, and SO₂ would not produce this change.
D. Wrong. This is the test for H₂S, which has a rotten-egg smell.
Tip: Remember the SO₂ tests as "dichromate goes green, permanganate goes colourless." Remember H₂S as "lead(II) ethanoate or nitrate paper goes black."
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45.
What is the partial pressure of oxygen at s.t.p. in a gaseous mixture containing 100 cm3 of oxygen and 900 cm3 of nitrogen gas?
1. A.1.0 atm
2. B.0.1 atm
3. C.9.0 atm
4. D.0.9 atm
Show answer
Answer: B. 0.1 atm
AI explanation
Analyze the question: You have a gas mixture at s.t.p. (total pressure = 1 atm, or 760 mmHg). You need the partial pressure of oxygen.
Concept: Dalton's Law of Partial Pressures says that each gas in a mixture exerts a pressure equal to its mole fraction multiplied by the total pressure.
Partial pressure of a gas = (volume or mole fraction of that gas) × total pressure
At constant temperature and pressure, volume is proportional to moles (Avogadro's law), so the volume fraction can be used in place of the mole fraction.
Solution:
Total volume = 100 + 900 = 1000 cm³
Mole fraction of O₂ = 100 / 1000 = 0.1
Total pressure at s.t.p. = 1 atm
Partial pressure of O₂ = 0.1 × 1 atm = 0.1 atm
As a check, the nitrogen's partial pressure is 0.9 × 1 = 0.9 atm. Then 0.1 + 0.9 = 1.0 atm, which equals the total pressure.
Answer: The correct option is B. 0.1 atm.
Tip: At s.t.p. the total pressure is always 1 atm, so the partial pressure is simply the fraction of the gas in the mixture. Oxygen is 100 out of 1000, which is 1/10, or 0.1 atm. Watch for option D (0.9 atm), which is the nitrogen's partial pressure and a common trap.
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46.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.
47.
Determine the quantity of electricity used when a current of 0.20 amperes is passed through an electrolytic cell for 60 minutes.
1. A.12C
2. B.120C
3. C.360C
4. D.720C
Show answer
Answer: D. 720C
AI explanation
Analyze the question: You are asked for the quantity of electricity (charge, Q) that flows when a current of 0.20 A passes for 60 minutes.
Concept: Quantity of electricity is current multiplied by time:
Q = I × t
Q is measured in coulombs (C)
I is measured in amperes (A)
t must be in seconds (this is where many students lose marks!)
Solution:
Convert time to seconds: t = 60 minutes × 60 s/min = 3600 s
Substitute into the formula: Q = 0.20 × 3600 Q = 720 C
Common mistakes:
Using 60 directly gives 0.20 × 60 = 12 C (option A, a trap).
Using 30 minutes by mistake gives 360 C (option C).
Tip: In every electrolysis question, convert minutes or hours to seconds first. Remember "Q = It, time in seconds". Also, 1 Faraday = 96,500 C, so you can later use Q to find the mass of a substance deposited with m = (Q × M)/(n × 96500).
Answer: D. 720 C
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48.
Ionization energy increases across the period in the periodic table because
1. A.atomic number increases
2. B.effective nuclear charge increases
3. C.mass number decreases
4. D.screening effect decreases
Show answer
Answer: B. effective nuclear charge increases
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49.
Which of the following statements about the standard hydrogen electrode is not correct?
1. A.The hydrogen gas is at a pressure of 1 atmosphere
2. B.A solution containing 1 mol $dm^{-3}$ of H+ (aq) ions is used
3. C.A platinum electrode is used
4. D.The temperature is kept at 20 °C
Show answer
Answer: D. The temperature is kept at 20 °C
AI explanation
Analyze the question: We need the statement about the standard hydrogen electrode (SHE) that is not correct.
Concept: The SHE is the reference electrode used to measure standard electrode potentials. By definition its potential is exactly 0.00 V. The standard conditions are:
Hydrogen gas at a pressure of 1 atm (101.3 kPa)
H⁺(aq) solution at a concentration of 1 mol dm⁻³
An inert platinum electrode, coated with platinum black, which adsorbs the hydrogen gas and conducts electrons
A temperature of 25 °C (298 K)
Check each option:
A. Hydrogen at 1 atm is correct.
B. 1 mol dm⁻³ H⁺(aq) is correct.
C. A platinum electrode is correct. It is inert and provides a surface for the reaction 2H⁺ + 2e⁻ ⇌ H₂.
D. A temperature of 20 °C is not correct. The standard temperature is 25 °C (298 K).
Tip: Remember the SHE as "1, 1, Pt, 25": 1 atm, 1 mol dm⁻³, platinum, 25 °C. If an option gives a different temperature, such as 20 °C, it is probably the false statement.
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50.
The maximum number of covalent bonds formed by nitrogen is
1. A.1
2. B.2
3. C.3
4. D.4
Show answer
Answer: D. 4
AI explanation
Analyze the question: It asks for the maximum number of covalent bonds a nitrogen atom can form. The word "maximum" matters here.
Concept: Nitrogen has atomic number 7, so its electronic configuration is 1s² 2s² 2p³. It has 5 valence electrons and needs 3 more to complete its octet.
Normally, it forms 3 covalent bonds by sharing its 3 unpaired electrons. Examples are NH₃ (ammonia) and N₂ (triple bond).
Nitrogen also has a lone pair. It can donate this pair to an electron-deficient species and form a dative (coordinate) covalent bond. This gives a fourth bond.
Example: NH₃ + H⁺ → NH₄⁺ (ammonium ion). Here nitrogen has 3 normal covalent bonds plus 1 dative covalent bond, so 4 bonds in total.
Why not more than 4? Nitrogen is in period 2, so it has no d-orbitals available. It can only use the 2s and 2p orbitals, which hold at most 8 electrons (4 pairs). That limits it to 4 bonds.
Solution:
Normal covalent bonds: 3
Dative bond from the lone pair: +1
Maximum = 4
Answer: The maximum number of covalent bonds nitrogen can form is 4, which is option D.
Exam tip: If a question says "maximum" for nitrogen or oxygen, think of the dative bond. Nitrogen gives NH₄⁺ (4 bonds), and oxygen gives H₃O⁺ (3 bonds). If the question says "usual" or "normally", the answer for nitrogen is 3.
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2022 WAEC (WASSCE) Chemistry Past Questions and Answers

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