WAEC Chemistry 2020 past questions

Checked by the Skolra question team · Last updated 15 October 2026

Ten questions from the 2020 WAEC Chemistry paper, with answers and explanations. We hold 55 questions from this paper, and the Skolra app has all of them.

WAEC Chemistry 2020 questions and answers

1. 1.

   A balanced chemical equation is based on thelaw of?

   1. A
      Constant composition
   2. B
      Multiple proportions
   3. C
      Conservation of mass
      Correct
   4. D
      Definite proportions
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   Answer: C

   A balanced chemical equation ensures that the number of atoms of each element is the same on both the reactant and product sides of the equation. This reflects the law of conservation of mass (Lavoisier's law), which states that matter is neither created nor destroyed in a chemical reaction.

   The total mass of the reactants equals the total mass of the products. Balancing an equation is the mathematical expression of this principle.

2. 2.

   The oxidation state of chlorine in NaClO$_3$ is?

   1. A
      −1
   2. B
      +7
   3. C
      +5
      Correct
   4. D
      +3
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   Answer: C

   To find the oxidation state of chlorine in NaClO₃, use the rule that the sum of all oxidation states in a neutral compound equals zero.

   Assigned oxidation states:

   • Na = +1
   • O = −2 each; three oxygen atoms contribute 3 × (−2) = −6
   • Let Cl = x

   Setting up the equation:
   (+1) + x + (−6) = 0
   x − 5 = 0
   x = +5

   The oxidation state of chlorine in NaClO₃ (sodium trioxochlorate(V)) is +5.

3. 3.

   If 20 cm$^3$ of sodium hydroxide was neutralized by 20 cm$^3$ of 0.01 mol dm$^{-3}$ tetraoxosulphate(VI) acid, what is the concentration of the solution?

   1. A
      0.04 mol dm⁻³
   2. B
      0.10 mol dm⁻³
   3. C
      0.02 mol dm⁻³
      Correct
   4. D
      0.01 mol dm⁻³
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   Answer: C

   The balanced equation is: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

   Mole ratio H₂SO₄ : NaOH = 1 : 2

   Moles of H₂SO₄ = 0.01 mol dm⁻³ × (20/1000) dm³ = 2.0 × 10⁻⁴ mol

   Moles of NaOH = 2 × 2.0 × 10⁻⁴ = 4.0 × 10⁻⁴ mol

   Concentration of NaOH = 4.0 × 10⁻⁴ mol ÷ (20/1000) dm³

   = 4.0 × 10⁻⁴ ÷ 0.020 = 0.02 mol dm⁻³

4. 4.

   Which of the following compounds would release hydrogen when reacted with sodium metal?

   I.CH$_3$COOH

   II.CH$_3$CH$_2$OH

   III.CH$_3$COOCH$_3$

   1. A
      I, II and III
   2. B
      I only
   3. C
      I and II only
      Correct
   4. D
      III only
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   Answer: C

   Sodium reacts with compounds that have an active O–H hydrogen to produce hydrogen gas:

   • CH₃COOH (ethanoic acid): has an acidic –COOH group. 2CH₃COOH + 2Na → 2CH₃COONa + H₂↑
   • CH₃CH₂OH (ethanol): has a hydroxyl –OH group. 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂↑
   • CH₃COOCH₃ (methyl ethanoate, an ester): has no free –OH or –COOH group; its oxygen atoms are fully bonded within the ester linkage and do not react with sodium to release hydrogen.

   Therefore compounds I and II release hydrogen with sodium.

5. 5.

   An example of a biodegradable pollutant is?

   1. A
      Mercury compounds
   2. B
      DDT pesticide
   3. C
      Polythene bags
   4. D
      Sewage
      Correct
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   Answer: D

   A biodegradable pollutant is one that can be broken down by microorganisms (bacteria, fungi) into harmless or simpler substances within a reasonable time.

   Sewage consists mainly of organic waste that microorganisms decompose — it is therefore biodegradable, though it can cause oxygen depletion and eutrophication if discharged untreated.

   Polythene bags, DDT, and mercury compounds are non-biodegradable: they persist in the environment for very long periods and accumulate in food chains (bioaccumulation/biomagnification).

6. 6.

   What volume of oxygen at s.t.p is required to burn completely 7.5 dm$^3$of methane according to the following equation?

   CH$_4$$_g$ + 20$_2$$_g$CO$_2$$_g$ +H$_2$O$_g$

   1. A
      22.4 dm³
   2. B
      7.5 dm³
   3. C
      15.0 dm³
      Correct
   4. D
      30.0 dm³
   Show answerHide answer

   Answer: C

   The balanced equation is: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

   By Gay-Lussac's law, at constant temperature and pressure, volumes of reacting gases are in the same ratio as their stoichiometric coefficients.

   Mole (volume) ratio: CH₄ : O₂ = 1 : 2

   Volume of O₂ = 2 × 7.5 dm³ = 15.0 dm³

7. 7.

   The boiling point of pentane is higher than that of propane because?

   1. A
      Pentane molecules have a higher polarity than propane molecules
   2. B
      Pentane has a larger molecular mass and stronger van der Waals forces
      Correct
   3. C
      Pentane has stronger ionic attractions between its molecules
   4. D
      Pentane molecules are more closely packed in the liquid state
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   Answer: B

   Both pentane (C₅H₁₂) and propane (C₃H₈) are non-polar alkanes with no dipole–dipole interactions or hydrogen bonding. Their boiling points are determined entirely by London (van der Waals) dispersion forces.

   Pentane has a larger molecular mass (72 g mol⁻¹) and more electrons than propane (44 g mol⁻¹). More electrons mean a larger, more polarisable electron cloud, producing stronger temporary dipoles and therefore stronger dispersion forces between molecules.

   Greater intermolecular forces require more energy to overcome, resulting in a higher boiling point for pentane (36°C) compared to propane (−42°C).

8. 8.

   How many electrons does ${31}_{15}$P$^3-$ contain?

   1. A
      21
   2. B
      12
   3. C
      18
      Correct
   4. D
      15
   Show answerHide answer

   Answer: C

   Phosphorus (P) has atomic number 15, so a neutral phosphorus atom has 15 electrons.

   The ion ³¹₁₅P³⁻ carries a charge of 3−, meaning it has gained 3 electrons relative to the neutral atom.

   Total electrons = 15 + 3 = 18

   This gives ³¹₁₅P³⁻ the same electron configuration as argon (2, 8, 8), making it isoelectronic with the noble gas argon.

9. 9.

   The metallic bond in magnesium is stronger than that in calcium because magnesium has a

   1. A
      Greater number of valence electrons
   2. B
      Lower melting point
   3. C
      Smaller atomic radius and higher charge density
      Correct
   4. D
      Larger atomic radius and lower charge density
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   Answer: C

   The strength of a metallic bond depends on the attraction between the delocalised electrons (electron sea) and the positively charged metal ions (cations).

   Both Mg (group 2, period 3) and Ca (group 2, period 4) contribute two valence electrons to the delocalised sea. The key difference is atomic size:

   • Mg²⁺ has a smaller ionic radius (72 pm) than Ca²⁺ (100 pm).
   • A smaller ion with the same charge has a higher charge density, so it attracts the electron sea more strongly.
10. 10.

    What is the mass of silver deposited when 24,125 C of electricity is passed through a solution or silversalt.

    [Ag = 108, IF = 96,500 C ]

    1. A
       27.0 g
       Correct
    2. B
       54.0 g
    3. C
       13.5 g
    4. D
       108.0 g
    Show answerHide answer

    Answer: A

    At the cathode: Ag⁺ + e⁻ → Ag

    One mole of silver requires one Faraday (96,500 C) of charge.

    Step 1 — moles of electrons:
    n(e⁻) = charge / F = 24,125 / 96,500 = 0.25 mol

    Step 2 — moles of silver:
    From the equation, 1 mol e⁻ deposits 1 mol Ag
    n(Ag) = 0.25 mol

    Step 3 — mass of silver:
    mass = n × M = 0.25 × 108 = 27.0 g

The questions come from past JAMB, WAEC and NECO papers. We use software to help write the answer options and explanations, and every answer is checked twice before it goes on the site. How we check

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