WAEC Chemistry 2021 past questions

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

Ten questions from the 2021 WAEC Chemistry paper, with answers and explanations. We hold 39 questions from this paper, and the Skolra app has all of them.

WAEC Chemistry 2021 questions and answers

1.
Protons and electrons are called fundamental particles because they----------
1. A
  have equal and opposite charges
2. B
  have the same mass as each other
3. C
  are found only in the nucleus of the atom
4. D
  cannot be further subdivided into simpler particles
  Correct
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Answer: D
Fundamental particles are defined as particles that cannot be broken down into smaller, simpler constituents. In the context of basic chemistry, protons and electrons (along with neutrons) are regarded as the indivisible building blocks of all matter — they are not made up of smaller chemical particles.
Electrons are found in the electron shells outside the nucleus, not only in the nucleus. Protons carry a positive charge and electrons carry a negative charge of equal magnitude, but this is a consequence of their nature, not the reason they are called fundamental. A proton has a mass approximately 1836 times greater than an electron, so they do not have the same mass.
2.
When CuSO $_{4(aq)}$ is added to Pb(NO $_{3}$ ) $_{2(aq)}$ --------
1. A
  a white precipitate of lead(II) sulfate is formed
  Correct
2. B
  a blue precipitate of copper(II) hydroxide is formed
3. C
  no visible change occurs as all products are soluble
4. D
  a red-brown precipitate of lead(II) oxide is formed
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Answer: A
When aqueous copper(II) sulfate is mixed with aqueous lead(II) nitrate, a double decomposition (precipitation) reaction takes place:
CuSO₄(aq) + Pb(NO₃)₂(aq) → PbSO₄(s) + Cu(NO₃)₂(aq)
- Lead(II) sulfate (PbSO₄) is insoluble in water — it appears as a dense white precipitate.
- Copper(II) nitrate remains in solution, giving the solution a pale blue colour.
The solubility rules confirm this: sulfates of lead, barium, and calcium are insoluble; most nitrates are soluble. The ionic equation is:
Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s)
3.What is the concentration of a solution which contains0.28g of potassium hydroxide in 100 cm $^{3}$ of solution [KOH = 56]
1. A
  0.28 mol dm⁻³
2. B
  0.05 mol dm⁻³
  Correct
3. C
  0.50 mol dm⁻³
4. D
  0.025 mol dm⁻³
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Answer: B
Step 1: Calculate the number of moles of KOH.
n = mass ÷ molar mass = 0.28 ÷ 56 = 0.005 mol
Step 2: Convert volume to dm³.
V = 100 cm³ = 100 ÷ 1000 = 0.1 dm³
Step 3: Calculate concentration.
c = n ÷ V = 0.005 ÷ 0.1 = 0.05 mol dm⁻³
4.
Which of the following oxide causes acid rain?
1. A
  Magnesium oxide
2. B
  Calcium oxide
3. C
  Sodium oxide
4. D
  Sulfur dioxide
  Correct
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Answer: D
Acid rain is precipitation (rain, snow, fog) with a pH below 5.6, caused by acidic oxides dissolving in atmospheric water.
5.
The most common method of preparing insoluble salts is by
1. A
  direct combination of the constituent elements
2. B
  action of dilute acid on a reactive metal
3. C
  evaporation of a saturated solution to dryness
4. D
  precipitation (mixing two aqueous solutions)
  Correct
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Answer: D
Insoluble salts cannot be prepared by neutralisation (because the salt cannot be collected from solution by evaporation without decomposing) or by direct synthesis in many cases. The standard laboratory method is precipitation:
- Two soluble salts (or a soluble salt and a soluble base/acid) are dissolved separately in water.
- The solutions are mixed; the desired insoluble salt immediately forms as a precipitate.
- The precipitate is filtered off, washed with distilled water, and dried.
Example: To prepare insoluble barium sulfate:
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)
The ionic equation is: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
6.
Which of the following statements is correct about ionization energy? It
1. A
  is the same for all elements in the same period
2. B
  decreases across a period from left to right
3. C
  increases across a period from left to right
  Correct
4. D
  increases down a group
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Answer: C
Ionization energy is the energy required to remove the outermost electron from a gaseous atom. Across a period from left to right, the nuclear charge (number of protons) increases while electrons are being added to the same energy level. The increased nuclear charge pulls the electrons closer and holds them more tightly, so more energy is needed to remove an outer electron — ionization energy increases.
Down a group, successive electrons occupy higher energy levels further from the nucleus, and the shielding effect of inner electrons also increases, so the outer electron is held less tightly. Ionization energy therefore decreases down a group.
7.
The ratio of carbon atoms of hydrogen atoms in a hydrocarbon is 1:2. If its molecular mass is 56, what is its molecular formula?
1. A
  C₅H₁₀
2. B
  C₂H₄
3. C
  C₄H₈
  Correct
4. D
  C₃H₆
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Answer: C
The ratio of carbon to hydrogen atoms is 1:2, so the empirical formula is CH₂.
Calculate the empirical formula mass:
- C: 12.0 g mol⁻¹
- H₂: 2 × 1.0 = 2.0 g mol⁻¹
- Empirical formula mass = 12.0 + 2.0 = 14 g mol⁻¹
Find the multiplier n:
n = Molecular mass ÷ Empirical formula mass = 56 ÷ 14 = 4
Therefore the molecular formula is:
(CH₂)₄ = C₄H₈
C₄H₈ corresponds to a hydrocarbon with one degree of unsaturation (either an alkene such as but-1-ene or but-2-ene, or a cycloalkane such as cyclobutane).
8.
An element, Q, contains 69% of $^{63}$ Q and 31% of $^{65}$ Q. What is the relative atomic mass of Q?
1. A
  64.0
2. B
  63.6
  Correct
3. C
  63.0
4. D
  65.0
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Answer: B
The relative atomic mass is the weighted average of the masses of all isotopes, taking their percentage abundances into account:
RAM = (69/100 × 63) + (31/100 × 65)
= 0.69 × 63 + 0.31 × 65
= 43.47 + 20.15
= 63.62 ≈ 63.6
This element is copper (Cu), which has two naturally occurring isotopes, ⁶³Cu (≈69%) and ⁶⁵Cu (≈31%), giving a standard atomic weight of approximately 63.5.
9.
The following ions have the same electron configuration except
$_{8}$ O, $_{12}$ Mg, $_{13}$ AI, $_{17}$ CI
1. A
  O²⁻
2. B
  Al³⁺
3. C
  Cl⁻
  Correct
4. D
  Mg²⁺
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Answer: C
To compare electron configurations, identify the ion each element commonly forms and count its electrons:
- O²⁻: 8 + 2 = 10 electrons → configuration 2, 8 (neon-like)
- Mg²⁺: 12 − 2 = 10 electrons → configuration 2, 8 (neon-like)
- Al³⁺: 13 − 3 = 10 electrons → configuration 2, 8 (neon-like)
- Cl⁻: 17 + 1 = 18 electrons → configuration 2, 8, 8 (argon-like)
O²⁻, Mg²⁺, and Al³⁺ are isoelectronic — they all have 10 electrons arranged like neon. Cl⁻ has 18 electrons and the electron configuration of argon, which is different from the other three.
10.
Under which conditions of pressure (p) and temperature (T) would the volume of an inflated balloon increase?
1. A
  Decreased pressure and increased temperature
  Correct
2. B
  Increased pressure and increased temperature
3. C
  Decreased pressure and decreased temperature
4. D
  Increased pressure and decreased temperature
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Answer: A
The combined gas law states:
V ∝ T / P (at constant amount of gas)
Therefore, to increase the volume of the gas in a balloon:
- Decrease pressure (P): At lower external pressure, the balloon walls expand outward because there is less force compressing the gas. By Boyle's Law, V increases as P decreases.
- Increase temperature (T): Higher temperature means gas molecules move faster and exert greater force on the walls, causing the balloon to expand. By Charles' Law, V increases as T increases.

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WAEC Chemistry 2021 past questions

WAEC Chemistry 2021 questions and answers

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