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WAEC Chemistry 2018 past questions

Ten questions from the 2018 WAEC Chemistry paper, with answers and explanations. We hold 47 questions from this paper, and the Skolra app has all of them.

WAEC Chemistry 2018 questions and answers

  1. 1.

    Consider the following reaction equation: Br2+2KI→2KBr+I2Br_{2} + 2KI \to 2KBr + I_{2}. Bromine is acting as

    1. A
      a catalyst
    2. B
      a precipitating agent
    3. C
      a reducing agent
    4. D
      an oxidising agent
      Correct
    Show answerHide answer

    Answer: D

    In this displacement reaction, iodide ions (I−) lose electrons and are oxidised to molecular iodine (I2): oxidation state changes from −1 to 0. Bromine molecules simultaneously gain those electrons and are reduced to bromide ions (Br−): oxidation state changes from 0 to −1. The species that causes oxidation of another is the oxidising agent, so bromine is the oxidising agent. Bromine is not itself oxidised, ruling out reducing agent.

  2. 2.

    When chlorine is passed through a sample of water, the pH of the water sample would be

    1. A
      greater than 7
    2. B
      exactly 14
    3. C
      less than 7
      Correct
    4. D
      equal to 7
    Show answerHide answer

    Answer: C

    Chlorine reacts with water according to: Cl2 + H2O → HCl + HClO. Both hydrochloric acid (HCl) and hypochlorous acid (HClO) are formed, releasing H+ ions into solution. This makes the water acidic, so the pH drops below 7.

  3. 3.

    How many atoms are contained in 0.2moles of nitrogen? [NA=6.02×1023][N_{A} = 6.02 \times 10^{23}]

    1. A
      2.408 × 10²³
      Correct
    2. B
      1.204 × 10²³
    3. C
      1.204 × 10²⁴
    4. D
      6.02 × 10²²
    Show answerHide answer

    Answer: A

    Nitrogen exists as a diatomic molecule, N2. Each molecule contains 2 atoms, so 0.2 moles of N2 contains 0.2 × 2 = 0.4 moles of nitrogen atoms. Multiplying by Avogadro's number: 0.4 × 6.02 × 1023 = 2.408 × 1023 atoms.

  4. 4.

    In which of the following compounds does hydrogen form ionic compounds?

    1. A
      HCl
    2. B
      H₂O
    3. C
      NH₃
    4. D
      NaH
      Correct
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    Answer: D

    In sodium hydride (NaH), hydrogen bonds ionically as the hydride ion, H− (oxidation state −1), while sodium exists as Na+. This ionic character arises because sodium's low electronegativity allows it to donate an electron to hydrogen. In HCl, H2O, and NH3, hydrogen is bonded covalently (polar or non-polar); none of these involves the transfer of electrons to form discrete ions.

  5. 5.

    What is the mass of solute in 500cm3cm^{3} of 0.005moldm−3moldm^{-3}H2SO4H_{2}SO_{4}? ( H = 1, S = 32.0, O=16.0)

    1. A
      0.245 g
      Correct
    2. B
      0.490 g
    3. C
      0.980 g
    4. D
      0.0245 g
    Show answerHide answer

    Answer: A

    Molar mass of H2SO4: 2(1) + 32 + 4(16) = 98 g mol−1.
    Volume in dm3: 500 cm3 = 0.5 dm3.
    Moles of H2SO4: 0.005 × 0.5 = 0.0025 mol.
    Mass: 0.0025 × 98 = 0.245 g.

  6. 6.

    What is the relative molecular mass of a compound which has empirical formula CH2OCH_{2}O? [C= 12, H= 1, O= 16]

    1. A
      120
    2. B
      30
      Correct
    3. C
      90
    4. D
      60
    Show answerHide answer

    Answer: B

    The formula mass of one CH₂O unit is: C (12) + 2 × H (1) + O (16) = 30 g mol⁻¹. Without additional experimental data specifying a larger molecular mass, the simplest compound with this empirical formula has a relative molecular mass of 30. This corresponds to methanal (formaldehyde, HCHO).

  7. 7.

    The most suitable substance for putting out petrol fire is

    1. A
      dilute hydrochloric acid
    2. B
      dilute sodium hydroxide solution
    3. C
      CO₂ gas
      Correct
    4. D
      water
    Show answerHide answer

    Answer: C

    Petrol fire is a Class B (flammable liquid) fire. Water must never be used because petrol is less dense than water and floats on it, spreading the fire. Carbon dioxide (CO2) is ideal: it is non-combustible, denser than air, and smothers the fire by displacing oxygen without leaving residue or reacting with petrol. Acids and alkalis are chemically reactive and dangerous near flames, and neither extinguishes a hydrocarbon fire.

  8. 8.

    The valence electrons of 12Mg_{12}Mg are in the

    1. A
      3p subshell
    2. B
      3s subshell
      Correct
    3. C
      2s subshell
    4. D
      2p subshell
    Show answerHide answer

    Answer: B

    Magnesium (atomic number 12) has the electronic configuration 1s2 2s2 2p6 3s2. The two outermost (valence) electrons occupy the 3s subshell in the third principal energy level. The 2s and 2p subshells belong to the second shell, which is fully filled and does not contain valence electrons. There is no 3p electron in ground-state magnesium.

  9. 9.

    The following factors would contribute to environmental pollution except

    1. A
      discharge of industrial effluents into rivers
    2. B
      combustion of fossil fuels
    3. C
      use of solar energy
      Correct
    4. D
      indiscriminate burning of refuse
    Show answerHide answer

    Answer: C

    Combustion of fossil fuels releases CO2, SO2, and nitrogen oxides — all pollutants. Burning refuse emits particulates and toxic gases. Industrial effluents introduce heavy metals and organic pollutants into waterways. Solar energy, by contrast, is a clean and renewable energy source that generates electricity without combustion or chemical emissions and therefore does not contribute to environmental pollution.

  10. 10.

    Consider the following reaction equation: 2HCl+Ca(OH)2→CaCl2+H2O2HCl + Ca(OH)_{2} \to CaCl_{2} + H_{2}O. What is the volume of 0.1moldm−3moldm^{-3} HCl that would completely neutralize 25cm3cm^{3} of 0.3moldm−3moldm^{-3} Ca(OH)2_{2}?

    1. A
      75 cm³
    2. B
      100 cm³
    3. C
      50 cm³
    4. D
      150 cm³
      Correct
    Show answerHide answer

    Answer: D

    Step 1 — moles of Ca(OH)2: 0.3 mol dm−3 × 0.025 dm3 = 0.0075 mol.
    Step 2 — moles of HCl needed: the equation shows a 2 : 1 molar ratio (HCl : Ca(OH)2), so 2 × 0.0075 = 0.015 mol HCl.
    Step 3 — volume of HCl: 0.015 mol ÷ 0.1 mol dm−3 = 0.15 dm3 = 150 cm3.

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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