WAEC Chemistry 2016 past questions

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

Ten questions from the 2016 WAEC Chemistry paper, with answers and explanations. We hold 43 questions from this paper, and the Skolra app has all of them.

WAEC Chemistry 2016 questions and answers

1. 1.On evaporation to dryness, 350 cm³ of saturated solution of salt Z gave 55.5g of salt. What is the solubility of the salt? [Z = 101]
   1. A
      1.57 mol dm⁻³
      Correct
   2. B
      0.55 mol dm⁻³
   3. C
      3.15 mol dm⁻³
   4. D
      2.14 mol dm⁻³
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   Answer: A

   Moles of salt Z = ^55.5/₁₀₁ = 0.5495 mol.

   Volume of solution = 350 cm³ = 0.350 dm³.

   Solubility = ^0.5495/_0.350 ≈ 1.57 mol dm⁻³.

2. 2.The refreshing characteristic taste of fizzy drinks is due to the presence of
   1. A
      Sucrose dissolved under pressure
   2. B
      Carbon dioxide dissolved under pressure
      Correct
   3. C
      Oxygen dissolved at low temperature
   4. D
      Citric acid added as a preservative
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   Answer: B

   Fizzy drinks are carbonated by dissolving carbon dioxide gas under high pressure. When the drink is opened, the pressure drops and CO₂ escapes as bubbles. In solution, CO₂ reacts with water to form carbonic acid (H₂CO₃), which produces the characteristic sharp, refreshing, slightly tangy taste of carbonated beverages.

3. 3.

   Consider the following ionic equation:

   Cr₂O₇^2- + 14H⁺ + ne^- $\to$ 2Cr³⁺ + 7H₂O.

   The value of n in the equation is_______?

   1. A
      4
   2. B
      8
   3. C
      10
   4. D
      6
      Correct
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   Answer: D

   Balance by charge. Left-hand side charge: (−2) + 14(+1) + (−n) = 12 − n. Right-hand side charge: 2(+3) + 0 = +6.

   Setting both sides equal: 12 − n = 6, so n = 6.

   This is consistent with the chromium atoms each being reduced from +6 (in dichromate) to +3, a gain of 3 electrons per chromium atom, giving 2 × 3 = 6 electrons in total.

4. 4.The compound that makes palm wine taste sour after exposure to the air for few days is
   1. A
      Ethanoic acid
      Correct
   2. B
      Propanoic acid
   3. C
      Ethanol
   4. D
      Methanoic acid
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   Answer: A

   Palm wine contains ethanol produced by fermentation of palm sap. On prolonged exposure to air, bacteria of the genus Acetobacter oxidise the ethanol using atmospheric oxygen:

   CH₃CH₂OH + O₂ → CH₃COOH + H₂O

   The product, ethanoic acid (acetic acid), is responsible for the sour taste. This is the same acid found in vinegar.

5. 5.Ripening of fruits is hastened by using
   1. A
      Propane gas
   2. B
      Ethylene (ethene) gas
      Correct
   3. C
      Methane gas
   4. D
      Carbon dioxide gas
   Show answerHide answer

   Answer: B

   Ethylene (ethene, C₂H₄) is a natural plant hormone that triggers and accelerates the ripening process in fruits. It activates enzymes that break down cell walls, convert starches to sugars, and develop colour. Commercially, fruits are exposed to controlled concentrations of ethylene gas in ripening rooms to hasten uniform ripening. Fruits such as bananas are often transported unripe and then ripened with ethylene just before sale.

6. 6.Which of the following arrangements shows increasing order of reactivity of the halogens?
   1. A
      Cl < Br < I < F
   2. B
      F < Cl < Br < I
   3. C
      Br < I < F < Cl
   4. D
      I < Br < Cl < F
      Correct
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   Answer: D

   Halogen reactivity decreases down Group VII (Group 17) of the periodic table. As the atomic radius increases down the group, the incoming electron is added to a shell farther from the nucleus, where the attraction is weaker and shielding by inner electrons is greater. Fluorine is therefore the most reactive and iodine the least reactive. The increasing order of reactivity is: I < Br < Cl < F.

7. 7.The shape of the water molecule is_______?
   1. A
      Linear
   2. B
      Bent (V-shaped)
      Correct
   3. C
      Tetrahedral
   4. D
      Trigonal planar
   Show answerHide answer

   Answer: B

   The oxygen atom in a water molecule has four electron pairs: two bond pairs (one to each hydrogen atom) and two lone pairs. According to the VSEPR model these four pairs adopt a roughly tetrahedral arrangement around the oxygen, but because only the two bonding pairs define the molecular geometry, the shape of the molecule is described as bent (or V-shaped), with an H–O–H bond angle of approximately 104.5°.

8. 8.Propane and carbon(IV) oxide diffuse at the same rate because [H = 1.00, C = 12.0, O = 16.0]
   1. A
      They are both non-polar molecules
   2. B
      They have the same number of atoms per molecule
   3. C
      They have the same molar mass
      Correct
   4. D
      They have the same boiling point
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   Answer: C

   Graham's law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Two gases with identical molar masses therefore diffuse at identical rates.

   Propane (C₃H₈): molar mass = 3(12.0) + 8(1.00) = 36.0 + 8.0 = 44.0 g mol⁻¹.

   Carbon(IV) oxide (CO₂): molar mass = 12.0 + 2(16.0) = 12.0 + 32.0 = 44.0 g mol⁻¹.

   Because both gases have a molar mass of 44.0 g mol⁻¹, they diffuse at the same rate under the same conditions.

9. 9.The bond formed between H₂O and H₊ to form the hydroxonium H₃O₊ is
   1. A
      ionic bond
   2. B
      dative (coordinate) covalent bond
      Correct
   3. C
      ordinary covalent bond
   4. D
      hydrogen bond
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   Answer: B

   When H₂O reacts with H⁺, the oxygen atom donates both electrons of one of its lone pairs to the proton, which has no electrons of its own. Because both electrons in the new O–H bond originate from the same atom (oxygen), the bond formed is a dative (coordinate) covalent bond. The resulting ion, H₃O⁺, is called the hydroxonium (or hydronium) ion.

10. 10.

    Which of the following compounds is/are secondary alkanols?i) CH₃CH(CH₃)CH₃OH

    Ii) CH₃CH₂CH(OH)CH₃

    Iii) CH₃C(OH)(CH₃)CH₃

    1. A
       ii only
       Correct
    2. B
       i only
    3. C
       i and ii only
    4. D
       iii only
    Show answerHide answer

    Answer: A

    A secondary alkanol has the –OH group on a carbon that is bonded to exactly two other carbon atoms.

    Compound i, CH₃CH(CH₃)CH₂OH (2-methylpropan-1-ol): the –OH is on a terminal CH₂ bonded to only one carbon — this is a primary alkanol.

    Compound ii, CH₃CH₂CH(OH)CH₃ (butan-2-ol): the –OH is on the carbon bearing both an ethyl group and a methyl group (two carbon neighbours) — this is a secondary alkanol.

    Compound iii, CH₃C(OH)(CH₃)CH₃ (2-methylpropan-2-ol): the –OH carbon is bonded to three methyl groups — this is a tertiary alkanol.

    Therefore only compound ii is a secondary alkanol.

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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