JAMB Chemistry 2016 past questions

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

Ten questions from the 2016 JAMB Chemistry paper, with answers and explanations. We hold 38 questions from this paper, and the Skolra app has all of them.

JAMB Chemistry 2016 questions and answers

1. 1.The industrial preparation of hydrogen gas from water gas is referred to as
   1. A
      the Bosch process
      Correct
   2. B
      the Contact process
   3. C
      the Solvay process
   4. D
      the Haber process
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   Answer: A

   Water gas (a mixture of carbon(II) oxide and hydrogen) is reacted with more steam over a catalyst to convert the carbon(II) oxide to carbon(IV) oxide while releasing extra hydrogen: CO + H₂O → CO₂ + H₂. This industrial route to hydrogen gas is called the Bosch process.

   The Haber process instead combines hydrogen with nitrogen to manufacture ammonia. The Contact process is the industrial route to tetraoxosulphate(VI) acid, and the Solvay process manufactures sodium trioxocarbonate(IV) — neither produces hydrogen from water gas.

2. 2.Which of the following compounds is used as a gaseous fuel?
   1. A
      petrol
   2. B
      diesel
   3. C
      kerosene
   4. D
      methane
      Correct
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   Answer: D

   Methane (natural gas) is a gas at room temperature and pressure and is burned directly as a fuel, for example piped into homes and industry.

   Kerosene, petrol and diesel are all liquid hydrocarbon fuels at room temperature, refined from crude oil, not gaseous fuels.

3. 3.What volume of 0.5 mol dm^-3 H₂SO₄ will exactly neutralize 20cm³ of 0.1 mol dm^-3 NaOH solution?
   1. A
      2cm³
      Correct
   2. B
      8cm³
   3. C
      10cm³
   4. D
      4cm³
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   Answer: A

   H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, so 1 mole of the acid neutralizes 2 moles of the base.

   Moles of NaOH = 0.1 × 20/1000 = 0.002 mol, so moles of H₂SO₄ needed = 0.002/2 = 0.001 mol, giving a volume of 0.001/0.5 = 0.002 dm³ = 2cm³.

   4cm³ comes from ignoring the acid's dibasic nature and treating the ratio as 1:1. 8cm³ comes from applying the ratio the wrong way round (doubling instead of halving), and 10cm³ is what you get from a simple M₁V₁=M₂V₂ slip that ignores the mole ratio entirely.

4. 4.The monomer of nylon is
   1. A
      styrene
   2. B
      tetrafluoroethene
   3. C
      hexanedioic acid (adipic acid) and 1,6-diaminohexane (hexamethylenediamine)
      Correct
   4. D
      ethane-1,2-diol and benzene-1,4-dicarboxylic acid
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   Answer: C

   Nylon-6,6 is a condensation polymer built from two monomers: hexanedioic acid and 1,6-diaminohexane. They join by repeated amide (peptide-type) linkages, releasing water at every step, to build up the long polyamide chain.

   Styrene is the monomer of polystyrene, and tetrafluoroethene is the monomer of PTFE (Teflon) — both addition polymers, not nylon. Ethane-1,2-diol and benzene-1,4-dicarboxylic acid are instead the monomers of the polyester Terylene.

5. 5.The pollutant that contributes to the depletion of the ozone layer is
   1. A
      chlorofluorocarbons (CFCs)
      Correct
   2. B
      carbon(IV) oxide
   3. C
      sulphur(IV) oxide
   4. D
      methane
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   Answer: A

   Chlorofluorocarbons, once widely used as refrigerants and aerosol propellants, release chlorine atoms in the upper atmosphere that catalytically break down ozone molecules, thinning the ozone layer.

   Carbon(IV) oxide and methane are chiefly linked to global warming through the greenhouse effect rather than ozone breakdown, and sulphur(IV) oxide is mainly associated with acid rain, not ozone depletion.

6. 6.Which of the following statements is correct about the periodic table?
   1. A
      metallic character increases across a period from left to right
   2. B
      all elements in the same period have the same number of valence electrons
   3. C
      elements in the same group have similar chemical properties because they have the same number of valence electrons
      Correct
   4. D
      atomic radius increases across a period from left to right
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   Answer: C

   Elements in the same group share the same number of electrons in their outermost shell, and since chemical behaviour is governed largely by these valence electrons, elements in a group show similar chemical properties.

   Atomic radius actually decreases across a period, as increasing nuclear charge pulls the electron cloud in more tightly. The number of valence electrons increases by one across each period rather than staying constant, and metallic character decreases, not increases, across a period as elements become more non-metallic.

7. 7.The drying agent suitable for drying ammonia is
   1. A
      concentrated tetraoxosulphate(VI) acid
   2. B
      calcium oxide
      Correct
   3. C
      fused calcium chloride
   4. D
      phosphorus(V) oxide
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   Answer: B

   Calcium oxide (quicklime) is basic and unreactive towards ammonia, so it can absorb the water vapour from moist ammonia gas without reacting with the ammonia itself.

   Concentrated tetraoxosulphate(VI) acid and phosphorus(V) oxide are both strongly acidic drying agents that would react with and absorb the alkaline ammonia gas instead of merely drying it. Fused calcium chloride also fails for ammonia because it forms a solid addition compound, CaCl₂.8NH₃, with the gas.

8. 8.

   n monosaccharide $\underset{Q}{\stackrel{P}{\rightleftharpoons}}$ polysaccharide -n water.

   In the process above, P and Q respectively represent

   1. A
      fermentation and hydrolysis, respectively
   2. B
      condensation and hydrolysis, respectively
      Correct
   3. C
      hydrolysis and condensation, respectively
   4. D
      oxidation and reduction, respectively
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   Answer: B

   Going forward (P), many monosaccharide units join up and release water at each linkage to build the polysaccharide — this joining-with-loss-of-water is condensation (polymerization). Going in reverse (Q), the polysaccharide is broken back down into monosaccharide units by reacting with water — this is hydrolysis.

   Reversing the two names describes the two arrows the wrong way round. Oxidation and reduction do not apply, since no electron transfer or change in oxidation state is involved in linking sugar units. Fermentation converts monosaccharides into ethanol and carbon(IV) oxide, not into a polysaccharide, so it does not fit the forward arrow here.

9. 9.An example of an acidic oxide is
   1. A
      carbon(IV) oxide
      Correct
   2. B
      sodium oxide
   3. C
      calcium oxide
   4. D
      carbon(II) oxide
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   Answer: A

   Carbon(IV) oxide (CO₂) is a non-metallic oxide that dissolves in water to give an acidic solution and reacts with bases to form a salt and water, making it an acidic oxide.

   Calcium oxide and sodium oxide are metallic oxides that are basic, reacting with acids rather than bases. Carbon(II) oxide (CO) is neutral — it shows neither acidic nor basic behaviour — despite also being a compound of carbon and oxygen.

10. 10.Ethanol is soluble in water due to the presence of a
    1. A
       carbonyl group present in its structure
    2. B
       carboxyl group present in its structure
    3. C
       long non-polar hydrocarbon chain
    4. D
       hydroxyl group, which forms hydrogen bonds with water
       Correct
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    Answer: D

    Ethanol's -OH (hydroxyl) group can form hydrogen bonds with water molecules, just as water molecules hydrogen-bond with each other, allowing ethanol to mix freely with water.

    Ethanol has neither a carbonyl group (found in aldehydes and ketones) nor a carboxyl group (found in carboxylic acids) in its structure. Its hydrocarbon chain is in fact non-polar and does not dissolve in water on its own — it is the hydroxyl group that overcomes this and accounts for the solubility.

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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