JAMB Chemistry 2023 past questions and answers

17 questions. Try each one, then open the answer to check yourself and read the explanation.

Question 1

Separation of Mixtures

Sodium chloride can be obtained from brine through

  1. A.decantation
  2. B.titration
  3. C.evaporation
  4. D.distillation
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Answer: C (evaporation)

Explanation

Brine is salt dissolved in water; evaporating the water leaves the solid sodium chloride behind.

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Question 2

Mole Concept and Stoichiometry

NH4NO2(s) → (heat) N2(g) + 2H2O(l)
In the above reaction, calculate the volume of N2(g) that would be produced at S.T.P from 3.20g of NH4NO2(s) [N=14, H=1, O = 16] Molar volume at s.t.p. = 22.4dm3

  1. A.2.24cm3
  2. B.2.24dm3
  3. C.1.12dm3
  4. D.1.12cm3
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Answer: C (1.12dm3)

Explanation

Molar mass of NH4NO2 = 64 g, so 3.20 g = 0.05 mol, giving 0.05 mol of N2. Volume = 0.05 × 22.4 = 1.12 dm3.

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Question 5

Electrolysis

In the above electrolyte cell, which species migrate to the anode?

Diagram for this question
  1. A.K+ and SO42−
  2. B.SO42− and OH−
  3. C.H3O+ and K+
  4. D.OH− and H3O+
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Answer: B (SO42− and OH−)

Explanation

The anode is positive, so the negative ions in K2SO4(aq), SO42− and OH−, move to it.

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Question 6

Air

What is the approximate volume of air containing 10cm3 of oxygen?

  1. A.25cm3
  2. B.20cm3
  3. C.100cm3
  4. D.50cm3
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Answer: D (50cm3)

Explanation

Air is about one-fifth (20%) oxygen, so 10 cm3 of oxygen is in about 5 × 10 = 50 cm3 of air.

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Question 11

Rates of Reaction

The above diagram illustrates the energy profile of the catalyst and uncatalysed reaction of A(g) + B(g) → AB(g). Calculate the activation energies in KJ of the catalysed and uncatalysed unreverse reactions.

Diagram for this question
  1. A.500, 300
  2. B.300, 500
  3. C.−500, −300
  4. D.−300, −500
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Answer: B (300, 500)

Explanation

Activation energy = peak energy − reactant energy. Catalysed: 400 − 100 = 300 kJ; uncatalysed: 600 − 100 = 500 kJ.

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Question 16

Solubility and Solutions

A property found in a colloidal dispersion which is absent in a true solution is

  1. A.homogeneity
  2. B.the Tyndale effect
  3. C.surface polarity
  4. D.Osmotic pressure
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Answer: B (the Tyndale effect)

Explanation

Colloid particles are big enough to scatter a beam of light (the Tyndall effect); a true solution does not.

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Question 17

Environmental Pollution

A gas when dissolved in water vapour, produced acid rain during rainfall. The gas is

  1. A.carbon (II) oxide
  2. B.oxygen
  3. C.sulphur (IV) oxide
  4. D.nitrogen
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Answer: C (sulphur (IV) oxide)

Explanation

SO2 dissolves in rainwater to form trioxosulphate(IV) acid, a major cause of acid rain.

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Question 18

Chemical Bonding

The carbon atoms in H2C=CH2 are

  1. A.Sp hybridized
  2. B.Sp3 hybridized
  3. C.not hybridized
  4. D.Sp2 hybridized
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Answer: D (Sp2 hybridized)

Explanation

Each carbon in ethene forms three σ bonds and one π bond, so it uses sp2 hybrid orbitals.

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Question 20

Electrolysis

Half cell reaction — Eθ
Cu2+(aq) + 2e− → Cu(s) +0.34V
Zn2+(aq) + 2e− → Zn(s) −0.76V
Fe2+(aq) + 2e− → Fe(s) −0.44V
Ba2+(aq) + 2e− → Ba(s) −2.90V
From the above data, it can be deduced that the most powerful reducing agent is?

  1. A.Fe
  2. B.Cu
  3. C.Zn
  4. D.Ba
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Answer: D (Ba)

Explanation

The most negative electrode potential (−2.90 V for Ba) means the metal loses electrons most easily, so barium is the strongest reducing agent.

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Question 24

Alkanoic Acids and Esters

CH3CH2OH + CH3COOH → (Conc H2SO4) Y + Z
Y and Z in the reaction above represent

  1. A.CH3CH2COCH3 and H2O
  2. B.CH3COCH3 and H2O
  3. C.CH3CH2CHO and CH4
  4. D.CH3COOCH2CH3 and H2O
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Answer: D (CH3COOCH2CH3 and H2O)

Explanation

An alkanol and an alkanoic acid react (esterification) to give an ester, ethyl ethanoate, and water.

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Question 30

Redox Reactions

Write out the increasing oxidation number of the transition metal ion for the compounds K2Cr2O7, V2O5 and KMnO4.

  1. A.K2Cr2O7 < KMnO4 < V2O5
  2. B.V2O5 < K2Cr2O7 < KMnO4
  3. C.KMnO4 < K2Cr2O7 < V2O5
  4. D.KMnO4 < V2O5 < K2Cr2O7
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Answer: B (V2O5 < K2Cr2O7 < KMnO4)

Explanation

V in V2O5 is +5, Cr in K2Cr2O7 is +6 and Mn in KMnO4 is +7.

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Question 32

Periodic Table

Given the distance between the nuclei of chlorine atoms in a chlorine molecule as 0.194mm, calculate its atomic radius.

  1. A.0.194mm
  2. B.0.097mm
  3. C.2.388mm
  4. D.0.388mm
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Answer: B (0.097mm)

Explanation

The nucleus-to-nucleus distance is two atomic radii, so r = 0.194/2 = 0.097.

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Question 35

Energy Changes

Which of these equations shows that a reaction is in equilibrium?

  1. A.ΔG = ΔH − TΔS
  2. B.ΔG < O
  3. C.ΔG = O
  4. D.ΔG > O
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Answer: C (ΔG = O)

Explanation

At equilibrium there is no driving force in either direction, so the free energy change ΔG is zero.

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Question 36

Alkanoic Acids and Esters

The formula CH2O for ethanoic acid represents its

  1. A.general formula
  2. B.molecular formula
  3. C.structural formula
  4. D.empirical formula
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Answer: D (empirical formula)

Explanation

Ethanoic acid is C2H4O2 = (CH2O)2; CH2O is the simplest whole-number ratio, the empirical formula.

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Question 38

Halogens

How is Cl2(g) prepared in the laboratory? by

  1. A.adding concentrated tetraoxosulphate (VI) acid to solid sodium chloride
  2. B.adding concentrated hydro chloric acid to solid manganese (IV) oxide
  3. C.dropping concentrated hydrochloric acid unto potassium tetraoxomanganate (VI) crystals and heating the mixture
  4. D.dropping concentrated hydrochloric acid unto potassium tetraoxomanganate(VI) crystals
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Answer: D (dropping concentrated hydrochloric acid unto potassium tetraoxomanganate(VI) crystals)

Explanation

Concentrated HCl is oxidised to chlorine by potassium permanganate crystals without heating (option A gives HCl gas, not chlorine).

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Question 39

Introduction to Organic Chemistry

Which of the following is an isomer of C5H12?

  1. A.2-methylbutane
  2. B.butane
  3. C.2-ethylbutane
  4. D.2-methylpropane
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Answer: A (2-methylbutane)

Explanation

2-methylbutane has five carbons and twelve hydrogens, the same formula as pentane but a branched chain.

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Question 40

Gas Laws

According to Charles's law, at what temperature does the volume of gas becomes zero?

  1. A.−100°C
  2. B.0°C
  3. C.373°C
  4. D.−273°C
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Answer: D (−273°C)

Explanation

Extending the volume-temperature line to zero volume gives absolute zero, −273 °C (0 K).

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