JAMB Chemistry 1984

26 reviewed questions with answers and explanations.

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Question 1

Sodium chloride may be obtained from brine by

  1. titration
  2. decantation
  3. distillation
  4. evaporation
  5. sublimation
Answer and explanation

D: evaporation

Evaporation removes water from brine. The dissolved sodium chloride remains and crystallises as enough water is lost.

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Question 3

NH₄NO₂ → N₂ + 2H₂O. What volume of nitrogen is produced from 3.20 g of ammonium nitrite at STP? Use a molar gas volume of 22.4 dm³ mol⁻¹. [N = 14, O = 16, H = 1]

  1. 2.24 dm³
  2. 2.24 cm³
  3. 1.12 cm³
  4. 1.12 dm³
  5. 4.48 dm³
Answer and explanation

D: 1.12 dm³

NH₄NO₂ has formula mass 2×14 + 4×1 + 2×16 = 64. Its amount is 3.20/64 = 0.050 mol. The equation produces one mole of N₂ per mole of NH₄NO₂, so the nitrogen volume is 0.050×22.4 = 1.12 dm³.

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Question 6

Which of these elements will not react with either water or steam? 1. Carbon 2. Oxygen 3. Copper 4. Bromine 5. Zinc

  1. 1 and 2
  2. 2 and 3
  3. 3 and 4
  4. 1, 2 and 3
  5. 2, 3 and 5
Answer and explanation

B: 2 and 3

Oxygen and copper do not react with water or steam in this comparison. Hot carbon and zinc react with steam. Bromine reacts reversibly with water to form HBr and HOBr, so copper and bromine cannot be the correct pair.

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Question 8

Naphthalene melts at 354 K (81 °C). At this temperature its molecules

  1. decompose into smaller molecules
  2. change their shape
  3. are oxidised by atmospheric oxygen
  4. contract
  5. become mobile as intermolecular forces holding the solid are overcome
Answer and explanation

E: become mobile as intermolecular forces holding the solid are overcome

On melting, naphthalene molecules gain enough energy to move out of their fixed crystal positions. The molecules remain chemically intact; it is the intermolecular attractions holding the solid structure that are overcome.

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Question 9

What is the ratio of the number of molecules in 2 g of hydrogen gas to that in 16 g of oxygen gas? [H = 1, O = 16]

  1. 2:1
  2. 1:1
  3. 1:2
  4. 1:4
  5. 1:8
Answer and explanation

A: 2:1

Hydrogen is H₂: 2 g represents 2/2 = 1 mol of molecules. Oxygen is O₂: 16 g represents 16/32 = 0.5 mol. Molecule counts are proportional to moles, giving 1:0.5 = 2:1.

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Question 14

In which reaction does hydrogen peroxide act as a reducing agent?

  1. H₂S + H₂O₂ → S + 2H₂O
  2. PbSO₃ + H₂O₂ → PbSO₄ + H₂O
  3. 2I⁻ + 2H⁺ + H₂O₂ → I₂ + 2H₂O
  4. PbO₂ + 2HNO₃ + H₂O₂ → Pb(NO₃)₂ + 2H₂O + O₂
  5. SO₂ + H₂O₂ → H₂SO₄
Answer and explanation

D: PbO₂ + 2HNO₃ + H₂O₂ → Pb(NO₃)₂ + 2H₂O + O₂

A reducing agent is itself oxidised. In D, oxygen in H₂O₂ changes from −1 to 0 in O₂, while lead changes from +4 in PbO₂ to +2 in Pb(NO₃)₂. In the other reactions peroxide is reduced to water.

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Question 15

For the reaction 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, which statement describes what happens?

  1. Fe is oxidised to Fe³⁺
  2. Fe³⁺ is oxidised to Fe²⁺
  3. I⁻ is oxidised to I₂
  4. I⁻ is reduced to I₂
  5. I⁻ removes an electron from Fe³⁺
Answer and explanation

C: I⁻ is oxidised to I₂

Iodide loses electrons: 2I⁻ → I₂ + 2e⁻. Its oxidation number rises from −1 to 0, so it is oxidised. Each Fe³⁺ ion gains an electron to become Fe²⁺, which is reduction.

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Question 16

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The energy profile represents A + B → C + D. Which description of the forward reaction follows from it?

  1. Spontaneous
  2. Isothermal
  3. Adiabatic
  4. Exothermic
  5. Endothermic

Question 17

In dilute solution, NaOH + HCl → NaCl + H₂O has an enthalpy change of −57.3 kJ for the equation as written. Using the same dilute strong-acid/strong-base neutralisation approximation, what is the enthalpy change for 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O?

  1. +28.65 kJ
  2. −28.65 kJ
  3. +57.3 kJ
  4. −114.6 kJ
  5. −229.2 kJ
Answer and explanation

D: −114.6 kJ

The reference reaction forms one mole of water and releases 57.3 kJ. The second equation forms two moles of water. In the stated dilute neutralisation approximation, ΔH = 2 × (−57.3) = −114.6 kJ. The negative sign indicates heat released.

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Question 20

Limewater, used to detect carbon dioxide in the laboratory, is an aqueous solution of

  1. Ca(OH)₂
  2. CaCO₃
  3. Ca(HCO₃)₂
  4. CaSO₄
  5. Na₂CO₃
Answer and explanation

A: Ca(OH)₂

Limewater is a clear aqueous solution of calcium hydroxide, Ca(OH)₂. Carbon dioxide produces suspended calcium carbonate: CO₂ + Ca(OH)₂ → CaCO₃ + H₂O, making it milky.

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Question 23

Which statement is not true of carbon monoxide under ordinary uncatalysed conditions?

  1. CO is poisonous
  2. CO is readily oxidised by air at room temperature to form CO₂
  3. CO can be prepared by passing CO₂ over coke heated to about 1000 °C
  4. CO can be prepared by heating charcoal with a limited supply of oxygen
  5. CO is a good reducing agent
Answer and explanation

B: CO is readily oxidised by air at room temperature to form CO₂

CO does not readily combine with air at room temperature without suitable activation or catalysis. It can burn to CO₂ when ignited. It is poisonous and can act as a reducing agent, so B is the false statement.

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Question 24

From ZnO + Na₂O → Na₂ZnO₂ and ZnO + CO₂ → ZnCO₃, zinc oxide may be classified as

  1. neutral
  2. basic
  3. acidic
  4. amphoteric
  5. a mixture
Answer and explanation

D: amphoteric

Zinc oxide reacts with a basic oxide, Na₂O, and with an acidic oxide, CO₂. An oxide that can show both acidic and basic behaviour is amphoteric.

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Question 25

An example of a neutral oxide is

  1. Al₂O₃
  2. NO₂
  3. CO₂
  4. CO
  5. SO₂
Answer and explanation

D: CO

Carbon monoxide is classified as a neutral oxide in ordinary acid-base chemistry. CO₂, SO₂ and NO₂ are acidic oxides, while Al₂O₃ is amphoteric.

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Question 26

In 3Cl₂ + 2NH₃ → N₂ + 6HCl, ammonia acts as

  1. a reducing agent
  2. an oxidizing agent
  3. an acid
  4. a catalyst
  5. a drying agent
Answer and explanation

A: a reducing agent

Nitrogen changes from −3 in NH₃ to 0 in N₂, so ammonia is oxidised. The substance that is oxidised reduces the other reactant and is therefore the reducing agent.

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Question 27

In the Haber process for manufacturing ammonia, finely divided iron is used as

  1. an ionizing agent
  2. a reducing agent
  3. a catalyst
  4. a dehydrating agent
  5. an oxidizing agent.
Answer and explanation

C: a catalyst

Iron provides a catalytic surface that speeds up the reaction of nitrogen with hydrogen to make ammonia. It changes the rate of reaching equilibrium, not the equilibrium composition.

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Question 28

A compound has vapour density 56.5 relative to hydrogen and composition C 53.1%, N 12.4%, O 28.3%, H 6.2%. What is its molecular formula? [C = 12, N = 14, O = 16, H = 1]

  1. C₃H₆O₂N
  2. C₅H₆O₂N
  3. (C₅H₇O₂N)½
  4. C₅H₇O₂N
  5. (C₅H₇ON)₂
Answer and explanation

D: C₅H₇O₂N

Vapour density relative to hydrogen gives molecular mass 2×56.5 = 113. For 100 g, the approximate mole amounts are C:53.1/12, H:6.2/1, O:28.3/16 and N:12.4/14. Dividing by the smallest gives C:H:O:N ≈ 5:7:2:1. C₅H₇O₂N has formula mass 113, so it is also the molecular formula.

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Question 30

When a kerosene fraction from petroleum is heated strongly and converted into a lower-boiling liquid, the process is known as

  1. polymerisation
  2. refining
  3. hydrogenation
  4. cracking
  5. fractional distillation
Answer and explanation

D: cracking

Cracking uses heat, often with a catalyst, to break larger hydrocarbon molecules into smaller ones. The smaller products generally have lower boiling points.

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Question 31

The compound CH₃CH₂C(=O)OH is

  1. acetic acid
  2. propanal
  3. propanol
  4. ethanoic acid
  5. propanoic acid
Answer and explanation

E: propanoic acid

CH₃CH₂C(=O)OH contains a three-carbon chain ending in the carboxyl group, −COOH. The three-carbon carboxylic acid is propanoic acid, not propanol.

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Question 32

Alkaline hydrolysis of naturally occurring fats and oils yields.

  1. fats and acids
  2. soaps and glycerol
  3. margarine and butter
  4. esters
  5. detergents.
Answer and explanation

B: soaps and glycerol

Alkaline hydrolysis breaks the ester bonds in fats and oils. It produces glycerol and the sodium or potassium salts of fatty acids, which are soaps.

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Question 33

Which formula represents a carboxylic acid? R denotes an organic group.

  1. R–C(=O)–OH
  2. R–C(=O)–OR
  3. H₂SO₄
  4. R–C(=O)–O–C(=O)–R
  5. R–C(=O)–H
Answer and explanation

A: R–C(=O)–OH

A carboxylic acid contains the −C(=O)OH group. A has this group; B is an ester, D an acid anhydride, and E an aldehyde. Sulfuric acid in C is not a carboxylic acid.

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Question 36

The electricity needed to deposit 1 g of magnesium costs ₦5.00. At the same cost per unit charge and 100% current efficiency, what does it cost to deposit 10 g of aluminium? [Al = 27, Mg = 24]

  1. ₦10.00
  2. ₦27.00
  3. ₦44.44
  4. ₦66.67
  5. ₦33.33
Answer and explanation

D: ₦66.67

Charge per gram is proportional to ionic charge divided by relative atomic mass. For 1 g Mg it is 2/24 mol of electrons; for 10 g Al it is 10×3/27 mol. The charge ratio is (30/27)/(2/24) = 13⅓. At the same cost per unit charge, the cost is ₦5×13⅓ = ₦66.67.

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Question 39

Nitrogen can best be obtained from a mixture of oxygen and nitrogen by passing the mixture over

  1. potassium hydroxide
  2. heated gold
  3. heated magnesium
  4. heated phosphorus
  5. calcium chloride.
Answer and explanation

D: heated phosphorus

Heated phosphorus removes oxygen by forming phosphorus oxide, leaving nitrogen. Magnesium is unsuitable because hot magnesium reacts with nitrogen as well as oxygen.

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Question 40

Water is said to be ‘hard’ if it

  1. easily forms ice
  2. must be warmed before sodium chloride dissolves
  3. forms an insoluble scum with soap
  4. contains nitrates
  5. contains sodium ions
Answer and explanation

C: forms an insoluble scum with soap

Calcium and magnesium ions in hard water react with soap ions to form insoluble salts, seen as scum. This also makes it harder to obtain a lather.

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Question 48

An aqueous metal salt M gives a white precipitate with NaOH, soluble in excess NaOH. It also gives a white precipitate with aqueous ammonia, soluble in excess ammonia. What is the cation in M?

  1. Zn²⁺
  2. Ca²⁺
  3. Al³⁺
  4. Pb²⁺
  5. Cu²⁺
Answer and explanation

A: Zn²⁺

Zinc ions form white Zn(OH)₂ with either reagent. The precipitate dissolves in excess NaOH because it is amphoteric, and in excess ammonia because zinc forms a soluble ammine complex. These two observations identify Zn²⁺.

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Question 49

What is the IUPAC name of CH₃–CH(CH₃)–CH₂–CH₃?

  1. isopropylethene
  2. acetylene
  3. 3-methylbutane
  4. 2-methylbutane
  5. 5-methylpentane
Answer and explanation

D: 2-methylbutane

The longest continuous carbon chain has four carbon atoms, so the parent is butane. Numbering from the nearer end puts the methyl branch on carbon 2, giving 2-methylbutane.

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Question 50

At STP, what volume of hydrogen is formed when 500 cm³ of 0.5 mol dm⁻³ H₂SO₄ reacts completely with excess zinc? Use a molar gas volume of 22.4 dm³ mol⁻¹.

  1. 22.4 dm³
  2. 11.2 dm³
  3. 6.5 dm³
  4. 5.6 dm³
  5. 0.00 dm³
Answer and explanation

D: 5.6 dm³

Zn + H₂SO₄ → ZnSO₄ + H₂. The acid amount is 0.5×500/1000 = 0.25 mol, producing 0.25 mol H₂ because zinc is in excess. At the given molar volume, 0.25×22.4 = 5.6 dm³ of hydrogen is formed.

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