All 41 questions from the Joint Admissions and Matriculation Board (JAMB) Chemistry 2021 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
The sequence of separation techniques required to separate a mixture of sand, sodium chloride and iodine is
A. Filtration → sublimation → dissolution → evaporation
B. Dissolution → sublimation → filtration → evaporation
C. Dissolution → filtration → sublimation → evaporation
D. Sublimation → dissolution → filtration → evaporationCorrect
Explanation
First: sublimation (heat to remove iodine as vapour). Then: dissolve residue in water (dissolution). Then: filter to remove sand. Then: evaporate filtrate to recover NaCl.
Which of the following gases cannot be collected by downward delivery?
A. Cl₂
B. NH₃Correct
C. SO₂
D. H₂S
Explanation
Downward delivery collects gases denser than air. NH₃ (M.M. = 17) is less dense than air (M.M. ≈ 29), so it is collected by upward delivery. Cl₂, SO₂ and H₂S are all denser than air.
KCl is a normal salt (all H⁺ replaced). NaHSO₄ is an acid salt (one H⁺ remaining). Mg(OH)Cl is a basic salt (one OH⁻ remaining). KAl(SO₄)₂·12H₂O is a double salt.
Which of the following is true about the bleaching action of chlorine and sulphur(iv) oxide?
A. Chlorine bleaches by reduction while SO₂ bleaches by oxidation
B. SO₂ bleaches by oxidation while chlorine bleaches by reduction
C. Chlorine bleaches by oxidation while SO₂ bleaches by reductionCorrect
D. Water is not required while bleaching with Cl₂
Explanation
Cl₂ bleaches by oxidation (oxidises coloured matter). SO₂ bleaches by reduction. Both require water. Cl₂ bleaching is permanent; SO₂ bleaching can be reversed by atmospheric oxidation.
The process by which a salt loses its water of crystallisation on exposure to the atmosphere is known as
A. Hygroscopy
B. EfflorescenceCorrect
C. Deliquescence
D. Efferhexane
Explanation
Efflorescence: loss of water of crystallisation in air (e.g. Na₂CO₃·10H₂O → Na₂CO₃·H₂O). Deliquescence: absorbing moisture until dissolving. Hygroscopy: absorbing moisture without dissolving.
The electronic configuration of an element is 1s²2s²2p⁶3s²3p⁵. It can be deduced that the element (i) is a non-metal (ii) it is a metal (iii) is a p-block element (iv) is a transition element (v) has five electrons in its outermost shell
A. i & iii
B. ii, iii & iv
C. i, iii & vCorrect
D. ii & iv
Explanation
Configuration = Cl (chlorine). Last orbital is 3p → p-block element. Chlorine is a non-metal. Outermost shell (n=3): 3s²3p⁵ = 7 electrons total in outer shell. Not a transition element (no d orbitals).
A gas occupies a volume of 500 cm³ at 30°C. At what temperature will its volume become double this initial value, provided that the pressure is kept constant?
2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = −ve. The yield of SO₃ can be improved by
A. Increasing the temperature
B. Increasing the pressureCorrect
C. Decreasing the pressure
D. Adding a catalyst
Explanation
Reaction has fewer moles of gas on right (2 vs 3). By Le Chatelier's principle, increasing pressure favours the side with fewer moles of gas (right), increasing SO₃ yield. Temperature increase would decrease yield (exothermic reaction).
A redox reaction requires BOTH oxidation AND reduction. Fe²⁺ → Fe³⁺ + e⁻ shows only oxidation (Fe²⁺ is oxidised to Fe³⁺). There is no corresponding reduction half-reaction shown.
What mass of silver will be deposited at the cathode when a current of 5 A is passed through a solution of AgNO₃ for 1 hour, 1 minute and 1 second? [Ag = 108, 1F = 96500 C]
A. 20.5 gCorrect
B. 10.25 g
C. 41 g
D. 5 g
Explanation
t = 1×3600 + 1×60 + 1 = 3661 s. Q = It = 5 × 3661 = 18305 C. m = (M × Q)/(n × F) = (108 × 18305)/(1 × 96500) = 20.5 g.
Which is the petroleum fraction used in jet engines?
A. Lubricating oil
B. Petrol
C. Diesel
D. KeroseneCorrect
Explanation
Kerosene (aviation fuel/jet fuel) is the petroleum fraction used in jet engines. Diesel is used in diesel engines; petrol in petrol engines; lubricating oil reduces friction.
In ¹H (protium), if an electron is lost, only one proton remains. Why is the hydrogen ion regarded as a proton?
A. Because it has one proton and one electron
B. Because losing one electron leaves only one protonCorrect
C. Because it is the lightest element
D. Because it forms covalent bonds
Explanation
¹H (protium) has 1 proton and 1 electron. When it loses its one electron (ionisation), only the proton remains. H⁺ = proton. This is why the hydrogen ion is regarded as simply a proton.
When gases are mixed, the chance of finding a molecule of any of those gases at its original position drastically reduces. This is an example of increase in
A. Enthalpy
B. EntropyCorrect
C. Free energy
D. Activation energy
Explanation
Entropy (S) is a measure of disorder/randomness. When gases mix, disorder increases (more possible positions for each molecule), so entropy increases.
The paper will turn black due to the formation of lead(ii) sulphide. A piece of filter paper moistened with lead(ii) ethanoate turns black when dropped into which gas?
A. SO₂
B. H₂SCorrect
C. SO₃
D. HCl
Explanation
H₂S + Pb(CH₃COO)₂ → PbS↓ (black) + 2CH₃COOH. The black precipitate of lead(ii) sulphide forms when lead ethanoate paper contacts hydrogen sulphide gas.
The two factors to be considered before choosing a method for collecting a gas are
A. Solubility and densityCorrect
B. Boiling point and density
C. Solubility and melting point
D. Colour and density
Explanation
Two factors for gas collection: (i) Solubility — a gas that is very soluble in water cannot be collected over water (e.g. NH₃, HCl, SO₂); (ii) Density — determines upward or downward delivery.
A metal X forms soluble trioxosulphate(iv) salt. The metal X is
A. Aluminium
B. Barium
C. PotassiumCorrect
D. Manganese
Explanation
Generally, all sodium, potassium and ammonium salts are soluble. Potassium sulphite (K₂SO₃) is soluble. BaSO₃ is insoluble; Al₂(SO₃)₃ hydrolyses; MnSO₃ is insoluble.
Which of the following statements is correct? (a) Atomic size decreases down the group (b) Cations are smaller than the parent atoms (c) Anions are larger than the parent atoms (d) Atomic size increases across the period
A. Atomic size decreases down the group
B. Cations are smaller than the parent atoms
C. Anions are larger than the parent atomsCorrect
D. Atomic size increases across the period
Explanation
Anions gain electrons → increased electron-electron repulsion → larger radius than parent atom. Cations lose electrons → smaller radius. Atomic size increases DOWN the group (more shells). Atomic size decreases ACROSS the period.
At room temperature, fluorine is a gas, chlorine is also gaseous, bromine is a liquid while iodine is solid. Which of the following halogens is liquid at room temperature?
A. Fluorine
B. Iodine
C. Chlorine
D. BromineCorrect
Explanation
At room temperature (~25°C): F₂ and Cl₂ are gases; Br₂ is a liquid; I₂ is a solid. Bromine is the only halogen that is liquid at standard room temperature.
The diatomic elements are nitrogen, oxygen, fluorine, hydrogen, chlorine, bromine and iodine. Which of the following elements is diatomic?
A. Neon
B. OxygenCorrect
C. Iron
D. Sodium
Explanation
The diatomic elements are H₂, N₂, O₂, F₂, Cl₂, Br₂ and I₂. Oxygen (O₂) is diatomic. Noble gases (Ne, Ar) are monoatomic; metals like Fe and Na are not diatomic.
The main function of limestone in the blast furnace is to
A. Melt the ore
B. Act as a catalyst
C. Act as reducing agent
D. Remove impurityCorrect
Explanation
Limestone (CaCO₃) decomposes to CaO in the blast furnace: CaCO₃ → CaO + CO₂. CaO then reacts with silica impurities (SiO₂) to form slag (CaSiO₃), removing impurities.
Which of the following statements about thermoplastic material is correct?
A. They do not melt on heating
B. They melt on heating
C. They soften and melt on heatingCorrect
D. They decompose on heating
Explanation
Thermoplastics soften and melt reversibly on heating. Thermosets do not soften — they decompose or char. Examples of thermoplastics: PVC, polyethene, perspex.
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