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JAMB Biology 2015 Objective Past Questions

All 50 questions from the Joint Admissions and Matriculation Board (JAMB) Biology 2015 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

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Biology 2015 Objective — Question 1

Pollution of water by crude oil can lead to

  • A. a decrease in carbon(IV)oxide content
  • B. an increase in oxygen content
  • C. a decrease in oxygen contentCorrect
  • D. an increase in growth of aquatic organisms

Explanation

Crude oil is a major pollutant of water; it blocks oxygen exchange at the surface, leading to a decrease in the oxygen content of the water.

Biology 2015 Objective — Question 2

When a few drops of barium chloride is added to an unknown sample acidified with hydrochloric acid, a white precipitate insoluble in excess of the acid was obtained. The likely anion in the sample is

  • A. SO₄²⁻Correct
  • B. Cl⁻
  • C. S²⁻
  • D. CO₃²⁻

Explanation

A white precipitate that remains undisclosed (insoluble) even after adding excess HCl confirms the presence of SO₄²⁻, since CO₃²⁻ and SO₃²⁻ would dissolve with effervescence.

Biology 2015 Objective — Question 3

Oxygen in air can be removed using

  • A. lime water or caustic soda solution
  • B. caustic soda solution
  • C. alkaline pyrogallol solutionCorrect
  • D. slaked lime

Explanation

Alkaline pyrogallol solution is commonly used to absorb/remove oxygen from air in gas analysis.

Biology 2015 Objective — Question 4

Calculate the pH of a solution of 0.0001moldm⁻³ hydrochloric acid.

  • A. 2
  • B. 4Correct
  • C. 2
  • D. 1

Explanation

[H⁺] = 0.0001 = 10⁻⁴ moldm⁻³. pH = −log₁₀(10⁻⁴) = 4.

Biology 2015 Objective — Question 5

Which of the following drying agents is NOT suitable for drying hydrogen sulphide?

  • A. CaOCorrect
  • B. P₄O₁₀
  • C. CaCl₂
  • D. H₂SO₄

Explanation

CaO (a basic drying agent) would react with the acidic gas H₂S, so it is unsuitable for drying it.

Biology 2015 Objective — Question 6

Diamond is used in making jewelry due to its

  • A. high refractive indexCorrect
  • B. transparency
  • C. hardness
  • D. high melting point

Explanation

Diamond's high refractive index makes it shine brilliantly when polished, making it a good ornament.

Biology 2015 Objective — Question 7

The presence of impurities in a solid will make the melting point to

  • A. decreaseCorrect
  • B. increase
  • C. remain unchanged
  • D. be zero

Explanation

Impurities generally cause the melting point of a solid to decrease (melting point depression), while they raise the boiling point of a liquid.

Biology 2015 Objective — Question 8

The discovery that protons and neutrons are concentrated at the nucleus of an atom was postulated by

  • A. E. RutherfordCorrect
  • B. J. Dalton
  • C. J.J. Thomson
  • D. N. Bohr

Explanation

Rutherford's gold-foil experiment led to the discovery that the mass/protons and neutrons of an atom are concentrated in a small dense nucleus.

Biology 2015 Objective — Question 9

Which of the following is a hydroscopic substance?

  • A. NaNO₃Correct
  • B. MgCl₂
  • C. FeCl₂
  • D. CaCl₂

Explanation

Hygroscopic substances absorb moisture from the atmosphere; NaNO₃ is a common example, alongside CuO and H₂SO₄.

Biology 2015 Objective — Question 10

In a chemical reaction, when the energy of the colliding reactant particles is less than the activation energy, the reaction will

  • A. be spontaneous
  • B. not occurCorrect
  • C. occur
  • D. be slow

Explanation

For a reaction to occur, colliding particles must have energy greater than or equal to the activation energy; if less, the reaction will not occur.

Biology 2015 Objective — Question 11

The monomer of natural rubber is

  • A. 2-methylbuta-1,3-dieneCorrect
  • B. 1-buten-3-yne
  • C. buta-1,3-diene
  • D. 2-chlorobuta-1,3-diene

Explanation

Natural rubber is a polymer of the diolefin monomer 2-methylbuta-1,3-diene (isoprene).

Biology 2015 Objective — Question 12

The monomer of natural rubber is

  • A. 2-methylbuta-1,3-dieneCorrect
  • B. 1-buten-3-yne
  • C. buta-1,3-diene
  • D. 2-chlorobuta-1,3-diene

Explanation

As with question 11, the monomer of natural rubber is 2-methylbuta-1,3-diene (this question is repeated in the original source).

Biology 2015 Objective — Question 13

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), ΔH = +ve. In the reaction above, the forward reaction is favoured by

  • A. increasing the size of the containing vessel
  • B. increasing the pressure
  • C. adding a catalyst
  • D. increasing the temperatureCorrect

Explanation

Since the forward reaction is endothermic (ΔH = +ve), it is favoured by an increase in temperature.

Biology 2015 Objective — Question 14

Which of the following halogens is the most reactive?

  • A. Cl₂
  • B. I₂
  • C. Br₂
  • D. F₂Correct

Explanation

Fluorine is the most electronegative element and the most reactive halogen; reactivity decreases down the group: F > Cl > Br > I.

Biology 2015 Objective — Question 15

A gas that forms a black precipitate with lead(II) ethanoate is

  • A. Cl₂
  • B. NH₃
  • C. SO₂
  • D. H₂SCorrect

Explanation

H₂S reacts with lead(II) ethanoate to produce lead(II) sulphide, an insoluble black precipitate.

Biology 2015 Objective — Question 16

The common ore of iron is

  • A. galena
  • B. cassiterite
  • C. bauxite
  • D. magnetiteCorrect

Explanation

Magnetite is a common ore of iron; galena is an ore of lead, cassiterite an ore of tin, and bauxite an ore of aluminium.

Biology 2015 Objective — Question 17

A property common to the two structures shown (normal pentane and 2-methylbutane) is that they have the same

Diagram for question 17
  • A. structural and molecular formulae
  • B. structural formula but different molecular formulae
  • C. number of carbon but different molecular mass
  • D. molecular formula but different structural formulaCorrect

Explanation

Normal pentane and 2-methylbutane are isomers — they share the same molecular formula (C₅H₁₂) but have different structural formulae.

Biology 2015 Objective — Question 18

The type of salt that ionizes to produce three different ions in solution is

  • A. acidic
  • B. complex
  • C. doubleCorrect
  • D. basic

Explanation

A double salt ionizes to produce three different ions in solution — two metallic ions and a negative acid radical, or a metallic ion, an ammonium ion, and a negative acid radical (e.g. potash alum).

Biology 2015 Objective — Question 19

H₂S(g) + Br₂(g) → 2HBr(g) + S(s). In the reaction above, the oxidizing agent is

  • A. H₂S(g)
  • B. S(s)
  • C. Br₂(g)Correct
  • D. HBr(g)

Explanation

The oxidation number of bromine decreases from 0 in Br₂ to −1 in HBr (reduction), so Br₂ is the oxidizing agent, while H₂S is the reducing agent.

Biology 2015 Objective — Question 20

The process whereby a gaseous body loses some of its kinetic energy to a colder body is referred to as

  • A. condensationCorrect
  • B. melting
  • C. evaporation
  • D. freezing

Explanation

When a gaseous substance loses kinetic energy, it may change into liquid form; this process is known as condensation.

Biology 2015 Objective — Question 21

Which of the quantum numbers determines the energy sub-level in an orbital?

  • A. magnetic
  • B. principal
  • C. azimuthalCorrect
  • D. spin

Explanation

The azimuthal (angular/secondary/subsidiary) quantum number determines the energy sub-level in an orbital.

Biology 2015 Objective — Question 22

Fehling's solution can be used to distinguish between

  • A. CH₃CHO and C₂H₅OH
  • B. CH₃COCH₃ and CH₃COOCH₃
  • C. CH₃CHO and CH₃COCH₃Correct
  • D. C₂H₅OH and CH₂CH₂COOH

Explanation

Alkanals (e.g. CH₃CHO, ethanal) reduce Fehling's solution while alkanones (e.g. CH₃COCH₃, propanone) do not, so Fehling's test distinguishes them.

Biology 2015 Objective — Question 23

What is the molecular formula of a compound with empirical formula CH₂ and vapour density 42? [C=12, H=1]

  • A. C₆H₁₂Correct
  • B. C₃H₆
  • C. C₄H₈
  • D. C₅H₁₀

Explanation

Molar mass = 2×vapour density = 84. Empirical formula mass (CH₂) = 14. n = 84/14 = 6, so molecular formula = C₆H₁₂.

Biology 2015 Objective — Question 24

The separation technique that is based on the principle of solubility of a solid in two miscible liquids is

  • A. fractional distillation
  • B. distillation
  • C. precipitationCorrect
  • D. filtration

Explanation

Fractional/solvent extraction (precipitation as listed) exploits differing solubility of a solid in two miscible liquids to separate it.

Biology 2015 Objective — Question 25

Hydrogen bond can be found in

  • A. hydrogen sulphide
  • B. hydrogen bromide
  • C. hydrogen fluorideCorrect
  • D. hydrogen chloride

Explanation

Hydrogen bonding occurs in compounds containing hydrogen bonded to a very small, highly electronegative element such as fluorine; hydrogen fluoride exhibits hydrogen bonding.

Biology 2015 Objective — Question 26

24g of magnesium reacts with dilute hydrochloric acid. Calculate the volume of hydrogen gas evolved at s.t.p. [Mg=24, Molar volume of gas at s.t.p. = 22.4dm³]

  • A. 22.4dm³Correct
  • B. 5.6dm³
  • C. 11.2dm³
  • D. 44.8dm³

Explanation

Moles of Mg = 24/24 = 1 mol. Mg + 2HCl → MgCl₂ + H₂, so 1 mol Mg produces 1 mol H₂ = 22.4dm³ at s.t.p.

Biology 2015 Objective — Question 27

Aluminium is usually used as electrical cables due to its

  • A. good conductivityCorrect
  • B. malleability
  • C. silvery-white appearance
  • D. high melting point

Explanation

Aluminium's good electrical conductivity is why it is used for electrical cables.

Biology 2015 Objective — Question 28

A chemical reaction with entropy change of +20 Jk⁻¹ and free energy change +55000J occurred at −23°C. Calculate the enthalpy change.

  • A. 55460J
  • B. 44540J
  • C. 50000J
  • D. 60000JCorrect

Explanation

ΔG = ΔH − TΔS ⟹ ΔH = ΔG + TΔS = 55000 + (250×20) = 55000+5000 = 60000J.

Biology 2015 Objective — Question 29

The enzyme invertase converts sucrose into

  • A. glucose and glucose
  • B. glucose and galactose
  • C. glucose and fructoseCorrect
  • D. fructose and galactose

Explanation

Invertase hydrolyses sucrose into glucose and fructose.

Biology 2015 Objective — Question 30

The number of lone pairs of electrons in a water molecule is

  • A. 2Correct
  • B. 4
  • C. 3
  • D. 1

Explanation

The oxygen atom in a water molecule has two lone pairs of electrons.

Biology 2015 Objective — Question 31

n monosaccharide (−x→) polysaccharide (−y→) + n water. In the illustration above, x and y respectively represent

  • A. condensation and hydrolysisCorrect
  • B. hydrolysis and condensation
  • C. fermentation and condensation
  • D. fermentation and hydration

Explanation

The conversion of monosaccharide to polysaccharide is condensation (with loss of water); the reverse process (polysaccharide back to monosaccharide) is hydrolysis.

Biology 2015 Objective — Question 32

The main product of the reaction between phosphorus(V) chloride and ethanol is

  • A. chloroethaneCorrect
  • B. chloromethane
  • C. dichloromethane
  • D. chloroethene

Explanation

Phosphorus(V) chloride reacts with ethanol (C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl) to give chloroethane as the main organic product.

Biology 2015 Objective — Question 33

The bond-breaking energies in an endothermic reaction is usually

  • A. equal to the bond-forming energies
  • B. more than the bond-forming energiesCorrect
  • C. less than the bond-forming energies
  • D. double the bond-forming energies

Explanation

In an endothermic reaction, the bond-breaking energy is usually more than the bond-forming energy (the reverse is true for exothermic reactions).

Biology 2015 Objective — Question 34

²³⁴₉₀Th → ˣ_yPa + β. In the reaction above, x and y respectively are

  • A. 234 and 91Correct
  • B. 235 and 90
  • C. 233 and 89
  • D. 230 and 86

Explanation

Beta decay: mass number is conserved (x=234) while atomic number increases by 1 (y=91).

Biology 2015 Objective — Question 35

The compound of calcium used to make refractory furnace linings is

  • A. calcium oxideCorrect
  • B. calcium trioxonitrate(V)
  • C. calcium hydroxide
  • D. calcium chloride

Explanation

Calcium oxide (quicklime), due to its extremely high melting point (about 2600°C), is used for making refractory furnace linings.

Biology 2015 Objective — Question 36

The mass of copper deposited when a current of 2.0A is passed through a solution of a copper salt for 60 minutes is [Cu=64, F=96500 Cmol⁻¹]

  • A. 4.78g
  • B. 0.04g
  • C. 2.39gCorrect
  • D. 9.55g

Explanation

Q=It=2×3600=7200C. For 1 mole Cu (64g), 2F (193000C) is required, so mass deposited = (64/193000)×7200 ≈ 2.39g.

Biology 2015 Objective — Question 37

Which of the following compounds will liberate carbon(IV) oxide on reacting with sodium trioxocarbonate(IV)?

  • A. CH₃COOC₂H₅
  • B. CH₃COOHCorrect
  • C. CH₃CHO
  • D. CH₃COCH₃

Explanation

Acids (like ethanoic acid, CH₃COOH) react with carbonates to liberate carbon(IV) oxide.

Biology 2015 Objective — Question 38

Kerosene is a petroleum fraction that distils over in the temperature range of

  • A. 50°C – 500°C
  • B. 40°C – 200°C
  • C. 200°C – 250°CCorrect
  • D. 25°C and above

Explanation

Kerosene is a mixture of hydrocarbons containing C₁₀–C₁₆ carbon atoms, boiling between 200°C and 250°C.

Biology 2015 Objective — Question 39

The preference in the usage of asbestos support to platinum block as a catalyst can be attributed to its

  • A. increased surface areaCorrect
  • B. high concentration
  • C. low concentration
  • D. decreased surface area

Explanation

A catalyst's effectiveness is higher with increased surface area; asbestos support provides this over a plain platinum block.

Biology 2015 Objective — Question 40

In the deduction of ethyl ethanoate with LiAlH₄, the product obtained is

  • A. ethanolCorrect
  • B. ethane
  • C. ethanoic acid
  • D. ethane

Explanation

Ethyl ethanoate is reduced to ethanol using reducing agents like lithium aluminium hydride (LiAlH₄).

Biology 2015 Objective — Question 41

A sample of a salt was heated on a Bunsen flame. If the colour of the flame turned bluish-green, the metal present is likely to be

  • A. potassium
  • B. copperCorrect
  • C. sodium
  • D. iron

Explanation

Copper produces a bluish-green flame colouration in a flame test.

Biology 2015 Objective — Question 42

A salt that loses moisture on exposure to the atmosphere is said to be

  • A. effervescent
  • B. hygroscopic
  • C. deliquescent
  • D. efflorescentCorrect

Explanation

A salt that loses its water of crystallization to become anhydrous on exposure to the atmosphere is said to be efflorescent.

Biology 2015 Objective — Question 43

In the graph, at what temperature is the solubility of KHCO₃ and NaCl the same?

Diagram for question 43
  • A. 35°C
  • B. 42°CCorrect
  • C. 37°C
  • D. 18°C

Explanation

KHCO₃ and NaCl have the same solubility at the point where their solubility curves intersect, which corresponds to about 42°C.

Biology 2015 Objective — Question 44

1050cm³ of helium gas was produced at −33°C and 700mmHg. What volume would the gas occupy at 23°C and 750mmHg?

  • A. 1208.67cm³Correct
  • B. 794.59cm³
  • C. 912.16cm³
  • D. 1387.50cm³

Explanation

Using the combined gas law: V₂ = P₁V₁T₂/(P₂T₁) = (700×1050×296)/(750×240) ≈ 1208.67cm³.

Biology 2015 Objective — Question 45

¹⁴₇N + X → ¹⁷₈O + ¹₁H. In the reaction above, X is a(n)

  • A. lithium atom
  • B. neutron
  • C. helium atomCorrect
  • D. deuterium atom

Explanation

Balancing mass and atomic numbers (14+4=17+1; 7+2=8+1) shows X is a helium atom (alpha particle).

Biology 2015 Objective — Question 46

The colour of phenolphthalein in an alkaline medium is

  • A. yellow
  • B. colourless
  • C. pink/redCorrect
  • D. orange

Explanation

Phenolphthalein is colourless in acidic/neutral medium but turns pink/red in alkaline medium.

Biology 2015 Objective — Question 47

Calculate the molecular formula of a compound with empirical formula CH₂O and relative molecular mass 120. [C=12, O=16, H=1]

  • A. C₂H₄O₂
  • B. C₄H₈O₄Correct
  • C. C₆H₁₆O₆
  • D. C₈H₁₆O₈

Explanation

Empirical formula mass = 12+2+16 = 30. n = 120/30 = 4, so molecular formula = (CH₂O)₄ = C₄H₈O₄.

Biology 2015 Objective — Question 48

MnO₂(s) + 4HCl(aq) → MnCl₂(aq) + Cl₂(g) + 2H₂O(l). In the reaction above, MnO₂ acts as a(n)

  • A. catalytic agent
  • B. oxidizing agentCorrect
  • C. bleaching agent
  • D. reducing agent

Explanation

The oxidation state of manganese decreases from +4 in MnO₂ to +2 in MnCl₂ (reduction), so MnO₂ is the oxidizing agent.

Biology 2015 Objective — Question 49

Calculate the mass of magnesium that would be deposited by 8 faradays of electricity. [Mg=24]

  • A. 96gCorrect
  • B. 48g
  • C. 72g
  • D. 120g

Explanation

Mg²⁺ + 2e⁻ → Mg, so 2F deposits 1 mole of Mg (24g). 8F will deposit 4 moles = 96g.

Biology 2015 Objective — Question 50

Which of the following hydrocarbons will NOT undergo polymerization reaction?

  • A. C₄H₆
  • B. C₃H₄
  • C. C₃H₆
  • D. C₄H₁₀Correct

Explanation

C₄H₁₀ (butane) is an alkane (saturated hydrocarbon); alkanes do not undergo polymerization reactions.

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