JAMB Past Questions

JAMB Chemistry 2003
Questions & Answers

20 questions · Correct answers highlighted · 20 with explanations

Explanations reviewed August 2026

This 2003 JAMB Chemistry examination paper covers a balanced selection of theoretical and numerical concepts designed to test secondary-level chemical competency. Several recurring themes emerge from the question set, notably acid-base neutralization titrations, electrochemistry involving standard electrode potentials, and organic reaction pathways including esterification and addition reactions. Engaging thoroughly with these items will help solidify your understanding of core chemical principles. When practicing on this platform, remember to write out balancing equations and check your unit conversions carefully rather than just guessing solutions.

20 Total Questions
2003 Exam Year
20 With Explanations

Topics in this paper

Functional Groups and Reactions (7) Redox Reactions (5) Solutions and Solubility (4) Stoichiometric Calculations (3) Rates of Reaction (3) Electrochemistry (3) Periodic Table Trends (3) Organic Nomenclature (3) Chemical Equilibrium (3) Atomic Structure (2) Acids Bases Salts (2) Thermochemistry and Enthalpy (1) Gas Laws (1)
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1
Question 1 of 20
JAMB · Chemistry · 2003

The quantum numbers for the last electron in the ground state of sodium (Na, atomic number 11) are

A. n=3, l=0, m=0, s=+1/2
B. n=2, l=1, m=0, s=-1/2
C. n=3, l=1, m=1, s=+1/2
D. n=2, l=0, m=0, s=+1/2
Explanation

One might mistakenly pick a configuration with different principal energy levels or spin values by confusing orbital designations. For sodium with atomic number 11, the ground state electron configuration is 1s²2s²2p⁶3s¹; the final valence electron resides in the 3s orbital, where the principal quantum number n equals 3, the azimuthal quantum number l equals 0 corresponding to the s subshell, the magnetic quantum number m is 0, and the spin quantum number s is +1/2. Common mistake: Forgetting that l equals 0 for an s orbital.

2
Question 2 of 20
JAMB · Chemistry · 2003

The volume of 0.5 mol/dm³ H₂SO₄ required to neutralize 25 cm³ of 0.2 mol/dm³ KOH is [1 cm³ = 0.001 dm³]

A. 5.0 cm³
B. 10.0 cm³
C. 12.5 cm³
D. 25.0 cm³
Explanation

An incorrect choice like 25.0 cm³ might be selected by overlooking the stoichiometric ratio between the acid and base in the neutralization equation. The balanced reaction is H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O; multiplying the concentration 0.2 mol/dm³ by the volume 0.025 dm³ gives 0.005 moles of KOH, which requires half that amount, 0.0025 moles of H₂SO₄, and dividing this by the acid concentration 0.5 mol/dm³ yields a volume of 0.005 dm³, equivalent to 10 cm³. Common mistake: Neglecting the stoichiometric coefficient of 2 for potassium hydroxide.

3
Question 3 of 20
JAMB · Chemistry · 2003

The type of hybridization in the carbon atom of ethanoic acid (CH₃COOH) in the carboxyl group is

A. sp
B. sp²
C. sp³
D. dsp²
Explanation

Students could mistakenly choose sp³ due to the presence of multiple single bonds around the carbon framework. The carboxyl carbon forms one double bond and two single bonds, which are C=O, C-O, and C-C, thus requiring sp² hybridization. Common mistake: Counting only single bonds and failing to account for the double-bonded oxygen atom.

4
Question 4 of 20
JAMB · Chemistry · 2003

The rate of decomposition of hydrogen peroxide is increased by adding

A. sodium chloride
B. manganese(IV) oxide
C. calcium carbonate
D. potassium nitrate
Explanation

A student might guess common salts like sodium chloride or calcium carbonate, assuming any foreign substance accelerates reactions. Manganese(IV) oxide acts as a catalyst, effectively lowering the activation energy for the breakdown of hydrogen peroxide into water and oxygen. Common mistake: Assuming all solid additives function as catalysts in peroxide reactions.

5
Question 5 of 20
JAMB · Chemistry · 2003

The enthalpy change for the reaction C(s) + O₂(g) → CO₂(g) is -393.5 kJ/mol. The formation of 4.4 g of CO₂ releases [C = 12, O = 16]

A. 39.35 kJ
B. 78.70 kJ
C. 393.5 kJ
D. 1574 kJ
Explanation

An incorrect option such as 393.5 kJ might be picked by forgetting to divide the mass of carbon dioxide by its molar mass. The molar mass of CO₂ is 44 g/mol, meaning that 4.4 g represents 0.1 moles; multiplying this mole value by the given enthalpy change of -393.5 kJ/mol shows that 39.35 kJ is released. Common mistake: Using the full molar enthalpy value directly without scaling for the given mass.

6
Question 6 of 20
JAMB · Chemistry · 2003

In the electrolysis of dilute CuSO₄ solution using copper electrodes, the mass of the cathode increases because

A. Cu²⁺ ions are reduced to Cu
B. SO₄²⁻ ions are oxidized
C. Cu atoms are oxidized to Cu²⁺
D. H⁺ ions are discharged
Explanation

One might incorrectly select the oxidation of sulfate ions or hydrogen ion discharge, confusing the roles of anode and cathode in electrolytic cells. At the cathode, Cu²⁺ ions undergo reduction via the half-reaction Cu²⁺ + 2e⁻ → Cu, resulting in copper deposition that increases the mass of the cathode. Common mistake: Mixing up the reactions occurring at the anode and the cathode during electrolysis.

7
Question 7 of 20
JAMB · Chemistry · 2003

The pH of a solution formed by mixing 50 cm³ of 0.1 M HCl with 50 cm³ of 0.1 M NaOH is

A. 1
B. 7
C. 10
D. 14
Explanation

A student might choose 1 or 14 by assuming the strong acid or base dominates without checking stoichiometry. The reaction is HCl + NaOH → NaCl + H₂O; multiplying 50 cm³ by 0.1 M gives 0.005 moles of both HCl and NaOH, and their complete neutralization creates a neutral solution with a pH of 7. Common mistake: Forgetting that equal volumes and concentrations of a strong acid and strong base yield a neutral pH.

8
Question 8 of 20
JAMB · Chemistry · 2003

The element with the electron configuration 1s²2s²2p⁶3s²3p³ belongs to

A. Group 3
B. Group 5
C. Group 7
D. Group 8
Explanation

An incorrect option like Group 3 or Group 7 might be selected by misinterpreting the outermost orbital numbers. The valence shell configuration 3s²3p³ contains a total of 5 electrons, placing the element in Group 5, as seen with phosphorus. Common mistake: Counting only the p-subshell electrons instead of summing all electrons in the highest principal energy level.

9
Question 9 of 20
JAMB · Chemistry · 2003

The compound formed when aluminium reacts with oxygen is

A. AlO
B. Al₂O₃
C. Al₃O₂
D. Al₂O
Explanation

One might mistakenly pick AlO or Al₂O based on random valency guessing. The balanced chemical equation is 4Al + 3O₂ → 2Al₂O₃, demonstrating that aluminium oxide has the correct formula Al₂O₃. Common mistake: Balancing the valencies incorrectly to form a 1:1 metal-to-oxygen ratio.

10
Question 10 of 20
JAMB · Chemistry · 2003

The IUPAC name of the compound CH₃CH(OH)CH₂CH₃ is

A. butan-1-ol
B. butan-2-ol
C. 2-methylpropan-2-ol
D. propan-2-ol
Explanation

A student could choose butan-1-ol by ignoring the exact position of the hydroxyl group on the carbon backbone. The four-carbon chain has the -OH group attached to carbon 2, giving the IUPAC name butan-2-ol. Common mistake: Failing to locate the lowest possible locant for the functional group.

11
Question 11 of 20
JAMB · Chemistry · 2003

The process used to separate a mixture of liquids with close boiling points is

A. simple distillation
B. fractional distillation
C. evaporation
D. chromatography
Explanation

An incorrect choice like simple distillation might be picked by assuming all liquid mixtures can be separated by a single boiling-and-condensation step. Fractional distillation separates liquids based on slight differences in boiling points. Common mistake: Confusing simple and fractional distillation when dealing with close boiling points.

12
Question 12 of 20
JAMB · Chemistry · 2003

The standard electrode potential of Zn²⁺/Zn is -0.76 V and Cu²⁺/Cu is +0.34 V. The EMF of the cell Zn|Zn²⁺||Cu²⁺|Cu is

A. 0.42 V
B. 1.10 V
C. 1.52 V
D. -1.10 V
Explanation

One might mistakenly pick -1.10 V by reversing the subtraction order or misapplying the electrode values. The cell EMF is calculated as E_cathode minus E_anode, which equals 0.34 minus (-0.76), giving 0.34 + 0.76 which equals 1.10 V. Common mistake: Subtracting the cathode potential from the anode potential instead of the reverse.

13
Question 13 of 20
JAMB · Chemistry · 2003

The gas that forms a red-brown vapour when heated is

A. chlorine
B. nitrogen dioxide
C. sulphur dioxide
D. hydrogen sulphide
Explanation

A student might choose chlorine or sulphur dioxide, overlooking the distinct color descriptions of gaseous oxides. Nitrogen dioxide forms a red-brown vapour, especially when heated. Common mistake: Confusing the green-yellow gas of chlorine with the brown fumes of nitrogen dioxide.

14
Question 14 of 20
JAMB · Chemistry · 2003

The functional group in CH₃COCH₃ is

A. alkene
B. alcohol
C. ketone
D. carboxylic acid
Explanation

An incorrect option like carboxylic acid or alcohol might be selected by misidentifying the carbonyl placement within the carbon chain. The compound CH₃COCH₃, known as propanone, contains a C=O group located between two alkyl groups, which is characteristic of ketones. Common mistake: Confusing ketones with aldehydes or carboxylic acids due to the shared carbonyl group.

15
Question 15 of 20
JAMB · Chemistry · 2003

The effect of adding a catalyst to a reversible reaction at equilibrium is

A. shifting the equilibrium to the right
B. increasing the equilibrium constant
C. increasing the rate of attaining equilibrium
D. decreasing the activation energy of the reverse reaction only
Explanation

A student might think a catalyst shifts the position of equilibrium or alters the equilibrium constant. A catalyst speeds up both forward and reverse reactions equally, reducing the time to reach equilibrium without shifting it. Common mistake: Assuming a catalyst can increase product yield in a reversible reaction.

16
Question 16 of 20
JAMB · Chemistry · 2003

The monomer used to produce polyethene is

A. ethene
B. ethane
C. chloroethene
D. propene
Explanation

One might mistakenly choose ethane or propene by confusing saturated alkanes with unsaturated alkene monomers. Polyethene is formed by polymerizing ethene, having the formula C₂H₄. Common mistake: Selecting the alkane instead of the corresponding alkene for addition polymerization.

17
Question 17 of 20
JAMB · Chemistry · 2003

The compound responsible for the smell of rotten eggs is

A. sulphur dioxide
B. hydrogen sulphide
C. carbon disulphide
D. sulphur trioxide
Explanation

An incorrect choice like sulphur dioxide might be picked by confusing different sulfur-containing atmospheric gases. Hydrogen sulphide has a characteristic rotten egg smell. Common mistake: Mixing up the pungent choking odor of SO₂ with the rotten egg smell of H₂S.

18
Question 18 of 20
JAMB · Chemistry · 2003

The mass of silver deposited when a current of 2 A is passed through a silver nitrate solution for 965 seconds is [Ag = 108, 1 F = 96500 C]

A. 1.08 g
B. 2.16 g
C. 4.32 g
D. 10.8 g
Explanation

A student could mistakenly choose a different value by miscalculating the total charge or Faraday constant conversion. The total charge Q equals current times time, 2 multiplied by 965, giving 1930 C; dividing by 96500 C/F yields 0.02 moles of silver, and multiplying by the molar mass of 108 gives a mass of 2.16 g. Common mistake: Forgetting to divide the charge by the Faraday constant.

19
Question 19 of 20
JAMB · Chemistry · 2003

The property of elements that increases across a period from left to right is

A. atomic radius
B. electronegativity
C. metallic character
D. ionization energy
Explanation

An incorrect option like atomic radius or metallic character might be chosen since these follow the reverse trend. Electronegativity increases across a period due to increasing nuclear charge, attracting electrons more strongly. Common mistake: Confusing atomic size trends with electronegativity trends.

20
Question 20 of 20
JAMB · Chemistry · 2003

The industrial process used to extract iron from its ore is

A. electrolysis
B. fractional distillation
C. blast furnace reduction
D. catalytic cracking
Explanation

One might mistakenly select electrolysis or fractional distillation, which apply to different classes of raw materials. Iron is extracted from hematite in a blast furnace using carbon monoxide as a reducing agent. Common mistake: Confusing the blast furnace reduction method with electrolytic extraction methods.

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