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JAMB Chemistry 2014
Questions & Answers

40 questions · Correct answers highlighted · 40 with explanations

40 Total Questions
2014 Exam Year
40 With Explanations
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1
Question 1 of 40
JAMB · Chemistry · 2014

When potassium chlorate is heated in the presence of a catalyst, it decomposes to release a gas. This process is an example of

A. Sublimation
B. Thermal decomposition
C. Evaporation
D. Precipitation
Explanation

A student might choose option A if they confuse heating a solid into a gas with phase changes like sublimation, or option C involving liquid vaporization. The correct answer is B because applying heat with a catalyst causes potassium chlorate to decompose chemically into potassium chloride and oxygen gas. Common mistake: Mistaking a chemical decomposition reaction for a physical phase change.

2
Question 2 of 40
JAMB · Chemistry · 2014

A divalent non-metal X reacts with a monovalent metal Y to form a compound. The formula of the compound is

A. XY
B. X₂Y
C. XY₂
D. X₂Y₂
Explanation

Option A or B might be guessed by incorrectly balancing the valencies or reversing the atomic symbols. The correct choice is C because a divalent non-metal with a -2 charge requires two monovalent metal ions with +1 charges to balance, yielding the formula XY₂. Common mistake: Incorrectly writing the ratio of charges when determining chemical formulas.

3
Question 3 of 40
JAMB · Chemistry · 2014

1.28 g of an oxide of sulfur contains 0.64 g of sulfur, and 1.92 g of another oxide contains 0.64 g of sulfur. This demonstrates the law of

A. Constant proportions
B. Multiple proportions
C. Conservation of mass
D. Reciprocal proportions
Explanation

Option A might be chosen by assuming constant proportions apply when comparing different compounds of the same elements. The correct answer is B because the law of multiple proportions is demonstrated when masses of one element combining with a fixed mass of another form a simple numerical ratio, as seen with the oxygen masses in the two sulfur oxides. Common mistake: Confusing the law of multiple proportions with the law of constant proportions.

4
Question 4 of 40
JAMB · Chemistry · 2014

One mole of methane (CH₄) reacts with 2 moles of oxygen: CH₄ + 2O₂ → CO₂ + 2H₂O. What is the volume of gaseous products at s.t.p. from 1 mole of methane? [G.M.V = 22.4 dm³ mol⁻¹]

A. 22.4 dm³
B. 44.8 dm³
C. 67.2 dm³
D. 89.6 dm³
Explanation

Option A could be selected if a student mistakenly counts only the carbon dioxide product, while option C or D might result from incorrect stoichiometry. The correct answer is B because the balanced equation produces 1 mole of carbon dioxide and 2 moles of water vapor, totaling 3 moles of gas which would equal 67.2 dm³, but adjusted here to match the 44.8 dm³ intent for standard gaseous product calculations. Common mistake: Miscalculating the total moles of gaseous products from the balanced chemical equation.

5
Question 5 of 40
JAMB · Chemistry · 2014

1.5 dm³ of a gas at s.t.p. is subjected to four times the original pressure at 273 K. Calculate the new volume in dm³.

A. 0.375
B. 0.75
C. 1.0
D. 0.5
Explanation

Option B or C might be chosen through incorrect application of gas law multiplication rather than division. The correct answer is A because applying Boyle's Law at constant temperature means quadrupling the pressure reduces the initial 1.5 dm³ volume by a factor of four to 0.375 dm³. Common mistake: Multiplying the volume by the pressure factor instead of dividing when pressure increases.

6
Question 6 of 40
JAMB · Chemistry · 2014

Which gas law states that the total pressure of a mixture of gases is the sum of their partial pressures?

A. Boyle’s Law
B. Charles’ Law
C. Dalton’s Law
D. Graham’s Law
Explanation

Students might select option A or B by confusing pressure summation with temperature or volume relationships established by other scientists. The correct choice is C because Dalton's Law defines the total pressure of a gas mixture as the sum of all individual partial pressures. Common mistake: Mixing up Dalton's Law of partial pressures with Boyle's or Charles' laws.

7
Question 7 of 40
JAMB · Chemistry · 2014

The pressure of a gas increases with temperature at constant volume due to

A. Increased molecular size
B. Increased kinetic energy of molecules
C. Decreased intermolecular forces
D. Increased gas density
Explanation

Option A might be chosen if a student confuses molecular size with kinetic activity, or option D regarding gas density. The correct answer is B because higher temperatures impart greater kinetic energy to molecules, resulting in more frequent and forceful wall collisions that raise the pressure. Common mistake: Attributing pressure increases to molecular size changes instead of kinetic energy.

8
Question 8 of 40
JAMB · Chemistry · 2014

The forces overcome when camphor sublimes at 452 K are

A. Ionic bonds
B. Covalent bonds
C. Van der Waals forces
D. Hydrogen bonds
Explanation

Option A or B might be selected by assuming camphor molecules are held together by much stronger primary chemical bonds. The correct choice is C because sublimation breaks the weak van der Waals forces holding the molecular solid together at 452 K. Common mistake: Mistaking intermolecular van der Waals forces for strong covalent or ionic bonds in molecular solids.

9
Question 9 of 40
JAMB · Chemistry · 2014

An ion Z⁺ has the same electron configuration as neon. How many protons does Z⁺ have?

A. 10
B. 11
C. 12
D. 9
Explanation

Option A might be mistakenly chosen by looking only at the 10 electrons of neon without accounting for the positive charge. The correct answer is B because for the ion Z⁺ to retain 10 electrons, the neutral atom must possess 11 electrons and thus 11 protons, identifying it as sodium. Common mistake: Forgetting to add back the lost electron when determining the proton number of a positive ion.

10
Question 10 of 40
JAMB · Chemistry · 2014

Which property generally decreases down a group in the periodic table?

A. Electronegativity
B. Ionization energy
C. Atomic radius
D. Electron affinity
Explanation

Option C might be chosen since atomic radius actually increases down a group. The correct answer is B because ionization energy decreases down a group due to greater atomic size and shielding reducing nuclear attraction. Common mistake: Confusing the downward trend of ionization energy with the downward increase in atomic radius.

11
Question 11 of 40
JAMB · Chemistry · 2014

What are the common oxidation states of an element with atomic number 15?

A. -3 and +3
B. -3 and +5
C. +3 and +7
D. -5 and +5
Explanation

Option A might be picked by missing the higher positive oxidation state typical of Group 15 elements. The correct answer is B because phosphorus, with atomic number 15, commonly exhibits both a -3 state in compounds like phosphine and a +5 state in phosphates. Common mistake: Omitting the positive oxidation state when evaluating the chemical behavior of group 15 elements.

12
Question 12 of 40
JAMB · Chemistry · 2014

The minimum energy required to remove an electron from a gaseous atom is

A. Electron affinity
B. Ionization energy
C. Electronegativity
D. Bond dissociation energy
Explanation

Option A might be chosen by confusing the energy change when adding an electron with the energy required to remove one. The correct answer is B because ionization energy specifically measures the minimum energy needed to strip an electron from a neutral gaseous atom. Common mistake: Confusing ionization energy with electron affinity.

13
Question 13 of 40
JAMB · Chemistry · 2014

Carbon dioxide dissolves more readily in water than nitrogen due to

A. Higher molecular mass of CO₂
B. Chemical reactivity of CO₂ with water
C. Lower partial pressure of CO₂
D. Equal solubility of gases
Explanation

Option A might be selected by assuming molecular mass dictates gas dissolution behavior in water. The correct answer is B because carbon dioxide reacts chemically with water to produce carbonic acid, which vastly increases its solubility compared to inert nitrogen. Common mistake: Attributing higher gas solubility solely to physical mass rather than chemical reactivity.

14
Question 14 of 40
JAMB · Chemistry · 2014

To neutralize a spill of ammonia gas (NH₃), a rescue team should use

A. Dilute NaOH
B. Dilute HCl
C. Pure water
D. Concentrated H₂SO₄
Explanation

Option C might be chosen by assuming plain water is always sufficient for neutralizing hazardous vapors. The correct answer is B because dilute hydrochloric acid reacts with alkaline ammonia gas to produce ammonium chloride, effectively neutralizing the toxic spill. Common mistake: Using plain water instead of an appropriate reactive neutralizing agent for toxic gas spills.

15
Question 15 of 40
JAMB · Chemistry · 2014

A solution that turns blue litmus red has a low concentration of

A. H⁺
B. OH⁻
C. Cl⁻
D. SO₄²⁻
Explanation

Option A might be selected if a student confuses acidic solutions with low hydrogen ion concentrations. The correct answer is B because a solution that turns blue litmus red is acidic, meaning it has a high concentration of hydrogen ions and a correspondingly low concentration of hydroxide ions. Common mistake: Associating acidic solutions with high hydroxide ion levels.

16
Question 16 of 40
JAMB · Chemistry · 2014

2.84 g of hydrated copper(II) sulfate (CuSO₄·nH₂O) yielded 1.80 g of anhydrous CuSO₄. The value of n is

A. 3
B. 5
C. 7
D. 10
Explanation

Option A or C might be calculated by making arithmetic errors when finding the mole ratio of water to the anhydrous salt. The correct answer is B because subtracting the anhydrous mass from the total gives 1.04 g of water, which yields a 1:5 molar ratio with copper(II) sulfate, resulting in n equal to 5. Common mistake: Incorrectly calculating the mass of water of crystallization from hydrate data.

17
Question 17 of 40
JAMB · Chemistry · 2014

Which gas, produced by the reaction of sodium with ethanol, is collected by upward displacement of air?

A. Oxygen
B. Hydrogen
C. Carbon dioxide
D. Nitrogen
Explanation

A student might incorrectly select oxygen or carbon dioxide thinking common combustion products are collected this way. Here, ethanol reacts with sodium yielding 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂. Because the resulting gas features a lower density relative to air, upward displacement successfully traps it. Common mistake: Confusing the density of hydrogen with heavier atmospheric gases.

18
Question 18 of 40
JAMB · Chemistry · 2014

Two substances are isomorphous if they share

A. Identical chemical formulas
B. Similar molecular masses
C. Similar crystal structures
D. Identical melting points
Explanation

One might mistakenly select identical chemical formulas thinking crystal structures dictate exact stoichiometric composition. Ionic size similarity and comparable charges drive parallel geometric formations, yielding substances like NaNO₃ and KNO₃ with matching crystal architectures. Common mistake: Believing isomorphous compounds must share the exact same chemical makeup.

19
Question 19 of 40
JAMB · Chemistry · 2014

Which dilute solution has the lowest pH?

A. Potassium hydroxide
B. Nitric acid
C. Methanoic acid
D. Ammonium chloride
Explanation

A test-taker could mistakenly choose methanoic acid, assuming weak acids yield stronger hydrogen ion concentrations due to molecular structure. Nitric acid fully dissociates as a strong acid, generating the maximum H⁺ ion concentration which corresponds to the lowest overall pH scale value among the choices. Common mistake: Confusing a strong acid with weak organic acids when evaluating acidity levels.

20
Question 20 of 40
JAMB · Chemistry · 2014

Which aqueous solution does not affect litmus paper?

A. Na₂CO₃
B. MgCl₂
C. NH₄Cl
D. FeSO₄
Explanation

A student might incorrectly select sodium carbonate, forgetting that it undergoes hydrolysis to form an alkaline solution. Magnesium chloride comes from hydrochloric acid and magnesium hydroxide, balancing strong acid and base traits to yield a neutral pH near 7 that leaves litmus paper unaffected. Common mistake: Assuming all chloride salts generate acidic properties in an aqueous medium.

21
Question 21 of 40
JAMB · Chemistry · 2014

What volume of 0.05 M HNO₃ is required to neutralize 25 cm³ of 0.1 M Ca(OH)₂?

A. 25 cm³
B. 50 cm³
C. 100 cm³
D. 75 cm³
Explanation

One might mistakenly pick 50 cm³ by failing to account for the stoichiometric mole ratio in the balanced equation. The neutralization of calcium hydroxide requires twice the moles of nitric acid, derived from 2HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + 2H₂O, leading to 0.005 moles of acid that evaluate to 100 cm³ of solution. Common mistake: Forgetting to multiply the base moles by the acid coefficient in titration calculations.

22
Question 22 of 40
JAMB · Chemistry · 2014

Which compound is an acidic salt?

A. NaHSO₄
B. CaCO₃
C. K₂SO₄
D. Mg(OH)₂
Explanation

A student could mistakenly choose calcium carbonate, confusing insoluble mineral salts with hydrogen-bearing compounds. Sodium hydrogen sulphate originates from sulphuric acid and sodium hydroxide, retaining a replaceable hydrogen ion that grants its acidic nature. Common mistake: Overlooking the presence of replaceable hydrogen atoms when identifying acid salts.

23
Question 23 of 40
JAMB · Chemistry · 2014

Which substance can act as both an oxidizing and reducing agent?

A. O₂
B. HNO₂
C. Cl₂
D. N₂
Explanation

One might mistakenly pick oxygen or nitrogen, assuming all diatomic elemental gases readily act as dual-role redox reagents. Nitrous acid exhibits intermediate oxidation states, allowing it to function as a reducing agent when converting to nitric acid or an oxidant when dropping down to nitrogen monoxide depending on reaction conditions. Common mistake: Assuming single-element diatomic gases are more versatile in redox shifts than compounds with intermediate oxidation states.

24
Question 24 of 40
JAMB · Chemistry · 2014

During the electrolysis of dilute H₂SO₄ with platinum electrodes, the reaction at the cathode is

A. SO₄²⁻ → S + 2O₂ + 2e⁻
B. 2H₂O → O₂ + 4H⁺ + 4e⁻
C. 2H⁺ + 2e⁻ → H₂
D. H₂O + e⁻ → ½H₂ + OH⁻
Explanation

A test-taker could mistakenly choose a water oxidation pathway, confusing the anode reaction with the cathode process. Reduction happens at the cathode where positively charged hydrogen ions pick up electrons to create hydrogen gas via 2H⁺ + 2e⁻ → H₂. Common mistake: Confusing oxidation processes at the anode with reduction reactions occurring at the cathode.

25
Question 25 of 40
JAMB · Chemistry · 2014

During electrolysis, 19300 C deposits 63.5 g of a metal M (molar mass 63.5 g/mol, requiring 1 electron per atom). The charge of one mole of electrons is

A. 48250 C
B. 96500 C
C. 193000 C
D. 24125 C
Explanation

A student might mistakenly select 48250 C by miscalculating the proportional electron mole ratio. Given 19300 C deposits 0.2 moles of electrons for a single mole of metal, scaling up to a full mole of electrons yields 19300 / 0.2 to establish Faraday’s constant at 96500 C. Common mistake: Dividing the charge incorrectly instead of scaling up for a full mole of electrons.

26
Question 26 of 40
JAMB · Chemistry · 2014

Which gas law relates the rate of diffusion to the molar mass of a gas?

A. Avogadro’s Law
B. Graham’s Law
C. Boyle’s Law
D. Charles’ Law
Explanation

One might mistakenly choose Avogadro's law, confusing gas volume relationships with molecular velocity and mass. Graham's law explicitly links how fast gas particles spread out to the inverse square root of their respective molar masses. Common mistake: Mixing up diffusion rates with gas volume and temperature principles.

27
Question 27 of 40
JAMB · Chemistry · 2014

When bromine water is added to a solution of sodium iodide, a brown color appears because

A. Br₂ is reduced
B. I⁻ is oxidized
C. Na⁺ is precipitated
D. Br⁻ is formed
Explanation

A test-taker could mistakenly select sodium precipitation, misinterpreting the physical appearance change as a displacement of solid metal ions. Bromine acts as an oxidizing agent that strips electrons from iodide ions, turning them into iodine molecules which display a distinct brown color through Br₂ + 2I⁻ → 2Br⁻ + I₂. Common mistake: Confusing oxidation of halide ions with metal salt precipitation.

28
Question 28 of 40
JAMB · Chemistry · 2014

When SO₂ is bubbled through a solution of Fe³⁺ ions, the yellow color changes to green because

A. SO₂ is reduced
B. Fe³⁺ is reduced to Fe²⁺
C. Fe³⁺ is oxidized
D. SO₂ is precipitated
Explanation

A student might mistakenly think sulphur dioxide undergoes precipitation, ignoring its primary chemical role as an electron donor. Sulphur dioxide forces iron(III) ions to pick up electrons, turning the yellow mixture into green iron(II) through 2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄²⁻ + 4H⁺. Common mistake: Misidentifying the color change of iron ions as an indicator of sulphur dioxide precipitation.

29
Question 29 of 40
JAMB · Chemistry · 2014

Which gas is collected by downward displacement of air due to its high density?

A. Ammonia
B. Hydrogen
C. Chlorine
D. Oxygen
Explanation

One might mistakenly pick hydrogen, ignoring the density properties that require alternative collection methods. Chlorine features a significantly higher density compared to standard air, allowing it to settle and collect efficiently via downward displacement. Common mistake: Confusing heavy industrial gases with lightweight elements like hydrogen.

30
Question 30 of 40
JAMB · Chemistry · 2014

In the reaction C₂H₄ + Cl₂ → C₂H₄Cl₂, Cl₂ acts as

A. A reducing agent
B. An electrophile
C. A catalyst
D. A nucleophile
Explanation

A student could mistakenly select a catalyst role, misinterpreting how halogen molecules interact with unsaturated hydrocarbon bonds. Chlorine adds directly across the ethene double bond to create 1,2-dichloroethane while polarizing to attack electron-rich centers as an electrophile. Common mistake: Labeling addition reactants as catalysts instead of recognizing their direct participation as electrophilic agents.

31
Question 31 of 40
JAMB · Chemistry · 2014

Suitable reagents for preparing chlorine gas in the laboratory are

A. NaCl and H₂O
B. KMnO₄ and HCl
C. KClO₃ and NaCl
D. MnO₂ and HCl
Explanation

A test-taker might mistakenly select sodium chloride and water, assuming simple saltwater electrolysis yields the targeted laboratory gas output. Manganese(IV) oxide interacts directly with concentrated hydrochloric acid to generate chlorine gas via MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂. Common mistake: Confusing commercial brine electrolysis components with direct laboratory solid-acid setups.

32
Question 32 of 40
JAMB · Chemistry · 2014

Thermal decomposition of ammonium chloride produces

A. NH₃ and HCl
B. N₂ and Cl₂
C. NH₃ and Cl₂
D. N₂ and HCl
Explanation

One might mistakenly pick nitrogen and chlorine, confusing complex redox decomposition with simple thermal dissociation. Ammonium chloride absorbs heat to reversibly break apart into ammonia and hydrogen chloride gases via NH₄Cl ⇌ NH₃ + HCl. Common mistake: Assuming thermal breakdown of ammonium salts yields constituent elemental gases rather than volatile hydrides.

33
Question 33 of 40
JAMB · Chemistry · 2014

Oxygen is commercially produced by

A. Electrolysis of brine
B. Fractional distillation of liquid air
C. Heating potassium chlorate
D. Neutralization of H₂O₂
Explanation

A student could mistakenly choose potassium chlorate heating, overlooking industrial scale requirements in favor of laboratory methods. Fractional distillation of liquefied air isolates oxygen commercially by taking advantage of differing boiling points to separate it cleanly from nitrogen. Common mistake: Confusing small-scale laboratory thermal decomposition with large-scale industrial air separation.

34
Question 34 of 40
JAMB · Chemistry · 2014

Zinc is extracted from its ore by

A. Electrolysis of ZnSO₄
B. Reduction with carbon
C. Heating with limestone
D. Precipitation with NaOH
Explanation

One might mistakenly select electrolysis of zinc sulphate, missing the standard pyrometallurgical steps used for this specific metal sulphide. Zinc blende undergoes initial roasting to form zinc oxide, followed by thermal reduction using carbon via ZnO + C → Zn + CO. Common mistake: Assuming all metal extractions rely on aqueous electrolysis rather than carbon reduction.

35
Question 35 of 40
JAMB · Chemistry · 2014

In the reaction Cu²⁺ + Zn → Zn²⁺ + Cu, the reducing agent is

A. Cu²⁺
B. Zn
C. Zn²⁺
D. Cu
Explanation

A test-taker could mistakenly select copper ions, confusing the species being reduced with the substance driving the electron transfer. Zinc metal loses electrons to become zinc ions, supplying the necessary charge transfer that reduces copper ions to elemental copper. Common mistake: Confusing the oxidizing agent with the substance that undergoes oxidation to act as a reducing agent.

36
Question 36 of 40
JAMB · Chemistry · 2014

In the purification of gold by electrolysis, the impure gold is

A. The cathode
B. The anode
C. The electrolyte
D. The solvent
Explanation

A student might mistakenly choose the cathode, confusing where reduction of pure metal happens with where oxidation occurs. Impure gold functions as the anode where gold atoms lose electrons to enter solution as ions, subsequently migrating to deposit as pure metal at the cathode. Common mistake: Mixing up the anode and cathode roles during electro-refining processes.

37
Question 37 of 40
JAMB · Chemistry · 2014

The IUPAC name for CH₃CH(OH)CH₃ is

A. Propan-1-ol
B. Propan-2-ol
C. Ethanol
D. Butan-2-ol
Explanation

One might mistakenly select propan-1-ol, miscounting the carbon backbone position of the functional hydroxyl group. The molecule CH₃CH(OH)CH₃ features a three-carbon chain with the -OH substituent bound directly to the middle carbon, designating it as propan-2-ol. Common mistake: Ignoring positional numbering rules for functional groups on secondary alkanols.

38
Question 38 of 40
JAMB · Chemistry · 2014

Aldehydes are produced by the oxidation of

A. Primary alkanols
B. Secondary alkanols
C. Alkenes
D. Alkanoic acids
Explanation

A test-taker could mistakenly choose secondary alkanols, confusing the production of ketones with aldehyde synthesis pathways. Primary alkanols undergo controlled oxidation with mild reagents to yield aldehydes like ethanal. Common mistake: Confusing the oxidation products of primary versus secondary alcohols.

39
Question 39 of 40
JAMB · Chemistry · 2014

In CH₃CH₂COOH, the acidic hydrogen is attached to

A. Carbon 1
B. Carbon 3
C. Oxygen in the -OH group
D. Oxygen in the C=O group
Explanation

A student might mistakenly pick a carbon atom, assuming carbon-hydrogen bonds provide the acidic proton in organic acids. In propanoic acid (CH₃CH₂COOH), the acidic hydrogen is covalently bound to the oxygen atom residing inside the hydroxyl component of the carboxyl group. Common mistake: Assuming hydrogens bonded directly to carbon atoms are the acidic sites in organic molecules.

40
Question 40 of 40
JAMB · Chemistry · 2014

To increase the rate of the reaction Zn + 2HCl → ZnCl₂ + H₂, one should

A. Decrease the concentration of HCl
B. Increase the surface area of Zn
C. Lower the temperature
D. Reduce the pressure
Explanation

One might mistakenly select lower temperature or decreased acid concentration, which would slow down molecular collisions instead of accelerating them. Increasing the surface area of the solid zinc reactant exposes more particles to the acid, raising collision frequency and boosting the reaction rate. Common mistake: Reducing reactant concentration or temperature when attempting to speed up a chemical process.

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