Back to Course

JAMB: CHEMISTRY

0% Complete
0/0 Steps
  1. Free Practice

    JAMB Chemistry Free Exam Questions
    1 Quiz
  2. Syllabus Practice
    Part A — Experimental & Quantitative Chemistry
    2 Topics
  3. Part B — States of Matter, Gas Laws & Solutions
    2 Topics
  4. Part C — Atomic Structure & Periodicity
    2 Topics
  5. Part D — Air, Water & Environment
    2 Topics
  6. Part E — Acids, Bases & Electrochemistry
    2 Topics
  7. Part F — Energy, Rates & Equilibrium
    2 Topics
  8. Part G — Elements & Their Compounds
    2 Topics
  9. Part H — Organic Chemistry
    2 Topics
  10. Part I — Chemistry & Industry
    2 Topics
  11. Practice by Topic
    1 Topic
  12. Mock Test
    Full Mock Test
    1 Topic
  13. Past Questions (Year-Based)
    Past Questions
    18 Quizzes
  • excellence
  • Follow

Lesson 12, Topic 1
In Progress

Full Mock Test (40 Questions)

Lesson Progress
0% Complete

Test Guide:

This timed mock test covers the complete JAMB Chemistry syllabus (Parts A–I) with a mix of questions from all topics. It simulates real UTME exam conditions: 40 questions, timed, with instant scoring and explanations after submission.

Because questions are randomly selected from the full syllabus bank, you can retake this mock to practise with new question sets and improve overall exam performance.

For best results, attempt this test like a real JAMB CBT: avoid checking answers during the test, submit before time runs out, then review your wrong questions using the “Review wrong” button.


Chemistry Full Mock Leaderboard
Ranked by highest score, then fastest time
#1
Esther OyedoyinEsther Oyedoyin@Estdoy
Score65/100
Raw26/40
Time17:03
Attempts7
JAMB Chemistry Full Mock
40 questions • 30 mins
The timer starts immediately after you click the button.
JAMB Chemistry Full Mock
Auto-built from Chemistry topic groups • 40 questions
0:00
Answered: 0/40
Score: /100
JAMB: CHEM - Boiling and Melting Points
1

For iodine crystals to sublime on heating, the particles must acquire energy that is

Explanation:

Sublimation is the direct change of a substance from solid to vapour without passing through the liquid state.

For iodine crystals to sublime, the particles must gain enough energy to overcome the intermolecular forces holding them in the solid lattice.

Once these forces are overcome, the particles escape directly into the gaseous state.

Therefore, the particles must acquire energy greater than the forces of attraction in the solid.

Need more practice in JAMB: CHEM - Boiling and Melting Points?
Practice this topic
JAMB: CHEM - Relative Atomic Mass
2

What is the percentage of sulphur in sulphur(IV) oxide? [S = 32, O = 16]

Solution:

relative molecular weight of SO2 = S + ([O2])

Given:

  • S = 32
  • O = 16

SO2 = 32 + 2(16) = 32 + 32

SO2 = 64

Relative percentage of sulphur = \(\Large \frac{32}{64}\times \frac{100}{1}\normalsize = 50\%\)

Since sulphur = 32

% of sulphur = 50%

Need more practice in JAMB: CHEM - Relative Atomic Mass?
Practice this topic
JAMB: CHEM - Mole Concept and Avogadro Number
3

How many atoms are in 6.0 g of magnesium?

[Mg = 24, NA = 6.02 × 1023 mol-1]

Solution:

Number of atom = number of mole × Avogadro’s constant

Number of moles = \(\Large\frac{mass}{molar\;mass}\)

= \(\Large\frac{6}{24} = \normalsize 0.25mol\)

Number of atom = 0.25 × 6.02 × 1023

= 1.51 × 1023

Need more practice in JAMB: CHEM - Mole Concept and Avogadro Number?
Practice this topic
JAMB: CHEM - Chemical Symbols, Formulae and Equations
4

The electronic configuration of element Z is 1s22s22p63s23p1. What is the formula of the compound formed between Z and tetraoxosulphate(VI) ion?

Explanation:  Z = 1s22s23p63s23p1

We have Z3+  and \(\normalsize SO_4^{-2}\)

The reaction is: \(\normalsize Z^{3} \: + \: SO^{2-}_4 \rightarrow Z_{2}\left ( SO_{4} \right )_{3}\)

= Z2(SO4)3

Need more practice in JAMB: CHEM - Chemical Symbols, Formulae and Equations?
Practice this topic
JAMB: CHEM - Gas Laws
5

To what temperature must a gas at 273 K be heated in order to double both its volume and pressure?

Explanation:

Using the combined gas law,

⇒ \(\Large \frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}\).

If P₂ = 2P₁ and V₂ = 2V₁,

then \(\normalsize T_2 = \Large \frac{(2P_1)(2V_1)T_1}{P_1V_1} \\ = \normalsize 4T_1 \\ =\normalsize 4 \times 273 = 1092 \: K\)

Need more practice in JAMB: CHEM - Gas Laws?
Practice this topic
JAMB: CHEM - Solutions and Solubility
6

The solubilities of different solutes in a given solvent can be compared by

Explanation: Plotting solubility curves of different substances on the same axes allows for a visual comparison of their solubility behavior at various temperatures, showing which substance is more soluble at a given temperature.

Need more practice in JAMB: CHEM - Solutions and Solubility?
Practice this topic
JAMB: CHEM - Kinetic Theory of Matter
7

Which of the following is not a basic assumption of the kinetic theory of gases?

Explanation: Collisions between gas molecules are perfectly elastic, not inelastic. This means that there is no loss of kinetic energy during the collisions

Need more practice in JAMB: CHEM - Kinetic Theory of Matter?
Practice this topic
JAMB: CHEM - Solutions and Solubility
8

The solubility in mol dm⁻³ of 20.2 g of potassium trioxonitrate(V) dissolved in 100 cm³ of water at room temperature is:

[K = 39, O = 16, N = 14]

Explanation:

Solutbility = \(\Large \frac{Mass}{Molar\:mass} \: \times \: \frac{1000}{Volume}\)

Molar mass KNO₃ = 39 + 14 + 3(16) = 101 g/mol.

Solubility (mol/dm³) = \(\Large \frac{g}{g/mole} \: \times \: \frac{1000 \: cm^3}{cm^3}\)

= \(\Large \frac{20.2}{101} \: \times \: \frac{1000}{100} \\ = \normalsize 2.00 \: mol/dm^3\)

Need more practice in JAMB: CHEM - Solutions and Solubility?
Practice this topic
JAMB: CHEM - Atomic Structure
9

The arrangement of particles in crystal lattices can be studied using

Explanation:

X-rays are used to study the arrangement of particles (atoms, molecules) within a crystal lattice through a technique called X-ray crystallography, by measuring the angles and intensities of scattered X-rays.

This technique exploits the fact that X-rays have wavelengths comparable to the spacing between atoms in a crystal. When a beam of X-rays interacts with a crystal, the regular arrangement of atoms causes the X-rays to diffract (scatter) in specific directions.

Need more practice in JAMB: CHEM - Atomic Structure?
Practice this topic
JAMB: CHEM - Periodic Properties
10

Which of the following physical properties decrease across the periodic table?

Explanation:

Going across the period, electrons are added to the same outer shell with a corresponding increase in the number of the positively charged protons in the nucleus, which produces an increased effective nuclear charge. The increased nuclear charge means a stronger pull exerted by the nucleus on the electrons of the outer shell, hence the contraction of the atoms, moving across the period.

Need more practice in JAMB: CHEM - Periodic Properties?
Practice this topic
JAMB: CHEM - Electron Configuration
11

An element, E with the electronic configuration of 1s2 2s2 2p63s2 3p3, the reaction of E with a halogen Y can give

Explanation:

The electronic configuration 1s2 2s2 2p6 3s2 3p3 corresponds to atomic number 15, which is phosphorus.

Phosphorus has five valence electrons and commonly exhibits valencies of 3 and 5.

Therefore, when phosphorus reacts with a halogen Y, it can form the halides EY3 and EY5.

Need more practice in JAMB: CHEM - Electron Configuration?
Practice this topic
JAMB: CHEM - Chemical Bonding
12

What type of bond exists between an element X with atomic number 12 and Y with atomic number 17?

Explanation: The bond between magnesium and chlorine in magnesium chloride (MgCl2) is ionic (electrovalent), formed through the transfer of electrons from magnesium to chlorine atoms.

Screenshot 2025 03 12 at 20.43.12

Need more practice in JAMB: CHEM - Chemical Bonding?
Practice this topic
JAMB: CHEM - Hardness of Water
13

200 cm3 of 0.50 mol/dm3 solution of calcium hydrogen tetraoxocarbonate(IV) is heated. The maximum weight of solid precipitated is

Solution:

Equation: ⇒ \(\normalsize Ca(HCO_3)_2\rightarrow CaCO_3\;{precipiate}+H_2O+CO_2\)

{in the presence of heat}.

Volume = 200 cm3 to dm3 = \(\Large\frac{200}{1000} =\normalsize 0.2\: dm^{3}\)

⇒ \(\normalsize C =\Large\frac{0.5\:mol}{dm^{3}}=\normalsize 0.5\: M\)

Number of moles = CV

= 0.5 × 0.2

= 0.1 mole

From the above equation:

1 mole of Ca(HCO3)2  gives 100 g of CaCO3

0.1 mole of Ca(HCO3)2 will give x g of CaCO3

⇒ \(\Large \frac{1}{0.10} = \frac{100}{x}\)

⇒ \(\normalsize x =\Large\frac{0.1\times 100}{1}\)

= 10 g

Need more practice in JAMB: CHEM - Hardness of Water?
Practice this topic
JAMB: CHEM - Hardness of Water
14

Water is said to be temporarily hard when it contains

Explanation:

Temporary hardness is the hardness of water which can be removed or softened by boiling. Temporary hardness is caused by hydrogen trioxocarbonate (IV) of calcium or magnesium, i.e. Ca(HCO3)2 or Mg(HCO3)2.

Need more practice in JAMB: CHEM - Hardness of Water?
Practice this topic
JAMB: CHEM - Air and Atmosphere
15

Helium is often used in observation balloons because it is:

Explanation: Helium is a noble gas (chemically inert), non‑combustible, and less dense than air, so it provides lift safely.

Need more practice in JAMB: CHEM - Air and Atmosphere?
Practice this topic
JAMB: CHEM - pH and Indicators
16

The colour of phenolphithalein in an alkaline medium is

Explanation: The pink colour of phenolphthalein in an alkaline medium is because it is a weak acid which can lose H+ ions in solution. The phenolphthalein molecule is colourless, and the ion is pink.

Need more practice in JAMB: CHEM - pH and Indicators?
Practice this topic
JAMB: CHEM - Electrolysis
17

What current will deposit 3.25 g of zinc in 2 hrs?

[Zn = 65, F = 96500 mol-1]

Explanation:

At the cathode:

Zn2+ + 2e- → Zn(s)

2F (i.e. 2 × 96500 C) deposits 65 g of Zn.

So, charge needed to deposit 3.25 g of Zn is:

⇒ \(\normalsize Q = \Large \frac{3.25 \times 2 \times 96500}{65} \\ = \normalsize 9650\ \text{C}\)

Time = 2 hours = 2 × 60 × 60 = 7200 s

Using Q = It:

⇒ \(\normalsize I= \Large \frac{Q}{t} \\ = \Large\frac{9650}{7200} \\ = \nromalsize 1.34\ \text{A}\)

Need more practice in JAMB: CHEM - Electrolysis?
Practice this topic
JAMB: CHEM - Electrolysis
18

What quantity of aluminum is deposited when a current of 10 A is passed through a solution of an aluminum salt for 1930 s?

( Al = 27, F = 96 500 C mol-1)

Solution

Al3+(aq)  + 3e-   →  Al(s)

3F                          27g

3F   =   ( 3 × 96500) C   =   289500 C

Q    =  It   =   (10 × 1930) C    =  19300 C

From the  reaction,

289500 C  =       27g   of Al

19300 C  =  \(\Large \frac {19300 \: \times \: 27} {289500}\)

= 1.8 g

Need more practice in JAMB: CHEM - Electrolysis?
Practice this topic
JAMB: CHEM - Oxidation and Reduction
19

What is the oxidation state of chlorine in HClO4?

Explanation:

HClO4 = 0

+1 + Cl + (-2 × 4) = 0

+1 + Cl – 8 = 0

Cl = 8 - 1

Cl = +7

Need more practice in JAMB: CHEM - Oxidation and Reduction?
Practice this topic
JAMB: CHEM - Oxidation and Reduction
20

H2S(g) +  Br2(g) \(\rightarrow\) 2HBr(g)  +  S(s)

In the above reaction, the oxidation agent is

Explanation: Br2 is the oxidizing agent, and H2S is the reducing agent. Oxidation is the loss of electrons, gain of oxygen, or loss of hydrogen.

Need more practice in JAMB: CHEM - Oxidation and Reduction?
Practice this topic
JAMB: CHEM - Salts
21

On exposure to the atmosphere, a hydrated salt loses its water of crystallization to become anhydrous. This phenomenon is referred to as

Explanation:

Efflorescence is the process where a hydrated salt loses its water of crystallization when exposed to the air, often resulting in a powdery appearance.

Examples of Efflorescent salts are:

  • Na2CO3  .  10H2O
  • Na2SO4  . 10H2O
Need more practice in JAMB: CHEM - Salts?
Practice this topic
JAMB: CHEM - Le Chatelier’s Principle
22

A certain industrial process is represented by:

2A(g) → C(g) + 3D(g)  ΔH = +X kJ mol⁻¹

Which of the following conditions will favour the yield of the products?

Explanation:

The reaction produces more moles of gas (2 mol on the left, 4 mol on the right). So decreasing pressure favours the side with more gas molecules (products).

Since ΔH is positive (endothermic), increasing temperature favours the forward reaction. Therefore, increase temperature and decrease pressure will favour the yield of products.

Need more practice in JAMB: CHEM - Le Chatelier’s Principle?
Practice this topic
JAMB: CHEM - Catalysis
23

 

Screenshot 2025 03 13 at 20.12.05

In the graph above, the activation energy of the catalyzed reaction is

Explanation:

Screenshot 2025 03 13 at 20.21.03

Activation energy is the minimum energy required for a reaction to occur.
On an energy profile diagram, it is the vertical distance between the energy of the reactants and the peak of the curve.

From the diagram:
– Energy of reactants ≈ 100 kJ
– Peak of catalyzed curve ≈ 200 kJ

Therefore,
Activation energy (catalyzed reaction) = 200 − 100 = 100 kJ

A catalyst lowers the activation energy of a reaction by providing an alternative pathway with lower energy. Hence, the activation energy of the catalyzed reaction is 100 kJ.

Need more practice in JAMB: CHEM - Catalysis?
Practice this topic
JAMB: CHEM - Entropy and Free Energy
24

\(\scriptsize X_{(g)} \: + \: 3Y_{(g)} \leftrightharpoons 2Z_{(g)} \; \Delta H = +ve\)

If the reaction above takes place at room temperature, the ΔG will be

Explanation:

The reaction is endothermic, so ΔH > 0. Also, the number of gas molecules decreases from 4 moles (reactants) to 2 moles (products), so ΔS < 0.

Therefore, ΔG = ΔH - TΔS becomes: (+) − T(−) = (+) + (+),

so ΔG is positive at room temperature (and in fact at all temperatures).

Need more practice in JAMB: CHEM - Entropy and Free Energy?
Practice this topic
JAMB: CHEM - Halogens
25

When chlorine water is exposed to bright sunlight, the following products are formed.

Explanation: Exposing chlorine water to sunlight causes it to decompose to HCl gas and oxygen. When chlorine water is exposed to strong sunlight, it produces an odourless and colourless gas called oxygen.

Need more practice in JAMB: CHEM - Halogens?
Practice this topic
JAMB: CHEM - Alloys
26

The constituent common to duralumin and alnico is

Explanation:

Constituents of Duralumin are: Al, Cu, Mg, and Mn.

Constituents of Alnico are: Al, Ni, and Co.

Need more practice in JAMB: CHEM - Alloys?
Practice this topic
JAMB: CHEM - Oxygen and Ozone
27

The pollutant that contributes to the depletion of the ozone layer is

Explanation: Chlorofluorocarbons (CFCs) are a major contributor to the depletion of the ozone layer, as they release chlorine atoms into the stratosphere that react with and destroy ozone molecules.

Need more practice in JAMB: CHEM - Oxygen and Ozone?
Practice this topic
JAMB: CHEM - Transition Metals
28

The coloured nature of transition metal ions are associated with their partially filled

Explanation: Transition metal ions with partially filled d orbitals exhibit characteristic properties like forming colored compounds and acting as catalysts due to the ease with which electrons can be removed or added to these orbitals.

Need more practice in JAMB: CHEM - Transition Metals?
Practice this topic
JAMB: CHEM - Nitrogen and its Compounds
29

Which of the following methods may be used to prepare tritoxonitrate(V) acid (nitric acid) in the laboratory?

Explanation:

NaNO3 + H2SO4 → NAHSO4 + HNO3

Need more practice in JAMB: CHEM - Nitrogen and its Compounds?
Practice this topic
JAMB: CHEM - General Properties of Metals
30

The metal whose ore can be concentrated by passing it through a magnetic separator is

Explanation:

Magnetic separation: This method involves passing magnetic ores, e.g. copper ore, through a magnetic separator.

magnetic separation

The ore is finely crushed and passed over the magnetic roller, where one is magnetic, and the other is nonmagnetic. Ores with a high metal content will attract and attach to the magnetic roller and get deflected into a pile, and the non-magnetic gangue particles will repel and fall into the pile from the conveyor belt.

Need more practice in JAMB: CHEM - General Properties of Metals?
Practice this topic
JAMB: CHEM - Aluminium
31

Aluminum is extracted commercially from its ore by

Explanation:

Alumina, dissolved in molten cryolite, is electrolyzed to obtain pure aluminium.

Need more practice in JAMB: CHEM - Aluminium?
Practice this topic
JAMB: CHEM - Carbon and its Compounds
32

When steam is passed through red hot coke, which of the following is produced?

Explanation:

Water gas, a mixture of hydrogen and carbon(II) oxide is prepared by passing steam over red hot coke. Water gas is a gaseous fuel produced from coke.

Need more practice in JAMB: CHEM - Carbon and its Compounds?
Practice this topic
JAMB: CHEM - Sulphur and its Compounds
33

Sulphur exists in several forms in the solid state. This property is known as:

Explanation: Allotropy is the existence of an element in more than one form in the same physical state (e.g., rhombic and monoclinic sulphur).

Need more practice in JAMB: CHEM - Sulphur and its Compounds?
Practice this topic
JAMB: CHEM - Nitrogen and its Compounds
34

The drying agent suitable for drying ammonia is

Explanation: The suitable drying agent for ammonia (NH3) is calcium oxide (CaO), also known as quicklime.

Need more practice in JAMB: CHEM - Nitrogen and its Compounds?
Practice this topic
JAMB: CHEM - Alkanes
35

The IUPAC name of the compound CF3CHBrCl is

Explanation: The parent chain is ethane (2 carbons). Number the chain to give the lowest set of locants: placing the CF3 group at carbon 1 gives 1,1,1-trifluoro and both halogens (Br and Cl) at carbon 2. Substituents are listed alphabetically, so bromo comes before chloro. Therefore the name is 2-bromo-2-chloro-1,1,1-trifluoroethane.

Need more practice in JAMB: CHEM - Alkanes?
Practice this topic
JAMB: CHEM - Aromatic Hydrocarbons
36

Ethyne is passed through a hot tube containing organo-nickel catalyst to produce

Explanation: Benzene is prepared from ethyne by the process of cyclic polymerization. In this process, ethyne is passed through a red-hot iron tube at 873 K. The ethyne molecules then undergo cyclic polymerization to form benzene.

Need more practice in JAMB: CHEM - Aromatic Hydrocarbons?
Practice this topic
JAMB: CHEM - Carbohydrates
37

Which of the following is not correct of glucose and fructose? They

Explanation:

Hydrolysis of starch yields 2 molecules of glucose.

While the hydrolysis of starch primarily yields glucose, fructose can be produced from starch through enzymatic isomerization of glucose, which is a subsequent step in the process of creating high-fructose syrups.

Need more practice in JAMB: CHEM - Carbohydrates?
Practice this topic
JAMB: CHEM - Alkenes
38

Which of these is an unsaturated hydrocarbon?

Explanation:

An unsaturated hydrocarbon contains at least one carbon–carbon double or triple bond.

Propene (C3H6) contains a carbon–carbon double bond and is therefore unsaturated.

Alkanes such as pentane contain only single bonds and are saturated.

Need more practice in JAMB: CHEM - Alkenes?
Practice this topic
JAMB: CHEM - Alkanes
39

Which of the following statements is true about 2-methylpropane and butane

Explanation: Both butane and 2-methylpropane are members of the same homologous series, specifically the alkane series (CnH2n + 2), and they are structural isomers with the same molecular formula (C4H10) but different structural arrangements.

Need more practice in JAMB: CHEM - Alkanes?
Practice this topic
JAMB: CHEM - Aromatic Hydrocarbons
40

The derivative of benzene that can be used in making explosives is

Explanation: Toluene is found naturally in crude oil and is used in oil refining and the manufacturing of paints, lacquers, explosives (TNT) and glues.

Need more practice in JAMB: CHEM - Aromatic Hydrocarbons?
Practice this topic

Like this content

Subscribe
Notify of
guest
0 Comments
Oldest
Newest
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x
×
×