JAMB Chemistry 2022
Questions & Answers

Common mistake: Students might invert the ratio or miscalculate by failing to set up the formula correctly. According to Charles’ law, which states that V divided by T is constant at constant pressure (V₁/T₁ = V₂/T₂), the ratio of the final absolute temperature to the initial absolute temperature is found by dividing the final volume by the initial volume, giving 975 over 32.5, which simplifies to a 30:1 ratio.

Common mistake: Test-takers might try to calculate complex molar masses or use incorrect gas laws, assuming different gases require different math. Avogadro’s law solves this by establishing that equal volumes of gases at identical temperatures and pressures contain the exact same number of moles, meaning cylinder B holds the same 5.0 moles of nitrogen as cylinder A holds of oxygen.

Common mistake: A student might mistakenly think boiling depends on reaching a specific temperature like 100°C under all conditions or merely on volume changes. A liquid actually begins to boil when its vapour pressure equals the surrounding atmospheric pressure, which allows vapor bubbles to freely form within the liquid.

Common mistake: Learners might forget to account for the electrical charge when the number of electrons differs from protons. With 8 protons, the element is oxygen, and its mass number is calculated by adding protons and neutrons (8 + 9 = 17); because there are 7 electrons, which is one less than the proton count, it carries a +1 charge and is written as ¹⁷O⁺.

Common mistake: Students might misread the periodic table coordinates and select transition metals or metalloids. Alkali metals belong to Group 1, represented here by letter G, while noble gases are located in Group 18, represented by letter L.

Common mistake: Examinees might choose alkali metals or noble gases, forgetting how bonding works. Covalent compounds form exclusively between non-metals, making R—likely carbon in Group 14—the correct choice, whereas alkali metals, typical metals, and noble gases do not fit this bonding category.

Common mistake: Students might confuse gaseous non-metals or metallic elements with solid non-metals. R represents a non-metal that remains solid at room temperature, such as carbon from Group 14.

Common mistake: Test-takers frequently confuse Millikan's oil drop experiment with J.J. Thomson's work and mistakenly choose the charge-to-mass ratio or the mass of the electron. Millikan specifically determined the fundamental charge of the electron by balancing gravitational and electric forces on charged oil droplets.

Common mistake: Learners might guess electron pair sharing or ionization potentials by confusing ionic bonding with covalent bonding. The stability of ionic solids is primarily due to lattice energy, which is the energy released when gaseous ions come together to form a stable solid crystal lattice.

Distractor check: A student might choose options A, B, or C because they describe true characteristics of isotopes, leading them to miss that D is factually incorrect. Reasoning to the answer: Isotopes share an identical count of protons and possess matching numbers of electrons in neutral atoms, yet they differ in their neutron counts and atomic masses. This makes option D false since isotopes do not have different numbers of electrons. Common mistake: Forgetting that neutral atoms of isotopes maintain the same electron configuration despite varying neutron numbers.

Distractor check: A student could mistakenly select option A or C, thinking that combustion properties matter less than lightness, or option D because they confuse the density of gases. Reasoning to the answer: Hydrogen is dangerously flammable, which prompted the preference for a safer alternative. Helium provides the necessary buoyancy because it is lighter than air, and it remains non-combustible, ensuring safety for observation balloons. Common mistake: Confusing helium with hydrogen regarding their flammability hazards in lighter-than-air applications.

Distractor check: A student might lean toward option A, B, or D if they mistakenly assume combustion of chlorinated compounds breaks down into pure elemental gases or unrelated hydrocarbon fragments. Reasoning to the answer: When burning chloromethane (CH₃Cl), the presence of chlorine within its chemical structure forces the release of hydrogen chloride (HCl) gas into the open atmosphere. Common mistake: Overlooking the chlorine atom's role in the molecular formula which directly dictates the generation of hydrogen chloride upon burning.

Distractor check: A student might select efflorescent (A) or anhydrous (B) due to confusion regarding terms describing water interactions in chemistry. Reasoning to the answer: Deliquescent substances take up moisture from the atmosphere and ultimately dissolve in it. This specific behavior is shared with hygroscopic substances, although deliquescence represents a more extreme version of moisture absorption. Common mistake: Mixing up deliquescent compounds with efflorescent ones, which lose water of crystallization to the air.

Distractor check: A student might pick options A, B, or C because they misattribute filtration traits or light-scattering behavior to suspensions or colloids interchangeably. Reasoning to the answer: Suspensions tend to settle out over time and may fail to scatter light effectively, whereas colloids successfully scatter light via the Tyndall effect and remain suspended without settling. Common mistake: Assuming that colloidal particles will eventually settle out of solution just like suspension particles do.

Distractor check: A student could choose options A, C, or D by incorrectly assuming that increased kinetic energy from heat causes dissociation rather than focusing on the thermodynamic heat requirement of the process. Reasoning to the answer: Raising the temperature provides the necessary energy to break solute bonds because the dissolution of most solids in water is endothermic, thereby favoring solubility under heated conditions. Common mistake: Forgetting that endothermic dissolution processes absorb heat, making high temperatures favorable for dissolving more solute.

Distractor check: A student might pick A, B, or D by confusing salts derived from strong acids and strong bases with those undergoing specific hydrolysis. Reasoning to the answer: Ammonium chloride (NH₄Cl) is produced from a weak base (NH₃) and a strong acid (HCl). The resulting solution becomes acidic, yielding a pH less than 7 because of the active hydrolysis of the NH₄⁺ ion. Common mistake: Assuming all chloride salts form neutral solutions regardless of their cationic parent base.

Distractor check: A student might select 3.0 (A) or 5.0 (B) if they mistake the concentration directly for pH or miscalculate the logarithmic conversion. Reasoning to the answer: Sodium hydroxide is a strong base, meaning the concentration of hydroxide ions equals the molarity given, [OH⁻] = 2.0 × 10⁻³ M. Calculating the pOH yields -log(2.0 × 10⁻³) ≈ 2.7, and subtracting this from 14 gives a pH of 11.3. Common mistake: Forgetting to subtract the calculated pOH value from 14 to find the final pH of a basic solution.

Distractor check: A student might incorrectly choose the anode (A) by confusing electroplating deposition sites with oxidation sites. Reasoning to the answer: During the silver-plating procedure, metal M functions as the cathode. At this specific electrode, silver ions (Ag⁺) undergo reduction to deposit solid silver onto the metal surface. Common mistake: Mixing up the anode and cathode roles in electroplating, placing deposition at the wrong electrode.

Distractor check: A student might choose 1.0 (B) or 3.0 (D) by misinterpreting the stoichiometric ratio of electrons required per copper ion. Reasoning to the answer: The reduction half-reaction is Cu²⁺ + 2e⁻ → Cu, which shows that 2 Faradays (2F) are needed to deposit 1 mole of copper. Passing 3F of electricity therefore deposits 3 divided by 2, resulting in 1.5 moles. Common mistake: Failing to account for the divalent charge of the copper ion (Cu²⁺) requiring two electrons per deposited atom.

Distractor check: A student might select A or B by focusing solely on either the iron or the acid component rather than recognizing the full scope of electron transfer. Reasoning to the answer: Solid iron is oxidized to iron(II) while hydrogen ions are simultaneously reduced to hydrogen gas, which means the overall chemical event represents a classic redox reaction. Common mistake: Labelling the entire process by only one half-reaction instead of acknowledging both simultaneous oxidation and reduction.