WAEC Chemistry 2023
Questions & Answers

A student might mistakenly choose option B, C, or D thinking that kinetic theory applies to fewer gas properties than it actually does. The kinetic theory of gases explains gas behavior through molecular motion, successfully accounting for pressure-volume relations in Boyle's law, volume-temperature relations in Charles' law, and pressure-temperature relations in Gay-Lussac's law. Dalton's atomic theory, however, addresses the composition of matter into indivisible atoms, a concept not directly derived from gas kinetics. Common mistake: grouping atomic composition theories together with gas laws.

One might mistakenly select option A, B, or D due to familiarity with various short-chain hydrocarbons present in petroleum fractions. Natural gas primarily consists of methane, making up about 70 to 90 percent of its composition, alongside smaller quantities of ethane, propane, and butane. Because methane acts as the main fuel source, it establishes natural gas as a clean-burning fossil fuel. Common mistake: assuming the minor hydrocarbon constituents represent the primary bulk of natural gas.

A student might mistakenly select option B or C believing that cracking releases heat or yields exclusively light alkanes. Thermal cracking breaks down large alkane molecules into smaller hydrocarbons through heat-induced decomposition, creating a mixture of alkenes and smaller alkanes utilized for fuels like petrol. This process is endothermic and does not require hydrogen, distinguishing it from catalytic methods. Common mistake: confusing thermal cracking's endothermic nature with exothermic combustion reactions.

A test-taker might mistakenly select option B, assuming that because carbon is a constituent of fuel molecules, it naturally deposits directly in the exhaust. Carbon is deposited in the exhaust due to the incomplete combustion of petrol under low-oxygen conditions, which produces carbon monoxide and soot instead of clean carbon dioxide and water vapor. Common mistake: confusing normal fuel composition with incomplete combustion byproducts.

A student might mistakenly pick option B or A by failing to account for the ionic charge or the stoichiometry of the reduction half-reaction. One faraday represents 1 mole of electrons, and the reduction at the cathode proceeds via Cu²⁺ + 2e⁻ → Cu, meaning 2 faradays of electricity are required to deposit 1 mole of copper. Consequently, passing 1 faraday deposits half that amount, yielding 0.5 moles. Common mistake: forgetting to balance the electron stoichiometry for divalent metal ions during electrolysis.

One might mistakenly select option A or C by confusing alkene production pathways with fat saturation or general polymerization. Alkenes are manufactured via the cracking of hydrocarbons, which breaks carbon-carbon bonds in larger alkanes to yield smaller alkanes and alkenes, such as converting butane into ethene and ethane. Hydrogenation turns oils into fats, polymerization links monomers, and combustion oxidizes compounds into carbon dioxide and water. Common mistake: confusing the thermal breaking of alkanes with the addition reactions used to saturate vegetable oils.

A student might mistakenly choose option A or C assuming that any large-scale chemical processing falls under pure research or biochemistry. Petrochemistry is an example of applied chemistry because it utilizes chemical principles to transform petroleum into practical industrial products like plastics and fuels rather than investigating fundamental theories. Common mistake: confusing industrial application fields with pure or biological branches of science.

A student might mistakenly select option A or C by confusing transient species with final products or initial starting materials. Intermediates are transient chemical species that form during a reaction pathway and are subsequently consumed, meaning they do not appear in the overall net equation, such as carbocations in SN1 reactions. Products represent the final outcome, reactants are initial materials, and inhibitors function to slow down reactions. Common mistake: listing reaction intermediates as final products in a net chemical equation.

A student might mistakenly pick option B, C, or D due to misconceptions about homologous series formulas or boiling point trends. Alkanes follow the general formula CₙH₂ₙ₊₂, meaning an alkane with 49 carbon atoms correctly contains 100 hydrogen atoms. Alkanes undergo substitution rather than addition with chlorine, higher molecular weight alkanes like butane have higher boiling points than propane due to increased electrons, and pentane has 3 isomers rather than five. Common mistake: misapplying general molecular formulas for aliphatic hydrocarbons.

A student might mistakenly choose option A or B because methyl red and methyl orange are common acid-base indicators. For the titration of ethanoic acid, which is a weak acid, with sodium hydroxide, a strong base, the equivalence point occurs at a basic pH around 8.7. Phenolphthalein is the best indicator because its color change range of 8.2 to 10 closely matches this endpoint, whereas methyl red and methyl orange change too early in the acidic range. Common mistake: using strong-acid indicators for weak-acid strong-base titrations.

A student might mistakenly select option D by choosing the oxidation half-reaction instead of the reduction process. The reduction half-equation represents the gain of electrons at the cathode, where copper ions gain two electrons to form solid copper according to Cu²⁺(aq) + 2e⁻ → Cu(s). Option A describes salt dissociation, and option C uses incorrect electron stoichiometry. Common mistake: confusing the oxidation half-reaction of the zinc electrode with the reduction of copper ions.

A student might mistakenly select option A, B, or D through calculation errors involving molar mass or volume conversions. To find the solubility, first calculate the moles of solute by dividing 10.5 g by the molar mass of 142 g/mol, which equals 0.074 moles. Then, divide this mole value by the volume of 0.1 dm³ (converted from 100 cm³) to yield a solubility of 0.74 mol dm⁻³. Common mistake: omitting the volume conversion from cubic centimeters to cubic decimeters.

A student might mistakenly pick option B, C, or D because hydrated crystal forms are common among many common salts. Potassium trioxonitrate(V), or KNO₃, is an anhydrous crystalline substance that does not possess water of crystallization in its chemical formula. In contrast, sodium carbonate, iron(II) sulfate, and sodium sulfate often exist in hydrated forms containing water molecules. Common mistake: assuming all crystalline salts contain water of crystallization.

One might mistakenly select option B or C, assuming that all reactions involving acids automatically yield acidic solutions or that neutralization creates neutral conditions. The salt solution formed from ethanoic acid, a weak acid, and sodium hydroxide, a strong base, is basic because the resulting salt undergoes hydrolysis where the acetate ion reacts with water to produce hydroxide ions. Common mistake: assuming neutral pH for salts derived from weak acids and strong bases.

A student might mistakenly select option A, which represents esterification rather than hydrolysis. The correct representation for the hydrolysis of an alkanoate is CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, where the acetate ion reacts with water to yield ethanoic acid and hydroxide ions, thereby creating a basic solution. Common mistake: confusing the reverse esterification reaction with salt hydrolysis.

A student might mistakenly select option A or D by assuming that every molecular impact leads to a chemical change. Collision theory states that the rate of a reaction is directly proportional to the number of effective collisions, which require both sufficient activation energy and correct spatial orientation. Ineffective collisions simply fail to bring about a chemical reaction. Common mistake: assuming all physical collisions between reactant molecules result in product formation.

A student might mistakenly choose option A, assuming that strong drying agents are avoided simply because they are corrosive. Sulfuric acid and calcium chloride are unsuitable for drying ammonia gas specifically because they react chemically with the gas, forming stable addition compounds such as ammonium sulfate and calcium chloride adducts. Common mistake: attributing the unsuitability of a desiccant to its corrosive nature rather than its chemical reactivity with the target gas.

Distractor check: A student might mistakenly select option B or D due to their familiarity as standard industrial preparations and bleaching reactions, or choose A since chlorine acts as an oxidizing agent there. Reasoning to the answer: Fluorine is more electronegative and reactive than chlorine, meaning chlorine cannot force fluorine out from sodium fluoride. Therefore, option C fails to describe a valid chemical displacement process, whereas the other choices correctly demonstrate chlorine bleaching hydrogen sulfide gas and creating sodium hypochlorite or bleaching powder. Common mistake: Assuming all halogens can displace one another indiscriminately regardless of their position in the electrochemical series.

Distractor check: A student might pick options A, B, C, or D by incorrectly configuring neutral iron or miscalculating the electrons lost during oxidation. Reasoning to the answer: A neutral iron atom possesses the electronic configuration 3d⁶4s². Removing three electrons to form the Fe³⁺ ion strips both 4s electrons and one 3d electron, leaving a 3d⁵ configuration. Distributing these remaining five electrons across five distinct orbitals according to Hund's rule results in five unpaired electrons, aligning with option E. Common mistake: Forgetting to remove electrons from the outermost 4s shell before taking them from the 3d sublevel when ionizing transition metals.

Distractor check: A student might mistakenly pick option A, C, or D assuming that atomic properties universally decrease down a group just like atomic size increases. Reasoning to the answer: Moving downward through group II, the atomic radii expand significantly, which brings an increase in the number of delocalized electrons participating in metallic binding. This enhanced metallic bonding causes melting points to rise overall, even though there are minor irregularities across the group. Common mistake: Confusing the trends of atomic shielding, ionization energy, and electronegativity—which decrease down a group—with the behavior of metallic melting points.