JAMB Chemistry 2013
Questions & Answers

A student might look at options such as the solid can be ground to a fine powder or the density of the solid is 2.25 g dm⁻³ and assume these indicate chemical purity or structural properties. The correct conclusion is reached by analyzing phase change characteristics: pure substances melt at a sharp, single temperature, whereas a combination of different components melts across a temperature span, making a melting range of 300°C to 309°C indicative of a mixture. Common mistake: Believing that any physical property like density or grinding capability can prove a solid is a mixture.

Options like C₂H₄O₂ might be guessed if a student miscalculates the molar mass or misinterprets the vapour data. By dividing the given volume of vapour by the molar volume (0.032 dm³ / 22.4 dm³ = 0.00143 mol) and dividing the mass by this value (0.127 g / 0.00143 mol ≈ 89 g/mol), we find the approximate molar mass. The empirical formula derived from the C:H:O ratio of 3:6:3 is CH₂O with a mass of 30 g/mol, and multiplying this unit by 3 yields (CH₂O)₃ = C₃H₆O₃, which gives a molar mass of 90 g/mol, closely matching the experimental value. Common mistake: Incorrectly matching the empirical formula multiple to the experimental molar mass.

Density or solubility in water might appear to be viable ways to tell liquids apart because different substances have distinct physical characteristics. However, separating or distinguishing liquids based on their volatility and boiling points relies directly on their tendency to vaporize; specifically, liquids with different boiling points will exhibit different vapour pressures at a given temperature, with the more volatile liquid showing a higher vapour pressure. Common mistake: Using solubility or density instead of vapour pressure when relating physical properties directly to boiling points.

A student might mistakenly select 0.10 or another value by miscalculating the stoichiometry or neglecting the titration volume ratios. To find the correct concentration, first determine the moles of anhydrous Na₂CO₃ using its mass and molar mass: (0.53 / 106) × (20 / 100) = 0.001 mol. According to the balanced equation (Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂), the acid reacts in a 1:1 mole ratio, meaning 0.001 mol of H₂SO₄ is neutralized, which yields a concentration of 0.001 / (25 / 1000) = 0.05 mol/dm³. Common mistake: Forgetting to account for the volume fraction of the carbonate solution used in the titration calculation.

Distractors like 25.0 cm³ could be chosen if a student simply adds the reactant volumes together without balancing their reaction stoichiometry. The correct minimum volume is calculated by looking at the individual combustion reactions: carbon monoxide reacts with oxygen in a 2:1 volume ratio (10 cm³ of CO requires 5 cm³ of O₂), and hydrogen also reacts with oxygen in a 2:1 ratio (15 cm³ of H₂ requires 7.5 cm³ of O₂), giving a combined total of 5 + 7.5 = 12.5 cm³ of oxygen. Common mistake: Adding reactant volumes directly instead of using stoichiometric mole-volume ratios for combustion.

A student might select 760 mmHg by forgetting to subtract the water vapour pressure, or choose 783 mmHg by mistakenly adding it. Dalton's law of partial pressures dictates that the total atmospheric pressure is the sum of the dry gas pressure and the water vapour pressure; therefore, subtracting the saturation vapour pressure of water (23 mmHg) from standard atmospheric pressure (760 mmHg) yields a partial pressure of 737 mmHg for hydrogen. Common mistake: Adding the water vapour pressure to the total atmospheric pressure instead of subtracting it.

Options stating that electropositive character decreases or electronegativity increases might seem plausible because periodic table trends often involve these properties. The correct explanation focuses on spatial dimensions: since lithium, sodium, and potassium all belong to Group 1, moving down the column adds successive principal electron shells, which naturally increases atomic radii. Common mistake: Confusing periodic group trends in electron shells with electronegativity changes down a group.

Charles’ Law or Avogadro’s Law might be mistakenly chosen because they are also fundamental gas laws dealing with proportional relationships. We arrive at the correct law by identifying the specific inverse relationship where pressure varies inversely with volume under constant temperature conditions, which defines Boyle’s Law. Common mistake: Confusing Boyle's Law (pressure-volume inverse relationship) with Charles' Law (temperature-volume direct relationship).

A student might guess X₂Y or XY due to miscounting valence electrons or improper balancing of ionic charges. Element X has 2 valence electrons, forming an X²⁺ ion (magnesium), while element Y has 7 valence electrons, forming a Y⁻ ion (chlorine); combining these ions requires two chlorine atoms for every one magnesium atom, resulting in the chemical formula XY₂. Common mistake: Incorrectly writing the empirical formula by failing to balance the valencies of the combining ions.

Options listing 78 protons and 55 neutrons are common traps for students who confuse atomic numbers with mass numbers. The atomic number specifies the count of protons (55), and subtracting this atomic number from the mass number (133 - 55) reveals the number of neutrons (78) present in the nucleus. Common mistake: Swapping the values for protons and neutrons by confusing atomic number with mass number.

Elements like P (atomic number 4) or S (atomic number 17) might be guessed if a student is unfamiliar with electronic configurations of group-specific elements. The correct classification is determined by looking for a stable octet or full outer shell: element Q has an atomic number of 10, corresponding to neon, which is a noble gas. Common mistake: Misidentifying halogen or alkaline earth elements as noble gases based on arbitrary atomic number order.

A student might answer 5 by looking only at neutral phosphorus or 3 by focusing on the ionic charge without considering core configurations. Neutral phosphorus (atomic number 15) has an electron configuration of 1s²2s²2p⁶3s²3p³, giving it 5 valence electrons in its outer shell; when it forms the P³⁻ ion, it gains 3 additional electrons to achieve a stable octet containing 8 valence electrons. Common mistake: Confusing the valence electron count of a neutral atom with that of its charged ion.

Distractors like 7.45 g could be selected if a student attempts complex fractional or percentage calculations instead of straightforward subtraction. The mass of crystals deposited from a solution when cooled is simply determined by finding the difference between the initial solubility and the final solubility per unit volume, which is 80 g minus 65 g, giving 15 g per dm³. Common mistake: Overcomplicating solubility deposition problems with unnecessary mathematical formulas instead of using simple subtraction.

Options like 100.0 cm³ might be chosen if a student uses the initial volume of the acid without testing for limitation. The stoichiometry of the reaction (KOH + H₂SO₄ → KHSO₄ + H₂O) shows a 1:1 molar ratio; calculating the moles of potassium hydroxide gives 1 M × 0.05 dm³ = 0.05 mol, which means exactly 0.05 moles of acid are required, translating to 50.0 cm³ of 1 M H₂SO₄. Common mistake: Failing to identify the limiting reagent and using the total volume of acid provided instead of the exact stoichiometric amount.

A student might mistakenly choose 0.4, 0.8 by failing to divide the mass concentration by the correct molar mass of the compound. The molarity of calcium bromide is calculated by dividing its concentration in grams per dm³ by its molar mass (20 / 200 = 0.1 mol/dm³); because each molecule dissociates to release two bromide ions for every one calcium ion, the concentration of bromide ions is doubled to 0.2 mol/dm³. Common mistake: Forgetting to multiply the anion concentration by the stoichiometric dissociation factor of the salt.

A student might mistakenly pick a peroxide or a dioxide because they describe chemical compositions rather than acid-base reaction behavior. The correct term is derived from the substance's ability to react with both acids and bases to yield soluble products, which defines an amphoteric oxide. Common mistake: Confusing amphoteric oxides (which react with both acids and bases) with neutral or basic oxides.

1) Distractor check: A student might mistakenly select option C, assuming that opposite charges define a conjugate pair because one species is often an ion and the other might be neutral or oppositely charged, or option D, assuming both species must carry a neutral charge. 2) Reasoning to the answer: A conjugate acid-base pair is related strictly by the transfer of a single hydrogen ion. When an acid loses a proton or a base gains one, the resulting pair differs precisely by one proton, such as hydrochloric acid and the chloride ion. 3) Common mistake: Confusing structural charge differences with the precise proton-count difference required for conjugate pairs.

1) Distractor check: A student might select option A or C by improperly balancing the electrode half-equations or incorrectly applying Faraday's laws of electrolysis. 2) Reasoning to the answer: First calculate the moles of copper deposited by taking the given mass and dividing by its molar mass, yielding 0.0157 moles based on the reduction half-reaction where copper ions accept two electrons. Since silver ions require only one electron per atom, twice as many moles of silver will be deposited, resulting in 0.0314 moles. Multiplying this amount by the molar mass of silver gives a final deposited mass of 3.4 grams. 3) Common mistake: Forgetting to account for the different valencies and electron stoichiometry between copper ions and silver ions.

1) Distractor check: A student might mistakenly pick option B or D, thinking strong oxidizing agents or strong bases are universally used for gas detection. 2) Reasoning to the answer: When carbon dioxide gas is introduced into an aqueous solution of calcium hydroxide, a white precipitate of calcium carbonate forms along with water, causing the liquid to turn milky. 3) Common mistake: Forgetting the specific precipitation reaction that occurs between carbon dioxide and lime water.

1) Distractor check: A student might choose option A or D, confusing the industrial production of chlorine or aluminium with the processing of sulfide-bearing ores. 2) Reasoning to the answer: The smelting of copper heavily utilizes sulfide minerals such as copper(I) sulfide, which undergo roasting in the presence of oxygen to yield copper(I) oxide and release sulfur(IV) oxide gas as a byproduct. 3) Common mistake: Associating all major industrial gas pollutants with brine electrolysis or bauxite processing instead of metallurgical smelting.