JAMB Chemistry 2004 Past Questions & Answers

Sublimation is the process where a substance changes directly from a solid state to a gaseous state without passing through the liquid state. The question describes exactly this phenomenon: a solid disappears to become a gas upon heating. Common examples of substances that sublime are iodine, camphor, and solid carbon dioxide (dry ice).

Common mistake: Students often mistakenly choose evaporation because it also involves a substance turning into a gas, but evaporation specifically describes the change from a liquid to a gas.

A chemical reaction, by its very definition, is a process that leads to the chemical transformation of one set of substances (reactants) into another set of substances (products). This means that the most fundamental and constant outcome of any chemical reaction is the formation of new substances with different identities and properties. While other changes like volume or energy might occur, they are not as defining as the creation of new materials.

Common mistake: Students might mistakenly choose option C, 'A change in the nature of the reactants,' because it sounds correct, but 'the formation of new substances' is the more precise and fundamental definition of a chemical reaction.

First, write the balanced chemical equation for the reaction: H₂SO₄ + CuO → CuSO₄ + H₂O. This shows that 1 mole of acid reacts with 1 mole of copper(II) oxide. Next, calculate the number of moles of H₂SO₄ given, which is mass divided by molar mass (4.8g / 98 g/mol) = 0.049 moles. Since the mole ratio is 1:1, we also need 0.049 moles of CuO. Finally, find the mass of CuO required by multiplying its moles by its molar mass (0.049 mol × 80 g/mol), which gives approximately 4.0g.

Common mistake: A common mistake is to incorrectly assume that you need one full mole of Copper(II) oxide, leading students to choose its molar mass, 80.0g, as the answer without performing the stoichiometric calculation.

First, find the number of moles of calcium trioxocarbonate(IV) used, which is mass divided by molar mass (2g / 100g/mol = 0.02 mol). The balanced chemical equation, CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂, shows that 1 mole of CaCO₃ produces 1 mole of CO₂ gas. Therefore, 0.02 moles of CaCO₃ will produce 0.02 moles of CO₂. To find the volume, multiply the moles of CO₂ by the molar volume at s.t.p (0.02 mol x 22.4 dm³), which gives 0.448 dm³, and then convert to cm³ by multiplying by 1000 to get 448 cm³.

Common mistake: Students might mistakenly calculate the moles of CaCO₃ as 0.01 instead of 0.02, which leads to half the correct volume, 224cm³.

Charles' law states that the volume of a fixed mass of gas is directly proportional to its absolute temperature in Kelvin. This implies that if you keep cooling a gas, its volume will keep reducing until it theoretically becomes zero. This theoretical point is called absolute zero, which is 0 on the Kelvin temperature scale. To convert 0 Kelvin to Celsius, you subtract 273, which gives -273°C.

Common mistake: Students often mistakenly choose 0°C because it is a familiar reference point (the freezing point of water), but they forget that the gas laws are based on the absolute Kelvin scale, not the Celsius scale.

According to the Kinetic Theory of Matter, particles in a gas are far apart and have very weak forces of attraction between them. This lack of significant intermolecular forces allows the molecules to move randomly and rapidly in all directions. Because they are not held together in a structure, it is difficult for them to form an orderly arrangement.

Common mistake: Students may incorrectly choose option C, as collisions are a key feature of gases, but these collisions are a result of the random motion, which is primarily caused by the weak forces of attraction.

This question is solved using Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. First, calculate the molar mass of methane (CH₄) as 16 g/mol and sulphur(IV) oxide (SO₂) as 64 g/mol. The formula relating time and molar mass is t₂/t₁ = $√(M₂/M₁)$, so t(SO₂)/20s = $√(64/16)$, which simplifies to t(SO₂)/20s = $√4$ = 2. Therefore, the time for SO₂ to diffuse is 20s × 2 = 40s.

Common mistake: The most common mistake is inverting the molar mass ratio in Graham's Law, which would incorrectly suggest the heavier gas diffuses faster and lead to a wrong answer.

Halogens are elements in Group 17 of the periodic table, meaning they have 7 electrons in their outermost shell (valence electrons). To become stable, atoms try to get a full outer shell, which is usually 8 electrons. When a halogen atom gains one electron, it completes its outer shell, forming a halide ion with a stable configuration of 8 valence electrons.

Common mistake: Students often mistakenly choose 7 valence electrons because that is the number of valence electrons in the neutral halogen atom, not the halide ion which is formed after gaining an electron.

Atomic orbitals are three-dimensional regions around the nucleus where there is a high probability of finding an electron. The s-orbital is the simplest orbital and has a non-directional, spherical shape, like a ball. This means the probability of finding the electron is the same in all directions at a given distance from the nucleus.

Common mistake: Students might choose 'Circular' because they confuse the three-dimensional shape of an orbital with the two-dimensional circular path of an electron in the older Bohr model of the atom.

Electronegativity is the ability of an atom to attract shared electrons in a chemical bond. In hydrogen chloride (HCl), chlorine is much more electronegative than hydrogen, so it pulls the bonding electrons strongly towards itself, creating a polar bond with significant ionic character. In a chlorine molecule (Cl₂), both atoms have the same electronegativity, so they share the electrons equally, resulting in a non-polar covalent bond with no ionic character.

Common mistake: Students often confuse electronegativity with electron affinity, but electron affinity refers to the energy change when an isolated atom gains an electron, not its ability to attract electrons within a chemical bond.

The ratio of the isotopes with mass numbers 35 and 37 is given as 3:1. This means the total parts in the ratio is 3 + 1 = 4. To find the relative abundance of the isotope with mass number 37, we take its part of the ratio (1) and divide by the total parts (4), then multiply by 100 to get a percentage. Therefore, the calculation is $(1/4)$ x 100%, which equals 25%.

Common mistake: Students often make the mistake of calculating the abundance for the other isotope (mass number 35), which would be $(3/4)$ * 100 = 75%, leading them to choose option D.

Burning in a flame is a chemical reaction called combustion, which requires a flammable substance. Among the gases listed in the options, only hydrogen gas is flammable. The other gases—neon, helium, and nitrogen—are all non-flammable, with neon and helium being noble gases that are chemically inert. Therefore, the mixture of neon and hydrogen is the only one that will burn because it contains a flammable component.

Common mistake: Students might mistakenly think nitrogen is flammable because it is a major component of air, but it is actually oxygen in the air that supports combustion, while nitrogen itself is unreactive.

Hardness of water is a property caused by the presence of dissolved mineral ions that react with soap to form scum. The principal ions responsible for this are divalent cations, specifically calcium ions (Ca²⁺) and magnesium ions (Mg²⁺). When these ions are present in water, they prevent soap from lathering properly, which is the main characteristic of hard water.

Common mistake: Students may mistakenly choose an option containing sodium because sodium compounds are very common and soluble in water, but they forget that it is specifically calcium and magnesium ions that cause hardness and prevent soap from lathering.

High acidity in soil, often caused by acid rain, makes it easier for many metal compounds to dissolve. Many metal compounds are basic and will react with the acid in the soil to form soluble salts. These soluble salts are then washed down into the underground water by rain, carrying the harmful metal ions with them and causing pollution.

Common mistake: Students may mistakenly choose 'Alkalinity' because they associate both strong acids and strong alkalis with being corrosive, but in this context, high alkalinity (basic conditions) actually makes most heavy metal compounds less soluble.

A carcinogenic substance is a substance that is capable of causing cancer in living tissue. Asbestos dust consists of tiny mineral fibres that, when inhaled, can get trapped in the lungs for a long time. Over many years, these fibres can cause inflammation and genetic damage that leads to serious diseases, including lung cancer and mesothelioma.

Common mistake: Students may confuse a generally toxic or poisonous substance with a carcinogenic one, often choosing Carbon (II) oxide (carbon monoxide) because it is a well-known poison, but it causes harm by asphyxiation, not by causing cancer.

First, calculate the molar mass of CuSO4: 64 + 32 + $(4 * 16)$ = 160 g/mol. Next, find the number of moles of CuSO4 dissolved, which is mass divided by molar mass: 20g / 160 g/mol = 0.125 moles. The volume of the solvent (100g of water) is 100cm3, which is 0.1 dm3 $(since 1000cm3 = 1dm3)$. Finally, solubility in mol dm-3 is calculated by dividing the moles by the volume in dm3: 0.125 mol / 0.1 dm3 = 1.25 mol dm-3.

Common mistake: Students often correctly calculate the number of moles (0.125 mol) but forget to divide by the volume of the solvent in dm3, leading them to mistakenly choose option A.

Calcium Tetraoxosulphate(VI), which is CaSO₄, is a sparingly soluble salt in water. This means only a very small amount dissolves, while the rest remains as very tiny, microscopic particles spread throughout the water. This type of mixture, where fine particles are dispersed in a medium but do not settle down, is defined as a colloid.

Common mistake: Students might mistakenly choose 'Suspension' because it also involves undissolved particles, but suspensions have larger particles that settle over time, unlike the very fine particles in a colloid.

A normal salt is formed when all the replaceable hydrogen ions of an acid are completely replaced by a metal ion. Sodium carbonate (Na2CO3) is formed from carbonic acid (H2CO3), and in this salt, both hydrogen ions have been completely replaced by two sodium ions. In contrast, options like NaHCO3 and NaHS are acid salts because they still contain a replaceable hydrogen atom from the original acid.

Common mistake: Students often mistakenly choose NaHCO3 because it is a very common salt, but they forget that it is an acid salt since it still contains a replaceable hydrogen atom from the parent acid (H2CO3).

First, write the balanced equation for the neutralization reaction: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. This shows that the mole ratio of acid to base (na/nb) is $1/2$. Using the titration formula CaVa/CbVb = na/nb, we substitute the given values: (0.5 × Va) / $(0.1 × 20)$ = $1/2$. Solving for the volume of the acid (Va) gives Va = $(0.1 × 20)$ / $(0.5 × 2)$ = 2.0cm³.

Common mistake: Students often forget to write a balanced chemical equation and mistakenly assume a 1:1 mole ratio between the acid and the base, which leads to an incorrect calculation.

To find the oxidation state of chlorine (Cl) in perchloric acid (HClO4), we use the known oxidation states of hydrogen and oxygen. The oxidation state of hydrogen (H) is +1, and the oxidation state of oxygen (O) is -2. Since the overall charge of the compound is zero, the sum of the oxidation states must be equal to zero. So, we can set up an equation: (+1) + (Cl) + 4(-2) = 0, which simplifies to 1 + Cl - 8 = 0, and therefore Cl = +7.

Common mistake: Students often mistakenly choose -1 because it is the common oxidation state of chlorine in simple ionic compounds like NaCl, but they forget to calculate it for this specific polyatomic compound.

Reduction is a chemical process that involves a decrease in oxidation number. In the reactant, zinc oxide (ZnO), the oxidation number of zinc is +2. In the product, elemental zinc (Zn), the oxidation number is 0. Since the oxidation number of zinc decreased from +2 to 0, the zinc oxide is said to be reduced.

Common mistake: Students often confuse oxidation with reduction and may choose 'Oxidized' by mistake, or they might look at the carbon monoxide which is being oxidized and incorrectly apply that to the zinc oxide.

A substance can act as both a reducing and an oxidizing agent if its central element is in an intermediate oxidation state. In Sulphur (IV) oxide (SO2), the oxidation state of sulphur is +4. Since the oxidation state of sulphur can range from -2 (in H2S) to +6 (in SO3), the +4 state is intermediate, allowing it to be either oxidized to +6 (acting as a reducing agent) or reduced to 0 (acting as an oxidizing agent).

Common mistake: Students often mistakenly choose H2S because it is a well-known reducing agent, but they forget that the sulphur in H2S is in its lowest oxidation state (-2) and can only be oxidized, not reduced.

Electrolysis is the process of using electricity to break down a substance. When brine (concentrated salt water) is electrolysed, chlorine gas is produced at the positive electrode, which is the anode. This chlorine gas is very reactive and corrosive, so the anode must be made of a material that will not react with it. Carbon, in the form of graphite, is used as the anode because it is inert (unreactive) and affordable for this industrial process.

Common mistake: Students might mistakenly choose Platinum because it is also an inert electrode used in electrolysis, but carbon (graphite) is the standard material used in the industrial electrolysis of brine due to its lower cost.

This question is an application of Faraday's first law of electrolysis. First, calculate the quantity of electricity (Q) passed using the formula Q = Current (I) × time (t), which gives 10A × 4830s = 48300 C. The reaction for silver deposition is Ag⁺ + e⁻ → Ag, meaning 1 mole of electrons (96500 C) deposits 1 mole of silver (108g). Therefore, the mass of silver deposited by 48300 C is calculated as (48300 C / 96500 C) × 108g, which equals 0.5 × 108g = 54.0g.

Common mistake: Students may mistakenly assume silver is a divalent ion (Ag²⁺), causing them to use 2 Faradays in their calculation, which leads to an incorrect answer of 27.0g.

To find the energy change of a reaction (ΔH), you calculate the total enthalpy of the products and subtract the total enthalpy of the reactants. The formula is ΔH = $[2 × ΔH_f(CO₂)]$ - $[2 × ΔH_f(CO)]$. Using the standard values $(and assuming ΔH_f(CO) is -110.4 kJ/mol)$, the calculation is [2 × (-393.0)] - [2 × (-110.4)] = -565.2 kJ. As this correct value is not an option, the question data or options are flawed; however, the question is designed to test your knowledge of reaction enthalpies.

Common mistake: Students often forget to use the stoichiometric coefficients, calculating for only one mole (i.e., -393.0 - (-110.4) = -282.6 kJ), which leads to the incorrect option B.