2013 JAMB CHEMISTRY PAST QUESTIONS AND ANSWERS

This is an administrative question to ensure students are using the correct question paper type. Paper Type I is the correct answer for this set.

Impurities disrupt the crystal lattice structure of a pure substance, causing a decrease in the melting point. This is known as melting point depression, a colligative property. Pure substances have definite melting points, while impure ones melt over a range of temperatures at lower points.

Following the equation $\text{C} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}$, for every 0.5 moles of $\text{O}_2$ used, 1 mole of CO is produced. Therefore, $10 \text{ dm}^3$ of $\text{O}_2$ will produce $20 \text{ dm}^3$ of CO.

In a PV diagram, an ideal gas follows a perfect hyperbolic curve (Boyle’s Law). Line M shows this ideal behavior, while other lines show deviations due to real gas behavior.

Using Graham’s Law: $\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$. Given $\frac{r_1}{r_2} = 2$ and $M_1 = 64$, solving gives $M_2 = 256$, then $\sqrt{M_2} = 16$.

Cobalt-60 is used in industrial radiography due to its high-energy gamma radiation, long half-life, and penetrating power, making it ideal for detecting flaws in metal welds and castings.

Fluorine’s electronic configuration is $1s^2 2s^2 2p^5$. In the p-orbital, there is one unpaired electron as the p-orbital has 5 electrons, with 4 paired and 1 unpaired.

Gamma rays (γ-rays) have the least ionizing power but greatest penetrating power. α-particles have highest ionization power, followed by β-particles, then γ-rays.

$\text{CO}_2$ has a linear shape due to its molecular geometry with two double bonds between carbon and oxygen atoms ($\text{O}=\text{C}=\text{O}$), resulting in $180°$ bond angles.

HF forms hydrogen bonds due to the high electronegativity of fluorine and small size of hydrogen. Other options lack the necessary electronegativity difference or proper elements for hydrogen bonding.

The first configuration is of magnesium (Group 2) and second is of oxygen (Group 6). Their electronegativity difference leads to electron transfer forming ionic bonds.

Nitrogen makes up about 78% of air and acts as a diluent, reducing the concentration of oxygen and other gases to safe levels. It is relatively inert in normal conditions.

Anhydrous cobalt(II) chloride is blue but turns pink when hydrated by steam. This is a common test for water presence and demonstrates the effect of hydration on transition metal compounds.

Copper(II) oxide (CuO) is hygroscopic, meaning it absorbs moisture from the air but doesn’t dissolve in it. This property is important in desiccation and dehydration processes.

Using formula: Solubility = $\frac{\text{mass of solute}}{\text{volume of solution in dm}^3}$. Volume ≈ 0.042 $\text{dm}^3$, therefore solubility = $\frac{24.4}{0.042} = 581 \text{ gdm}^{-3}$

Ammonia solution is effective for removing grease stains due to its ability to break down and emulsify fats and oils. It forms soluble compounds with fatty substances.

Excess sewage increases bacterial population, leading to increased oxygen consumption during decomposition. This reduces dissolved oxygen levels, harming aquatic life.

Using $\text{C}_1\text{V}_1 = \text{C}_2\text{V}_2$: $(2.0)(2.5) = \text{C}_2(12.5)$. Solving gives $\text{C}_2 = 0.4 \text{ moldm}^{-3}$

Adding an acid (HCl) to a solution will decrease its pH. Since the original solution was slightly acidic (pH 6.4), adding more acid will further lower the pH.

Aluminium oxide is an oxide, not a salt. It’s formed by direct combination of aluminum and oxygen, lacking the characteristic cation-anion structure of salts.

Double decomposition (precipitation) reactions produce insoluble salts when two soluble salts react to form one insoluble product. This is a common method in inorganic synthesis.

$\text{F}_2$ is reduced as it gains electrons, changing from 0 to -1 oxidation state in HF. Other species either oxidize or maintain their oxidation states.

$\text{CuSO}_4$ is the oxidizing agent as it accepts electrons from Zn, causing Cu²⁺ to be reduced to Cu while Zn is oxidized to Zn²⁺.

Hydrogen accumulation on electrodes causes polarization by creating a barrier that increases resistance and reduces current flow. This affects cell efficiency.

Using Faraday’s laws: Q = It = 5 × 193 = 965 C. Moles of electrons = 965/96500. Moles of $\text{O}_2$ = moles e⁻/4. Volume = 0.056 $\text{dm}^3$

For spontaneous reactions, ΔG must be negative. In endothermic reactions with entropy decrease, ΔG will always be positive, making the reaction non-spontaneous.

Since the reaction is exothermic (negative ΔH), decreasing temperature favors forward reaction according to Le Chatelier’s principle, increasing $\text{SO}_3$ concentration.

Activation energy is the minimum energy barrier that reactants must overcome to form products. It determines reaction rate and feasibility.

From the energy profile diagram, the catalyzed pathway shows a lower activation energy of 100KJ compared to the uncatalyzed pathway.

For heterogeneous equilibrium, only gas phase concentrations are included in K expression. K = $[\text{H}_2]^4/[\text{H}_2\text{O}]^4$

Carbon (II) oxide (CO) is a neutral oxide as it neither reacts with acids nor bases to form salts. It doesn’t affect the pH of water.

Slanting prevents condensed water from flowing back into the hot flask, which could cause the flask to crack due to thermal shock.

Nitrous oxide ($\text{N}_2\text{O}$), also known as laughing gas, is commonly used as an anesthetic in medical and dental procedures.

Potassium forms soluble trioxosulphate (IV) salts, while other metals listed form insoluble or less soluble sulfites. This is due to the ionic size and charge density of potassium.

As a transition element, copper is low in the reactivity series, making it easily displaced by more reactive metals. This is based on standard electrode potentials.

The characteristic colors of transition metal ions arise from d-d transitions in partially filled d-orbitals. This allows for electronic transitions that absorb specific wavelengths of light.

Aluminum’s low density makes it ideal for transport containers, reducing shipping costs while maintaining structural integrity.

2-methylbutan-2-ol has an OH group attached to a carbon bearing no hydrogen atoms (only methyl groups), making it a tertiary alcohol.

The reaction produces ethanol and ethanamide through nucleophilic acyl substitution. Ammonia replaces the ethoxy group of the ester.

Decarboxylation of ethanoic acid ($\text{CH}_3\text{COOH}$) produces $\text{CH}_4$ (methane) and $\text{CO}_2$. The $\text{-COOH}$ group is removed as $\text{CO}_2$.

Based on the carbonyl group (C=O) in the middle of the carbon chain, this compound is a ketone (alkanone).

Ethanoic acid ($\text{CH}_3\text{COOH}$) reacts with NaOH in a neutralization reaction to form sodium ethanoate and water.

Basic solutions turn red litmus paper blue. Among the options, only amines have basic properties due to the lone pair on nitrogen.

When ammonium salts of carboxylic acids are dehydrated, they form amides. The process involves loss of water, resulting in $\text{R-CONH}_2$ structure.

Lubricating oil, being a higher molecular weight fraction, is commonly used as raw material for cracking to produce smaller, more useful hydrocarbons.

Carboxylic acids ($\text{C}n\text{H}{2n+1}\text{COOH}$) typically have pleasant fruity smells and are common in fragrances and flavoring.

Primary amides have the formula $\text{RCONH}_2$, where R is an alkyl group and the nitrogen has two hydrogen atoms attached.

Based on the structure shown, the longest carbon chain with the double bond gives 4-methylpent-1-ene, following IUPAC naming rules.

Converting percentages to moles: C=5, H=13.3, O=1.67. Dividing by smallest number gives C:H:O ratio of 3:8:1, yielding $\text{C}_3\text{H}_8\text{O}$