JAMB UTME - Chemistry - 2025

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The equation is:

\[2\text{X} + 2\text{HCl} \rightarrow 2\text{XCl} + \text{H}_2\]

The product formed is XCl, which tells us that element X combines with chlorine in a 1:1 ratio. This means X has a valency of +1 and forms a monovalent chloride.

Examining the options:

• K (potassium): Group 1 element with a valency of +1. It forms KCl. The equation becomes \(2\text{K} + 2\text{HCl} \rightarrow 2\text{KCl} + \text{H}_2\), which balances correctly.
• Mg (magnesium): Group 2 element with a valency of +2. It forms MgCl₂, not MgCl. Does not match.
• Ca (calcium): Group 2 element with a valency of +2. It forms CaCl₂, not CaCl. Does not match.
• Ba (barium): Group 2 element with a valency of +2. It forms BaCl₂, not BaCl. Does not match.

Only potassium (K) has a valency of +1 and forms a chloride with the formula XCl, making it the correct identity of X.

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Brine is a concentrated solution of sodium chloride (NaCl) in water. When brine is electrolysed using inert (neutral) electrodes such as carbon or platinum, the ions present in solution are:

• Cations: Na⁺ and H⁺ (from water)
• Anions: Cl^- and OH^- (from water)

At the anode (positive electrode), anions migrate and are discharged. Both Cl^- and OH^- are present, but Cl^- is preferentially discharged because it is present in much higher concentration in the brine solution. Despite OH^- having a lower discharge potential, the high concentration of Cl^- gives it priority at the anode.

The half-equation at the anode is:

\[2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-\]

Chlorine gas (Cl₂) is released at the anode.

Na⁺ and H⁺ are cations and migrate to the cathode, not the anode. At the cathode, H⁺ is discharged (since Na⁺ has a very high discharge potential), producing hydrogen gas.

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Acidified potassium dichromate (K₂Cr₂O₇) is a strong oxidising agent. When it oxidises a substance, the orange dichromate ion (Cr₂O₇^2-, containing Cr⁶⁺) is reduced to the green chromium(III) ion (Cr³⁺). This colour change from orange to green is the characteristic indicator that an oxidation reaction has occurred.

Human breath can contain ethanol (an alkanol) if the person has consumed alcohol. Ethanol is readily oxidised by acidified dichromate:

\[\text{C}_2\text{H}_5\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CHO} + \text{H}_2\text{O}\]

and further:

\[\text{CH}_3\text{CHO} + [\text{O}] \rightarrow \text{CH}_3\text{COOH}\]

This is the principle behind the breathalyser test used to detect alcohol (alkanol) in a person's breath. The colour change from orange to green confirms the presence of an alkanol (alcohol).

Alkanones (ketones) are not easily oxidised by acidified dichromate under mild conditions, so they would not produce this colour change. Alkanals (aldehydes) and alkanoic acids (carboxylic acids) are not typically present in human breath in significant amounts.

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The compound [Cu(NH₃)₄]Cl₂ consists of two distinct parts:

• The complex cation: [Cu(NH₃)₄]²⁺ (copper(II) tetraamine ion)
• The anions: 2 Cl^- (chloride ions)

The question asks about the bond between the complex cation and the chloride ions. The complex cation carries a 2+ charge, and each chloride ion carries a 1- charge. The attraction between these oppositely charged ions is an ionic bond (electrostatic attraction).

It is important to distinguish this from the bonding within the complex ion. Inside [Cu(NH₃)₄]²⁺, each NH₃ molecule donates a lone pair of electrons from nitrogen to the Cu²⁺ ion, forming dative (coordinate) bonds. However, the question specifically asks about the bond between the complex and chlorine, which is ionic.

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The fountain experiment demonstrates the very high solubility of certain gases in water. A round-bottom flask is filled with the gas and inverted over a trough of water (often containing an indicator). When a small amount of water enters the flask and dissolves the gas, the pressure inside drops dramatically. Atmospheric pressure then forces water up into the flask in a spectacular fountain.

For this experiment to work, the gas must be extremely soluble in water so that it dissolves almost instantly on contact, creating a near-vacuum inside the flask.

Hydrogen chloride (HCl) is the classic gas used. It is one of the most soluble gases in water: about 450 volumes of HCl dissolve in one volume of water at room temperature, forming hydrochloric acid. Ammonia (NH₃) is also commonly used for the same experiment, but it is not among the given options.

Hydrogen sulphide (H₂S) is only moderately soluble and is extremely toxic, making it unsuitable. Dinitrogen(I) oxide (N₂O, nitrous oxide) and nitrogen(II) oxide (NO, nitric oxide) are both poorly soluble in water and would not produce the dramatic pressure drop needed for the fountain effect.

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Solubility is defined as the maximum amount of a solute that can dissolve in a given quantity of solvent at a particular temperature to form a saturated solution. Its importance lies directly in the fact that it determines the amount of solute that can dissolve in a given amount of solvent.

This knowledge is practically essential in:

• Preparing solutions of known concentration in the laboratory
• Predicting whether a precipitate will form when solutions are mixed
• Understanding purification techniques such as recrystallisation
• Pharmaceutical formulation (ensuring a drug dissolves sufficiently for absorption)

The other options describe colligative properties - effects that arise after a solute has been dissolved:

• Lowering the vapour pressure and raising the boiling point are consequences of having dissolved particles in solution, not descriptions of what solubility itself tells us.
• Knowing chemical reactions of a gas with a liquid relates to chemical reactivity, not solubility as a concept.

Solubility is fundamentally about how much dissolves, and that is its primary importance.

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The molecular mass (molar mass) of a substance is defined as the mass of one mole of that substance. The relationship is:

\[\text{Molar mass} = \frac{\text{Mass}}{\text{Number of moles}}\]

Given that 0.5 mole of the alkanoic acid weighs 44 g:

\[\text{Molar mass} = \frac{44\,\text{g}}{0.5\,\text{mol}} = 88\,\text{g/mol}\]

The molecular mass of the alkanoic acid is therefore 88. This corresponds to butanoic acid (CH₃CH₂CH₂COOH), which has the molecular formula C₄H₈O₂: (4 x 12) + (8 x 1) + (2 x 16) = 48 + 8 + 32 = 88.

A common error is to multiply mass by moles instead of dividing. Remember: if a fraction of a mole has a certain mass, the full mole must weigh proportionally more.

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Le Chatelier's principle states that if a system at equilibrium is subjected to a change in conditions, the equilibrium shifts in the direction that tends to counteract that change. For pressure changes, the key factor is the difference in the total number of moles of gas on each side of the equation.

If there is no change in volume during a gaseous reaction, this means the total number of moles of gaseous products equals the total number of moles of gaseous reactants. In such a case, increasing or decreasing the pressure gives the system no direction in which to shift, because neither the forward nor the backward reaction would reduce the total number of gas molecules.

Therefore, a change in pressure will have no effect on the equilibrium position when the reaction involves equal moles of gas on both sides.

For example, in the reaction:

\[ \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \]

there are 2 moles of gas on each side, so pressure changes do not shift the equilibrium.

Pressure only affects equilibrium when there is an unequal number of moles of gas on either side. In those cases, increasing pressure favours the side with fewer moles of gas, and decreasing pressure favours the side with more moles.

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A reaction is feasible (spontaneous) when the Gibbs free energy change is negative: \(\Delta G < 0\). The relationship between Gibbs free energy, enthalpy, and entropy is:

\[\Delta G = \Delta H - T\Delta S\]

The question states that \(\Delta G^\circ\) is negative. To determine which combination of \(\Delta H\) and \(\Delta S\) guarantees this, consider each option:

• If \(\Delta H\) is positive and \(\Delta S\) is positive: \(\Delta G = (+) - T(+)\). This is negative only at sufficiently high temperatures, so feasibility is not guaranteed at all temperatures.
• If \(\Delta H = 0\) and \(\Delta S\) is positive: \(\Delta G = 0 - T(+) = -T\Delta S\), which is always negative at any temperature above 0 K. This guarantees a negative \(\Delta G\).
• If \(\Delta H\) is positive and \(\Delta S\) is negative: \(\Delta G = (+) - T(-) = (+) + T|\Delta S|\), which is always positive. The reaction would never be feasible.
• If \(\Delta S = 0\): \(\Delta G = \Delta H\), and feasibility depends entirely on the sign of \(\Delta H\), which is not specified.

The only option that ensures \(\Delta G\) is negative under all conditions is \(\Delta H = 0\) and \(\Delta S\) is positive, because the \(-T\Delta S\) term is always negative when \(\Delta S > 0\).

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Brine is a concentrated solution of sodium chloride (NaCl) in water. During electrolysis of brine, the ions present are Na⁺, Cl^-, H⁺ (from water), and OH^- (from water).

At the cathode (negative electrode), reduction takes place. The two cations competing for discharge are Na⁺ and H⁺. Because hydrogen ions are much easier to reduce than sodium ions (sodium has a very negative standard electrode potential), H⁺ ions are preferentially discharged:

\[2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)\]

The gas produced at the cathode is therefore hydrogen.

At the anode (positive electrode), chloride ions are oxidised to produce chlorine gas. Sodium hydroxide remains in solution. Steam is not produced during electrolysis, and oxygen would only appear at the anode if a dilute solution were used instead of concentrated brine.

Exam tip: In electrolysis of brine, remember the three products: hydrogen at the cathode, chlorine at the anode, and sodium hydroxide in solution.

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In the electrolysis of concentrated sodium chloride solution (brine) using inert graphite electrodes, the products depend on the concentration of the solution and the electrode positions.

At the anode (positive electrode), negatively charged ions migrate and are discharged. In concentrated NaCl solution, both chloride ions (Cl^-) and hydroxide ions (OH^-) from water are present. However, because the chloride ion concentration is very high, chloride ions are preferentially discharged at the anode:

\[2\text{Cl}^{-}(aq) \rightarrow \text{Cl}_2(g) + 2e^{-}\]

This produces chlorine gas, which can be identified by its greenish-yellow colour and its ability to bleach damp litmus paper.

At the cathode, hydrogen gas is produced from the reduction of water (since Na⁺ ions are too electropositive to be discharged). Oxygen gas would be the anode product only in the electrolysis of dilute sodium chloride or dilute sulphuric acid, where hydroxide ions are discharged instead of chloride ions. Water vapour and hydrogen gas are not anode products in this process.

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Oxidation and reduction are defined in terms of electron transfer:

• Oxidation is the loss (removal) of electrons from a species.
• Reduction is the gain (addition) of electrons to a species.

A useful mnemonic is OIL RIG: Oxidation Is Loss, Reduction Is Gain.

For example, when iron is oxidised:

\[\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-\]

Iron loses two electrons, so its oxidation state increases from 0 to +2. The electrons are removed from the iron atom.

The other options are incorrect: "added" describes reduction (the opposite process), while "hydrolysed" (broken down by water) and "hydrated" (combined with water molecules) are unrelated to the electron-transfer definition of oxidation.

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The basicity of an acid is the number of replaceable hydrogen ions (\(\text{H}^+\)) that one molecule of the acid can donate in a reaction with a base.

The compound \(\text{C}_2\text{H}_2\text{O}_4\) is oxalic acid (also called ethanedioic acid). Its structural formula is:

\(\text{HOOC-COOH}\)

Oxalic acid contains two carboxyl groups (\(-\text{COOH}\)). Each carboxyl group carries one hydrogen atom that can be released as \(\text{H}^+\) during a neutralisation reaction. The remaining hydrogen atoms in the molecule are bonded to carbon and are not ionisable.

Since there are two replaceable hydrogen atoms, the basicity of oxalic acid is 2. This means it is a dibasic acid (also called a diprotic acid).

The neutralisation reaction with sodium hydroxide confirms this:

Two moles of NaOH are required to completely neutralise one mole of oxalic acid, confirming a basicity of 2.

When determining basicity, count only the hydrogen atoms bonded to oxygen in carboxyl or hydroxyl groups, not those bonded directly to carbon.

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The compound CH₃-CH₂-COOCH₂-CH₃ contains the ester functional group (-COO-). To identify the reactants that formed this ester, split the structure at the ester linkage (between the carbonyl carbon and the oxygen bonded to the alkyl group).

The ester bond in -COO- comes from two parts:

• The acid part (to the left of -COO-): CH₃-CH₂-COO-. This corresponds to the carboxylate portion of propanoic acid (CH₃CH₂COOH), a three-carbon carboxylic acid.
• The alcohol part (to the right of -COO-): -CH₂-CH₃. This corresponds to the alkyl group from ethanol (CH₃CH₂OH), a two-carbon alcohol.

The ester is therefore ethyl propanoate, formed from propanoic acid and ethanol:

\[\text{CH}_3\text{CH}_2\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}\]

The correct reactants are propanoic acid and ethanol.

Exam tip: To identify the parent acid and alcohol of an ester, break the molecule at the single-bond oxygen in the -COO- group. The fragment bonded to the carbonyl (C=O) gives the acid; the fragment bonded through the oxygen gives the alcohol.

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A physical change is a change in which no new substance is formed. The original substance can be recovered by simple physical methods (such as evaporation, filtration, or condensation), and no chemical bonds are broken or formed between different types of atoms.

Dissolving sodium chloride in water is a physical change. When NaCl dissolves, the ionic lattice breaks apart and Na⁺ and Cl^- ions become surrounded by water molecules (hydration). However, no new chemical substance is created. The sodium chloride can be fully recovered by evaporating the water. The process is reversible.

The other options all involve chemical changes:

• Addition of water to calcium oxide: CaO + H₂O → Ca(OH)₂. A new substance (calcium hydroxide) is formed, and considerable heat is released.
• Corrosion of iron: Iron reacts with oxygen and moisture to form iron(III) oxide (rust, Fe₂O₃). New substances are produced irreversibly.
• Burning of charcoal: C + O₂ → CO₂. Combustion produces a completely different substance (carbon dioxide) and is irreversible.

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Dilute nitric acid (HNO\(_3\)) is an oxidising acid, which means it usually oxidises the metal and is itself reduced to nitrogen oxides (such as NO or NO\(_2\)) rather than producing hydrogen gas. This is different from non-oxidising acids like dilute HCl or dilute H\(_2\)SO\(_4\), which readily liberate H\(_2\) with reactive metals.

However, magnesium (Mg) is an exception. Because magnesium is extremely reactive (high up in the electrochemical series), it reacts so vigorously with very dilute HNO\(_3\) that the reaction proceeds faster than the acid can act as an oxidising agent. The result is that hydrogen gas is liberated:

\[\text{Mg} + 2\text{HNO}_3\text{(very dilute)} \rightarrow \text{Mg(NO}_3\text{)}_2 + \text{H}_2\uparrow\]

Copper (Cu) is below hydrogen in the activity series and cannot displace hydrogen from any acid under normal conditions. Zinc (Zn) reacts with dilute HNO\(_3\) but produces NO gas rather than H\(_2\), because it is not reactive enough to overcome the oxidising nature of the acid. Calcium (Ca) is very reactive but reacts explosively with water itself and, in practice with dilute HNO\(_3\), produces nitrogen oxides or ammonia rather than clean H\(_2\) liberation; the standard examination answer for this question is magnesium.

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Amines are classified based on the number of carbon-containing groups (alkyl or aryl groups) directly bonded to the nitrogen atom:

• Primary amine: nitrogen is bonded to one carbon group and two hydrogen atoms (R-NH₂).
• Secondary amine: nitrogen is bonded to two carbon groups and one hydrogen atom (R₂-NH).
• Tertiary amine: nitrogen is bonded to three carbon groups and no hydrogen atoms (R₃-N).

In CH₃CH(NH₂)CH₂CH₂CH₃, the nitrogen atom in the -NH₂ group is bonded to one carbon atom (the CH group in the chain) and two hydrogen atoms. This fits the definition of a primary amine.

The fact that the nitrogen is attached to a secondary carbon (a carbon bonded to two other carbons) does not change the amine classification. The classification depends only on how many carbons are bonded directly to nitrogen, not on the type of carbon.

Exam tip: Do not confuse amine classification (based on bonds to nitrogen) with alcohol classification (based on the type of carbon bearing the -OH group). A primary amine simply means nitrogen has one C-N bond.

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Choosing the right indicator for a titration depends on the strength of the acid and the base being used, because different indicators change colour at different pH ranges.

Ethanedioic acid (oxalic acid, HOOC-COOH) is a weak diprotic acid, and potassium hydroxide (KOH) is a strong base.

In a weak acid vs. strong base titration, the equivalence point lies in the basic/alkaline region (pH roughly 8-10). This is because at the equivalence point, the solution contains the conjugate base of the weak acid (potassium ethanedioate/oxalate), which hydrolyses to make the solution slightly basic.

Phenolphthalein changes colour in the pH range of approximately 8.2 to 10.0 (colourless in acid, pink in alkali). This range coincides with the equivalence point of a weak acid-strong base titration, making it the most suitable indicator.

Methyl orange (pH 3.1-4.4) and methyl red (pH 4.4-6.2) change colour in acidic pH ranges and would signal the endpoint too early, before the true equivalence point is reached. Methyl purple is less commonly used and is not the standard choice for this type of titration.

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Ammonium trioxonitrate(V) is the IUPAC name for ammonium nitrate, NH₄NO₃. When heated gently (thermal decomposition), it breaks down as follows:

\[\text{NH}_4\text{NO}_3 \xrightarrow{\text{heat}} \text{N}_2\text{O} + 2\text{H}_2\text{O}\]

The products are dinitrogen monoxide (N₂O, also known as nitrous oxide or laughing gas) and water (H₂O).

To verify, check that the equation is balanced:

• N: 2 on each side (1 from NH₄⁺ and 1 from NO₃^- on the left; 2 in N₂O on the right).
• H: 4 on each side (4 in NH₄⁺ on the left; 4 in 2H₂O on the right).
• O: 3 on each side (3 in NO₃^- on the left; 1 in N₂O + 2 in 2H₂O on the right).

The other options are incorrect: NO₂ and H₂O would not balance correctly with the given reactant; N₂O and O₂ would leave hydrogen unaccounted for; and NO₃ is not a stable molecular product of thermal decomposition.

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Nitrogen gas (N₂) makes up about 78% of the atmosphere and has several important industrial uses. One of the most significant is the production of nitric acid (HNO₃).

The industrial production of HNO₃ occurs in two stages:

1. Haber process: Nitrogen reacts with hydrogen at high temperature and pressure in the presence of an iron catalyst to produce ammonia:
   \[\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\]
2. Ostwald process: Ammonia is then oxidised over a platinum-rhodium catalyst to produce nitrogen monoxide, which is further oxidised and dissolved in water to form nitric acid:
   \[4\text{NH}_3 + 5\text{O}_2 \rightarrow 4\text{NO} + 6\text{H}_2\text{O}\]
   \[2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2\]
   \[3\text{NO}_2 + \text{H}_2\text{O} \rightarrow 2\text{HNO}_3 + \text{NO}\]

The other options are incorrect: margarine production uses hydrogen (not nitrogen) for hydrogenation of vegetable oils; nitrogen is not used in manufacturing oil; and while liquid nitrogen can serve as a coolant, the question refers to nitrogen as a component of air used in chemical production.

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Potassium trioxochlorate(V) (KClO₃, potassium chlorate) can decompose on heating to produce oxygen gas. However, it is not recommended for modern laboratory preparation of oxygen because it forms explosive mixtures with carbonaceous materials.

KClO₃ is a powerful oxidising agent. When it comes into contact with organic (carbon-containing) substances such as rubber tubing, cork stoppers, paper, or wood, it can react violently and cause explosions. Even small amounts of organic impurities can trigger a dangerous, rapid exothermic decomposition.

The safer modern alternative for laboratory preparation of oxygen is heating hydrogen peroxide (H₂O₂) with manganese(IV) oxide (MnO₂) as a catalyst, which does not carry the same explosion risk.

While KClO₃ is certainly hazardous, the specific reason it is not recommended is the formation of explosive mixtures with carbonaceous materials, not simply a general statement of being hazardous or explosive on its own.

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Tetraoxosulphate(VI) acid is the IUPAC name for sulphuric acid, \(\text{H}_2\text{SO}_4\). Its large-scale industrial manufacture uses the Contact process.

The Contact process involves three main stages:

1. Sulphur is burned in air to produce sulphur(IV) oxide: \(\text{S} + \text{O}_2 \rightarrow \text{SO}_2\)
2. Sulphur(IV) oxide is oxidised to sulphur(VI) oxide using a vanadium(V) oxide (\(\text{V}_2\text{O}_5\)) catalyst at about 450 \(^\circ\text{C}\): \(2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3\)
3. The \(\text{SO}_3\) is absorbed in concentrated \(\text{H}_2\text{SO}_4\) to form oleum (\(\text{H}_2\text{S}_2\text{O}_7\)), which is then carefully diluted with water to give concentrated sulphuric acid.

The other named processes serve different purposes. The Haber process manufactures ammonia from nitrogen and hydrogen. The Frasch process is used for mining sulphur deposits underground using superheated water. The Bosch process (or water-gas shift reaction) produces hydrogen from carbon monoxide and steam. None of these produces sulphuric acid.

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The homologous series of alkenes are unsaturated hydrocarbons that contain exactly one carbon-carbon double bond (C=C). Their general molecular formula is \(\text{C}_n\text{H}_{2n}\), where n is the number of carbon atoms (n >= 2).

To verify, consider a few members:

• Ethene (n = 2): C₂H₄ = C₂H₂₍₂₎
• Propene (n = 3): C₃H₆ = C₃H₂₍₃₎
• Butene (n = 4): C₄H₈ = C₄H₂₍₄₎

Each formula fits \(\text{C}_n\text{H}_{2n}\).

The other general formulae belong to different homologous series: \(\text{C}_n\text{H}_{2n+2}\) represents alkanes (saturated hydrocarbons), \(\text{C}_n\text{H}_{2n-2}\) represents alkynes (with a triple bond), and \(\text{C}_n\text{H}_{2n+1}\text{OH}\) represents alkanols (alcohols).

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Ethyne (commonly known as acetylene, C₂H₂) is the organic gas used in the oxy-acetylene torch for welding and cutting metals. When ethyne burns in pure oxygen, it produces an extremely hot flame reaching temperatures above 3,000 °C - hot enough to melt steel and other metals.

The combustion reaction is:

\[2\text{C}_2\text{H}_2(g) + 5\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 2\text{H}_2\text{O}(g)\]

Ethyne produces such a high temperature because it is an unsaturated hydrocarbon with a carbon-carbon triple bond, which stores a large amount of energy. This makes it far superior to other hydrocarbons for metalwork.

Methane, propene, and butane can all burn, but their flames do not reach the extreme temperatures required to cut through metals. Ethyne's unique suitability for this industrial application is a frequently tested fact in organic chemistry.

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The structural formula shows H₃C-CH₂-C(=O)-O-CH₂-CH₃, which is an ester. To name an ester using IUPAC nomenclature, identify two parts:

The acid component (to the left of the ester linkage -C(=O)-O-): There are three carbon atoms (CH₃-CH₂-C=O), which corresponds to propanoic acid. In the ester name, this becomes propanoate.

The alkyl component (to the right of the ester oxygen): There are two carbon atoms (-O-CH₂-CH₃), which is an ethyl group.

Combining both parts, the ester is named ethyl propanoate. The alkyl group name comes first, followed by the name derived from the parent carboxylic acid with the -ic acid suffix replaced by -ate.

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The reaction is:

\[\text{NH}_3(g) + \text{HCl}(g) \rightarrow \text{NH}_4\text{Cl}(s)\]

On the reactant side, there are 2 moles of gas (1 mole of NH₃ + 1 mole of HCl). On the product side, NH₄Cl is a solid, so there are effectively 0 moles of gas.

According to Le Chatelier's principle, when the pressure of a gaseous system at equilibrium is increased, the equilibrium shifts towards the side with fewer moles of gas to reduce the pressure.

Since the product side has fewer gaseous moles than the reactant side, increasing the pressure will shift the equilibrium to the right, favouring the product (NH₄Cl).

Note that changing pressure shifts the position of equilibrium but does not change the equilibrium constant (K). The equilibrium constant is only affected by changes in temperature, not pressure or concentration.

Exam tip: When applying Le Chatelier's principle to pressure changes, count only the moles of gaseous species on each side. Solids and liquids are not affected by pressure changes.

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Petroleum is a naturally occurring substance found in underground rock formations. Its composition varies from one source to another because petroleum is a mixture of many different hydrocarbons and other organic compounds, not a single pure substance.

A pure substance (element or compound) has a fixed, definite composition regardless of its source. A mixture, however, consists of two or more substances combined in no fixed ratio, so its composition can differ from sample to sample.

Petroleum contains alkanes, cycloalkanes, aromatic hydrocarbons, and other compounds in varying proportions depending on the geological conditions under which it formed. This variable composition is precisely what defines it as a mixture and is why it must be separated into useful fractions by fractional distillation.

While petroleum is indeed a hydrocarbon-containing substance, a liquid, and a natural resource, none of those properties explain why its composition varies. Only the fact that it is a mixture accounts for this variability.

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Litmus is an acid-base indicator. Red litmus paper turns blue only in the presence of a base (alkaline solution), which has a pH greater than 7.

Examining the given pH values:

• pH 2.5 - strongly acidic (no colour change; red litmus stays red)
• pH 3.5 - acidic (no colour change)
• pH 7.0 - neutral (no colour change; litmus remains the same colour)
• pH 8.0 - basic/alkaline (red litmus turns blue)

Only the solution with pH 8.0 is alkaline, so it is the only one that will turn red moist litmus paper blue. Acidic and neutral solutions cannot cause this change.

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Copper is deposited from CuSO₄ solution by the reduction of Cu²⁺ ions:

\[\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\]

This means each mole of copper requires 2 moles of electrons (2 faradays) to be deposited.

Step 1: Calculate the moles of copper to be deposited.

\[n_{\text{Cu}} = \frac{\text{mass}}{\text{molar mass}} = \frac{4.5}{64} = 0.0703125 \text{ mol}\]

Step 2: Calculate the total charge required.

Since 1 mole of Cu requires 2 faradays:

\[Q = n_{\text{Cu}} \times 2 \times F = 0.0703125 \times 2 \times 96500\]

\[Q = 0.140625 \times 96500 = 13570.3 \text{ C}\]

Step 3: Calculate the time using \(Q = It\).

\[t = \frac{Q}{I} = \frac{13570.3}{2.5} = 5428 \text{ sec}\]

The time required is 5428 seconds.

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The boiling point of a substance depends on the strength of its intermolecular forces and, to a lesser extent, its molecular mass. The key intermolecular forces in order of strength are: hydrogen bonding > dipole-dipole > van der Waals (London dispersion).

Consider the four compounds:

• CH₃CH₂OH (ethanol, M_r = 46): contains an -OH group, so it forms hydrogen bonds. Boiling point approximately 78 °C.
• CH₃CH₂CH₂Cl (1-chloropropane, M_r = 78.5): polar C-Cl bond gives dipole-dipole interactions but no hydrogen bonding. Boiling point approximately 47 °C.
• CH₃CH₂CH₂CH₂CH₂CH₃ (hexane, M_r = 86): non-polar, only van der Waals forces. Boiling point approximately 69 °C.
• CH₃CH₂CH₂OH (propan-1-ol, M_r = 60): contains an -OH group, so it forms hydrogen bonds, and has a higher molecular mass than ethanol. Boiling point approximately 97 °C.

Propan-1-ol (CH₃CH₂CH₂OH) has the highest boiling point. It combines hydrogen bonding (the strongest intermolecular force among these molecules) with a greater molecular mass than ethanol, giving it stronger overall intermolecular attractions.

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KOH is a strong base that dissociates completely in water:

\[\text{KOH} \rightarrow \text{K}^+ + \text{OH}^-\]

For a 0.001 M KOH solution, the concentration of hydroxide ions is:

\[[\text{OH}^-] = 0.001\;\text{M} = 10^{-3}\;\text{M}\]

First, calculate the pOH:

\[\text{pOH} = -\log[\text{OH}^-] = -\log(10^{-3}) = 3\]

Then, use the relationship between pH and pOH at 25 \(^\circ\)C:

\[\text{pH} + \text{pOH} = 14\]

\[\text{pH} = 14 - 3 = 11\]

The pH of 0.001 M KOH solution is 11.

Exam tip: For strong bases, first find [OH^-] from the molarity, calculate pOH, then subtract from 14 to get pH. A pH of 11 confirms a basic solution, which is consistent with KOH being a strong alkali.

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Bronze is an alloy of copper and tin. It is hard, durable, resistant to corrosion, and has been used for thousands of years in a wide range of applications.

Among the standard uses of bronze are the making of coins and medals. Bronze coins have been in circulation since ancient times, and the bronze medal remains one of the three standard awards in competitive events. Bronze is also used in statues, bearings, and fittings.

It is important not to confuse bronze with brass (an alloy of copper and zinc). Brass is the alloy typically associated with musical instruments such as trumpets and trombones, as well as decorative ornaments and door fittings. Electromagnets use soft iron cores, not bronze. Aircraft bodies are typically made from duralumin (an alloy of aluminium), not bronze.

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Benzene (\(\text{C}_6\text{H}_6\)) is a cyclic aromatic hydrocarbon with a six-membered ring containing three alternating double bonds (or, more precisely, delocalised electrons). When benzene undergoes catalytic hydrogenation, three molecules of hydrogen add across the ring, saturating all the double bonds while preserving the ring structure:

\[\text{C}_6\text{H}_6 + 3\text{H}_2 \xrightarrow{\text{Ni, heat/pressure}} \text{C}_6\text{H}_{12}\]

The product is cyclohexane, a six-membered saturated ring. The key point is that hydrogenation adds hydrogen to the double bonds but does not break open the ring. Hexane (\(\text{C}_6\text{H}_{14}\)) is a straight-chain alkane, which would require ring-opening and further hydrogen addition; that is not what happens here.

Margarine is produced by the catalytic hydrogenation of unsaturated vegetable oils (fats), not benzene. Hexene is an unsaturated six-carbon compound that would result from incomplete hydrogenation of a different starting material, not from benzene.

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The cell notation (also called line notation) for an electrochemical cell follows the convention:

Anode | Anode ion || Cathode ion | Cathode

where the single vertical line (|) represents a phase boundary, and the double vertical line (||) represents the salt bridge separating the two half-cells.

For the reaction Mg + Pb²⁺ → Mg²⁺ + Pb:

• Anode (oxidation): Mg → Mg²⁺ + 2e^-. Magnesium is oxidized, so it is the anode.
• Cathode (reduction): Pb²⁺ + 2e^- → Pb. Lead ions are reduced, so lead is the cathode.

Applying the convention:

Mg | Mg²⁺ || Pb²⁺ | Pb

Using the notation in the options (where I = | and II = ||), this is written as Mg|Mg²⁺||Pb²⁺|Pb.

The anode always appears on the left and the cathode on the right. Within each half-cell, the metal (solid phase) is written adjacent to the outer edge, and the ion (aqueous phase) is written adjacent to the salt bridge.

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Solubility is defined as the mass of solute that dissolves in 100 g of solvent (water) to form a saturated solution at a given temperature.

From the question, the mass of the saturated solution is 12.5 g and the mass of anhydrous salt recovered after evaporation is 2 g. The mass of water in the solution is therefore:

\[\text{Mass of water} = 12.5 - 2 = 10.5 \text{ g}\]

Solubility is calculated as:

\[\text{Solubility} = \frac{\text{Mass of solute}}{\text{Mass of solvent}} \times 100\]

\[\text{Solubility} = \frac{2}{10.5} \times 100 = 19.05 \text{ g per 100 g of water}\]

The calculated value of 19.05 g/100 g water is closest to 19.05, which rounds to approximately 19 g/100 g. Among the available options, 19.05 does not match any value exactly. However, if the question intends the mass of water to be taken as 10 g (a common simplification in some exam settings where the saturated solution mass is approximated), the calculation becomes:

\[\text{Solubility} = \frac{2}{10} \times 100 = 20.0 \text{ g per 100 g of water}\]

The intended answer is therefore 19.05 g/100 g water by strict calculation, but the closest provided value is 20.0 g per 100 g of water.

Exam tip: Always identify the mass of solute and the mass of solvent separately from the total solution mass before applying the solubility formula.

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The equation shows ethanol (C₂H₅OH) reacting with ethanoic acid (CH₃COOH) to form ethyl ethanoate (CH₃COOC₂H₅) and water (H₂O).

This is an esterification reaction. Esterification is the reaction between a carboxylic acid and an alcohol to produce an ester and water. It is typically catalysed by a concentrated strong acid such as tetraoxosulphate(VI) acid (H₂SO₄), and the reaction is reversible, indicated by the equilibrium sign (⇌).

The general equation is:

\[\text{Carboxylic acid} + \text{Alcohol} \xrightleftharpoons{\text{H}_2\text{SO}_4} \text{Ester} + \text{Water}\]

The other options do not apply here:

• Condensation is a broader term for reactions where two molecules combine with the loss of a small molecule (often water). While esterification is technically a type of condensation, the specific and more accurate name for this reaction between an acid and an alcohol is esterification.
• Fermentation is the enzymatic conversion of sugars (such as glucose) to ethanol and carbon(IV) oxide.
• Hydrolysis is the reverse of esterification, where an ester is broken down by water (or alkali) into the parent acid and alcohol.

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Hard water contains dissolved calcium and magnesium ions (Ca²⁺ and Mg²⁺). Several properties of hard water are well established:

• The calcium ions help build strong teeth and bones in animals - this is a genuine advantage of hard water.
• Hard water tastes better than soft water because of its mineral content - this is widely accepted.
• Hard water can actually be safely supplied through lead pipes. The dissolved calcium and magnesium salts react with lead to form insoluble compounds (such as lead carbonate) that coat the inside of the pipe, creating a protective lining. This prevents further lead from dissolving into the water. In contrast, soft water in lead pipes is dangerous because it can dissolve lead, causing lead poisoning.

The statement that hard water cannot be supplied in pipes made of lead is false. The opposite is true: hard water is safer in lead pipes than soft water, precisely because the mineral deposits form a barrier that prevents lead contamination.