2024 JAMB Chemistry Past Questions and Answers

Questions 1–69

1. 1.

   The chemical formula for potassiumhexacyanoferrate(II) is

   1. A.[Fe(CN)$_6$]$^{4-}$
   2. B.K$_3$Fe(CN)$_6$
   3. C.Fe(CN)$_6$
   4. D.K$_4$Fe(CN)$_6$

   Show answer

   Answer: D. K$_4$Fe(CN)$_6$

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2. 2.

   The IUPAC nomenclature of the compound above is

   1. A.propan-1-ol
   2. B.butan-1-ol
   3. C.2-methyl propan-1-ol
   4. D.2-methyl propan-2-ol

   Show answer

   Answer: B. butan-1-ol

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3. 3.

   When n = 3, the quantum number of an element is

   1. A.3
   2. B.9
   3. C.18
   4. D.21

   Show answer

   Answer: C. 18

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4. 4.

   When a specie undergoes oxidation, its

   1. A.oxidation number increases
   2. B.oxidation number decreases
   3. C.oxidation number remains constant
   4. D.acts as oxidizing agent

   Show answer

   Answer: A. oxidation number increases

   AI explanation

   Analysis of the Question

   This question tests your understanding of the fundamental definitions in redox (reduction-oxidation) reactions. You are asked to identify what happens to the oxidation number of a chemical species when it undergoes oxidation.

   Concept Explanation

   In chemistry, redox reactions involve the transfer of electrons. We use the mnemonic OIL RIG to remember the definitions:

   • Oxidation Is Loss (of electrons)
   • Reduction Is Gain (of electrons)

   When an atom or ion loses electrons, it becomes more positive (or less negative). Because electrons carry a negative charge, losing them causes the oxidation number of the species to increase. Conversely, when a species gains electrons (reduction), its oxidation number decreases.

   Step-by-Step Solution

   1. Define Oxidation: Oxidation is the process where a species loses electrons.
   2. Analyze the effect on charge: Since electrons are negatively charged, removing them from a species makes the overall charge of that species more positive.
   3. Relate to Oxidation Number: The oxidation number is essentially the charge an atom would have if all bonds were ionic. Therefore, if a species loses electrons, its oxidation number must shift in the positive direction (increase). • Example: In the reaction $Zn \rightarrow Zn^{2+} + 2e^-$, the oxidation number of Zinc increases from $0$ to $+2$. This is oxidation.
   4. Evaluate the Options: • A: Oxidation number increases (Correct, as explained above). • B: Oxidation number decreases (This describes reduction). • C: Oxidation number remains constant (This describes a non-redox reaction). • D: Acts as an oxidizing agent (Incorrect; a species that undergoes oxidation is actually the reducing agent).

   Conclusion

   Exam Tip

   Always remember the "Number Line Rule":

   • If the oxidation number moves to the right (e.g., $-2$ to $0$, or $+1$ to $+3$), it is Oxidation.
   • If the oxidation number moves to the left (e.g., $+2$ to $0$, or $0$ to $-1$), it is Reduction.

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5. 5.

   The indicator used in a titration between strong acid and weak base is

   1. A.methyl orange
   2. B.phenolphthalein
   3. C.bromothymol blue
   4. D.methyl red

   Show answer

   Answer: A. methyl orange

   AI explanation

   Analysis

   The question asks for the appropriate chemical indicator to use when performing a titration between a strong acid and a weak base.

   Concept Explanation

   In acid-base titrations, the choice of indicator depends on the pH range at which the indicator changes color (the "pH range of the indicator") and the pH at the equivalence point of the reaction.

   1. Strong Acid + Strong Base: The equivalence point is at pH 7. Any indicator (like phenolphthalein or methyl orange) works.
   2. Strong Acid + Weak Base: The resulting salt is acidic because the conjugate acid of the weak base hydrolyzes in water. Therefore, the equivalence point occurs at a pH lower than 7 (typically between pH 3 and 5). We need an indicator that changes color in this acidic range.
   3. Weak Acid + Strong Base: The equivalence point is at a pH higher than 7 (typically between 8 and 10). Phenolphthalein is the standard choice here.

   Step-by-Step Solution

   • Step 1: Identify the nature of the reaction. A strong acid (e.g., HCl) reacting with a weak base (e.g., $NH_3$) produces an acidic salt.
   • Step 2: Determine the pH at the equivalence point. Because the acid is strong and the base is weak, the solution will be acidic at the endpoint (pH < 7).
   • Step 3: Evaluate the indicators: • Methyl orange: Changes color in the pH range of 3.1 – 4.4. This matches the acidic equivalence point of a strong acid-weak base titration. • Phenolphthalein: Changes color in the pH range of 8.2 – 10.0. This is used for weak acid-strong base titrations. • Bromothymol blue: Changes color around pH 6.0 – 7.6, which is better suited for strong acid-strong base titrations.

   Since methyl orange changes color in the acidic range, it is the correct choice for this titration.

   Exam Tip

   Remember the mnemonic "SAWB-MO":

   • Strong Acid + Weak Base = Methyl Orange.
   • Weak Acid + Strong Base = Phenolphthalein (WASP).

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6. 6.

   An oxide of nitrogen that can rekindle a glowing splint is

   1. A.nitrogen(I)oxide
   2. B.nitrogen(II)oxide
   3. C.nitrogen(IV)oxide
   4. D.dinitrogen tetraoxide

   Show answer

   Answer: A. nitrogen(I)oxide

   AI explanation

   Analysis

   The question asks for an oxide of nitrogen that has the ability to support combustion (rekindle a glowing splint). In chemistry, a substance that rekindles a glowing splint is typically an oxidizing agent that decomposes to release oxygen gas ($O_2$).

   Concept Explanation

   To solve this, we must look at the chemical properties of nitrogen oxides:

   1. Nitrogen(I) oxide ($N_2O$): Also known as nitrous oxide or "laughing gas." It is a neutral oxide. When heated, it decomposes into nitrogen and oxygen: $$2N_2O(g) \rightarrow 2N_2(g) + O_2(g)$$ Because it releases oxygen upon decomposition, it supports combustion and can rekindle a glowing splint.
   2. Nitrogen(II) oxide ($NO$): A neutral oxide that is colorless. It reacts readily with oxygen to form nitrogen(IV) oxide ($NO_2$) but does not spontaneously decompose to release oxygen in a way that supports a glowing splint.
   3. Nitrogen(IV) oxide ($NO_2$): A reddish-brown, acidic gas. It is a strong oxidizing agent, but it does not decompose to release oxygen as readily as $N_2O$ under the conditions required to rekindle a glowing splint.
   4. Dinitrogen tetraoxide ($N_2O_4$): This is the dimer of nitrogen(IV) oxide. It shares similar chemical properties with $NO_2$ and does not act as a source of oxygen for a glowing splint.

   Step-by-Step Solution

   • Step 1: Identify the requirement. A glowing splint rekindles in the presence of oxygen. Therefore, we are looking for an oxide that decomposes to produce oxygen gas.
   • Step 2: Evaluate the thermal stability of the options. Nitrogen(I) oxide ($N_2O$) is known to be thermally unstable at high temperatures.
   • Step 3: Write the decomposition equation. As shown above, $2N_2O \rightarrow 2N_2 + O_2$. The production of $O_2$ is the key factor here.

   Conclusion

   Nitrogen(I) oxide ($N_2O$) is the only oxide among the options that effectively decomposes to supply the oxygen necessary to rekindle a glowing splint.

   Exam Tip: Always remember the "Neutral Oxides" of nitrogen: $N_2O$ and $NO$. Among these, $N_2O$ is the one famous for its ability to support combustion, similar to how oxygen gas ($O_2$) behaves.

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7. 7.

   When Sulphur(IV)oxide is passed into solution of acidified tetraoxomanganate(VII), the colour changes from

   1. A.blue to green
   2. B.pink to yellow
   3. C.purple to colourless
   4. D.Orange to green

   Show answer

   Answer: C. purple to colourless

   AI explanation

   Analysis

   The question asks for the color change observed when Sulphur(IV) oxide ($\text{SO}_2$) reacts with an acidified solution of tetraoxomanganate(VII) ($\text{KMnO}_4$). This is a classic redox reaction frequently tested in JAMB Chemistry.

   Concept Explanation

   1. The Reactants: • Acidified potassium tetraoxomanganate(VII) ($\text{KMnO}_4$) is a powerful oxidizing agent. In its acidic state, the manganese ion is in the $+7$ oxidation state, which gives the solution a deep purple color. • Sulphur(IV) oxide ($\text{SO}_2$) acts as a reducing agent.

   2. The Reaction: When $\text{SO}_2$ is bubbled through the acidified $\text{KMnO}_4$ solution, the $\text{SO}_2$ is oxidized to sulphate ions ($\text{SO}_4^{2-}$), and the manganese(VII) ions are reduced to manganese(II) ions ($\text{Mn}^{2+}$).

   The balanced ionic equation is: 2\text{MnO}_4^-_{(aq)} + 5\text{SO}_{2(g)} + 2\text{H}_2\text{O}_{(l)} \rightarrow 2\text{Mn}^{2+}_{(aq)} + 5\text{SO}_4^{2-}_{(aq)} + 4\text{H}^+_{(aq)}

   3. The Color Change: • The $\text{Mn}^{2+}$ ion in aqueous solution is virtually colourless (or very pale pink, which is often considered colourless in the context of this specific reaction). • Therefore, the intense purple color of the permanganate solution disappears as the reaction proceeds.

   Step-by-Step Solution

   • Step 1: Identify the initial state. Acidified $\text{KMnO}_4$ is purple.
   • Step 2: Identify the reaction type. This is a redox reaction where $\text{KMnO}_4$ is reduced.
   • Step 3: Determine the final state. The reduction of $\text{Mn}^{7+}$ to $\text{Mn}^{2+}$ results in a loss of color.
   • Step 4: Match with options. The change is from purple to colourless.

   Conclusion

   The reaction causes the purple color of the permanganate to be discharged.

   Exam Tip: Always remember that acidified $\text{KMnO}_4$ is a "color-killer" for reducing agents. If you see it in a question, look for the option that involves the disappearance of the purple color!

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8. 8.

   The group VIII elements are the inert gases because they

   1. A.have lone pair of electrons on their valence shells
   2. B.all have completely filled valence shells
   3. C.are all solids
   4. D.are all polyatomic molecules

   Show answer

   Answer: B. all have completely filled valence shells

   AI explanation

   Analysis

   The question asks for the reason why Group VIII elements (also known as Group 0 or Noble Gases) are chemically inert, meaning they are generally unreactive under standard conditions.

   Concept Explanation

   The chemical reactivity of an element is primarily determined by its electron configuration, specifically the electrons in its outermost shell (the valence shell).

   • The Octet Rule: Atoms tend to gain, lose, or share electrons to achieve a stable electron configuration, which usually means having a full outer shell (typically 8 electrons, or 2 for Helium).
   • Noble Gases: Elements in Group VIII (Helium, Neon, Argon, Krypton, Xenon, and Radon) already possess a stable, "closed-shell" configuration. Their valence shells are completely filled with electrons. Because they are already stable, they have no "need" to gain, lose, or share electrons with other atoms, which is why they do not readily form chemical bonds.

   Step-by-Step Solution

   1. Examine Option A: While some noble gases can form compounds under extreme conditions (like Xenon fluorides), they do not have "lone pairs" in the way that makes them reactive in the context of standard bonding; their stability comes from the fullness of the shell.
   2. Examine Option B: This is the definition of their stability. Helium has a full $1s^2$ shell, and the others have a full $ns^2 np^6$ configuration. This complete filling creates a high ionization energy and low electron affinity, making them inert.
   3. Examine Option C: This is factually incorrect. Noble gases are gases at room temperature, not solids.
   4. Examine Option D: This is factually incorrect. Noble gases are monatomic (they exist as single atoms), not polyatomic.

   Conclusion

   The stability and inert nature of Group VIII elements are due to their completely filled valence shells, which provide a low-energy, stable state.

   Exam Tip: Whenever you see a question about "Noble Gases" or "Inert Gases," always look for keywords like "stable," "full valence shell," "octet configuration," or "monatomic." If you see options mentioning "solids" or "polyatomic," you can immediately eliminate them!

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9. 9.

   Aqueous solution of sodium hydroxide can be used to test for the presence of : I. Ca$^{2+}$ ,   II. Zn$^{2+}$ ,  III. Cu$^{2+}$

   1. A.I only
   2. B.I and III
   3. C.I, II and III
   4. D.I and II only

   Show answer

   Answer: C. I, II and III

   AI explanation

   Analysis of the Question

   The question asks us to identify which of the given metal ions ($Ca^{2+}$, $Zn^{2+}$, and $Cu^{2+}$) can be identified using an aqueous solution of sodium hydroxide ($NaOH$). In qualitative analysis, $NaOH$ is used as a reagent to form characteristic precipitates with various metal cations.

   Explaining the Concept

   When aqueous $NaOH$ is added to solutions containing these metal ions, they react to form metal hydroxides. We look for distinct physical changes, such as the color of the precipitate or its solubility in excess $NaOH$.

   1. $Ca^{2+}$ (Calcium ion): When $NaOH$ is added, it forms a white precipitate of calcium hydroxide: $Ca^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Ca(OH)_{2(s)}$ (White precipitate)
   2. $Zn^{2+}$ (Zinc ion): When $NaOH$ is added, it forms a white precipitate of zinc hydroxide. Crucially, this precipitate is amphoteric and dissolves in excess $NaOH$ to form a colorless solution of sodium tetrahydroxozincate(II): $Zn^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Zn(OH)_{2(s)}$ (White precipitate) $Zn(OH)_{2(s)} + 2OH^-_{(aq)} \rightarrow [Zn(OH)_4]^{2-}_{(aq)}$ (Colorless solution)
   3. $Cu^{2+}$ (Copper(II) ion): When $NaOH$ is added, it forms a characteristic pale blue precipitate of copper(II) hydroxide: $Cu^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Cu(OH)_{2(s)}$ (Pale blue precipitate)

   Step-by-Step Solution

   • Step 1: Test for $Ca^{2+}$: $NaOH$ produces a white precipitate. This confirms the presence of $Ca^{2+}$.
   • Step 2: Test for $Zn^{2+}$: $NaOH$ produces a white precipitate that dissolves in excess. This confirms the presence of $Zn^{2+}$.
   • Step 3: Test for $Cu^{2+}$: $NaOH$ produces a pale blue precipitate. This confirms the presence of $Cu^{2+}$.

   Since all three ions produce distinct, observable reactions with aqueous sodium hydroxide, $NaOH$ is a valid reagent for testing the presence of all three.

   Conclusion

   Therefore, the correct option is C.

   Exam Tip

   When dealing with qualitative analysis questions, always remember the "Amphoteric Trio": Zinc ($Zn^{2+}$), Lead ($Pb^{2+}$), and Aluminium ($Al^{3+}$). These three form white precipitates with $NaOH$ that dissolve in excess. Knowing this will help you quickly identify them in both JAMB and WASSCE practical-based theory questions!

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10. 10.

    The basicity of tetraoxophosphate(V) acid is

    1. A.1
    2. B.4
    3. C.2
    4. D.3

    Show answer

    Answer: D. 3

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11. 11.

    The electronic configuration of an atom of Nitrogen is 1s$^2$ 2s$^2$ 2p$^1_x$ 2p$^1_y$ 2p$^1_z$ because the atom is

    1. A.in an excited state
    2. B.undergoing energy changes
    3. C.unstable and ready to combine with another
    4. D.simply obeying rules governing electron configuration

    Show answer

    Answer: D. simply obeying rules governing electron configuration

    AI explanation

    Analysis

    The question asks why the electronic configuration of Nitrogen ($Z=7$) is written as $1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$ rather than filling one orbital completely before moving to the next.

    Concept: Hund’s Rule of Maximum Multiplicity

    To understand this, we look at Hund’s Rule. This rule states that for a given electron configuration, the term with the greatest multiplicity (the most unpaired electrons) has the lowest energy.

    In the $2p$ subshell, there are three degenerate orbitals ($2p_x, 2p_y, 2p_z$). According to Hund's Rule, electrons will occupy these orbitals singly first with parallel spins before pairing up. This minimizes electron-electron repulsion, making the atom more stable.

    Step-by-Step Solution

    1. Identify the Atomic Number: Nitrogen has 7 electrons.
    2. Distribute the electrons: • $1s$ orbital: 2 electrons ($1s^2$) • $2s$ orbital: 2 electrons ($2s^2$) • $2p$ subshell: 3 electrons remaining.
    3. Apply the Rules: • Aufbau Principle: Electrons fill lower energy levels first. • Pauli Exclusion Principle: No two electrons can have the same four quantum numbers. • Hund’s Rule: Electrons fill degenerate orbitals (orbitals with the same energy) singly before pairing.
    4. Conclusion: Writing the configuration as $2p_x^1 2p_y^1 2p_z^1$ is the standard, ground-state way of representing the distribution of electrons in the $p$-orbitals to satisfy Hund's Rule. It is not an excited state; it is the most stable arrangement for a Nitrogen atom. Therefore, the atom is simply following the established laws of quantum mechanics and electron configuration.

    Evaluation of Options

    • A & B: Incorrect. This is the ground state of Nitrogen, not an excited state.
    • C: Incorrect. This configuration is the most stable state for Nitrogen.
    • D: Correct. This notation is the standard way to represent the ground state of Nitrogen in accordance with Hund's Rule.

    Exam Tip

    Whenever you see $p$, $d$, or $f$ orbitals, remember the "Bus Seat Rule": Just like people entering an empty bus prefer to sit in an empty row before sitting next to someone else, electrons prefer to occupy empty orbitals before pairing up to avoid repulsion!

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12. 12.

    In the conductance of aqueous CuSO$_4$ solution, the current carriers are the

    1. A.mobile electrons
    2. B.metallic ions
    3. C.hydrated ions
    4. D.hydrated electrons

    Show answer

    Answer: C. hydrated ions

    AI explanation

    Analysis

    The question asks for the identity of the particles responsible for carrying electric current through an aqueous solution of Copper(II) tetraoxosulphate(VI) ($CuSO_4$).

    Concept Explanation

    In chemistry, electrical conductivity is determined by the presence of mobile charge carriers.

    1. In metals: Current is carried by delocalized electrons (mobile electrons).
    2. In aqueous solutions (electrolytes): Current is carried by ions. When an ionic compound like $CuSO_4$ dissolves in water, it dissociates into its constituent ions: $$CuSO_4(aq) \rightarrow Cu^{2+}(aq) + SO_4^{2-}(aq)$$
    3. Hydration: In an aqueous environment, these ions are surrounded by water molecules. This process is called hydration. The water molecules form a "shell" around the ions, making them hydrated ions. These hydrated ions are free to move toward the electrodes (cathode and anode) when a potential difference is applied, thereby conducting electricity.

    Step-by-Step Solution

    1. Identify the substance: $CuSO_4$ is an ionic salt.
    2. Dissociation: Upon dissolving in water, it breaks into $Cu^{2+}$ and $SO_4^{2-}$ ions.
    3. Interaction with solvent: Because the solvent is water, these ions become surrounded by water dipoles. They are no longer "naked" ions; they are hydrated ions.
    4. Movement: Under the influence of an electric field, these hydrated ions migrate through the solution, acting as the charge carriers.
    5. Evaluation of options: • A is incorrect because electrons carry current in wires/metals, not in aqueous solutions. • B is incomplete because it ignores the role of the solvent (water) in the solution phase. • C is correct because the ions are hydrated in the aqueous medium. • D is incorrect because "hydrated electrons" are specific, highly reactive species not involved in standard electrolytic conduction of salt solutions.

    Conclusion

    The current carriers in an aqueous solution of an electrolyte are the hydrated ions resulting from the dissociation of the solute.

    Exam Tip: Always remember the rule: Metals conduct via electrons; Electrolytes (solutions/molten salts) conduct via ions. If the question specifies "aqueous," look for the option mentioning ions!

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13. 13.

    C$_x$H$_y$O + 5O$_2$  →  4CO$_2$  +  4H$_2$O C$_x$H$_y$O in the equation is

    1. A.CH$_2$O
    2. B.C$_2$H$_6$O
    3. C.CH$_4$O
    4. D.C$_2$H$_4$O

    Show answer

    Answer: D. C$_2$H$_4$O

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14. 14.

    Determine the half-life of a first order reaction with constant 4.5 x 10$^{-3}$ sec$^{-1}$.

    1. A.154 s
    2. B.15.4 s
    3. C.1540 s
    4. D.1.54 s

    Show answer

    Answer: A. 154 s

    AI explanation

    Analysis

    To solve this problem, we need to understand the relationship between the rate constant ($k$) and the half-life ($t_{1/2}$) for a first-order reaction.

    Concept

    For any first-order reaction, the half-life is the time required for the concentration of a reactant to decrease to half of its initial value. The formula connecting the rate constant ($k$) and the half-life ($t_{1/2}$) is:

    $t_{1/2} = \frac{\ln(2)}{k}$

    Since $\ln(2) \approx 0.693$, the formula simplifies to:

    $t_{1/2} = \frac{0.693}{k}$

    Solution

    Step 1: Identify the given values.

    • Rate constant ($k$) = $4.5 \times 10^{-3} \text{ s}^{-1}$

    Step 2: Substitute the values into the formula.

    $t_{1/2} = \frac{0.693}{4.5 \times 10^{-3}}$

    Step 3: Perform the calculation.

    • First, handle the power of 10: $\frac{1}{10^{-3}} = 10^3 = 1000$
    • Now, calculate the division: $\frac{0.693}{4.5} \times 1000$
    • $0.693 \div 4.5 = 0.154$
    • $0.154 \times 1000 = 154 \text{ s}$

    The half-life of the reaction is 154 seconds.

    Conclusion

    Comparing our result with the provided options, we can see that it matches option A.

    Exam Tip

    When dealing with first-order kinetics in JAMB, always remember the constant 0.693. If you ever forget the formula, just remember that the half-life is inversely proportional to the rate constant. If $k$ is a very small number (like $10^{-3}$), your answer for $t_{1/2}$ should be a relatively large number. This helps you quickly eliminate impossible options!

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15. 15.

    The salt formed when strong acid reacts with strong base is

    1. A.normal
    2. B.double
    3. C.complex
    4. D.basic

    Show answer

    Answer: A. normal

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16. 16.

    A gas when mixed with oxygen, it produces a very hot and early controllable flame. What is the name of the flame and where is it used?

    1. A.Acetylene flame; miners' lamb
    2. B.Oxy-ethylene ; hunters' torch
    3. C.Oxy-ethylene flame; miners' lamb
    4. D.Oxy-ethylene flame ; cutting and welding metals

    Show answer

    Answer: D. Oxy-ethylene flame ; cutting and welding metals

    AI explanation

    Analysis

    The question asks for the identification of a specific gas mixture that produces a high-temperature, controllable flame, and its practical application in industry.

    Concept Explanation

    The flame in question is produced by the combustion of ethyne (commonly known as acetylene) in the presence of pure oxygen. When ethyne ($C_2H_2$) burns in oxygen, it produces an extremely high temperature (reaching up to 3000°C). This is known as an oxy-acetylene flame.

    Note: In some older or specific curriculum contexts, "oxy-ethylene" is occasionally used interchangeably or erroneously in textbooks, but the chemical process described—a very hot, controllable flame used for metalwork—refers to the oxy-acetylene process.

    Step-by-Step Solution

    1. Identify the gas: The gas is ethyne ($C_2H_2$). When mixed with oxygen, it undergoes complete combustion to produce a very hot flame.
    2. Identify the application: Because this flame produces enough heat to melt steel and iron, it is the standard tool used in welding (joining metals) and cutting (slicing through thick metal plates).
    3. Evaluate the options: • Miners' lamps (Options A and C) typically use acetylene, but they are designed to produce light, not the high-intensity heat required for cutting or welding. • The term "Oxy-ethylene" in the options is a common nomenclature variation found in some JAMB question banks for this specific question, referring to the oxy-acetylene process. • Option D correctly identifies the application as "cutting and welding metals," which is the primary industrial use of this flame.

    Conclusion

    The oxy-acetylene flame is essential in engineering and construction for metal fabrication. While the terminology in the options uses "oxy-ethylene," the functional description of "cutting and welding metals" makes Option D the only scientifically accurate application provided.

    Exam Tip: Whenever you see questions about "very hot flames" in Chemistry, immediately think of welding. If the question mentions "miners' lamp," look for the use of acetylene for illumination, not for heat.

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17. 17.

    Hydrogen chloride gas and ammonia can be used to demonstrate the fountain experiment because they are

    1. A.light
    2. B.volatile
    3. C.very soluble in water
    4. D.heavy gases

    Show answer

    Answer: C. very soluble in water

    AI explanation

    1. Analyze the question

    The question asks for the specific physical property of hydrogen chloride (HCl) gas and ammonia ($NH_3$) gas that makes them suitable for the "fountain experiment."

    2. Explain the concept

    The fountain experiment is a classic demonstration in chemistry used to show the high solubility of certain gases in water.

    • How it works: A round-bottom flask is filled with the gas (HCl or $NH_3$). A glass tube is inserted into the flask, and the other end is placed in a beaker of water (often containing an indicator like phenolphthalein).
    • The mechanism: When a small amount of water is introduced into the flask, the gas dissolves almost instantaneously. This creates a partial vacuum (a sudden drop in pressure) inside the flask. Because the atmospheric pressure outside is higher than the pressure inside the flask, water is forced up the tube into the flask, creating a "fountain."

    3. Show the solution

    • Hydrogen Chloride (HCl): This is a highly polar covalent compound. It dissolves extremely well in water to form hydrochloric acid.
    • Ammonia ($NH_3$): This gas is also highly polar and reacts with water to form ammonium hydroxide ($NH_4OH$). Its solubility is so high that one volume of water can dissolve hundreds of volumes of ammonia gas.
    • Evaluation of options: • A & D: Being "light" or "heavy" (density) does not cause the pressure drop required for the fountain effect. • B: While both gases are volatile, volatility refers to the ease of turning into a gas, which is not the primary reason for the fountain effect. • C: The high solubility is the critical factor. Without the gas dissolving rapidly into the water, the pressure inside the flask would not decrease, and the water would not be drawn up.

    4. Identify the answer

    The correct option is C. Both gases are chosen specifically because they are extremely soluble in water, which allows them to create the vacuum necessary for the experiment to work.

    5. Exam Tip

    Whenever you see the "fountain experiment" in an exam, immediately associate it with high solubility. Remember: the faster the gas dissolves, the more dramatic the fountain will be!

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18. 18.

    Hydrochloric acid is not suitable in the preparation of ethanoic acid because it

    1. A.is too acidic
    2. B.is too volatile
    3. C.can attack cork and rubber used as containers
    4. D.turns blue litmus red

    Show answer

    Answer: B. is too volatile

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19. 19.

    Kerosene is used as solvent for

    1. A.Paints
    2. B.Sulphur
    3. C.Gums
    4. D.Fats

    Show answer

    Answer: A. Paints

    AI explanation

    Analysis

    The question asks for the specific application of kerosene as a solvent. In chemistry, "like dissolves like"—non-polar substances dissolve in non-polar solvents. Kerosene is a non-polar hydrocarbon mixture derived from petroleum, making it an excellent solvent for other non-polar substances.

    Concept Explanation

    Kerosene is a mixture of hydrocarbons. Because it is non-polar, it is widely used in industrial and domestic settings to dissolve substances that do not dissolve in water (a polar solvent).

    • Paints: Oil-based paints use organic solvents (like turpentine or kerosene) to keep the pigments and binders in a liquid state.
    • Sulphur: Sulphur is typically dissolved in carbon disulfide ($CS_2$).
    • Gums: Natural gums are often complex polysaccharides and are generally insoluble in non-polar solvents like kerosene.
    • Fats: While kerosene can dissolve some fats, its primary commercial and industrial application as a solvent is specifically formulated for thinning oil-based paints and varnishes.

    Step-by-Step Solution

    1. Evaluate the options: • Option A (Paints): Kerosene is commonly used as a thinner for oil-based paints to adjust their viscosity for easier application. • Option B (Sulphur): Sulphur is not effectively dissolved by kerosene; it requires specific solvents like carbon disulfide. • Option C (Gums): Gums are generally insoluble in petroleum-based solvents. • Option D (Fats): While kerosene has some grease-cutting properties, "Paints" is the standard industrial classification for its use as a solvent.

    2. Conclusion: Among the given choices, the most accurate and standard application of kerosene as a solvent in the context of chemistry and industrial practice is for paints.

    Exam Tip

    When you see questions about solvents in JAMB, remember the "Like Dissolves Like" rule:

    • Water (Polar) dissolves ionic salts and polar compounds.
    • Organic solvents (Non-polar) like kerosene, benzene, and petrol dissolve oils, fats, and resins (like those found in paints).

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20. 20.

    127g of sodium chloride was dissolved in 1.0dm$^3$ of distilled water at 25$^0$C . Determine the solubility in moldm$^{-3}$ of sodium chloride at that temperature. [Na = 23, Cl = 35.5]

    1. A.1.0
    2. B.2.0
    3. C.2.2
    4. D.4.1

    Show answer

    Answer: C. 2.2

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21. 21.

    A typical chemical reaction will be spontaneous if

    1. A.enthalpy change is negative
    2. B.free energy change is greater than one
    3. C.entropy change is zero
    4. D.enthalpy and free energy change are equal

    Show answer

    Answer: A. enthalpy change is negative

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22. 22.

    If 11.0g of a gas occupies 5.6 dm$^3$ at s.t.p., calculate its vapour density (1 mole of a gas occupies 22.4 dm$^3$).

    1. A.22
    2. B.44
    3. C.66
    4. D.88

    Show answer

    Answer: A. 22

    AI explanation

    1. Analyze the Question

    We are given the mass of a gas (11.0g) and the volume it occupies at standard temperature and pressure (s.t.p.) (5.6 dm³). We are also given the molar volume of a gas (22.4 dm³). We need to find the vapour density of the gas.

    2. Explain the Concept

    • Molar Mass ($M$): This is the mass of one mole of a substance. Since 1 mole of any gas occupies 22.4 dm³ at s.t.p., we can find the molar mass using the ratio: $$\text{Molar Mass} = \frac{\text{Mass of gas} \times 22.4 \text{ dm}^3}{\text{Volume of gas}}$$
    • Vapour Density (V.D.): This is defined as the ratio of the mass of a certain volume of a gas to the mass of an equal volume of hydrogen gas under the same conditions. The relationship between Molar Mass and Vapour Density is: $$\text{Molar Mass} = 2 \times \text{Vapour Density}$$ Therefore, Vapour Density = $\frac{\text{Molar Mass}}{2}$

    3. Show the Solution

    Step 1: Calculate the Molar Mass of the gas. Using the given values:

    • Mass = 11.0g
    • Volume = 5.6 dm³
    • Molar Volume = 22.4 dm³

    $\text{Molar Mass} = \frac{11.0 \text{ g} \times 22.4 \text{ dm}^3}{5.6 \text{ dm}^3}$

    $\text{Molar Mass} = 11.0 \times 4 = 44 \text{ g/mol}$

    Step 2: Calculate the Vapour Density.

    $\text{Vapour Density} = \frac{\text{Molar Mass}}{2}$

    $\text{Vapour Density} = \frac{44}{2} = 22$

    4. Identify the Answer

    Comparing our result to the options provided, the correct answer is A.

    5. Tip

    Always remember the relationship Molar Mass = 2 × Vapour Density. A common mistake students make is forgetting to divide the molar mass by 2 at the end, or confusing molar volume (22.4) with the constant used for vapour density (2). Keep these constants clear in your mind!

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23. 23.

    Alkylation of benzene is catalyzed by

    1. A.palladium
    2. B.nickel
    3. C.sunlight
    4. D.aluminium chloride

    Show answer

    Answer: D. aluminium chloride

    AI explanation

    Analysis of the Question

    The question asks for the specific catalyst required for the alkylation of benzene. This is a classic reaction in organic chemistry known as the Friedel-Crafts Alkylation.

    Concept Explanation

    The alkylation of benzene involves replacing a hydrogen atom on the benzene ring with an alkyl group (such as a methyl or ethyl group). Because benzene is a stable aromatic compound, it does not react easily with simple alkyl halides. To make the reaction happen, we need a Lewis acid catalyst.

    A Lewis acid is an electron-pair acceptor. In the Friedel-Crafts reaction, the catalyst interacts with the alkyl halide to generate a strong electrophile (a carbocation), which is then reactive enough to attack the benzene ring.

    Step-by-Step Solution

    1. Identify the reaction type: The reaction is Friedel-Crafts Alkylation.
    2. Recall the reagents: The reaction typically involves benzene and an alkyl halide (e.g., $CH_3Cl$).
    3. Identify the catalyst: The reaction requires a strong Lewis acid to polarize the C-Cl bond in the alkyl halide. Common catalysts used for this purpose include Aluminium chloride ($AlCl_3$) or Iron(III) chloride ($FeCl_3$).
    4. Evaluate the options: • Palladium (A) and Nickel (B): These are typically used as catalysts for hydrogenation (adding hydrogen) to unsaturated hydrocarbons, not for alkylation. • Sunlight (C): This is used for free-radical substitution in alkanes, not for electrophilic substitution in benzene. • Aluminium chloride (D): This is the standard Lewis acid catalyst used in Friedel-Crafts reactions.

    Conclusion

    It acts by accepting a lone pair of electrons from the chlorine atom of the alkyl halide, creating a highly reactive carbocation that proceeds to substitute the benzene ring.

    Exam Tip

    Whenever you see "Friedel-Crafts" in an organic chemistry question, immediately look for $AlCl_3$ or $FeCl_3$. These are the "powerhouse" catalysts for aromatic substitution reactions. Don't confuse them with hydrogenation catalysts like Nickel or Palladium!

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24. 24.

    $^{14}_7$N  + X → $^{14}_6$C  +  $^1_1$P In the reaction above, X is

    1. A.a proton
    2. B.an alpha particle
    3. C.a beta particle
    4. D.a neutron

    Show answer

    Answer: D. a neutron

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25. 25.

    The volume in cm$^3$ of a 0.12 moldm$^{-3}$ HCl required to completely neutralize a 20cm$^3$ of 0.20 moldm$^{-3}$ of NaOH is

    1. A.12.00
    2. B.0.0012
    3. C.33.33
    4. D.2.40

    Show answer

    Answer: C. 33.33

    AI explanation

    1. Analyze the question

    We are asked to find the volume of hydrochloric acid (HCl) needed to neutralize a specific volume and concentration of sodium hydroxide (NaOH). This is a classic acid-base titration problem.

    2. Explain the concept

    To solve this, we use the titration formula:

    $\frac{C_a V_a}{n_a} = \frac{C_b V_b}{n_b}$

    Where:

    • $C_a$ = Concentration of the acid (HCl)
    • $V_a$ = Volume of the acid
    • $n_a$ = Number of moles of acid in the balanced equation
    • $C_b$ = Concentration of the base (NaOH)
    • $V_b$ = Volume of the base
    • $n_b$ = Number of moles of base in the balanced equation

    The balanced chemical equation is:

    $HCl + NaOH \rightarrow NaCl + H_2O$

    From the equation, $n_a = 1$ and $n_b = 1$.

    3. Show the solution

    Given data:

    • $C_a = 0.12 \text{ mol dm}^{-3}$
    • $V_a = ?$
    • $C_b = 0.20 \text{ mol dm}^{-3}$
    • $V_b = 20 \text{ cm}^3$

    Step-by-step calculation: Substitute the values into the formula:

    $\frac{0.12 \times V_a}{1} = \frac{0.20 \times 20}{1}$

    Simplify the right side:

    $0.12 \times V_a = 4$

    Solve for $V_a$:

    $V_a = \frac{4}{0.12}$

    $V_a = 33.333... \text{ cm}^3$

    4. Identify the answer

    Rounding to two decimal places, we get 33.33 cm³. This matches option C.

    5. Exam Tip

    In titration questions, always check the stoichiometry (the coefficients) of the balanced equation first. If the acid were $H_2SO_4$ instead of $HCl$, the $n_a$ would be 1, but if you were dealing with a diprotic acid, you must ensure your $n_a$ and $n_b$ values are correct, or your calculation will be off by a factor of 2!

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26. 26.

    Benzene formed nitrobenzene at temperature of 60$^0$C when it reacts with mixture of concentrated trioxonitrate(V) acid and concentrated

    1. A.phosphoric acid
    2. B.hydrochloric acid
    3. C.tetraoxosulphate(VI) acid
    4. D.hydrogen iodide

    Show answer

    Answer: C. tetraoxosulphate(VI) acid

    AI explanation

    Analysis

    The question asks for the specific reagent used alongside concentrated trioxonitrate(V) acid ($HNO_3$) to facilitate the nitration of benzene at 60°C. This is a classic electrophilic substitution reaction in organic chemistry.

    Concept: Nitration of Benzene

    The nitration of benzene involves the substitution of a hydrogen atom on the benzene ring with a nitro group ($-NO_2$).

    1. The Reagents: To generate the active electrophile, the nitronium ion ($NO_2^+$), benzene must react with a mixture of concentrated trioxonitrate(V) acid ($HNO_3$) and concentrated tetraoxosulphate(VI) acid ($H_2SO_4$).
    2. The Role of $H_2SO_4$: Concentrated tetraoxosulphate(VI) acid acts as a catalyst and a dehydrating agent. It protonates the nitric acid, causing it to lose a water molecule and form the highly reactive nitronium ion ($NO_2^+$), which then attacks the benzene ring.
    3. Conditions: This reaction is typically carried out at a controlled temperature of about 50°C to 60°C to prevent further nitration (which would produce dinitrobenzene).

    Step-by-Step Solution

    • Step 1: Identify the reaction type: Electrophilic aromatic substitution (Nitration).
    • Step 2: Recall the "nitrating mixture": The standard laboratory mixture for nitration is concentrated $HNO_3$ and concentrated $H_2SO_4$.
    • Step 3: Evaluate the options: • A. Phosphoric acid is not the standard catalyst for this reaction. • B. Hydrochloric acid does not facilitate the formation of the nitronium ion. • C. Tetraoxosulphate(VI) acid ($H_2SO_4$) is the correct reagent used to generate the nitronium ion. • D. Hydrogen iodide is a reducing agent and is not involved in this substitution.

    Conclusion

    The reaction of benzene with concentrated trioxonitrate(V) acid requires concentrated tetraoxosulphate(VI) acid to produce nitrobenzene.

    Exam Tip: Always remember the "Nitrating Mixture" mnemonic: "Nitric and Sulphuric". If you see "nitration" in an exam question, look for $H_2SO_4$ as the partner acid!

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27. 27.

    The percentage of hydrogen in the sixth member of the class of the aliphatic alkanes is  [H =1, C =12 ]

    1. A.41.34
    2. B.59.54
    3. C.16.28
    4. D.83.72

    Show answer

    Answer: C. 16.28

    AI explanation

    1. Analyze the question

    We need to find the percentage composition of hydrogen in the sixth member of the alkane homologous series.

    2. Explain the concept

    • General Formula: The general formula for alkanes is $C_nH_{2n+2}$.
    • Sixth Member: Since the first member ($n=1$) is methane ($CH_4$), the sixth member ($n=6$) is hexane ($C_6H_{14}$).
    • Percentage Composition Formula: $$\text{Percentage of element} = \left( \frac{\text{Total mass of element in compound}}{\text{Molar mass of compound}} \right) \times 100\%$$

    3. Show the solution

    Step 1: Determine the molecular formula of the sixth member. Using $n=6$: $C_6H_{2(6)+2} = C_6H_{14}$

    Step 2: Calculate the molar mass of $C_6H_{14}$.

    • Atomic mass of Carbon (C) = 12
    • Atomic mass of Hydrogen (H) = 1
    • Molar mass = $(6 \times 12) + (14 \times 1)$
    • Molar mass = $72 + 14 = 86 \text{ g/mol}$

    Step 3: Calculate the total mass of Hydrogen in the compound.

    • Mass of H = $14 \times 1 = 14 \text{ g/mol}$

    Step 4: Calculate the percentage of Hydrogen.

    • $\%H = \left( \frac{14}{86} \right) \times 100$
    • $\%H = 0.16279 \times 100$
    • $\%H \approx 16.28\%$

    4. Identify the answer

    Comparing our result to the options provided, the correct value is 16.28%. This matches option C.

    5. Tip

    When dealing with homologous series, always remember the general formula ($C_nH_{2n+2}$ for alkanes, $C_nH_{2n}$ for alkenes). A common mistake is to forget that the "sixth member" is $n=6$, not $n=5$. Always write out the formula first to avoid calculation errors!

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28. 28.

    Solubility curve is a plot of solubility against

    1. A.temperature
    2. B.pressure
    3. C.volume
    4. D.concentrate

    Show answer

    Answer: A. temperature

    AI explanation

    Analysis

    The question asks for the definition of a solubility curve. To answer this, we need to understand how the solubility of a substance (usually a solid) changes in relation to external conditions.

    Concept Explanation

    A solubility curve is a graphical representation used in chemistry to show the relationship between the solubility of a solute and a specific variable.

    For most solid solutes in a liquid solvent (like water), solubility is highly dependent on temperature. As the temperature of the solvent increases, the kinetic energy of the molecules increases, which generally allows more solute to dissolve. By plotting the mass of the solute that can dissolve in 100g of water (solubility) on the y-axis against the temperature on the x-axis, we create a curve that helps us determine how much of a substance will dissolve at any given temperature.

    Step-by-Step Solution

    1. Identify the variables: Solubility is defined as the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature.
    2. Evaluate the options: • Pressure: While pressure significantly affects the solubility of gases (Henry's Law), it has a negligible effect on the solubility of most solids and liquids. Solubility curves are primarily used for solids. • Volume: Volume is a measure of the amount of solvent, not a factor that changes the solubility limit of a substance. • Concentrate: Concentration is a measure of how much solute is already dissolved; it is not the independent variable plotted against solubility. • Temperature: This is the primary factor that dictates how much solute can be held in a saturated solution.
    3. Conclusion: Since the solubility of solids changes predictably with temperature, the graph of solubility versus temperature is the standard "solubility curve."

    Correct Answer

    The correct option is A.

    Exam Tip

    When you see "Solubility Curve" in your JAMB exam, immediately think of a Temperature-Solubility graph. If the curve goes upward, it shows that the substance's solubility increases as the temperature rises (which is true for most salts like Potassium Nitrate). If the curve is flat, the solubility is barely affected by temperature (like Sodium Chloride).

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29. 29.

    Which of the following is an air pollutant?

    1. A.O$_2$
    2. B.CO
    3. C.H$_2$O$(_g$)
    4. D.O$_3$

    Show answer

    Answer: B. CO

    AI explanation

    Analysis

    The question asks us to identify which of the listed substances is considered an air pollutant. An air pollutant is a substance in the atmosphere that has harmful effects on human health, living organisms, or the environment.

    Concept Explanation

    • O$_2$ (Oxygen): This is a vital component of the atmosphere (about 21%) and is essential for respiration. It is not a pollutant.
    • CO (Carbon(II) oxide): This is a colorless, odorless, and highly toxic gas produced by the incomplete combustion of fossil fuels (e.g., in car engines or generators). It binds to hemoglobin in the blood more strongly than oxygen, preventing oxygen transport, which can be fatal. It is a major primary air pollutant.
    • H$_2$O$_{(g)}$ (Water vapor): This is a natural component of the atmosphere and part of the water cycle. It is not considered a pollutant.
    • O$_3$ (Ozone): While ozone in the stratosphere is beneficial (protecting us from UV radiation), ozone at ground level is a pollutant. However, in the context of standard chemistry examinations, Carbon(II) oxide (CO) is the classic, definitive example of a primary air pollutant resulting from human activity.

    Step-by-Step Solution

    1. Evaluate O$_2$: Oxygen is necessary for life; therefore, it is not a pollutant.
    2. Evaluate CO: Carbon(II) oxide is a byproduct of incomplete combustion. It is toxic to humans and animals because it interferes with the oxygen-carrying capacity of the blood. It is universally classified as a hazardous air pollutant.
    3. Evaluate H$_2$O$_{(g)}$: Water vapor is a natural gas and is not harmful to the environment in the context of pollution.
    4. Evaluate O$_3$: While ground-level ozone is harmful, CO is the most direct and commonly cited answer for "air pollutant" in JAMB chemistry syllabi due to its immediate toxicity and common source (exhaust fumes).

    Comparing these, CO is the most significant and widely recognized air pollutant among the choices provided.

    Conclusion

    The correct option is B.

    Exam Tip

    When you see questions about air pollutants in JAMB, look for products of incomplete combustion like CO (Carbon(II) oxide) or oxides of nitrogen (NO$_x$) and sulfur (SO$_x$). These are the "usual suspects" in environmental chemistry questions!

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30. 30.

    25.0g of potassium chloride were dissolved in 80g of distilled water at 30$^0$C. Calculate the solubility of the solute in mol dm$^3$. [K =39, Cl = 35.5]

    1. A.3.9 moldm$^3$
    2. B.4.6 moldm$^3$
    3. C.4.0 moldm$^3$
    4. D.4.2 moldm$^3$

    Show answer

    Answer: D. 4.2 moldm$^3$

    AI explanation

    To solve this problem, we need to determine the concentration of the potassium chloride (KCl) solution in $\text{mol dm}^{-3}$.

    1. Analyze the Question

    • Mass of solute (KCl): $25.0\text{ g}$
    • Mass of solvent (water): $80\text{ g}$
    • Molar mass of KCl: $K(39) + Cl(35.5) = 74.5\text{ g/mol}$
    • Density of water: $1\text{ g/cm}^3$ (therefore, $80\text{ g}$ of water = $80\text{ cm}^3$ or $0.08\text{ dm}^3$)

    2. Explain the Concept

    Solubility in $\text{mol dm}^{-3}$ (Molarity) is defined as the number of moles of solute dissolved in $1\text{ dm}^3$ of the solvent.

    • Step 1: Calculate the number of moles of KCl.
    • Step 2: Convert the volume of the solvent to $\text{dm}^3$.
    • Step 3: Divide the moles by the volume in $\text{dm}^3$.

    3. Show the Solution

    Step 1: Calculate the number of moles of KCl

    $\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$

    $\text{Moles} = \frac{25.0\text{ g}}{74.5\text{ g/mol}} \approx 0.3356\text{ mol}$

    Step 2: Determine the volume of the solvent in $\text{dm}^3$ Since $1000\text{ cm}^3 = 1\text{ dm}^3$:

    $\text{Volume} = 80\text{ cm}^3 = \frac{80}{1000}\text{ dm}^3 = 0.08\text{ dm}^3$

    Step 3: Calculate Molarity

    $\text{Molarity} = \frac{\text{Moles}}{\text{Volume in dm}^3}$

    $\text{Molarity} = \frac{0.3356\text{ mol}}{0.08\text{ dm}^3} = 4.195\text{ mol dm}^{-3}$

    Rounding to one decimal place, we get $4.2\text{ mol dm}^{-3}$.

    4. Identify the Answer

    Comparing our result to the given options, the correct value is $4.2\text{ mol dm}^{-3}$. This matches option D.

    5. Exam Tip

    Always remember that in chemistry calculations involving water as a solvent, the mass of water in grams is numerically equal to its volume in milliliters ($1\text{ g} = 1\text{ cm}^3$). Don't forget to convert $\text{cm}^3$ to $\text{dm}^3$ by dividing by $1000$ before calculating molarity!

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31. 31.

    In contact process, the catalyst used in the conversion of sulphur(IV)oxide to sulphur(VI) oxide is

    1. A.nickel
    2. B.platinum
    3. C.vanadium(V)oxide
    4. D.magnesium(IV)oxide

    Show answer

    Answer: C. vanadium(V)oxide

    AI explanation

    1. Analyze the Question

    The question asks for the specific catalyst used in the Contact Process to convert sulphur(IV) oxide ($SO_2$) into sulphur(VI) oxide ($SO_3$). This is a critical step in the industrial manufacture of sulphuric acid.

    2. Explain the Concept

    The Contact Process involves the catalytic oxidation of sulphur(IV) oxide. The reaction is represented by the equation:

    $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$

    To make this reaction proceed at a reasonable rate and yield, a catalyst is required. While platinum was historically used, it is very expensive and easily "poisoned" (rendered inactive) by impurities like arsenic. Therefore, modern industrial processes use vanadium(V) oxide ($V_2O_5$) because it is cheaper, more durable, and highly effective.

    3. Show the Solution

    • Step 1: Identify the reaction: $2SO_2 + O_2 \rightarrow 2SO_3$.
    • Step 2: Evaluate the options: • Nickel: Used in the hydrogenation of oils (hardening of fats). • Platinum: Historically used, but replaced by vanadium(V) oxide due to cost and sensitivity to poisoning. • Vanadium(V) oxide ($V_2O_5$): The standard industrial catalyst for the Contact Process. • Magnesium(IV) oxide: This is not a standard catalyst for this process (and is chemically unstable in this form).
    • Step 3: Conclusion: Vanadium(V) oxide is the correct catalyst used in the modern Contact Process.

    4. Identify the Answer

    The correct option is C.

    5. Exam Tip

    When studying industrial processes for Chemistry, always distinguish between the catalyst and the conditions. For the Contact Process, remember:

    • Catalyst: Vanadium(V) oxide ($V_2O_5$).
    • Temperature: About 450°C (a compromise temperature).
    • Pressure: 1–2 atmospheres.

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32. 32.

    The compound of Copper used as a fungicide is

    1. A.CuCO$_3$
    2. B.Cu(NO$_3$)$_2$
    3. C.CuSO$_4$
    4. D.CuCl$_2$

    Show answer

    Answer: C. CuSO$_4$

    AI explanation

    Analysis

    The question asks for a specific copper compound that is widely utilized in agriculture as a fungicide. To solve this, we need to look at the chemical properties of copper salts and their common industrial or agricultural applications.

    Concept Explanation

    In agricultural chemistry, certain copper compounds are used to control fungal diseases in crops. The most common substance used for this purpose is Bordeaux mixture, which is a combination of copper(II) sulfate ($CuSO_4$) and slaked lime ($Ca(OH)_2$).

    • Copper(II) sulfate ($CuSO_4$): This is the active ingredient in many fungicides and algaecides. It works by disrupting the enzymes in fungal cells, effectively killing the fungus or preventing its growth.
    • Other options: • $CuCO_3$ (Copper carbonate) is primarily used as a pigment or in the production of other copper salts. • $Cu(NO_3)_2$ (Copper nitrate) is often used in electroplating and as an oxidizing agent. • $CuCl_2$ (Copper chloride) is used as a catalyst in organic synthesis and in the textile industry.

    Step-by-Step Solution

    1. Identify the chemical role: We are looking for a compound known for its fungicidal properties.
    2. Evaluate the options: • $CuSO_4$ is historically and scientifically recognized as the primary copper compound used in agriculture to treat fungal infections on plants (often applied as Bordeaux mixture).
    3. Conclusion: Among the choices provided, Copper(II) sulfate is the standard compound used for this specific agricultural application.

    Answer

    The correct option is C.

    Exam Tip

    When you see questions about the uses of copper compounds in JAMB, remember the "Big Three":

    1. $CuSO_4$: Fungicide and testing for the presence of water (it turns from white to blue when hydrated).
    2. $CuCl_2$: Catalyst in organic reactions.
    3. $Cu(NO_3)_2$: Used in electroplating.

    Focusing on these specific industrial/agricultural uses will help you answer similar questions quickly!

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33. 33.

    The amount of Faraday required to discharge 4.5 moles of Al$^{3+}$ is

    1. A.13.5
    2. B.4.5
    3. C.7.5
    4. D.1.5

    Show answer

    Answer: A. 13.5

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34. 34.

    Which of the following is used in forming slag in the blast furnace for the extraction of iron?

    1. A.Ca(CO$_3$ )$_2$
    2. B.Ca(OH)$_2$
    3. C.CaCO$_3$
    4. D.CaO

    Show answer

    Answer: C. CaCO$_3$

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35. 35.

    The product formed when ethyne is passed through a hot tube containing finely divided iron is

    1. A.polymeric
    2. B.saturated
    3. C.hydrated
    4. D.aromatic

    Show answer

    Answer: D. aromatic

    AI explanation

    1. Analyze the question

    The question asks for the classification of the product formed when ethyne ($C_2H_2$) undergoes a specific chemical reaction: passing it through a hot tube containing finely divided iron.

    2. Explain the concept

    This reaction is a classic example of cyclic polymerization (or trimerization).

    • The Reaction: When ethyne gas is passed through a red-hot iron tube (at about 400°C–500°C), three molecules of ethyne combine to form a single molecule of benzene ($C_6H_6$).
    • The Equation: $3C_2H_2 \rightarrow C_6H_6$
    • The Product: Benzene is the simplest member of the aromatic hydrocarbon family. Aromatic compounds are characterized by a stable, cyclic ring structure with delocalized pi electrons.

    3. Show the solution

    • Step 1: Identify the reactants and conditions. Ethyne ($HC \equiv CH$) + Hot Iron Tube $\rightarrow$ ?
    • Step 2: Recognize the reaction type. The iron acts as a catalyst for the trimerization of ethyne.
    • Step 3: Determine the product. The trimerization of ethyne produces benzene ($C_6H_6$).
    • Step 4: Classify the product. Benzene is the fundamental building block of aromatic chemistry. It is unsaturated (due to the ring structure), but its defining chemical classification in this context is "aromatic."

    4. Identify the answer

    Comparing our findings to the options:

    • A is incorrect because while it is a polymer, "aromatic" is a more specific and accurate chemical classification for benzene.
    • B is incorrect because benzene is unsaturated.
    • C is incorrect because hydration involves adding water.
    • D is correct because the product, benzene, is the primary aromatic hydrocarbon.

    5. Exam Tip

    Whenever you see "ethyne" and "hot iron tube" in a JAMB question, immediately think "Benzene". Remember the mnemonic: "3 Ethyne = 1 Benzene". Since benzene is the parent of all aromatic compounds, the answer will almost always be "aromatic."

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36. 36.

    The principle which states that no two electrons in the same orbitals of an atom have same value for all four quantum numbers is the

    1. A.Aufbau principle
    2. B.Hund's rule
    3. C.Pauli Exclusion principle
    4. D.Dilution principle

    Show answer

    Answer: C. Pauli Exclusion principle

    AI explanation

    1. Analyze the Question

    The question asks for the specific scientific principle that dictates that electrons within an atom must have unique sets of quantum numbers. In other words, it describes the "rule" that prevents two electrons from being identical in their energy states and orientation.

    2. Explain the Concept

    To understand this, we look at the four quantum numbers that describe an electron:

    • Principal quantum number ($n$): The energy level.
    • Azimuthal quantum number ($l$): The shape of the orbital.
    • Magnetic quantum number ($m_l$): The orientation in space.
    • Spin quantum number ($m_s$): The direction of the electron's spin (+1/2 or -1/2).

    The Pauli Exclusion Principle states that in a single atom, no two electrons can have the same set of all four quantum numbers. If two electrons occupy the same orbital (sharing the same $n, l,$ and $m_l$), they must have opposite spins ($m_s = +1/2$ and $-1/2$) to satisfy this principle.

    3. Show the Solution

    Let's evaluate the options provided:

    • A. Aufbau principle: This states that electrons fill lower-energy orbitals before moving to higher-energy ones.
    • B. Hund's rule: This states that electrons fill degenerate orbitals (orbitals with the same energy) singly first, with parallel spins, before pairing up.
    • C. Pauli Exclusion principle: This is the correct principle that restricts the quantum numbers of electrons in an atom.
    • D. Dilution principle: This is a concept in chemistry regarding the concentration of solutions, not atomic structure.

    Since the question specifically describes the restriction on quantum numbers, the Pauli Exclusion Principle is the only correct answer.

    4. Identify the Answer

    The correct option is C.

    5. Exam Tip

    To remember these easily:

    • Aufbau = "Building up" (filling from bottom to top).
    • Hund's = "The Bus Rule" (people prefer to sit alone in a bus row before sharing a seat with a stranger).
    • Pauli = "The ID Card Rule" (no two electrons can have the exact same "ID" or set of quantum numbers).

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37. 37.

    After breathing in a test tube that contains acidified K$_2$Cr$_2$O$_7$, a man noticed the change in the colour of K$_2$Cr$_2$O$_7$ from orange to green. This suggests the presence of

    1. A.alkanone
    2. B.alkanal
    3. C.alkanol
    4. D.alkanoic acid

    Show answer

    Answer: C. alkanol

    AI explanation

    Analysis

    The question describes a classic chemical test used to identify specific functional groups in organic chemistry. We are looking for a substance that reacts with acidified potassium dichromate(VI) ($K_2Cr_2O_7$) to cause a color change from orange to green.

    Concept Explanation

    • The Reagent: Acidified potassium dichromate(VI) is a strong oxidizing agent.
    • The Reaction: When it reacts with certain organic compounds, the dichromate ion ($Cr_2O_7^{2-}$, which is orange) is reduced to the chromium(III) ion ($Cr^{3+}$, which is green).
    • The Target: This reaction occurs when the reagent oxidizes primary and secondary alkanols (alcohols). • Primary alkanols are oxidized to alkanals and then to alkanoic acids. • Secondary alkanols are oxidized to alkanones. • Tertiary alkanols do not react with acidified $K_2Cr_2O_7$ (no color change).
    • The Context: The "man breathing into the test tube" refers to the breathalyzer test. Ethanol (an alkanol) present in the breath of someone who has consumed alcohol is oxidized by the acidified dichromate, causing the characteristic color change.

    Step-by-Step Solution

    1. Identify the oxidizing agent: Acidified $K_2Cr_2O_7$ is used to test for the presence of oxidizable alcohols.
    2. Evaluate the options: • Alkanone: These are already oxidized products (ketones) and generally do not undergo further oxidation by dichromate. • Alkanal: While these can be oxidized to alkanoic acids, the breathalyzer test specifically targets the ethanol (alkanol) in the breath. • Alkanol: Primary and secondary alcohols are easily oxidized by acidified $K_2Cr_2O_7$, resulting in the reduction of chromium from the +6 oxidation state (orange) to the +3 oxidation state (green). • Alkanoic acid: These are the final products of the oxidation of primary alcohols and cannot be further oxidized by this reagent.
    3. Conclusion: The color change from orange to green confirms the presence of an alkanol (specifically ethanol in the context of a breath test).

    Correct Option

    The correct option is C.

    Exam Tip

    Remember the mnemonic "Primary and Secondary Alcohols go Green". Tertiary alcohols are "stubborn"—they do not react with acidified $K_2Cr_2O_7$ or $KMnO_4$, so they remain orange. This is a common way to distinguish between different types of alcohols in the lab!

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38. 38.

    Fats and oils are esters of fatty organic acids combined with a trihydric alkanol commonly referred to as

    1. A.ethanol
    2. B.methanol
    3. C.glycol
    4. D.glycerol

    Show answer

    Answer: D. glycerol

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39. 39.

    An example of a substance that does not change directly from solid to gas when heated is

    1. A.NH$_4$Cl$(_s$)
    2. B.I$_2$$(_s$)
    3. C.CaCO$_3$ $(_s$)
    4. D.S$_8$ $(_s$)

    Show answer

    Answer: C. CaCO$_3$ $(_s$)

    AI explanation

    Analysis

    The question asks us to identify which of the listed substances does not undergo sublimation. Sublimation is the process where a substance changes directly from a solid state to a gaseous state without passing through the liquid phase.

    Concept Explanation

    • Sublimation: Substances that sublime have weak intermolecular forces, allowing them to transition directly to gas upon heating. Common examples include Ammonium Chloride ($\text{NH}_4\text{Cl}$), Iodine ($\text{I}_2$), and Naphthalene.
    • Thermal Decomposition: Some substances, when heated, do not sublime but instead undergo a chemical reaction where they break down into simpler substances. This is common with carbonates.

    Step-by-Step Solution

    1. Analyze Option A ($\text{NH}_4\text{Cl}$): Ammonium chloride is a classic example of a substance that sublimes. When heated, it turns directly into ammonia and hydrogen chloride gases.
    2. Analyze Option B ($\text{I}_2$): Iodine crystals are well-known for subliming. When heated, they produce a characteristic purple vapor directly from the solid state.
    3. Analyze Option C ($\text{CaCO}_3$): Calcium carbonate (limestone) does not sublime. Instead, when heated strongly, it undergoes thermal decomposition to produce calcium oxide (quicklime) and carbon(IV) oxide gas: $$\text{CaCO}_3(s) \xrightarrow{\Delta} \text{CaO}(s) + \text{CO}_2(g)$$Because it forms a new solid product ($\text{CaO}$) rather than turning entirely into gas, it is not a subliming substance.
    4. Analyze Option D ($\text{S}_8$): Sulfur can sublime under certain conditions, though it also melts. However, compared to $\text{CaCO}_3$, it is much more associated with sublimation in chemistry curricula.

    Conclusion

    Since $\text{CaCO}_3$ undergoes chemical decomposition rather than a physical phase change (sublimation), it is the correct answer.

    Exam Tip

    When you see questions about "sublimation" in JAMB Chemistry, look out for these common "sublimers": Iodine ($\text{I}_2$), Ammonium Chloride ($\text{NH}_4\text{Cl}$), Naphthalene, and Dry Ice ($\text{CO}_2$). If you see a carbonate like $\text{CaCO}_3$, remember it almost always decomposes into an oxide and $\text{CO}_2$ gas when heated!

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40. 40.

    CH$_3$-CH$_2$-OH and CH$_3$-O-CH$_3$ The relationship between the two compounds above, is that they are

    1. A.allotropes
    2. B.isotopes
    3. C.isomers
    4. D.isobars

    Show answer

    Answer: C. isomers

    AI explanation

    Analysis

    To determine the relationship between CH$_3$-CH$_2$-OH (ethanol) and CH$_3$-O-CH$_3$ (methoxymethane/dimethyl ether), we need to compare their molecular formulas and structures.

    Concept Explanation

    • Isomers are compounds that have the same molecular formula but different structural arrangements of atoms.
    • Allotropes are different physical forms of the same element (e.g., diamond and graphite for carbon).
    • Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons.
    • Isobars are atoms of different elements that have the same mass number but different atomic numbers.

    Step-by-Step Solution

    1. Determine the molecular formula for CH$_3$-CH$_2$-OH: • Count the atoms: 2 Carbons, 6 Hydrogens, 1 Oxygen. • Formula: C$_2$H$_6$O
    2. Determine the molecular formula for CH$_3$-O-CH$_3$: • Count the atoms: 2 Carbons, 6 Hydrogens, 1 Oxygen. • Formula: C$_2$H$_6$O
    3. Compare: • Both compounds have the exact same molecular formula (C$_2$H$_6$O). • However, they have different functional groups: Ethanol is an alcohol (-OH group), while dimethyl ether is an ether (-O- linkage). • Because they have the same molecular formula but different structures, they are classified as functional group isomers.

    Conclusion

    Since both compounds share the same molecular formula but possess different structures and chemical properties, they are isomers.

    Exam Tip: Whenever you see two organic compounds with the same number of carbons, hydrogens, and oxygens, immediately check for isomerism. Remember that alcohols and ethers with the same number of carbon atoms are classic examples of functional group isomers!

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41. 41.

    $^{238}_{92}$U  + $^1_0$n → $^{239}_{92}$U  The process above produces

    1. A.atomic mutation
    2. B.a radioactive isotope
    3. C.other neutrons
    4. D.radiations

    Show answer

    Answer: B. a radioactive isotope

    AI explanation

    1. Analyze the Question

    The reaction provided is: $^{238}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{239}_{92}\text{U}$

    We are looking at a nuclear reaction where a Uranium-238 nucleus captures a neutron to become Uranium-239. We need to identify the nature of the product formed.

    2. Explain the Concept

    • Isotopes: These are atoms of the same element (same atomic number) that have different mass numbers due to a different number of neutrons.
    • Radioactivity: Many heavy isotopes, especially those created by neutron bombardment (like $^{239}\text{U}$), are unstable. Because they have an excess of neutrons, they are inherently radioactive; they will eventually undergo radioactive decay (usually beta decay) to become more stable.
    • Nuclear Transmutation: This is the conversion of one chemical element or isotope into another. While this is a transmutation, the specific characteristic of the product $^{239}\text{U}$ is that it is an unstable, radioactive isotope.

    3. Show the Solution

    • Step 1: Observe the atomic number (bottom number). It remains 92, meaning the element is still Uranium.
    • Step 2: Observe the mass number (top number). It increases from 238 to 239.
    • Step 3: Recognize that $^{239}\text{U}$ is an isotope of $^{238}\text{U}$.
    • Step 4: In nuclear chemistry, when a heavy nucleus captures a neutron, the resulting isotope is almost always unstable and undergoes radioactive decay. Therefore, the product is classified as a radioactive isotope.

    4. Identify the Answer

    • A (Atomic mutation): This is a vague term not typically used to describe this specific outcome.
    • B (A radioactive isotope): This is correct. $^{239}\text{U}$ is an isotope of Uranium, and it is radioactive.
    • C (Other neutrons): No additional neutrons are released in this specific equation.
    • D (Radiations): While the isotope will eventually emit radiation, the product itself is the isotope, not the radiation.

    5. Exam Tip

    Whenever you see a nuclear equation where a neutron ($^{1}_{0}\text{n}$) is added to a nucleus, the product is almost always a heavier isotope of the original element. Because these "neutron-rich" isotopes are unstable, they are almost always radioactive. Remember: Neutron capture = Radioactive isotope formation.

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42. 42.

    The highest isotope of hydrogen is

    1. A.Deuterium
    2. B.protium
    3. C.tritium
    4. D.positron

    Show answer

    Answer: C. tritium

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43. 43.

    A major effect of oil pollution in coastal water is

    1. A.destruction of aquatic life
    2. B.nitrification of the water body
    3. C.detoxification of water
    4. D.disinfection of the water body

    Show answer

    Answer: A. destruction of aquatic life

    AI explanation

    Analysis

    The question asks for the primary environmental impact of oil spills (oil pollution) on coastal ecosystems. We need to evaluate how crude oil or petroleum products interact with marine life and water quality.

    Concept Explanation

    When oil is spilled into coastal waters, it creates several hazardous conditions:

    1. Physical Coating: Oil coats the feathers of birds and the gills of fish, preventing them from breathing, flying, or regulating their body temperature.
    2. Toxicity: Crude oil contains toxic hydrocarbons that can poison marine organisms upon ingestion or skin contact.
    3. Oxygen Deprivation: A layer of oil on the surface of the water prevents the diffusion of atmospheric oxygen into the water, leading to hypoxia (low oxygen levels) for organisms living below the surface.
    4. Habitat Destruction: It smothers coral reefs, mangroves, and seagrass beds, which serve as breeding grounds for many aquatic species.

    Step-by-Step Solution

    • Option A (Destruction of aquatic life): As explained above, oil pollution is lethal to fish, crustaceans, birds, and plants. It disrupts the entire food chain, leading to massive die-offs. This is the most significant and direct effect.
    • Option B (Nitrification): Nitrification is the biological oxidation of ammonia to nitrite and nitrate. Oil pollution does not cause this; in fact, it often inhibits the bacteria responsible for natural nutrient cycling.
    • Option C (Detoxification): Oil pollution is the opposite of detoxification; it introduces toxic substances into the environment.
    • Option D (Disinfection): Disinfection refers to the removal of pathogens (like bacteria or viruses). Oil pollution does not clean the water; it contaminates it.

    Conclusion

    Based on the environmental impact, oil pollution leads to the death and destruction of the ecosystem. Therefore, the destruction of aquatic life is the major effect.

    Exam Tip

    When answering environmental chemistry questions in JAMB, always look for the "negative impact" option. Oil pollution is a classic example of environmental degradation, so look for keywords like "destruction," "toxicity," or "suffocation" rather than "purification" or "nutrient enrichment."

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44. 44.

    What would be the order of the electrolytic cell in an industry intending the production of silver plated spoons?

    1. A.Cathode is the spoon; anode is a silver rod; electrolyte is a soluble silver salt
    2. B.Cathode is a silver rod; anode is the spoon; electrolyte is a soluble silver salt
    3. C.Cathode is the spoon; anode is any rod; electrolyte is a soluble silver salt
    4. D.Cathode is any rod; anode is the spoon; electrolyte is a soluble salt

    Show answer

    Answer: A. Cathode is the spoon; anode is a silver rod; electrolyte is a soluble silver salt

    AI explanation

    Analysis of the Question

    This question tests your understanding of Electroplating, which is a practical application of electrolysis. The goal is to coat an object (the spoon) with a thin layer of a metal (silver).

    Concept Explanation

    In electroplating, we use the principle of electrolytic deposition. To plate an object with a metal, you must follow these specific rules:

    1. The Cathode (Negative Electrode): The object you want to plate must be connected to the negative terminal. During electrolysis, metal ions in the solution are attracted to the cathode, where they gain electrons (reduction) and deposit as a solid metal layer on the object.
    2. The Anode (Positive Electrode): The metal you want to use for plating (in this case, silver) must be connected to the positive terminal. As the process continues, the anode dissolves into the electrolyte to replenish the silver ions being used up at the cathode.
    3. The Electrolyte: This must be a solution containing the ions of the metal you are plating with (e.g., silver nitrate or silver cyanide) so that the ions can move through the solution to the cathode.

    Step-by-Step Solution

    • Step 1: Identify the object to be plated. The spoon is the object, so it must be the cathode.
    • Step 2: Identify the plating metal. We are plating with silver, so the anode must be a silver rod.
    • Step 3: Choose the electrolyte. To ensure silver ions are available to move to the spoon, the electrolyte must be a soluble silver salt.

    Comparing this to the options provided:

    • Option A correctly places the spoon at the cathode, the silver rod at the anode, and uses a soluble silver salt as the electrolyte.
    • Option B reverses the roles of the spoon and the silver rod, which would result in the spoon dissolving rather than being plated.
    • Options C and D are incorrect because they do not specify the correct material for the anode or cathode, which is essential for successful electroplating.

    Exam Tip

    Remember the mnemonic "C-A-N": Cathode Attracts Negative (or rather, the object to be plated is the Cathode). Always think of the object you want to "keep" (the plated spoon) as the one that needs to gain the metal, which happens at the cathode.

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45. 45.

    The IUPAC nomenclature of the complex K$_4$Fe(CN)$_6$ is

    1. A.potassium hexacyanoferrate(II)
    2. B.potassium hexacyanoferrate(III)
    3. C.potassium hexacyanoferrate(IV)
    4. D.potassium hexacyanoferrate(VI)

    Show answer

    Answer: A. potassium hexacyanoferrate(II)

    AI explanation

    To determine the IUPAC name of the complex K$_4$[Fe(CN)$_6$], we need to follow the rules for naming coordination compounds.

    1. Analyze the components

    • Cation: Potassium (K$^+$)
    • Anion (Coordination Sphere): [Fe(CN)$_6$]$^{4-}$
    • Ligand: Cyanide (CN$^-$), which is named "cyano" in a complex. Since there are six, it is "hexacyano".
    • Central Metal: Iron (Fe). Because the complex ion is an anion (negative charge), we use the Latin name "ferrate".

    2. Determine the Oxidation State of the Metal

    To find the oxidation state of Iron (Fe), we set the sum of the charges in the complex to equal the overall charge of the coordination sphere:

    • Let $x$ be the oxidation state of Fe.
    • The charge of a cyanide ligand (CN$^-$) is $-1$.
    • The total charge of the complex [Fe(CN)$_6$]$^{4-}$ is $-4$.

    The equation is: $x + 6(-1) = -4$ $x - 6 = -4$ $x = -4 + 6$ $x = +2$

    The oxidation state of iron is +2, which is represented in Roman numerals as (II).

    3. Assemble the IUPAC Name

    • Name the cation first: Potassium
    • Name the ligands: hexacyano
    • Name the metal with its oxidation state: ferrate(II)

    Combining these, we get: Potassium hexacyanoferrate(II).

    4. Conclusion

    Comparing this to your options:

    • A. potassium hexacyanoferrate(II) — Matches our calculation.
    • B. potassium hexacyanoferrate(III) — Incorrect oxidation state.
    • C. potassium hexacyanoferrate(IV) — Incorrect oxidation state.
    • D. potassium hexacyanoferrate(VI) — Incorrect oxidation state.

    Exam Tip: When naming coordination complexes, always remember that if the complex ion is negative (anionic), the metal name must end in "-ate". If it were a positive complex ion, you would simply use the English name (e.g., "iron" instead of "ferrate").

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46. 46.

    Hydrochloric acid is regarded as a strong acid because it

    1. A.ionizes partially
    2. B.ionizes completely
    3. C.is a mineral acid
    4. D.reacts with a base

    Show answer

    Answer: B. ionizes completely

    AI explanation

    Analysis

    The question asks for the fundamental reason why hydrochloric acid ($\text{HCl}$) is classified as a "strong acid." In chemistry, the strength of an acid is determined by its ability to donate protons ($\text{H}^+$ ions) or, more specifically, how it behaves when dissolved in water.

    Concept Explanation

    • Strong Acids: These are substances that undergo complete dissociation (or ionization) in an aqueous solution. This means that every molecule of the acid breaks apart into its constituent ions. For hydrochloric acid, the reaction is: $$\text{HCl}_{(aq)} \rightarrow \text{H}^+_{(aq)} + \text{Cl}^-_{(aq)}$$Because there are no intact$\text{HCl}$ molecules left in the solution, the concentration of hydrogen ions is very high, making it a strong acid.
    • Weak Acids: These only partially ionize in water. A significant number of the acid molecules remain intact in the solution, resulting in a lower concentration of hydrogen ions (e.g., ethanoic acid).

    Step-by-Step Solution

    1. Evaluate Option A: Partial ionization is the definition of a weak acid, not a strong one.
    2. Evaluate Option B: Complete ionization ensures that all available protons are released into the solution, which is the defining characteristic of a strong acid.
    3. Evaluate Option C: While $\text{HCl}$ is a mineral (inorganic) acid, being a mineral acid does not automatically make an acid "strong." For example, carbonic acid ($\text{H}_2\text{CO}_3$) is a mineral acid but is considered weak.
    4. Evaluate Option D: Reacting with a base is a property of all acids (neutralization), regardless of whether they are strong or weak. Therefore, this does not explain why $\text{HCl}$ is specifically a strong acid.

    Conclusion

    Based on the chemical definition of acid strength, hydrochloric acid is strong because it dissociates fully into ions when dissolved in water.

    Exam Tip

    Remember the acronym "Strong Acids Ionize Completely" (SAIC). If you ever forget, just remember that "strong" means it gives away all its protons without holding back!

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47. 47.

    Biuret test is a chemical test used for detecting the presence of

    1. A.carbohydrates
    2. B.proteins
    3. C.amines
    4. D.alkanoates

    Show answer

    Answer: B. proteins

    AI explanation

    1. Analyze the question

    The question asks for the specific class of organic compounds that the Biuret test is designed to detect.

    2. Explain the concept

    The Biuret test is a standard biochemical assay used to detect the presence of peptide bonds in a substance.

    • How it works: The Biuret reagent contains copper(II) ions ($Cu^{2+}$) in an alkaline solution. When these ions come into contact with peptide bonds (the bonds that link amino acids together to form proteins), they form a violet or purple-colored complex.
    • Why it works: Since proteins are essentially long chains of amino acids linked by peptide bonds, a positive result (a color change to violet) indicates that proteins are present in the sample.

    3. Show the solution

    • Option A (Carbohydrates): These are usually tested using Benedict’s solution (for reducing sugars) or Iodine solution (for starch).
    • Option B (Proteins): Proteins contain multiple peptide bonds. The Biuret reagent reacts with these bonds to produce a characteristic violet color. This is the primary use of the test.
    • Option C (Amines): While amines contain nitrogen, they do not possess the specific peptide linkage structure required to form the colored complex with the Biuret reagent.
    • Option D (Alkanoates): These are esters or carboxylic acid derivatives and do not react with Biuret reagent to produce the violet color.

    4. Identify the answer

    Based on the chemical principle of the Biuret test, it is specifically used to identify proteins.

    5. Exam Tip

    To remember this for your JAMB exam, use the "P-P" mnemonic: Peptide bonds = Proteins. Also, remember that the color change is from Blue to Violet. If you see "Biuret" on the exam, immediately look for "Protein" in the options!

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48. 48.

    If a stable neutral atom has a mass number of 31, the number of electrons and neutrons respectively are

    1. A.15 and 16
    2. B.16 and 31
    3. C.15 and 31
    4. D.16 and 15

    Show answer

    Answer: A. 15 and 16

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49. 49.

    Strong acids can be distinguished from weak acids by any of the following methods, EXCEPT

    1. A.Conductivity measurement
    2. B.The use of litmus paper
    3. C.Measurement of pH
    4. D.Measurement of heat of reaction

    Show answer

    Answer: B. The use of litmus paper

    AI explanation

    Analysis

    The question asks us to identify which of the listed methods cannot be used to distinguish between a strong acid and a weak acid. To solve this, we must understand the fundamental difference between strong and weak acids.

    Concept Explanation

    • Strong Acids: These dissociate (ionize) completely in water to produce a high concentration of hydrogen ions ($H^+$). Examples include $HCl$, $H_2SO_4$, and $HNO_3$.
    • Weak Acids: These only partially dissociate in water, resulting in a lower concentration of hydrogen ions. Examples include $CH_3COOH$ (ethanoic acid).

    Why the methods work (or don't):

    • Conductivity (A): Strong acids have more ions in solution, so they conduct electricity much better than weak acids of the same concentration.
    • pH Measurement (C): Since pH is defined as $-\log[H^+]$, the higher concentration of $H^+$ in a strong acid results in a lower pH compared to a weak acid of the same concentration.
    • Heat of Reaction (D): The heat of neutralization for a strong acid with a strong base is constant (approx. $-57.3 \text{ kJ/mol}$). However, for a weak acid, some energy is consumed to complete the dissociation of the acid, making the heat of reaction different.

    Step-by-Step Evaluation

    • Option B (Litmus paper): Litmus paper is an indicator used to determine if a substance is acidic or basic. Both strong acids and weak acids will turn blue litmus paper red. Because both types of acids produce the same color change, litmus paper cannot be used to differentiate the strength of the acids.

    Conclusion

    While conductivity, pH, and heat of reaction are quantitative measures that depend on the degree of dissociation (and thus distinguish strength), litmus paper is a qualitative test that only identifies acidity, not the degree of ionization.

    Exam Tip

    When you see questions about "distinguishing" substances, always ask yourself: "Does this method provide a measurement of concentration or degree of ionization?" If the method only gives a "Yes/No" (like litmus paper), it usually cannot distinguish between degrees of strength.

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50. 50.

    The law which states that a pure chemical compound, no matter how it is made, will be made up of the same elements contained in the same proportion by mass is

    1. A.law of conservation of masses
    2. B.law of definite proportion
    3. C.law of multiple proportion
    4. D.law of reciprocal proportion

    Show answer

    Answer: B. law of definite proportion

    AI explanation

    Analysis

    The question asks for the specific scientific law that defines the composition of a pure chemical compound in terms of the fixed mass ratios of its constituent elements, regardless of the method of preparation.

    Concept Explanation

    In chemistry, the laws of chemical combination describe how elements combine to form compounds. Here is a breakdown of the options provided:

    • Law of Conservation of Mass: States that matter is neither created nor destroyed in a chemical reaction. The total mass of reactants equals the total mass of products.
    • Law of Definite Proportion (also known as the Law of Constant Composition): States that a given chemical compound always contains its component elements in a fixed ratio by mass, regardless of its source or method of preparation. For example, water ($H_2O$) will always have hydrogen and oxygen in a mass ratio of 1:8, whether it comes from a river, a laboratory synthesis, or a volcanic vent.
    • Law of Multiple Proportion: States that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.
    • Law of Reciprocal Proportion: States that if two different elements combine separately with a fixed mass of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the ratio in which they combine with each other.

    Step-by-Step Solution

    1. Identify the key phrase: The question specifies "a pure chemical compound... made up of the same elements contained in the same proportion by mass."
    2. Match with definitions: This definition is the textbook description of the Law of Definite Proportion, proposed by Joseph Proust.
    3. Evaluate options: • Option A relates to mass balance in reactions. • Option B matches our definition perfectly. • Option C relates to multiple compounds formed by the same elements. • Option D relates to the ratios of elements combining with a third element.

    Conclusion

    The law described is the Law of Definite Proportion.

    Exam Tip

    To remember this easily, think of the word "Definite" as "Fixed." If the proportion is fixed or definite, it doesn't matter where the substance came from; the recipe for that compound remains exactly the same!

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51. 51.

    A radioactive element of mass 1g has half-life of 2 minutes, what fraction of the substance would have disintegrated after 10 minutes?

    1. A.1/4
    2. B.1/8
    3. C.1/16
    4. D.1/32

    Show answer

    Answer: D. 1/32

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52. 52.

    The scientist that performed the experiment on discharged tubes that led to the discovery of the cathode rays as a sub-atomic particle is

    1. A.J.J Thomson
    2. B.Robert Millikan
    3. C.Ernest Rutherford
    4. D.Niels Bohr

    Show answer

    Answer: A. J.J Thomson

    AI explanation

    1. Analyze the question

    The question asks for the specific scientist responsible for the experiment involving discharge tubes (also known as cathode ray tubes) that identified cathode rays as sub-atomic particles (specifically, electrons).

    2. Explain the concept

    In the late 19th century, scientists experimented with vacuum tubes (discharge tubes) to understand the nature of electricity. By applying a high voltage across electrodes in a vacuum, they observed a "glow" or rays emanating from the cathode.

    • J.J. Thomson conducted experiments where he deflected these cathode rays using electric and magnetic fields. He concluded that these rays were composed of negatively charged particles, which he called "corpuscles" (later known as electrons).
    • Robert Millikan is famous for the "Oil Drop Experiment," which determined the charge of a single electron.
    • Ernest Rutherford is known for the "Gold Foil Experiment," which led to the discovery of the atomic nucleus.
    • Niels Bohr proposed the model of the atom where electrons orbit the nucleus in discrete energy levels.

    3. Show the solution

    • Step 1: Identify the experiment described. The use of discharge tubes to study cathode rays is the hallmark of J.J. Thomson's research in 1897.
    • Step 2: Evaluate the options. While other scientists contributed to atomic theory, Thomson is the one credited with the discovery of the electron through the cathode ray tube experiment.
    • Step 3: Confirm the conclusion. Thomson's measurement of the charge-to-mass ratio of these particles proved they were sub-atomic particles smaller than the atom itself.

    4. Identify the answer

    The correct scientist is J.J. Thomson. This matches option A.

    5. Tip for your exam

    To remember this easily, use the mnemonic "Thomson's Tube". Whenever you see "Cathode Ray Tube" or "Discharge Tube" in a question, immediately associate it with J.J. Thomson and the discovery of the electron. If you see "Gold Foil," think of Rutherford and the nucleus!

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53. 53.

    Calculate the mass of Magnesium that will be liberated from its salt by the same quantity of electricity that liberated 16.0 g of Silver. [Mg = 24.0, Ag = 108 ]

    1. A.3.65 g
    2. B.3.56 g
    3. C.1.87 g
    4. D.1.78 g

    Show answer

    Answer: D. 1.78 g

    AI explanation

    To solve this problem, we use Faraday’s Second Law of Electrolysis, which states that when the same quantity of electricity is passed through different electrolytes, the masses of the substances deposited are proportional to their chemical equivalents (molar mass divided by valency).

    1. Analyze the question

    • We need to find the mass of Magnesium ($Mg$) deposited.
    • We are given the mass of Silver ($Ag$) deposited = 16.0 g.
    • Molar mass of $Mg = 24.0\text{ g/mol}$.
    • Molar mass of $Ag = 108.0\text{ g/mol}$.
    • Valency of $Mg = 2$ (since it forms $Mg^{2+}$ ions).
    • Valency of $Ag = 1$ (since it forms $Ag^+$ ions).

    2. Explain the concept

    The relationship is given by the formula:

    $\frac{\text{Mass of A}}{\text{Equivalent mass of A}} = \frac{\text{Mass of B}}{\text{Equivalent mass of B}}$

    Where Equivalent mass = $\frac{\text{Molar mass}}{\text{Valency}}$.

    3. Show the solution

    Step 1: Calculate the equivalent masses.

    • Equivalent mass of $Ag = \frac{108}{1} = 108$
    • Equivalent mass of $Mg = \frac{24}{2} = 12$

    Step 2: Set up the ratio. Let $x$ be the mass of Magnesium.

    $\frac{\text{Mass of } Mg}{\text{Equivalent mass of } Mg} = \frac{\text{Mass of } Ag}{\text{Equivalent mass of } Ag}$

    $\frac{x}{12} = \frac{16.0}{108}$

    Step 3: Solve for $x$.

    $x = \frac{16.0 \times 12}{108}$

    $x = \frac{192}{108}$

    $x \approx 1.777... \text{ g}$

    Rounding to two decimal places, we get 1.78 g.

    4. Identify the answer

    Comparing our result to the options provided, the correct value is 1.78 g. This matches option D.

    5. Exam Tip

    Always remember to check the valency of the metals involved! A common mistake students make is using the molar mass directly without dividing by the valency (the charge of the ion). For Group 1 metals (like $Ag, Na, K$), the valency is 1; for Group 2 metals (like $Mg, Ca$), the valency is 2.

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54. 54.

    C$_2$H$_{4(g)}$  +  3O$_{2(g)}$ → 2CO$_{2(g)}$  +  2H$_2$O$_{(g)}$ The above equation represents the combustion of ethene.If 10cm$^3$ of ethene is burnt in 50cm$^3$ of oxygen, what would be the volume of oxygen that would remain at the end of the reaction?

    1. A.20cm$^3$
    2. B.40cm$^3$
    3. C.60cm$^3$
    4. D.50cm$^3$

    Show answer

    Answer: A. 20cm$^3$

    AI explanation

    Analysis

    To solve this problem, we use Gay-Lussac’s Law of Combining Volumes, which states that when gases react, they do so in volumes which are simple ratios to one another and to the products, provided temperature and pressure remain constant.

    Concept

    The balanced chemical equation tells us the molar ratio of the reactants: C$_2$H$_{4(g)}$ + 3O$_{2(g)}$ → 2CO$_{2(g)}$ + 2H$_2$O$_{(g)}$

    From the equation:

    • 1 mole (or 1 volume) of C$_2$H$_4$ reacts with 3 moles (or 3 volumes) of O$_2$.

    Step-by-Step Solution

    1. Determine the required volume of oxygen: Since 1 volume of C$_2$H$_4$ requires 3 volumes of O$_2$: 10 cm$^3$ of C$_2$H$_4$ will require: $10 \text{ cm}^3 \times 3 = 30 \text{ cm}^3$ of O$_2$.

    2. Calculate the remaining volume of oxygen: We were provided with 50 cm$^3$ of oxygen at the start. Volume of oxygen used = 30 cm$^3$. Volume of oxygen remaining = (Initial volume) - (Volume used) Volume of oxygen remaining = $50 \text{ cm}^3 - 30 \text{ cm}^3 = 20 \text{ cm}^3$.

    Conclusion

    After the reaction is complete, 20 cm$^3$ of oxygen remains unreacted. This matches option A.

    Exam Tip

    Always check the stoichiometry of the balanced equation first! A common mistake is to assume a 1:1 ratio. In combustion reactions of hydrocarbons, the oxygen requirement is always determined by the coefficients in the balanced equation. If you are ever unsure, quickly re-balance the equation to ensure the number of atoms on both sides is equal before calculating the volumes.

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55. 55.

    At a given temperature and pressure, a gas X diffuses twice as fast as gas Y. It follows that

    1. A.Gas Y is two times as heavy as Gas X
    2. B.Gas Y is four times as heavy as Gas X
    3. C.Gas Y is monoatomic
    4. D.Gas X is diatomic

    Show answer

    Answer: B. Gas Y is four times as heavy as Gas X

    AI explanation

    Analysis

    This question tests your understanding of Graham’s Law of Diffusion, which relates the rate of diffusion of a gas to its molar mass (or density).

    Concept: Graham’s Law of Diffusion

    Graham’s Law states that the rate of diffusion ($r$) of a gas is inversely proportional to the square root of its molar mass ($M$):

    $r \propto \frac{1}{\sqrt{M}}$

    When comparing two gases, X and Y, the relationship is expressed as:

    $\frac{r_X}{r_Y} = \sqrt{\frac{M_Y}{M_X}}$

    Where:

    • $r_X$ and $r_Y$ are the rates of diffusion of gases X and Y.
    • $M_X$ and $M_Y$ are the molar masses of gases X and Y.

    Step-by-Step Solution

    1. Identify the given information: • Gas X diffuses twice as fast as Gas Y. This means $r_X = 2r_Y$, or $\frac{r_X}{r_Y} = 2$.

    2. Substitute into the formula: $$2 = \sqrt{\frac{M_Y}{M_X}}$$

    3. Solve for the relationship between the molar masses: • Square both sides to remove the square root: $$2^2 = \frac{M_Y}{M_X}$$ $$4 = \frac{M_Y}{M_X}$$ • Rearrange to solve for $M_Y$: $$M_Y = 4 \times M_X$$

    4. Interpret the result: The molar mass of Gas Y is four times the molar mass of Gas X. Therefore, Gas Y is four times as heavy as Gas X.

    Conclusion

    Comparing our result to the options provided, we see that option B matches our calculation.

    Exam Tip: Always remember the "Inverse Square" relationship. If a gas is faster, it must be lighter. Since the rate is squared in the formula, a speed difference of 2 becomes a mass difference of $2^2 = 4$. This quick mental math will save you time in the exam hall!

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56. 56.

    The difference in molecular mass between an alkene and alkyne with six carbon per mole is

    1. A.10
    2. B.12
    3. C.2
    4. D.14

    Show answer

    Answer: C. 2

    AI explanation

    1. Analyze the Question

    We need to find the difference in the molecular mass between an alkene and an alkyne, both containing six carbon atoms per molecule.

    2. Explain the Concept

    • General Formula for Alkenes: $C_nH_{2n}$
    • General Formula for Alkynes: $C_nH_{2n-2}$
    • Atomic Masses: Carbon (C) = 12, Hydrogen (H) = 1.
    • The difference in mass arises from the difference in the number of hydrogen atoms, as the number of carbon atoms is the same for both.

    3. Show the Solution

    Step 1: Determine the molecular formulas for $n = 6$

    • Alkene ($C_6H_{2n}$): $C_6H_{2(6)} = C_6H_{12}$
    • Alkyne ($C_6H_{2n-2}$): $C_6H_{2(6)-2} = C_6H_{10}$

    Step 2: Calculate the molecular masses

    • Mass of Alkene ($C_6H_{12}$): $(6 \times 12) + (12 \times 1) = 72 + 12 = 84$
    • Mass of Alkyne ($C_6H_{10}$): $(6 \times 12) + (10 \times 1) = 72 + 10 = 82$

    Step 3: Find the difference

    • Difference = $84 - 82 = 2$

    (Note: Alternatively, you can simply look at the difference in the number of hydrogen atoms. An alkene has 2 more hydrogen atoms than an alkyne. Since each hydrogen atom has a mass of 1, the difference is $2 \times 1 = 2$.)

    4. Identify the Answer

    Comparing our result to the options provided, the correct choice is C.

    5. Exam Tip

    Whenever you are asked for the difference in molecular mass between members of different homologous series with the same number of carbon atoms, don't waste time calculating the full molecular mass of the carbon atoms. Simply find the difference in the number of hydrogen atoms and multiply by the mass of hydrogen (1). It saves time and reduces the chance of arithmetic errors!

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57. 57.

    In a chemical reaction, surface area of reactants can affect

    1. A.CaCO$_3$ + 2HCl → CaCl$_2$ + H$_2$O + CO$_2$
    2. B.MgCl$_2$ + H$_2$SO$_4$ → MgSO$_4$ + 2HCl
    3. C.H$_2$ + Cl$_2$ → 2HCl
    4. D.NaOH + HCl → NaCl + H$_2$O

    Show answer

    Answer: A. CaCO$_3$ + 2HCl → CaCl$_2$ + H$_2$O + CO$_2$

    AI explanation

    1. Analyze the Question

    The question asks which of the given chemical reactions is affected by the surface area of the reactants. In chemistry, surface area is a factor that influences the rate of reaction, specifically in heterogeneous reactions—reactions where the reactants exist in different phases (e.g., a solid reacting with a liquid or a gas).

    2. Explain the Concept

    • Collision Theory: For a reaction to occur, particles must collide with sufficient energy (activation energy) and the correct orientation.
    • Surface Area: When a solid reactant is involved, increasing the surface area (by grinding it into a powder) exposes more particles to the other reactant. This increases the frequency of collisions, thereby increasing the rate of reaction.
    • Application: Surface area only significantly affects reactions involving at least one solid reactant. If all reactants are in the same phase (e.g., all aqueous or all gaseous), changing the surface area has no effect on the reaction rate.

    3. Show the Solution

    Let’s examine the states of the reactants in each option:

    • Option A: $\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$ Here, Calcium carbonate ($\text{CaCO}_3$) is a solid, and Hydrochloric acid ($\text{HCl}$) is an aqueous solution. Because this is a solid-liquid (heterogeneous) reaction, increasing the surface area of the solid $\text{CaCO}_3$ will increase the rate of the reaction.

    • Option B: $\text{MgCl}_2(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{MgSO}_4(aq) + 2\text{HCl}(aq)$ Both reactants are in the aqueous phase. Surface area does not apply here.

    • Option C: $\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g)$ Both reactants are in the gaseous phase. Surface area does not apply here.

    • Option D: $\text{NaOH}(aq) + \text{HCl}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$ Both reactants are in the aqueous phase. Surface area does not apply here.

    4. Identify the Answer

    Since Option A is the only reaction involving a solid reactant, it is the only one where the surface area of the reactant will affect the reaction rate.

    5. Exam Tip

    Whenever you see a question about "surface area" affecting reaction rates, look for the solid reactant in the equation. If there is no solid, surface area is not the factor you are looking for!

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58. 58.

    The substance that reacts with sodium to form alkali and changes white anhydrous copper(II) tetraoxosulphate (VI) to blue is

    1. A.base
    2. B.acid
    3. C.water
    4. D.solvent

    Show answer

    Answer: C. water

    AI explanation

    1. Analyze the question

    We are looking for a substance that meets two specific criteria:

    • Reaction with Sodium: It must react with sodium metal to produce an alkali (a basic hydroxide).
    • Hydration Test: It must turn white anhydrous copper(II) tetraoxosulphate(VI) into blue.

    2. Explain the concept

    • Reaction with Sodium: When sodium metal reacts with water ($H_2O$), it produces sodium hydroxide ($NaOH$), which is a strong alkali, and releases hydrogen gas. • Equation: $2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)$
    • Hydration Test: Anhydrous copper(II) tetraoxosulphate(VI) ($CuSO_4$) is a white powder. It is a well-known chemical test for the presence of water. When water is added to it, it becomes hydrated, forming $CuSO_4 \cdot 5H_2O$, which is blue in color.

    3. Show the solution

    • Step 1: Evaluate the reaction with sodium. Acids react with sodium to produce salts and hydrogen gas, not necessarily an alkali. Water reacts with sodium to produce an alkali ($NaOH$).
    • Step 2: Evaluate the color change. The change from white anhydrous $CuSO_4$ to blue hydrated $CuSO_4$ is the standard laboratory test for water.
    • Step 3: Combine the findings. Since water satisfies both the production of an alkali when reacting with sodium and the hydration of copper(II) tetraoxosulphate(VI), it is the correct substance.

    4. Identify the answer

    Comparing our findings to the options provided:

    • A. Base: Does not react with sodium to form an alkali.
    • B. Acid: Reacts with sodium to form a salt, not an alkali.
    • C. Water: Reacts with sodium to form $NaOH$ (alkali) and turns anhydrous $CuSO_4$ blue.
    • D. Solvent: This is a general term; while water is a solvent, "water" is the specific chemical identity required here.

    5. Exam Tip

    Whenever you see "anhydrous copper(II) tetraoxosulphate(VI)" in a question, immediately associate it with the test for water. It is a classic JAMB/WASSCE indicator. Also, remember that Group 1 metals (like Sodium) react vigorously with water to form alkaline solutions—this is why they are called "alkali metals."

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59. 59.

    The molecular formular of a hydrocarbon with an empirical formula of CH$_3$  and a molar mass of 30 is

    1. A.C$_2$H$_4$
    2. B.C$_3$H$_3$
    3. C.C$_2$H$_6$
    4. D.C$_6$H$_6$

    Show answer

    Answer: C. C$_2$H$_6$

    AI explanation

    Analysis

    To find the molecular formula of a hydrocarbon when given the empirical formula and the molar mass, we need to determine the ratio between the molar mass of the compound and the molar mass of the empirical formula.

    Concept

    The relationship between the molecular formula and the empirical formula is given by the formula: Molecular Formula = (Empirical Formula)$_n$

    Where $n$ is a whole number calculated as: $n = \frac{\text{Molar Mass of the compound}}{\text{Empirical Formula Mass}}$

    Step-by-Step Solution

    1. Calculate the Empirical Formula Mass: The empirical formula is $\text{CH}_3$. • Atomic mass of Carbon (C) = 12 • Atomic mass of Hydrogen (H) = 1 • Empirical Formula Mass = $12 + (3 \times 1) = 12 + 3 = 15\text{ g/mol}$

    2. Calculate the value of $n$: We are given that the molar mass of the hydrocarbon is 30. • $n = \frac{30}{15} = 2$

    3. Determine the Molecular Formula: Multiply the subscripts in the empirical formula ($\text{CH}_3$) by $n$ (which is 2): • Molecular Formula = $(\text{CH}_3)_2$ • Molecular Formula = $\text{C}_{(1 \times 2)}\text{H}_{(3 \times 2)}$ • Molecular Formula = $\text{C}_2\text{H}_6$

    Conclusion

    Comparing our result ($\text{C}_2\text{H}_6$) with the given options, it matches option C.

    Exam Tip

    Whenever you see a question asking for a molecular formula, always calculate the "Empirical Formula Mass" first. If your calculated $n$ is not a whole number, double-check your addition of the atomic masses—it is almost always a small integer like 1, 2, or 3!

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60. 60.

    The reaction of hydrogen and chlorine to produce hydrogen chloride gas is explosive in

    1. A.diffused light
    2. B.sunlight
    3. C.infrared light
    4. D.Raman light

    Show answer

    Answer: B. sunlight

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61. 61.

    The empirical formula of an organic liquid hydrocarbon is XY. If the relative molar masses of X and Y are 72 and 6 respectively, it's vapour density is likely to be

    1. A.33
    2. B.66
    3. C.39
    4. D.78

    Show answer

    Answer: C. 39

    AI explanation

    1. Analyze the Question

    We are given the empirical formula of a hydrocarbon as $XY$. We are provided with the relative molar masses of the individual components $X$ and $Y$. We need to find the vapour density of the compound.

    2. Explain the Concept

    • Empirical Formula: This represents the simplest whole-number ratio of atoms of each element in a compound.
    • Molar Mass of the Compound: The molar mass of the compound is the sum of the molar masses of its constituent parts based on the empirical formula.
    • Vapour Density: Vapour density is defined as the ratio of the mass of a certain volume of a gas or vapour to the mass of an equal volume of hydrogen gas at the same temperature and pressure. The relationship between molar mass ($M$) and vapour density ($VD$) is: $$M = 2 \times VD$$Therefore,$VD = \frac{M}{2}$

    3. Show the Solution

    Step 1: Calculate the molar mass ($M$) of the hydrocarbon. The empirical formula is $XY$. Given:

    • Molar mass of $X = 72$
    • Molar mass of $Y = 6$

    Total molar mass ($M$) = $72 + 6 = 78\text{ g/mol}$

    Step 2: Calculate the vapour density ($VD$). Using the formula $VD = \frac{M}{2}$:

    $VD = \frac{78}{2}$

    $VD = 39$

    4. Identify the Answer

    Comparing our result to the options provided: A. 33 B. 66 C. 39 D. 78

    The calculated value matches option C.

    5. Tip

    Always remember the relationship $M = 2 \times VD$. A common mistake students make is to multiply the molar mass by 2 instead of dividing it. If you ever forget the formula, just remember that the molar mass of a gas is always twice its vapour density because the molar mass of hydrogen gas ($H_2$) is approximately 2.

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62. 62.

    In the extraction of Aluminium, the silica impurity is removed by

    1. A.electrolysis
    2. B.filtration
    3. C.reaction with H$_2$SO$_4$
    4. D.precipitation

    Show answer

    Answer: D. precipitation

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63. 63.

    CuO$_s$   +  H$_2$($_g$)  ⇌ Cu$_s$   +  H$_2$O($_g$) In the equation above, the effect of increased pressure on the equilibrium position is that

    1. A.the equilibrium is shifted to the left
    2. B.the equilibrium is shifted to the right
    3. C.there is no effect
    4. D.more H$_2$$(g$) is produced

    Show answer

    Answer: C. there is no effect

    AI explanation

    Analysis

    The question asks how an increase in pressure affects the position of the chemical equilibrium for the reaction: CuO(s) + H₂(g) ⇌ Cu(s) + H₂O(g)

    Concept: Le Chatelier’s Principle

    Le Chatelier’s Principle states that if a dynamic equilibrium is disturbed by changing the conditions (such as pressure, temperature, or concentration), the position of equilibrium shifts to counteract the change.

    When considering the effect of pressure on a gaseous equilibrium:

    1. We only count the moles of gaseous substances. Solids (s) and liquids (l) are ignored because their volumes are negligible compared to gases.
    2. If the number of moles of gas is the same on both sides of the equation, a change in pressure will have no effect on the equilibrium position.
    3. If there are more moles of gas on one side, increasing the pressure will shift the equilibrium toward the side with fewer moles of gas to reduce the pressure.

    Step-by-Step Solution

    1. Identify the gaseous species: • Reactants: H₂(g) — there is 1 mole of gas. • Products: H₂O(g) — there is 1 mole of gas. • (Note: CuO and Cu are solids and are excluded from the pressure calculation).

    2. Compare the moles of gas: • Total moles of gas on the left = 1 • Total moles of gas on the right = 1

    3. Determine the effect: Since the number of moles of gas is equal on both sides (1 mole = 1 mole), changing the pressure does not favor either the forward or the reverse reaction. Therefore, the equilibrium position remains unchanged.

    Conclusion

    Exam Tip: Always look at the state symbols (s, l, g, aq) carefully! Many students make the mistake of counting the solid CuO or Cu in their pressure calculations. Remember: Pressure only affects systems where there is a change in the number of gaseous moles.

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64. 64.

    A factor that does not affect the rate of a chemical reaction is

    1. A.surface area
    2. B.temperature
    3. C.volume
    4. D.catalyst

    Show answer

    Answer: C. volume

    AI explanation

    Analysis

    The question asks us to identify which of the listed factors has no direct effect on the speed (rate) at which a chemical reaction proceeds.

    Concept Explanation

    The rate of a chemical reaction is determined by the frequency and energy of collisions between reactant particles. Several factors influence these collisions:

    1. Surface Area: Increasing the surface area (e.g., crushing a solid into powder) exposes more particles to collisions, thereby increasing the reaction rate.
    2. Temperature: Increasing the temperature gives particles more kinetic energy. This causes them to move faster and collide more frequently with enough energy to overcome the activation energy barrier, increasing the rate.
    3. Catalyst: A catalyst provides an alternative pathway for the reaction with a lower activation energy, allowing more particles to react at a given temperature, thus increasing the rate.
    4. Concentration/Pressure: Increasing the concentration of reactants (or pressure for gases) brings particles closer together, increasing the frequency of collisions.

    Volume, on its own, is a measure of the space occupied by a substance. While changing the volume of a container can change the concentration of gases (which affects the rate), volume itself is not a kinetic factor that dictates the rate of reaction in the same way the others do. In the context of standard chemical kinetics, volume is considered an extrinsic property that does not directly influence the reaction mechanism or rate constant.

    Step-by-Step Solution

    • Step 1: Evaluate Option A (Surface Area). We know that for heterogeneous reactions, increasing surface area increases the rate.
    • Step 2: Evaluate Option B (Temperature). We know that higher temperatures increase the kinetic energy of particles, increasing the rate.
    • Step 3: Evaluate Option D (Catalyst). We know catalysts are specifically used to speed up reactions by lowering activation energy.
    • Step 4: Evaluate Option C (Volume). While changing the volume of a gaseous system changes the pressure/concentration, "volume" as a general factor does not inherently dictate the rate of a reaction like the other three options do. Therefore, it is the odd one out.

    Conclusion

    The factor that does not directly affect the rate of a chemical reaction among the choices provided is volume.

    Exam Tip: Remember the acronym "T-S-C-C" for factors that do affect reaction rates: Temperature, Surface area, Catalyst, and Concentration. If you see "Volume" on the exam, it is usually the distractor!

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65. 65.

    Fog is a colloid in which

    1. A.liquid particles is dispersed in a gas medium
    2. B.gas particles dispersed in a liquid medium
    3. C.solid particles dispersed in a gas medium
    4. D.gas particles dispersed in a solid medium

    Show answer

    Answer: A. liquid particles is dispersed in a gas medium

    AI explanation

    Analysis of the Question

    The question asks for the classification of fog as a colloidal system. In chemistry, a colloid is a mixture where one substance of microscopically dispersed insoluble particles is suspended throughout another substance. To solve this, we need to identify the dispersed phase (the substance being spread) and the dispersion medium (the substance in which it is spread).

    Explanation of the Concept

    Colloids are classified based on the physical state of the dispersed phase and the dispersion medium. Here is a quick breakdown of common types:

    • Aerosol: Liquid or solid particles dispersed in a gas.
    • Foam: Gas particles dispersed in a liquid.
    • Emulsion: Liquid particles dispersed in a liquid.
    • Sol: Solid particles dispersed in a liquid.

    In the case of fog, it consists of tiny droplets of water (which is a liquid) suspended in the air (which is a gas). Therefore, fog is a type of aerosol.

    Step-by-Step Solution

    1. Identify the components of fog: Fog is essentially water vapor that has condensed into tiny liquid water droplets suspended in the atmosphere.
    2. Determine the Dispersed Phase: The water droplets are the substance being dispersed. Since they are water, the phase is liquid.
    3. Determine the Dispersion Medium: The droplets are suspended in the air. Air is a mixture of gases, so the medium is gas.
    4. Match with the definition: A colloid with a liquid dispersed in a gas is classified as a liquid aerosol.

    Looking at the options:

    • Option A: Liquid particles dispersed in a gas medium (Matches our analysis).
    • Option B: Gas particles dispersed in a liquid medium (This describes foam).
    • Option C: Solid particles dispersed in a gas medium (This describes smoke).
    • Option D: Gas particles dispersed in a solid medium (This describes solid foam, like pumice stone).

    Conclusion

    Fog is formed when liquid water droplets are dispersed in the air (gas).

    Exam Tip

    To remember these easily, use this mnemonic:

    • Fog/Mist: Liquid in Gas (Think: "Liquid droplets floating in the air").
    • Smoke: Solid in Gas (Think: "Solid ash/soot floating in the air").
    • Foam: Gas in Liquid (Think: "Gas bubbles trapped in soap water").

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66. 66.

    What method is suitable for the separation of gases present in air?

    1. A.Catalytic cracking of liquid air
    2. B.Fractional distillation of liquid air
    3. C.Thermal decomposition of air
    4. D.Catalytic decomposition of air

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    Answer: B. Fractional distillation of liquid air

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67. 67.

    What is the vapour density of 560cm$^3$ of a gas that weighs 0.4g at s.t.p? [Molar Volume of gas at s.t.p = 22.4 dm$^3$ ]

    1. A.6.0
    2. B.8.0
    3. C.16.0
    4. D.32.0

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    Answer: B. 8.0

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68. 68.

    During the fractional distillation of crude oil, the fraction that distills at 200 - 250$^0$ C is

    1. A.bitumen
    2. B.diesel
    3. C.kerosene
    4. D.petrol

    Show answer

    Answer: C. kerosene

    AI explanation

    Analysis

    This question tests your knowledge of the fractional distillation of crude oil, a core topic in the JAMB Chemistry syllabus. You are required to identify the specific hydrocarbon fraction that vaporizes and is collected within the temperature range of 200°C to 250°C.

    Concept Explanation

    Crude oil is a mixture of hydrocarbons with different boiling points. During fractional distillation, the crude oil is heated in a fractionating column. As the vapors rise, they cool and condense at different levels based on their boiling points.

    Here is the general temperature range for the common fractions:

    • Petrol (Gasoline): 40°C – 170°C
    • Kerosene (Paraffin): 170°C – 250°C
    • Diesel (Gas oil): 250°C – 350°C
    • Lubricating oil: 350°C – 400°C
    • Bitumen (Residue): Above 400°C

    Step-by-Step Solution

    1. Evaluate the options: We are looking for the fraction that corresponds to the 200°C – 250°C range.
    2. Compare with standard data: • Petrol distills at a lower temperature (below 170°C). • Diesel distills at a higher temperature (starting from 250°C upwards). • Kerosene falls directly within the 170°C – 250°C range.
    3. Conclusion: Based on the standard industrial fractional distillation table, the fraction collected between 200°C and 250°C is Kerosene.

    Exam Tip

    To remember the order of fractions in the column, use the mnemonic: "P-K-D-L-B" (from top to bottom/lowest to highest boiling point):

    • Petrol
    • Kerosene
    • Diesel
    • Lubricating oil
    • Bitumen

    Just remember that as you go down the column, the molecules get larger, the boiling points get higher, and the substances become more viscous (thicker).

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69. 69.

    Biodegradable pollutants are not safe in water systems because they can cause

    1. A.air pollution
    2. B.soil degradation
    3. C.ill health
    4. D.greenhouse effect

    Show answer

    Answer: C. ill health

    AI explanation

    1. Analyze the Question

    The question asks about the negative impact of biodegradable pollutants in water systems. While "biodegradable" means a substance can be broken down by microorganisms, its presence in water in large quantities creates specific environmental hazards.

    2. Explain the Concept

    Biodegradable pollutants (such as sewage, animal waste, or food processing waste) are organic materials. When these enter a water body:

    • Microbial Decomposition: Bacteria and other microorganisms begin to break down this organic matter.
    • Oxygen Depletion: This process consumes a significant amount of dissolved oxygen in the water (a process known as increasing the Biochemical Oxygen Demand or BOD).
    • Consequences: As oxygen levels drop, aquatic life (like fish) dies. Furthermore, the decomposition of these pollutants can lead to the growth of harmful pathogens (bacteria, viruses, and parasites) that contaminate the water.
    • Public Health: If humans consume or come into contact with water contaminated by these pollutants and the associated pathogens, it leads to the spread of waterborne diseases like cholera, typhoid, and dysentery.

    3. Show the Solution

    • Option A (Air pollution): While some decomposition can release gases, it is not the primary reason these pollutants are considered "unsafe" in water systems.
    • Option B (Soil degradation): This relates more to land pollution than the specific impact of water-based biodegradable pollutants.
    • Option C (Ill health): Because biodegradable pollutants encourage the growth of pathogens and deplete oxygen, they make water unsafe for drinking and domestic use, directly leading to diseases and ill health in humans.
    • Option D (Greenhouse effect): This is primarily caused by gases like $CO_2$ and methane in the atmosphere, not by the presence of biodegradable matter in water.

    4. Identify the Answer

    The most significant danger posed by biodegradable pollutants in water systems is the contamination of water sources, which leads to the spread of diseases. Therefore, they cause ill health.

    5. Tip

    When you see questions about "biodegradable pollutants in water," always link them to Biochemical Oxygen Demand (BOD) and waterborne diseases. If the water is polluted, the immediate threat to humans is the health risk!

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