2010 JAMB Chemistry Past Questions

Questions

1. 1.

   What quantity of aluminum is deposited when a current of 10A is passed through a solution of an aluminum salt for 1930s? [Al = 27, F = 96500 C mol -1]

   1. A.0.2 g
   2. B.1.8 g
   3. C.5.4 g
   4. D.14.2 g

   Show answer

   Answer: B. 1.8 g

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2. 2.

   Calculate the percentage by mass of nitrogen in calcium trioxonitrate (V) [Ca = 40, N = 14, O = 16]

   1. A.8.5%
   2. B.13.1%
   3. C.17.1%
   4. D.27.6%

   Show answer

   Answer: C. 17.1%

   AI explanation

   1. Analyze the question

   We need to find the percentage by mass of nitrogen in calcium trioxonitrate(V).

   • Chemical Formula: Calcium is $Ca^{2+}$ and trioxonitrate(V) is $NO_3^-$. Therefore, the formula is $Ca(NO_3)_2$.
   • Given Atomic Masses: $Ca = 40$, $N = 14$, $O = 16$.

   2. Explain the concept

   The percentage by mass of an element in a compound is calculated using the formula:

   $\text{Percentage} = \frac{\text{Total mass of the element in the compound}}{\text{Molar mass of the compound}} \times 100\%$

   3. Show the solution

   Step 1: Calculate the molar mass of $Ca(NO_3)_2$

   • $Ca = 1 \times 40 = 40$
   • $N = 2 \times 14 = 28$ (Note: there are two nitrogen atoms because of the subscript 2 outside the bracket)
   • $O = 6 \times 16 = 96$ (Note: $3 \times 2 = 6$ oxygen atoms)

   Total Molar Mass = $40 + 28 + 96 = 164 \text{ g/mol}$

   Step 2: Calculate the total mass of Nitrogen

   • Mass of Nitrogen = $2 \times 14 = 28 \text{ g/mol}$

   Step 3: Calculate the percentage

   $\text{Percentage of N} = \left( \frac{28}{164} \right) \times 100\%$

   $\text{Percentage of N} = 0.17073 \times 100\%$

   $\text{Percentage of N} \approx 17.1\%$

   4. Identify the answer

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

   5. Exam Tip

   Always pay close attention to the subscripts in brackets! Many students forget to multiply the atoms inside the bracket (like $NO_3$) by the subscript outside (the 2 in $Ca(NO_3)_2$). If you had calculated for $CaNO_3$ instead, you would have gotten the wrong answer. Always write out the full formula clearly before starting your calculation.

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3. 3.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.

4. 4.

   In which of the following is the entropy change positive?

   1. A.Thermal dissociation of ammonium chloride
   2. B.Reaction between an acid and a base
   3. C.Addition of concentrated acid to water
   4. D.Dissolution of sodium metal in water

   Show answer

   Answer: A. Thermal dissociation of ammonium chloride

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5. 5.

   Moving from left to right across a period, the general rise in the first ionization energy can be attributed to the?

   1. A.decrease in nuclear charge
   2. B.increase in nuclear charge
   3. C.decrease in screening effect
   4. D.increase in screening effect

   Show answer

   Answer: B. increase in nuclear charge

   AI explanation

   Analysis

   The question asks for the reason behind the general increase in first ionization energy as we move from left to right across a period in the Periodic Table.

   Concept Explanation

   Ionization Energy is the minimum amount of energy required to remove the most loosely held electron from an isolated gaseous atom.

   As you move from left to right across a period:

   1. Nuclear Charge: The number of protons in the nucleus increases. This creates a stronger positive pull on the electrons.
   2. Atomic Radius: Because the nucleus is pulling the electrons more strongly, the electron cloud is pulled closer to the nucleus, causing the atomic radius to decrease.
   3. Screening (Shielding) Effect: The number of inner electron shells remains the same across a period. Therefore, the screening effect remains relatively constant and does not significantly change.

   Because the nuclear charge increases while the distance between the nucleus and the valence electrons decreases (and the screening effect stays constant), the electrostatic force of attraction between the nucleus and the valence electrons becomes much stronger. Consequently, it requires more energy to remove an electron.

   Step-by-Step Solution

   1. Identify the trend: Across a period, the atomic number increases, meaning the number of protons (nuclear charge) increases.
   2. Relate to force: According to Coulomb’s Law, a higher nuclear charge exerts a stronger pull on the valence electrons.
   3. Relate to energy: A stronger pull means the electrons are held more tightly. Therefore, more energy is needed to overcome this attraction to remove an electron.
   4. Evaluate options: • A is incorrect because nuclear charge increases, not decreases. • B is correct because the increasing positive charge of the nucleus is the primary driver for the increased hold on electrons. • C and D are incorrect because the screening effect remains roughly constant across a period, so it is not the cause of the rise in ionization energy.

   Exam Tip

   Always remember: "Across a period, the nucleus gets stronger." If you ever get confused, think of the nucleus as a magnet. As you add more protons (moving right), the magnet gets stronger, making it harder to pull an electron away.

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6. 6.

   The final products of the reaction between methane and chlorine in the presence of ultraviolet light are hydrogen chloride and?

   1. A.tricloromethane
   2. B.dichloromethane
   3. C.tetrachloromethane
   4. D.chloromethane

   Show answer

   Answer: C. tetrachloromethane

   AI explanation

   Analysis of the Question

   This question tests your understanding of free-radical substitution reactions in alkanes. Specifically, it asks for the final product of the reaction between methane ($CH_4$) and chlorine ($Cl_2$) under ultraviolet (UV) light.

   Concept Explanation

   When an alkane like methane reacts with a halogen (like chlorine) in the presence of UV light, a series of substitution reactions occurs. The UV light provides the energy to break the chlorine molecule into highly reactive chlorine radicals. These radicals then replace the hydrogen atoms in the methane molecule one by one:

   1. $CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$ (Chloromethane)
   2. $CH_3Cl + Cl_2 \rightarrow CH_2Cl_2 + HCl$ (Dichloromethane)
   3. $CH_2Cl_2 + Cl_2 \rightarrow CHCl_3 + HCl$ (Trichloromethane)
   4. $CHCl_3 + Cl_2 \rightarrow CCl_4 + HCl$ (Tetrachloromethane)

   The reaction continues until all hydrogen atoms are replaced by chlorine atoms. The "final product" refers to the stage where no more hydrogen atoms remain to be substituted.

   Step-by-Step Solution

   1. Identify the reactants: Methane ($CH_4$) and Chlorine ($Cl_2$).
   2. Understand the process: The reaction is a chain reaction. As long as there is an excess of chlorine, the substitution will proceed until all four hydrogen atoms in the methane molecule are replaced by chlorine atoms.
   3. Determine the final state: After the fourth substitution, the molecule becomes $CCl_4$.
   4. Name the product: $CCl_4$ is known as tetrachloromethane (also commonly called carbon tetrachloride).

   Conclusion

   Since the reaction proceeds until all hydrogen atoms are replaced, the final product is tetrachloromethane.

   Exam Tip: Remember the sequence of substitution: Mono- (one Cl), Di- (two Cl), Tri- (three Cl), and Tetra- (four Cl). If the question asks for the "final product," always look for the option with the highest number of chlorine atoms (the "tetra" prefix).

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7. 7.

   2H2(g) + O2(g) ⇌ 2H2O(g) ΔH = -ve what happens to the equilibrium constant of the reaction above if the temperature is increased?

   1. A.it is unaffected
   2. B.it becomes zero
   3. C.it decrease
   4. D.it increase

   Show answer

   Answer: C. it decrease

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8. 8.

   The compound that is used as an anesthetic is?

   1. A.CCl4
   2. B.CH Cl3
   3. C.CH2Cl2
   4. D.CH3Cl

   Show answer

   Answer: B. CH Cl3

   AI explanation

   Analysis

   The question asks for the chemical compound among the given options that is historically and chemically recognized for its use as an anesthetic.

   Concept Explanation

   The compounds listed are all haloalkanes (specifically chloromethanes).

   • CCl₄ (Tetrachloromethane): Primarily used as a solvent and in fire extinguishers. It is toxic to the liver.
   • CHCl₃ (Trichloromethane): Commonly known as chloroform. In the 19th and early 20th centuries, it was widely used as an inhalational anesthetic during surgery. Although it has been largely replaced by safer alternatives (like halothane) due to its toxicity, it remains the classic answer in chemistry examinations regarding anesthetics.
   • CH₂Cl₂ (Dichloromethane): Used primarily as a paint stripper and solvent.
   • CH₃Cl (Chloromethane): Used as a refrigerant and in the production of silicones.

   Step-by-Step Solution

   1. Identify the chemical names: • A: Tetrachloromethane • B: Trichloromethane (Chloroform) • C: Dichloromethane • D: Chloromethane
   2. Evaluate the medical history: Among these, Trichloromethane (CHCl₃) is the only one historically documented and widely taught in the Nigerian secondary school chemistry curriculum as an anesthetic agent. 3.

   Conclusion

   The correct compound is Trichloromethane, which is represented by the formula CHCl₃.

   Exam Tip

   When you see questions about haloalkanes, remember their common names:

   • CHCl₃ = Chloroform (Anesthetic)
   • CCl₄ = Carbon tetrachloride (Solvent/Fire extinguisher)
   • CHI₃ = Triiodomethane (Iodoform - Antiseptic)

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9. 9.

   The residual solids from the fractional distillation of petroleum are used as?

   1. A.coatings of pipes
   2. B.raw materials for the cracking process
   3. C.fuel for the driving tractors
   4. D.fuel for jet engines

   Show answer

   Answer: A. coatings of pipes

   AI explanation

   Analysis

   This question tests your knowledge of the fractional distillation of crude oil (petroleum) and the specific uses of the various fractions obtained from the process. You are asked to identify the application of the "residual solids" (often called bitumen or asphalt) left at the bottom of the fractionating column.

   Concept: Fractional Distillation of Petroleum

   In a fractionating column, crude oil is heated, and components are separated based on their boiling points. The substances with the highest boiling points do not vaporize and remain at the bottom as a thick, viscous residue.

   • Refinery Gases: Used for cooking gas (LPG).
   • Petrol/Gasoline: Used for cars.
   • Kerosene: Used for jet engines and domestic stoves.
   • Diesel/Gas Oil: Used for heavy machinery and tractors.
   • Lubricating Oils/Waxes: Used for lubrication and candles.
   • Bitumen (Residue): This is the thick, black, sticky substance at the bottom. Because of its waterproof and adhesive properties, it is primarily used for road surfacing and as a protective coating for pipes to prevent corrosion.

   Step-by-Step Solution

   1. Identify the fraction: The "residual solids" mentioned are known as bitumen or asphalt.
   2. Evaluate the options: • A (Coatings of pipes): Bitumen is widely used for coating pipes and road surfaces because it is waterproof and durable. This fits the description. • B (Raw materials for cracking): Cracking usually involves heavier gas oils or naphtha, not the final solid residue. • C (Fuel for tractors): This is the role of diesel (gas oil). • D (Fuel for jet engines): This is the role of kerosene (specifically Jet A-1).
   3. Conclusion: Based on the properties of the residual fraction, it is used for surfacing and protective coatings.

   Exam Tip

   When studying the fractional distillation table, remember the order of boiling points: "Gases rise, solids sink." The higher the boiling point, the lower the fraction settles in the column. Bitumen is at the very bottom, so it is the "heaviest" and most viscous.

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10. 10.

    In the preparation of oxygen by heating KClO3 in the presence of MnO2, only moderate heat is needed because the catalyst acts by?

    1. A.lowering the pressure of the reaction
    2. B.increasing the surface area of the reactant
    3. C.increase the rate of the reaction
    4. D.lowering the energy barrier of the reaction

    Show answer

    Answer: D. lowering the energy barrier of the reaction

    AI explanation

    Analysis

    The question asks for the mechanism by which a catalyst (in this case, Manganese(IV) oxide, $MnO_2$) facilitates the decomposition of Potassium trioxochlorate(V) ($KClO_3$) at a lower temperature.

    Concept Explanation

    A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed in the process.

    In chemical kinetics, for a reaction to occur, reactant particles must collide with a minimum amount of energy known as the Activation Energy ($E_a$). This is the "energy barrier" that must be overcome for the reactants to transform into products.

    A catalyst provides an alternative reaction pathway that has a lower activation energy. Because the energy barrier is lower, a larger fraction of reactant molecules possess enough energy to react at a given temperature. Consequently, the reaction proceeds much faster at lower temperatures than it would without the catalyst.

    Step-by-Step Solution

    1. Identify the role of $MnO_2$: In the laboratory preparation of oxygen, $MnO_2$ is added to $KClO_3$ to speed up the decomposition.
    2. Evaluate the options: • A: Catalysts do not change the pressure of a reaction system. • B: While catalysts can be solid surfaces, their primary function is not just "increasing surface area" (that is the role of grinding or powdering reactants). • C: While it is true that a catalyst increases the rate of reaction, the question specifically asks how it allows the reaction to occur at "moderate heat." • D: By lowering the activation energy (the energy barrier), the catalyst ensures that the reaction can proceed efficiently at lower temperatures. This is the fundamental thermodynamic explanation for why "moderate heat" is sufficient.

    Conclusion

    While option C is a correct statement about what a catalyst does, option D provides the specific mechanism that explains why the heat requirement is reduced. In the context of JAMB chemistry, lowering the activation energy is the standard scientific explanation for this phenomenon.

    Exam Tip: Always remember the "Energy Profile Diagram." A catalyst creates a "shortcut" on the graph, lowering the peak (the activation energy) between the reactants and the products. This is a favorite topic for JAMB examiners!

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11. 11.

    Which of the following salts is slightly soluble in water?

    1. A.AgCl
    2. B.CaSO4
    3. C.Na2CO3
    4. D.PbCl2

    Show answer

    Answer: B. CaSO4

    AI explanation

    Analysis

    The question asks us to identify which of the given salts is "slightly soluble" in water. Solubility refers to the ability of a substance to dissolve in a solvent. In chemistry, we categorize salts based on their solubility rules.

    Concept Explanation

    To solve this, we apply the general solubility rules for inorganic salts:

    1. Sodium (Na⁺) salts: Almost all sodium salts (like $Na_2CO_3$) are highly soluble in water.
    2. Chlorides (Cl⁻): Most chlorides are soluble, but there are exceptions. Silver chloride ($AgCl$) and Lead(II) chloride ($PbCl_2$) are generally considered insoluble (they form precipitates).
    3. Sulfates ($SO_4^{2-}$): Most sulfates are soluble, but there are notable exceptions like Barium sulfate ($BaSO_4$), Lead sulfate ($PbSO_4$), and Calcium sulfate ($CaSO_4$).

    "Slightly soluble" refers to substances that do not dissolve completely but are not as insoluble as a precipitate like $AgCl$. $CaSO_4$ is a classic example of a salt that is sparingly or slightly soluble in water.

    Step-by-Step Solution

    • Option A ($AgCl$): This is an insoluble salt. It is used in qualitative analysis to test for chloride ions because it forms a white precipitate.
    • Option B ($CaSO_4$): Calcium sulfate is known to be sparingly (slightly) soluble in water. It is often used in the context of "temporary hardness" in water because it dissolves only to a small extent.
    • Option C ($Na_2CO_3$): Sodium carbonate is highly soluble in water. It is a common washing soda used in laboratories and homes.
    • Option D ($PbCl_2$): Lead(II) chloride is considered insoluble at room temperature, though its solubility increases significantly in hot water.

    Comparing these, $CaSO_4$ is the most accurate fit for the term "slightly soluble" in the context of standard JAMB chemistry curriculum.

    Conclusion

    While $PbCl_2$ has some solubility, $CaSO_4$ is the standard textbook example for a slightly soluble sulfate.

    Exam Tip

    Remember the acronym "SNAP" for solubility: Sodium, Nitrate, Ammonium, and Potassium salts are always soluble. When you see these in an option, you can usually rule them out if the question asks for an insoluble or slightly soluble salt!

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12. 12.

    Coffee stains can best be removed by?

    1. A.Kerosine
    2. B.turpentine
    3. C.a solution of borax in water
    4. D.ammonia solution

    Show answer

    Answer: C. a solution of borax in water

    AI explanation

    Analysis

    The question asks for the most effective substance to remove coffee stains from fabrics or surfaces. Coffee stains are primarily organic in nature, containing tannins and other plant-based pigments.

    Concept Explanation

    To remove stains, you must use a substance that can either dissolve the stain (a solvent) or chemically break it down (a cleaning agent).

    • Kerosene and Turpentine: These are non-polar organic solvents. They are excellent for removing oil-based stains like grease, paint, or wax, but they are ineffective against water-soluble stains like coffee.
    • Ammonia solution: While ammonia is a common household cleaner, it is highly alkaline and can damage certain fabrics or cause color bleeding.
    • Borax (Sodium tetraborate): Borax acts as a mild alkali and a bleaching agent. When dissolved in water, it helps to break down the complex organic molecules found in coffee, making them water-soluble and easier to wash away without damaging the fabric.

    Step-by-Step Solution

    1. Identify the nature of the stain: Coffee is a water-based, organic stain.
    2. Evaluate the solvents: • Kerosene and Turpentine are non-polar; they will not mix with the water-based coffee stain. • Ammonia is a strong base that is often too harsh for delicate fabrics.
    3. Evaluate the cleaning agent: Borax is a salt of boric acid. In water, it creates a slightly alkaline solution that effectively lifts organic pigments (tannins) from fibers. It is widely recognized in home economics and chemistry for its ability to remove stubborn organic stains like coffee, tea, and wine.

    Conclusion

    Based on chemical properties and standard cleaning practices, a solution of borax in water is the most effective and safe method for removing coffee stains among the given options.

    Exam Tip: When dealing with "stain removal" questions in Chemistry, always categorize the stain first:

    • Oil/Grease: Use organic solvents (Kerosene, Turpentine, Petrol).
    • Water-based/Organic (Coffee, Tea, Ink): Use mild alkalis or oxidizing agents (Borax, Hydrogen peroxide, or mild detergents).

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13. 13.

    Substance employed as drying agents are usually?

    1. A.amphoteric
    2. B.hygroscopic
    3. C.efflorescent
    4. D.acidic

    Show answer

    Answer: B. hygroscopic

    AI explanation

    Analysis of the Question

    The question asks for the property of substances that are used as "drying agents." In chemistry, a drying agent (or desiccant) is a substance that removes moisture from its surroundings, such as from gases or other chemical compounds.

    Explanation of Concepts

    To identify the correct answer, let's look at the definitions of the terms provided in the options:

    • Amphoteric: These are substances that can react as both an acid and a base (e.g., Aluminum oxide). This property does not determine whether a substance can absorb water.
    • Hygroscopic: These are substances that have the ability to absorb moisture from the atmosphere. Because they "soak up" water vapor, they are ideal for use as drying agents in the laboratory.
    • Efflorescent: These are hydrated salts that lose their water of crystallization to the atmosphere when exposed to air. They do the opposite of drying agents; they release water.
    • Acidic: This refers to the pH level of a substance. While some drying agents (like concentrated tetraoxosulphate(VI) acid) are acidic, being acidic is not the defining characteristic that makes a substance a drying agent.

    Step-by-Step Solution

    1. Define the role: A drying agent must be capable of attracting and holding water molecules from the environment.
    2. Evaluate the options: • An efflorescent substance loses water, so it cannot be a drying agent. • Amphoteric and acidic are chemical properties related to reactivity and pH, not necessarily water absorption. • A hygroscopic substance actively absorbs water from the air or from a substance it is in contact with. This is the exact definition of a drying agent.
    3. Conclusion: Since hygroscopic substances naturally attract and retain water, they are the standard materials used to dry gases or keep environments moisture-free.

    Identify the Answer

    The correct option is B.

    Exam Tip

    To remember this easily, think of the word "Hygro" (related to humidity/water) and "Scopic" (like a scope, reaching out to grab). A hygroscopic substance "reaches out" to grab water from the air. Conversely, remember that efflorescent starts with "E" for "Exit"—the water is exiting the substance!

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14. 14.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.

15. 15.

    The volume of a given gas is V cm$^3$ at a pressure P mm Hg. what is the new volume of the gas if the pressure is reduced to half at constant temperature?

    1. A.4 V cm3
    2. B.2 V cm3
    3. C.V/2 cm3
    4. D.V cm3

    Show answer

    Answer: B. 2 V cm3

    AI explanation

    Analysis

    The question asks for the relationship between the volume and pressure of a gas when the temperature is held constant. This is a classic application of the gas laws.

    Concept: Boyle's Law

    Boyle's Law states that for a fixed mass of an ideal gas, the volume ($V$) of the gas is inversely proportional to its pressure ($P$), provided the temperature remains constant.

    Mathematically, this is expressed as:

    $P_1V_1 = P_2V_2$

    Where:

    • $P_1$ = Initial pressure
    • $V_1$ = Initial volume
    • $P_2$ = Final pressure
    • $V_2$ = Final volume

    Step-by-Step Solution

    1. Identify the initial values: • Initial pressure ($P_1$) = $P$ • Initial volume ($V_1$) = $V$

    2. Identify the final values: • The pressure is reduced to half, so the new pressure ($P_2$) = $\frac{P}{2}$ • The new volume ($V_2$) = ?

    3. Apply Boyle's Law formula: $$P \times V = \left(\frac{P}{2}\right) \times V_2$$

    4. Solve for $V_2$: • Divide both sides by $P$: $$V = \frac{1}{2} \times V_2$$ • Multiply both sides by 2 to isolate $V_2$: $$2V = V_2$$

    The new volume is $2V$.

    Conclusion

    When the pressure of a gas is halved at a constant temperature, the volume must double to maintain the constant product of $PV$.

    Exam Tip: Whenever you see "constant temperature" in a gas law question, immediately think of Boyle's Law. Remember: Pressure and Volume are like a see-saw—if one goes down, the other must go up!

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16. 16.

    How many unpaired electron(s) are there in the nitrogen sub-levels?

    1. A.3
    2. B.2
    3. C.1
    4. D.none

    Show answer

    Answer: A. 3

    AI explanation

    1. Analyze the question

    The question asks for the number of unpaired electrons in a neutral nitrogen atom. To find this, we need to determine the electronic configuration of nitrogen and apply Hund's Rule of Maximum Multiplicity.

    2. Explain the concept

    • Atomic Number: Nitrogen (N) has an atomic number of 7. This means a neutral nitrogen atom has 7 electrons.
    • Electronic Configuration: Electrons fill orbitals in order of increasing energy: $1s^2, 2s^2, 2p^3$.
    • Hund’s Rule: This rule states that for a given electron sub-shell (like the $p$-subshell), electrons fill empty orbitals singly before they start pairing up. The $p$-subshell has three orbitals ($p_x, p_y, p_z$).

    3. Show the solution

    • Step 1: Write the full electronic configuration of Nitrogen ($Z=7$): $1s^2, 2s^2, 2p^3$
    • Step 2: Focus on the outermost sub-level, which is $2p^3$.
    • Step 3: Distribute the 3 electrons in the $2p$ sub-shell into the three available $p$-orbitals ($p_x, p_y, p_z$): • $p_x$ gets 1 electron • $p_y$ gets 1 electron • $p_z$ gets 1 electron
    • Step 4: Count the unpaired electrons. Since each of the three $p$-orbitals contains exactly one electron, there are 3 unpaired electrons.

    4. Identify the answer

    Based on the configuration, there are 3 unpaired electrons in the $2p$ sub-level. This matches option A.

    5. Tip

    Memory Trick: Remember the "Bus Seat Rule." Electrons are like people getting on a bus; they prefer to sit in empty seats (orbitals) individually before they are forced to share a seat (pair up). Since nitrogen has 3 electrons in the $p$-subshell and there are 3 "seats" available, each electron gets its own seat, leaving 3 unpaired!

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17. 17.

    Calculate the solubility in mol $dm^{-3}$ of 40g of CuSO4 dissolved in 100g of water at 120°C? [Cu = 64, S = 32, O = 16]

    1. A.4.00
    2. B.2.50
    3. C.0.40
    4. D.0.25

    Show answer

    Answer: B. 2.50

    AI explanation

    1. Analyze the question

    We are asked to find the solubility of Copper(II) tetraoxosulphate(VI) ($\text{CuSO}_4$) in $\text{mol dm}^{-3}$. We are given the mass of the solute (40g), the mass of the solvent (100g of water), and the atomic masses to calculate the molar mass.

    2. Explain the concept

    • Solubility in $\text{mol dm}^{-3}$: This is the molar concentration of a saturated solution. It is calculated as: $$\text{Solubility} = \frac{\text{Number of moles of solute}}{\text{Volume of solvent in dm}^3}$$
    • Molar Mass: The sum of the atomic masses of all atoms in the formula.
    • Density of Water: In these calculations, we assume the density of water is $1\text{g/cm}^3$ (or $1\text{g/ml}$). Therefore, $100\text{g}$ of water is equal to $100\text{cm}^3$, which is $0.1\text{dm}^3$.

    3. Show the solution

    Step 1: Calculate the molar mass of $\text{CuSO}_4$

    • $\text{Cu} = 64$
    • $\text{S} = 32$
    • $\text{O} = 16 \times 4 = 64$
    • $\text{Molar Mass} = 64 + 32 + 64 = 160\text{ g/mol}$

    Step 2: Calculate the number of moles of $\text{CuSO}_4$

    • $\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$
    • $\text{Moles} = \frac{40\text{g}}{160\text{g/mol}} = 0.25\text{ moles}$

    Step 3: Convert the volume of water to $\text{dm}^3$

    • $100\text{g of water} = 100\text{cm}^3$
    • $100\text{cm}^3 = \frac{100}{1000}\text{dm}^3 = 0.1\text{dm}^3$

    Step 4: Calculate the solubility

    • $\text{Solubility} = \frac{0.25\text{ moles}}{0.1\text{dm}^3} = 2.5\text{ mol dm}^{-3}$

    4. Identify the answer

    Comparing our result to the options provided: A. 4.00 B. 2.50 C. 0.40 D. 0.25

    The correct option is B.

    5. Tip

    When solving solubility questions, always check the units requested. If the question asks for $\text{mol dm}^{-3}$, ensure your volume is in $\text{dm}^3$ (liters) and your mass is converted to moles. A common mistake is forgetting to divide by the volume of the solvent—don't let that happen to you!

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18. 18.

    The maximum number of electrons in the L shell of an atom is?

    1. A.2
    2. B.8
    3. C.18
    4. D.32

    Show answer

    Answer: B. 8

    AI explanation

    Analysis

    The question asks for the maximum number of electrons that can be accommodated in the L shell of an atom. In atomic structure, electrons are arranged in shells (energy levels) designated by the principal quantum number, $n$.

    Concept

    The maximum number of electrons that any shell can hold is determined by the formula: Maximum electrons = $2n^2$

    Where $n$ is the shell number:

    • K shell: $n = 1$
    • L shell: $n = 2$
    • M shell: $n = 3$
    • N shell: $n = 4$

    Solution

    1. Identify the shell number for the L shell. The shells are ordered K, L, M, N... which correspond to $n = 1, 2, 3, 4...$ respectively. Therefore, for the L shell, $n = 2$.
    2. Apply the formula $2n^2$: • Maximum electrons = $2 \times (2)^2$ • Maximum electrons = $2 \times 4$ • Maximum electrons = 8

    Conclusion

    The L shell can hold a maximum of 8 electrons.

    Exam Tip: Remember the sequence 2, 8, 18, 32. These are the maximum capacities for the K, L, M, and N shells respectively. If you ever forget the formula, just remember that the shells fill up in this specific order!

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19. 19.

    Stainless steel is used for making?

    1. A.magnets
    2. B.tools
    3. C.coins and medals
    4. D.moving parts of clocks

    Show answer

    Answer: B. tools

    AI explanation

    1. Analyze the question

    The question asks for the primary application or use of stainless steel. We need to identify which of the provided options is the most common and practical use for this specific alloy.

    2. Explain the concept

    Stainless steel is an alloy of iron, carbon, and a significant amount of chromium (usually at least 10.5%). The addition of chromium is what gives stainless steel its defining property: corrosion resistance (it does not rust easily).

    Because it is strong, durable, and resistant to rust and staining, it is widely used in environments where hygiene and longevity are critical. Common uses include:

    • Cutlery and kitchenware: Knives, spoons, and forks.
    • Surgical instruments: Scalpels and medical tools.
    • Industrial tools: Wrenches, hammers, and other hardware that need to withstand moisture without rusting.

    3. Show the solution

    Let's evaluate the options based on the properties of stainless steel:

    • A. Magnets: Magnets are typically made from "hard" magnetic materials like Alnico (Aluminum, Nickel, Cobalt) or Ferrite. While some stainless steels are magnetic, they are not the primary material used to manufacture magnets.
    • B. Tools: Because stainless steel is hard, durable, and resistant to corrosion, it is the standard material for high-quality hand tools, surgical tools, and kitchen tools. This fits the properties perfectly.
    • C. Coins and medals: Coins are usually made from cupro-nickel or bronze alloys. While some modern coins use stainless steel, it is not its most defining or primary application compared to tools.
    • D. Moving parts of clocks: Clock parts are often made from brass or specialized steel alloys that are easy to machine and have low friction, but "tools" is a much broader and more accurate classification for the general use of stainless steel.

    4. Identify the answer

    Comparing the options, tools is the most accurate and widely recognized application for stainless steel in the context of chemistry and materials science.

    5. Tip

    Memory Trick: Think of the "S" in Stainless Steel as standing for Surgical and Shop tools. Whenever you see "stainless steel," remember it is the "rust-proof" metal used for things that need to stay clean and strong, like your kitchen knives or a mechanic's wrench!

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20. 20.

    A constituent common to bronze and solder is?

    1. A.lead
    2. B.silver
    3. C.copper
    4. D.tin

    Show answer

    Answer: D. tin

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21. 21.

    The droplets of water observed around a bottle of milk taken out of the refrigerator is due to the fact that the?

    1. A.water vapour in the air around the bottle gains some energy from the bottle
    2. B.temperature of the milk drops as it loses heat into the surroundings
    3. C.saturated vapour pressure of the milk is equal to the atmospheric pressure
    4. D.water vapour in the air around the bottle loses some of its energy to the bottle

    Show answer

    Answer: D. water vapour in the air around the bottle loses some of its energy to the bottle

    AI explanation

    Analysis

    This question tests your understanding of phase changes and the kinetic theory of matter, specifically regarding how water vapour in the atmosphere interacts with cold surfaces.

    Concept Explanation

    The phenomenon described is condensation.

    1. Water Vapour: The air around us contains invisible water vapour (gaseous water).
    2. Kinetic Energy: Molecules in a gas possess kinetic energy, which keeps them in a gaseous state.
    3. Heat Transfer: When this warm, moist air comes into contact with a cold surface (like a bottle of milk from the refrigerator), heat flows from the warmer air to the colder bottle.
    4. Phase Change: As the water vapour molecules lose kinetic energy to the cold bottle, they slow down. When they lose enough energy, the intermolecular forces become strong enough to pull them together, turning the gas into liquid droplets on the surface of the bottle.

    Step-by-Step Solution

    • Step 1: Identify the state of the water. It starts as vapour (gas) in the air.
    • Step 2: Identify the interaction. The cold bottle acts as a "heat sink," absorbing heat from the surrounding air.
    • Step 3: Apply the principle of energy transfer. Since the bottle is cold, the water vapour molecules hitting the bottle lose energy to it.
    • Step 4: Conclude the result. Losing energy causes the vapour to condense into liquid water droplets.

    Looking at the options:

    • A is incorrect because the vapour loses energy, it does not gain it.
    • B describes the milk, but it does not explain why water forms on the outside of the bottle.
    • C refers to boiling points, which is irrelevant here.
    • D correctly identifies that the water vapour loses energy to the cold bottle, leading to condensation.

    Conclusion

    The correct answer is D.

    Exam Tip

    Whenever you see questions about "droplets" or "mist" forming on a cold surface, always look for the option that mentions loss of energy or cooling of the surrounding vapour. Remember: Condensation = Cooling = Loss of Energy.

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22. 22.

    Which of the following is used as fuel in miners' lamp?

    1. A.Ethanal
    2. B.Ethyne
    3. C.Ethene
    4. D.Ethane

    Show answer

    Answer: B. Ethyne

    AI explanation

    Analysis

    The question asks for the specific hydrocarbon used as a fuel in a miner's lamp. This is a classic application of the properties of alkynes in industrial and practical settings.

    Concept Explanation

    • Ethyne ($C_2H_2$): Also known as acetylene, this is an alkyne. When ethyne burns in oxygen, it produces an exceptionally hot, bright, and luminous flame. Because of this high luminosity and heat, it was historically used in "carbide lamps" (or acetylene lamps) by miners to provide light in underground tunnels.
    • Ethanal ($CH_3CHO$): This is an aldehyde, not a fuel used for lighting.
    • Ethene ($C_2H_4$): This is an alkene used primarily in the production of plastics (like polythene) and for fruit ripening.
    • Ethane ($C_2H_6$): This is an alkane used mainly as a fuel for heating or as a feedstock in the petrochemical industry, but it does not produce the intense, bright light required for a miner's lamp.

    Step-by-Step Solution

    1. Identify the requirement: A miner's lamp requires a fuel that produces a bright, steady, and luminous flame to ensure visibility in dark, underground environments.
    2. Evaluate the options: • Ethyne ($C_2H_2$) undergoes combustion: $2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O + \text{Heat/Light}$. • The reaction releases a significant amount of energy and produces a very bright white light, which is ideal for illumination.
    3. Conclusion: Among the given options, Ethyne is the only one with the chemical properties suitable for use in a miner's lamp.

    Exam Tip

    Whenever you see questions about Ethyne (Acetylene), remember two key uses:

    1. Oxy-acetylene welding: Used for cutting and welding metals due to its high flame temperature.
    2. Miner's lamp: Used for its high luminosity (brightness).

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23. 23.

    The stability of the noble gases is due to the fact that they?

    1. A.have no electron in their outermost shells
    2. B.have duplet or octet electron configurations
    3. C.belong to group zero of the periodic table
    4. D.are volatile in nature

    Show answer

    Answer: B. have duplet or octet electron configurations

    AI explanation

    Analysis

    The question asks for the fundamental reason behind the chemical inertness (stability) of noble gases (Group 18 elements).

    Concept Explanation

    In chemistry, atoms tend to react and form bonds to achieve a stable electronic configuration. This is often referred to as the Octet Rule.

    • Octet configuration: Most atoms aim to have 8 electrons in their outermost shell (valence shell) to become stable, similar to the noble gas Neon, Argon, etc.
    • Duplet configuration: Helium, being the first noble gas, is stable with only 2 electrons in its outermost shell (the 1s orbital).

    Noble gases are already "satisfied" because their outermost shells are completely filled. They do not need to gain, lose, or share electrons with other atoms, which is why they are chemically inert (unreactive).

    Step-by-Step Solution

    1. Evaluate Option A: This is incorrect. Noble gases have electrons in their outermost shells; they are just fully occupied.
    2. Evaluate Option B: This is correct. Helium has a stable duplet (2 electrons), while Neon, Argon, Krypton, Xenon, and Radon have a stable octet (8 electrons). This full configuration prevents them from reacting under normal conditions.
    3. Evaluate Option C: While it is true that they belong to Group 0 (or Group 18), this is a classification, not the reason for their stability. The stability is the cause of their position in the table, not the other way around.
    4. Evaluate Option D: Volatility refers to how easily a substance turns into a gas. While noble gases are gases, their volatility is a physical property, not the reason for their chemical stability.

    Conclusion

    The chemical stability of noble gases is directly attributed to their electronic structure, which consists of either a duplet (for Helium) or an octet (for the others) of electrons in their valence shells.

    Exam Tip: Whenever you see a question about the reactivity of elements, always look for the option that mentions "valence electrons" or "octet/duplet configuration." These are the keys to understanding chemical bonding!

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24. 24.

    12D + 13T → 24He + 01n + energy The reaction above illustrates?

    1. A.alpha decay
    2. B.artificial transmutation
    3. C.nuclear fussion
    4. D.nuclear fission

    Show answer

    Answer: C. nuclear fussion

    AI explanation

    Analysis of the Question

    The equation provided is:

    $^2_1\text{D} + ^3_1\text{T} \rightarrow ^4_2\text{He} + ^1_0\text{n} + \text{energy}$

    This represents the reaction between Deuterium ($^2_1\text{D}$) and Tritium ($^3_1\text{T}$), which are isotopes of hydrogen, to form a Helium nucleus ($^4_2\text{He}$) and a neutron ($^1_0\text{n}$).

    Concept Explanation

    • Nuclear Fission: This is the process where a heavy, unstable nucleus (like Uranium-235) splits into two or more smaller, lighter nuclei, releasing a large amount of energy and neutrons.
    • Nuclear Fusion: This is the process where two light atomic nuclei combine (fuse) to form a single, heavier nucleus. This process releases a tremendous amount of energy. It is the same process that powers the sun and other stars.
    • Alpha Decay: This is a type of radioactive decay where an unstable nucleus emits an alpha particle ($^4_2\text{He}$ nucleus) to become more stable.
    • Artificial Transmutation: This is the conversion of one element into another by bombarding it with high-energy particles (like alpha particles or neutrons).

    Step-by-Step Solution

    1. Identify the reactants: We have two light nuclei (Deuterium and Tritium).
    2. Identify the product: They combine to form a heavier nucleus (Helium).
    3. Analyze the process: Since two light nuclei are joining together to form a heavier one, this is the definition of nuclear fusion.
    4. Compare with options: • Option A is incorrect because this is not a decay process. • Option B is incorrect because this is a spontaneous/natural-style reaction often used in thermonuclear weapons or stars, not a laboratory bombardment of a stable element. • Option C matches our analysis. • Option D is incorrect because the nuclei are joining, not splitting.

    Conclusion

    The reaction describes the fusion of two hydrogen isotopes into helium.

    Exam Tip: Remember the mnemonic: Fission = Fission (splitting/division), Fusion = Fuse (joining together). If you see small numbers (like hydrogen isotopes) becoming a larger one, it is always fusion!

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25. 25.

    One of the major uses of alkane is?

    1. A.as domestic and industrial fuel
    2. B.in the hydrogenation of oils
    3. C.in the textile industries
    4. D.in the production of plastics

    Show answer

    Answer: A. as domestic and industrial fuel

    AI explanation

    Analysis

    The question asks for the primary application of alkanes in everyday life and industry. To answer this, we must consider the chemical properties of alkanes, specifically their reactivity and energy content.

    Concept Explanation

    Alkanes are saturated hydrocarbons with the general formula $C_nH_{2n+2}$. They are characterized by strong carbon-carbon (C-C) and carbon-hydrogen (C-H) single bonds.

    1. Combustion: Because alkanes are rich in carbon and hydrogen, they undergo combustion reactions in the presence of oxygen to release a significant amount of heat energy. This makes them excellent fuels.
    2. Reactivity: Alkanes are relatively unreactive (often called paraffins, meaning "little affinity") compared to alkenes or alkynes. They do not easily undergo addition reactions, which rules out many industrial chemical synthesis processes like the hydrogenation of oils (which requires unsaturated hydrocarbons).

    Step-by-Step Solution

    • Option A (Domestic and industrial fuel): Alkanes like methane (natural gas), propane, and butane (LPG) are widely used for cooking and heating. Petrol (gasoline) and diesel, which are mixtures of alkanes, power internal combustion engines. This is their most significant use.
    • Option B (Hydrogenation of oils): This process involves adding hydrogen to unsaturated fats (alkenes) to make them saturated. Alkanes are already saturated, so they are not used for this.
    • Option C (Textile industries): While some petrochemical derivatives are used in textiles (like synthetic fibers), alkanes themselves are not the primary raw material for the textile industry.
    • Option D (Production of plastics): Plastics are typically produced from alkenes (like ethene to make polyethene) through polymerization. Alkanes are generally too stable to undergo polymerization easily.

    Conclusion

    Based on the chemical properties of alkanes, their high heat of combustion makes them the primary choice for energy production in both homes and factories.

    Exam Tip: Whenever you see a question about the uses of alkanes, always associate them with combustion and energy. If the question asks about plastics or hydrogenation, look for alkenes instead!

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26. 26.

    6$AgNO_{4(aq)} + PH_{3(g)} + 3H_2O_{(l)}$ → 6$Ag_{(s)} + H_3PO_{3(g)} + 6HNO_{3(aq)}$ In the above reaction, the reducing agent is?

    1. A.HNO3(aq)
    2. B.H2O(l)
    3. C.PH3(g)
    4. D.AgNO3(aq)

    Show answer

    Answer: C. PH3(g)

    AI explanation

    Analysis of the Question

    To identify the reducing agent in a redox reaction, we need to track the changes in the oxidation states of the elements involved. A reducing agent is the substance that gets oxidized (loses electrons or increases in oxidation number) during the reaction.

    Concept Explanation

    • Oxidation: An increase in oxidation number (loss of electrons).
    • Reduction: A decrease in oxidation number (gain of electrons).
    • Reducing Agent: The species that causes reduction by being oxidized itself.
    • Oxidizing Agent: The species that causes oxidation by being reduced itself.

    Step-by-Step Solution

    1. Assign oxidation numbers to the reactants and products: • In $AgNO_3$: $Ag$ is $+1$, $N$ is $+5$, $O$ is $-2$. • In $PH_3$: $P$ is $-3$, $H$ is $+1$. • In $Ag_{(s)}$: The oxidation state of an element in its standard state is $0$. • In $H_3PO_3$: $H$ is $+1$, $O$ is $-2$. Let $P = x$. $3(1) + x + 3(-2) = 0 \implies 3 + x - 6 = 0 \implies x = +3$.

    2. Track the changes: • Silver ($Ag$): Goes from $+1$ in $AgNO_3$ to $0$ in $Ag_{(s)}$. Since the oxidation number decreased, $AgNO_3$ is the oxidizing agent. • Phosphorus ($P$): Goes from $-3$ in $PH_3$ to $+3$ in $H_3PO_3$. Since the oxidation number increased (from $-3$ to $+3$), $PH_3$ has lost electrons.

    3. Conclusion: Because $PH_3$ increased in oxidation state, it is the substance being oxidized. Therefore, $PH_3$ is the reducing agent.

    Identifying the Answer

    Comparing our findings with the given options: A. $HNO_3$ (Product) B. $H_2O$ (Reactant, but not the reducing agent) C. $PH_3$ (The species being oxidized) D. $AgNO_3$ (The oxidizing agent)

    The correct option is C.

    Exam Tip

    Always remember the mnemonic OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons). If you see an element's oxidation number increase from the reactant side to the product side, that substance is your reducing agent!

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27. 27.

    Benzene reacts with hydrogen in the presence of nickel catalyst at 180ºC to give?

    1. A.xylene
    2. B.toluene
    3. C.cyclopentane
    4. D.cyclohexane

    Show answer

    Answer: D. cyclohexane

    AI explanation

    1. Analyze the question

    The question asks for the product formed when benzene ($C_6H_6$) undergoes a hydrogenation reaction (reaction with hydrogen) in the presence of a nickel catalyst at a specific temperature (180ºC).

    2. Explain the concept

    This is a classic addition reaction in organic chemistry. Benzene is an unsaturated aromatic hydrocarbon. Because it contains three double bonds within its ring structure, it can undergo addition reactions under specific conditions.

    When benzene is reacted with hydrogen gas ($H_2$) in the presence of a finely divided nickel catalyst at an elevated temperature (typically 150ºC – 200ºC), the hydrogen atoms add across the double bonds, breaking them and saturating the ring. This process converts the aromatic benzene into a saturated cyclic alkane.

    3. Show the solution

    The chemical equation for this reaction is:

    $C_6H_6 + 3H_2 \xrightarrow{Ni, 180ºC} C_6H_{12}$

    • Reactants: Benzene ($C_6H_6$) + 3 moles of Hydrogen ($3H_2$).
    • Conditions: Nickel catalyst ($Ni$) at 180ºC.
    • Product: The hydrogen atoms saturate the benzene ring, resulting in cyclohexane ($C_6H_{12}$), which is a cyclic alkane.

    4. Identify the answer

    Comparing our result ($C_6H_{12}$) with the given options:

    • A. xylene (a dimethylbenzene)
    • B. toluene (methylbenzene)
    • C. cyclopentane (a 5-carbon ring)
    • D. cyclohexane (a 6-carbon saturated ring)

    The correct option is D.

    5. Tip

    Remember that benzene is an aromatic ring, but when you "saturate" it completely with hydrogen, it loses its aromaticity and becomes a simple cycloalkane. A quick way to remember this is that the prefix "cyclo-" indicates a ring, and "-hexane" indicates six carbons, matching the six carbons in the original benzene ring.

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28. 28.

    Which of the following is an example of a mixture?

    1. A.Common salt
    2. B.Blood
    3. C.Sand
    4. D.Washing soda

    Show answer

    Answer: B. Blood

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29. 29.

    The colour of methyl orange in alkaline medium is?

    1. A.yellow
    2. B.pink
    3. C.orange
    4. D.red

    Show answer

    Answer: A. yellow

    AI explanation

    Analysis of the Question

    This question tests your knowledge of acid-base indicators. Indicators are substances that change color depending on the pH of the solution they are in. You are required to identify the specific color change of methyl orange when placed in an alkaline (basic) environment.

    Concept Explanation

    Methyl orange is a common pH indicator used in titrations. Its behavior is defined by its pH range:

    • Acidic medium (pH < 3.1): It appears red.
    • Transition range (pH 3.1 – 4.4): It appears orange.
    • Alkaline/Basic medium (pH > 4.4): It appears yellow.

    Because alkaline solutions have a pH greater than 7, methyl orange will consistently show its basic-range color in these environments.

    Step-by-Step Solution

    1. Identify the medium: The question specifies an "alkaline medium."
    2. Recall the indicator properties: Methyl orange is known to be red in acidic conditions and yellow in basic conditions.
    3. Evaluate the options: • Red is for acidic conditions. • Orange is the transition color (neutral/slightly acidic). • Yellow is for alkaline conditions.
    4. Conclusion: Since the medium is alkaline, the color must be yellow.

    Exam Tip

    A great way to remember this is the mnemonic "MY": Methyl orange is Yellow in alkaline solutions. Also, remember that methyl orange is often used for strong acid-weak base titrations, where the endpoint color change is from yellow to red as the acid is added.

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30. 30.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.

31. 31.

    If a reaction is exothermic and there is a great disorder, it means that?

    1. A.the reaction is static
    2. B.the reaction is in a state of equilibrium
    3. C.there will be a large increase in free energy
    4. D.there will be a large decrease in free energy

    Show answer

    Answer: D. there will be a large decrease in free energy

    AI explanation

    Analysis of the Question

    This question tests your understanding of the Gibbs Free Energy equation, which determines the spontaneity of a chemical reaction. You are asked to determine what happens to the free energy when a reaction is exothermic and leads to a great increase in disorder.

    Concept Explanation

    To solve this, we use the Gibbs Free Energy equation:

    $\Delta G = \Delta H - T\Delta S$

    Where:

    • $\Delta G$ is the change in Gibbs Free Energy.
    • $\Delta H$ is the change in Enthalpy (heat content).
    • $T$ is the absolute temperature (in Kelvin).
    • $\Delta S$ is the change in Entropy (disorder).

    Key Principles:

    1. Exothermic reaction: The system releases heat, so $\Delta H$ is negative ($-$).
    2. Great disorder: The system becomes more disordered, so $\Delta S$ is positive ($+$).
    3. Spontaneity: For a reaction to be spontaneous, $\Delta G$ must be negative. A large decrease in free energy indicates a highly spontaneous reaction.

    Step-by-Step Solution

    1. Substitute the signs into the equation: • $\Delta H$ is negative ($-$). • $\Delta S$ is positive ($+$).
    2. The equation becomes: $$\Delta G = (\text{negative}) - T(\text{positive})$$
    3. Since $T$ is always positive, $-T(\text{positive})$ results in a negative value.
    4. Therefore, $\Delta G = (\text{negative}) - (\text{positive}) = \text{a larger negative value}$.
    5. A large negative value for $\Delta G$ represents a large decrease in free energy. This confirms that the reaction is highly spontaneous.

    Conclusion

    When a reaction is exothermic ($\Delta H < 0$) and results in increased disorder ($\Delta S > 0$), the Gibbs Free Energy ($\Delta G$) becomes significantly negative. This is described as a large decrease in free energy.

    Exam Tip

    Remember the "Spontaneity Table":

    • If $\Delta H$ is negative and $\Delta S$ is positive, the reaction is spontaneous at all temperatures.
    • Think of "Free Energy" like a bank account: if a reaction is "spontaneous," it is "spending" its energy to happen, hence a "decrease" in free energy!

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32. 32.

    Which of the following is used in rocket fuels?

    1. A.HNO3
    2. B.CH3COOH
    3. C.H2SO4
    4. D.HCl

    Show answer

    Answer: A. HNO3

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33. 33.

    The IUPAC nomenclature of the compound LiAlH4 is?

    1. A.lithiumtetrahydridoaluminate (III)
    2. B.aluminium tetrahydrido lithium
    3. C.tetrahydrido lithium aluminate (III)
    4. D.lithium aluminium hydride

    Show answer

    Answer: A. lithiumtetrahydridoaluminate (III)

    AI explanation

    Analysis

    The question asks for the IUPAC (International Union of Pure and Applied Chemistry) name for the compound $\text{LiAlH}_4$. This compound is a well-known reducing agent in organic chemistry.

    Concept: IUPAC Nomenclature for Complex Hydrides

    To name complex inorganic compounds like $\text{LiAlH}_4$, we follow these rules:

    1. Identify the Cation: The metal outside the complex bracket (or the first part of the formula) is named first. Here, it is Lithium.
    2. Identify the Anion (Complex Ion): The complex part is $[\text{AlH}_4]^-$. • The ligand (the group attached to the central metal) is the hydride ion ($\text{H}^-$). In IUPAC nomenclature, when hydrogen acts as a ligand, it is called hydrido. Since there are four of them, we use the prefix tetra-. • The central metal is Aluminium. Because it is part of an anionic complex (the whole bracket has a negative charge), we add the suffix -ate to the metal name, making it aluminate.
    3. Determine Oxidation State: We calculate the oxidation state of the central metal (Aluminium). • $\text{Li} = +1$ • $\text{H} = -1$ (since it is a hydride) • Let $x$ be the oxidation state of $\text{Al}$: $(+1) + x + 4(-1) = 0$ $1 + x - 4 = 0$ $x - 3 = 0 \implies x = +3$ • The oxidation state is written in Roman numerals in parentheses: (III).

    Step-by-Step Solution

    1. Name the cation: Lithium.
    2. Name the ligands: Four hydrides = tetrahydrido.
    3. Name the central metal: Aluminium becomes aluminate.
    4. Combine: Put the parts together in the order: [Cation] + [Ligand prefix] + [Ligand name] + [Metal name] + [Oxidation state].
    5. Result: Lithium + tetrahydrido + aluminate + (III) = Lithiumtetrahydridoaluminate (III).

    Conclusion

    Comparing our result with the given options, we find that option A matches our systematic derivation. Note that while "Lithium aluminium hydride" is the common name used in laboratories, the formal IUPAC name is "lithiumtetrahydridoaluminate (III)".

    Exam Tip

    When naming complex ions, always remember that if the complex ion is an anion (negative), the central metal must end in -ate (e.g., aluminate, ferrate, cuprate). If it were a cation, the metal name would remain unchanged.

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34. 34.

    The correct order of increasing boiling points of the following compounds C3H7OH, C7H16 and C4H10 is?

    1. A.C3H7OH → C4H10 → C7H10
    2. B.C4H10 → C7H16 → C3H7OH
    3. C.C7H16 → C3H7OH → C4H10
    4. D.C4H10 → C3H7OH → C7H16

    Show answer

    Answer: D. C4H10 → C3H7OH → C7H16

    AI explanation

    Analysis of the Question

    To determine the order of increasing boiling points, we must look at the intermolecular forces present in each compound. The compounds are:

    1. $C_4H_{10}$ (Butane): An alkane.
    2. $C_3H_7OH$ (Propanol): An alcohol.
    3. $C_7H_{16}$ (Heptane): An alkane.

    Concept Explanation

    Boiling point is determined by the strength of the intermolecular forces holding the molecules together:

    • Van der Waals forces (London Dispersion Forces): These are the primary forces in alkanes. The strength of these forces increases as the molecular mass (or chain length) increases. Therefore, $C_7H_{16}$ (a larger molecule) has stronger Van der Waals forces than $C_4H_{10}$ (a smaller molecule).
    • Hydrogen Bonding: Alcohols ($R-OH$) possess a hydroxyl group, which allows for hydrogen bonding. Hydrogen bonds are significantly stronger than the Van der Waals forces found in simple alkanes of similar size.

    Comparing the compounds:

    • $C_4H_{10}$: Smallest alkane, weakest forces, lowest boiling point.
    • $C_3H_7OH$: Contains hydrogen bonding. Even though it is a smaller molecule than $C_7H_{16}$, the hydrogen bonding gives it a higher boiling point than the small alkane ($C_4H_{10}$), but it is generally lower than a long-chain alkane like $C_7H_{16}$ due to the significant increase in Van der Waals forces in the much larger $C_7$ chain.
    • $C_7H_{16}$: Largest molecule with the strongest Van der Waals forces, resulting in the highest boiling point.

    Step-by-Step Solution

    1. Identify the lowest: $C_4H_{10}$ is a small alkane with weak dispersion forces. It has the lowest boiling point.
    2. Identify the middle: $C_3H_7OH$ has hydrogen bonding, which makes its boiling point higher than the small alkane $C_4H_{10}$.
    3. Identify the highest: $C_7H_{16}$ is a much larger hydrocarbon. The cumulative effect of the Van der Waals forces in a 7-carbon chain is strong enough to exceed the boiling point of the 3-carbon alcohol.
    4. Order: $C_4H_{10} < C_3H_7OH < C_7H_{16}$.

    Conclusion

    Exam Tip

    When comparing boiling points of organic compounds, always check for Hydrogen Bonding first (Alcohols, Carboxylic acids). If no hydrogen bonding is present, simply look at the Molecular Mass—the heavier the molecule, the higher the boiling point!

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35. 35.

    The IUPAC nomenclature of the compound above is?

    1. A.4 - ethyloctane
    2. B.5 - ethyloctane
    3. C.5 - propylheptane
    4. D.3 - propylheptane

    Show answer

    Answer: A. 4 - ethyloctane

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36. 36.

    The acid that is used to remove rust is?

    1. A.boric
    2. B.hydrochloric
    3. C.trioxonitrate (V)
    4. D.tetraoxosulphate (VI)

    Show answer

    Answer: B. hydrochloric

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37. 37.

    Permanent hardness of water can be removed by?

    1. A.filtration
    2. B.adding slaked lime
    3. C.adding caustic soda
    4. D.boiling

    Show answer

    Answer: C. adding caustic soda

    AI explanation

    Analysis of the Question

    The question asks for a method to remove permanent hardness from water. It is important to distinguish between temporary hardness (which can be removed by boiling) and permanent hardness (which requires chemical treatment).

    Explanation of the Concept

    • Temporary Hardness: Caused by the presence of calcium hydrogen trioxocarbonate(IV) or magnesium hydrogen trioxocarbonate(IV). This is easily removed by boiling, which precipitates the minerals as carbonates.
    • Permanent Hardness: Caused by the presence of soluble calcium or magnesium chlorides or tetraoxosulphate(VI) salts. Because these salts do not decompose upon boiling, we must use chemical methods to precipitate the calcium or magnesium ions out of the solution.
    • Methods for removing permanent hardness: • Adding washing soda (sodium trioxocarbonate(IV)). • Using ion-exchange resins (Permutit process). • Adding caustic soda (sodium hydroxide).

    Step-by-Step Solution

    1. Evaluate Option A (Filtration): Filtration only removes suspended solid particles; it cannot remove dissolved mineral salts that cause hardness.
    2. Evaluate Option B (Adding slaked lime): Slaked lime ($Ca(OH)_2$) is used in the Clark’s process to remove temporary hardness, not permanent hardness.
    3. Evaluate Option C (Adding caustic soda): Caustic soda ($NaOH$) reacts with the calcium or magnesium ions in the hard water to form insoluble precipitates (like $Ca(OH)_2$ or $Mg(OH)_2$), effectively removing the hardness. This is a standard chemical method for treating permanent hardness.
    4. Evaluate Option D (Boiling): As established, boiling only removes temporary hardness. It has no effect on the chlorides or tetraoxosulphate(VI) salts responsible for permanent hardness.

    Conclusion

    Adding caustic soda is an effective chemical method for removing permanent hardness by precipitating the ions responsible for the hardness.

    Exam Tip: Always remember the "Boiling Rule": If the question asks about temporary hardness, think "Boiling." If it asks about permanent hardness, think "Chemicals" (like washing soda or caustic soda) or "Ion Exchange."

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38. 38.

    Elements in the same period in the periodic table have the same?

    1. A.number of shells
    2. B.atomic number
    3. C.chemical properties
    4. D.physical properties

    Show answer

    Answer: A. number of shells

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39. 39.

    A noble gas with a high power of fog penetration used in aerodrome beacons is?

    1. A.krypton
    2. B.argon
    3. C.helium
    4. D.neon

    Show answer

    Answer: D. neon

    AI explanation

    Analysis

    The question asks for a specific noble gas known for its ability to penetrate fog, making it ideal for use in airport (aerodrome) beacons and warning lights.

    Concept Explanation

    Noble gases (Group 18 elements) are known for being chemically inert. However, they have distinct applications based on their physical properties, particularly how they behave when an electric current is passed through them in a discharge tube.

    • Neon (Ne): When electricity passes through neon gas, it emits a bright, intense reddish-orange light. This light has a long wavelength, which allows it to scatter less and penetrate through fog, mist, and rain much better than white light. This property makes it the standard choice for high-visibility signage and airport landing beacons.
    • Argon (Ar): Often used in light bulbs to prevent the filament from burning out, but it does not have the same high-visibility penetration power as neon.
    • Krypton (Kr): Used in high-speed photography flashes and some specialized lamps, but not primarily for fog penetration in beacons.
    • Helium (He): Used in balloons and airships due to its low density, but it is not used for lighting beacons.

    Step-by-Step Solution

    1. Identify the requirement: We need a gas that produces a light capable of cutting through poor visibility conditions like fog.
    2. Evaluate the options: • Argon, Krypton, and Helium do not possess the specific high-intensity, long-wavelength emission characteristic required for fog-penetrating beacons. • Neon produces a characteristic red-orange glow that is highly visible even in adverse weather conditions.
    3. Conclusion: Because of its high visibility and ability to penetrate fog, neon is the gas used in aerodrome beacons.

    Exam Tip

    When you see questions about noble gases in JAMB, remember these common uses:

    • Neon: Advertising signs and airport beacons (high visibility).
    • Argon: Electric light bulbs (inert atmosphere).
    • Helium: Weather balloons and airships (low density).
    • Xenon: Strobe lights and high-intensity car headlights.

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40. 40.

    Calculate the volume of 0.5 mol dm$^{-3}$ H$_2$SO$_4$ that is neutralized by 25 cm$^3$ of 0.1 mol dm$^3$ NaOH?

    1. A.5.0 cm3
    2. B.2.5 cm3
    3. C.0.4 cm3
    4. D.0.1 cm3

    Show answer

    Answer: B. 2.5 cm3

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41. 41.

    Carbon (II) oxide is considered dangerous if inhaled mainly because it?

    1. A.can cause injury to the nervous system
    2. B.competes with oxygen in the blood
    3. C.competes with carbon (IV) oxide in the blood
    4. D.can cause lung cancer

    Show answer

    Answer: B. competes with oxygen in the blood

    AI explanation

    Analysis

    The question asks for the primary reason why Carbon (II) oxide (CO), also known as carbon monoxide, is toxic to humans when inhaled.

    Concept Explanation

    Carbon (II) oxide is a colorless, odorless gas produced by the incomplete combustion of carbon-containing fuels. Its toxicity is rooted in how it interacts with hemoglobin, the protein in red blood cells responsible for transporting oxygen throughout the body.

    1. Hemoglobin's Role: Normally, hemoglobin binds with oxygen ($O_2$) to form oxyhemoglobin, which carries oxygen to body tissues.
    2. The Affinity Factor: Hemoglobin has a significantly higher affinity (about 200–300 times greater) for Carbon (II) oxide than it does for oxygen.
    3. The Competition: When CO is inhaled, it binds to hemoglobin to form carboxyhemoglobin. Because CO binds so strongly, it effectively "kicks out" or prevents oxygen from binding to the hemoglobin.
    4. The Result: This deprives the body's tissues and vital organs (like the brain and heart) of the oxygen they need to function, leading to suffocation at the cellular level, which can be fatal.

    Step-by-Step Solution

    • Step 1: Evaluate Option A. While CO can cause brain damage due to lack of oxygen, it is not a direct neurotoxin that attacks the nervous system like a nerve agent would.
    • Step 2: Evaluate Option B. As explained above, CO binds to hemoglobin in place of oxygen. This is a direct competition for the same binding site on the hemoglobin molecule. This is the primary mechanism of CO poisoning.
    • Step 3: Evaluate Option C. Carbon (IV) oxide ($CO_2$) is a waste product of respiration. While high levels are dangerous, CO does not compete with $CO_2$ in the blood; it competes with $O_2$.
    • Step 4: Evaluate Option D. Carbon (II) oxide is not classified as a carcinogen (cancer-causing agent).

    Conclusion

    The danger of Carbon (II) oxide lies in its ability to bind to hemoglobin more effectively than oxygen, thereby preventing the transport of oxygen to the body's tissues.

    Exam Tip

    When you see questions about Carbon (II) oxide, always remember the keyword "Affinity." Because it has a higher affinity for hemoglobin than oxygen, it "steals" the spot meant for oxygen, leading to chemical asphyxiation.

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42. 42.

    Two hydrocarbons X and Y were treated with bromine water. X decolorised the solution and Y did not not. Which class of compound does Y belong?

    1. A.Benzene
    2. B.Alkynes
    3. C.Alkenes
    4. D.Alkanes

    Show answer

    Answer: D. Alkanes

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43. 43.

    When iron is exposed to moist air, it gradually rusts. This is due to the formation of?

    1. A.hydrate iron (III) oxide
    2. B.anhydrous iron (III) oxide
    3. C.anhydrous iron (II) oxide
    4. D.hydrate iron (II) oxide

    Show answer

    Answer: A. hydrate iron (III) oxide

    AI explanation

    Analysis

    The question asks for the chemical identity of the substance formed when iron undergoes the process of rusting in the presence of moisture and oxygen.

    Concept Explanation

    Rusting is a common electrochemical process known as corrosion. For iron to rust, it requires two essential components: oxygen and water (moisture).

    1. Chemical Reaction: When iron ($Fe$) reacts with oxygen ($O_2$) in the presence of water ($H_2O$), it forms a reddish-brown flaky substance.
    2. Oxidation State: During this process, iron is oxidized from its metallic state ($Fe^0$) to the Iron(III) state ($Fe^{3+}$).
    3. Hydration: The resulting compound is not just a simple oxide; it incorporates water molecules into its crystal lattice. This is why it is referred to as a "hydrated" oxide.
    4. Formula: The chemical formula for rust is $Fe_2O_3 \cdot xH_2O$. This is chemically named hydrated iron(III) oxide.

    Step-by-Step Solution

    • Step 1: Identify the oxidation state of iron in rust. Rust is an Iron(III) compound, which eliminates options C and D (which refer to Iron(II)).
    • Step 2: Determine if the compound is anhydrous or hydrated. Since the process requires "moist air," water is chemically bonded to the oxide. This eliminates option B (anhydrous means without water).
    • Step 3: Combine these facts. The substance is an Iron(III) oxide that contains water, making it hydrated iron(III) oxide.

    Conclusion

    The formation of rust is the chemical reaction of iron with oxygen and water to produce hydrated iron(III) oxide.

    Exam Tip

    Always remember that "anhydrous" means "without water" (think of a dry desert), while "hydrated" means "with water." Since rust only forms in the presence of moisture, it must be a hydrated compound!

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44. 44.

    The haloalkanes used in dry-cleaning industries are?

    1. A.trichloromethane and tetrachloromethane
    2. B.chloroethene and dichloroethene
    3. C.trichloroethene and tetrachloroethene
    4. D.chloroethane and dichloroethene

    Show answer

    Answer: C. trichloroethene and tetrachloroethene

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45. 45.

    A compound gives an orange-red color to non-luminous flame. This compound is likely to contain?

    1. A.Na+
    2. B.Ca2+
    3. C.Fe3+
    4. D.Fe2+

    Show answer

    Answer: B. Ca2+

    AI explanation

    1. Analyze the question

    The question asks you to identify a metal cation based on the characteristic color it imparts to a non-luminous Bunsen flame. This is a classic "Flame Test" question frequently tested in JAMB Chemistry.

    2. Explain the concept

    The Flame Test is an analytical procedure used in chemistry to detect the presence of certain metal ions based on the emission spectrum of the element. When a metal salt is heated in a non-luminous flame, the electrons in the metal atoms absorb energy and become "excited." As they return to their ground state, they release this energy as light of a specific wavelength (color).

    Each metal ion produces a unique flame color:

    • Sodium (Na⁺): Intense golden-yellow.
    • Calcium (Ca²⁺): Brick-red or orange-red.
    • Potassium (K⁺): Lilac (pale violet).
    • Copper (Cu²⁺): Blue-green.
    • Barium (Ba²⁺): Apple-green.

    3. Show the solution

    • Option A (Na⁺): Sodium ions produce a characteristic golden-yellow flame. This does not match the description.
    • Option B (Ca²⁺): Calcium ions are well-known in qualitative analysis for producing a brick-red or orange-red color in a non-luminous flame. This matches the description perfectly.
    • Option C (Fe³⁺) and D (Fe²⁺): Iron ions do not produce a distinct flame color that is typically used for identification in the standard flame test; they are usually identified by their reactions with reagents like NaOH or K₄[Fe(CN)₆].

    Comparing the options to the known flame test results, the orange-red color is the signature of Calcium (Ca²⁺).

    4. Identify the answer

    The correct option is B.

    5. Tip for your exam

    To remember these easily, use this simple mnemonic:

    • Na (Sodium) = Na-yellow (Golden-yellow).
    • Ca (Calcium) = Ca-brick (Brick-red/Orange-red).
    • K (Potassium) = K-lilac (Lilac).

    Always remember that the flame must be non-luminous (the blue flame) to get an accurate reading, as a luminous (yellow) flame would mask the color of the metal!

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46. 46.

    Iron can be protected from corrosion by coating the surface with?

    1. A.gold
    2. B.silver
    3. C.copper
    4. D.zinc

    Show answer

    Answer: D. zinc

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47. 47.

    To a solution of an unknown compound, a little dilute tetraoxosulphate (VI) acid was added with some freshly prepared iron (II) tetraoxosulphate (VI) solution. The brown ring observed after the addition of a stream of concentrated tetraoxosulphate (VI) acid confirmed the presence of?

    1. A.CO$^{2-}_{3}$
    2. B.Cl-
    3. C.SO$_{3}^{2-}$
    4. D.NO$^{2-}_{3}$

    Show answer

    Answer: D. NO$^{2-}_{3}$

    AI explanation

    Analysis of the Question

    This question tests your knowledge of qualitative analysis in chemistry, specifically the confirmatory test for a particular anion. You are asked to identify the anion that produces a "brown ring" when reacted with iron(II) tetraoxosulphate(VI) ($FeSO_4$) and concentrated tetraoxosulphate(VI) acid ($H_2SO_4$).

    Concept Explanation: The Brown Ring Test

    The Brown Ring Test is a classic laboratory procedure used to detect the presence of the nitrate ion ($NO_3^-$).

    1. The Reagents: The test requires the unknown solution, freshly prepared iron(II) sulphate ($FeSO_4$), and concentrated tetraoxosulphate(VI) acid ($H_2SO_4$).
    2. The Process: When $FeSO_4$ is added to the solution containing nitrate ions, and concentrated $H_2SO_4$ is carefully poured down the side of the test tube, a brown ring forms at the junction of the two layers.
    3. The Chemistry: The concentrated $H_2SO_4$ reduces the nitrate ion to nitrogen(II) oxide ($NO$). The $NO$ then reacts with the excess $Fe^{2+}$ ions to form a brown-colored complex compound known as nitrosonium iron(II) tetraoxosulphate(VI), which has the formula $[Fe(H_2O)_5(NO)]SO_4$.

    Step-by-Step Solution

    • Step 1: Identify the reagents provided in the question. The use of $FeSO_4$ and concentrated $H_2SO_4$ is the standard procedure for the Brown Ring Test.
    • Step 2: Evaluate the options provided: • A (Carbonate $CO_3^{2-}$): Tested by adding dilute acids, which causes effervescence (bubbles of $CO_2$). • B (Chloride $Cl^-$): Tested by adding silver trioxonitrate(V) ($AgNO_3$) solution, which forms a white precipitate. • C (Sulphite $SO_3^{2-}$): Tested by adding dilute acids, which releases $SO_2$ gas that turns acidified potassium tetraoxomanganate(VII) from purple to colorless. • D (Nitrate $NO_3^-$): This is the specific ion that undergoes the Brown Ring Test. (Note: The option in your question is written as $NO_3^-$, though it appears as $NO_3^2-$ in your text, it refers to the nitrate ion).

    Conclusion

    The formation of the brown ring is the definitive confirmatory test for the nitrate ion ($NO_3^-$). Therefore, the presence of the brown ring confirms the presence of $NO_3^-$.

    Exam Tip: Always remember that the Brown Ring Test must be performed with freshly prepared iron(II) sulphate. If the $FeSO_4$ is old, it will have oxidized to iron(III) ($Fe^{3+}$) and will not be able to form the brown complex, leading to a false negative result.

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48. 48.

    Which of the following organic compounds is very soluble in water?

    1. A.CH3COOH
    2. B.C2H2
    3. C.C2H4
    4. D.CH3COOC2H5

    Show answer

    Answer: A. CH3COOH

    AI explanation

    Analysis of the Question

    The question asks us to identify which of the given organic compounds is highly soluble in water. Solubility in water for organic compounds is generally determined by the ability of the molecule to form hydrogen bonds with water molecules.

    Concept Explanation

    1. Hydrogen Bonding: For an organic compound to be very soluble in water, it must be polar and capable of forming hydrogen bonds with water. Small carboxylic acids, alcohols, and amines are typically soluble because their polar functional groups (like -OH or -COOH) interact strongly with water.
    2. Hydrocarbons: Compounds like alkanes, alkenes (e.g., $C_2H_4$), and alkynes (e.g., $C_2H_2$) are non-polar. Because they cannot form hydrogen bonds with water, they are generally insoluble in water.
    3. Esters: While esters (like $CH_3COOC_2H_5$) have some polarity, they cannot form hydrogen bonds with themselves and have limited solubility in water compared to carboxylic acids.
    4. Carboxylic Acids: Ethanoic acid ($CH_3COOH$) has a polar carboxyl group (-COOH). The small size of the molecule allows the polar end to interact effectively with water, making it miscible (very soluble) in water.

    Step-by-Step Solution

    • Option A ($CH_3COOH$): This is Ethanoic acid. It is a polar molecule that forms strong hydrogen bonds with water. It is completely miscible in water.
    • Option B ($C_2H_2$): This is Ethyne. It is a non-polar hydrocarbon and is only sparingly soluble in water.
    • Option C ($C_2H_4$): This is Ethene. Like ethyne, it is a non-polar hydrocarbon and is insoluble in water.
    • Option D ($CH_3COOC_2H_5$): This is Ethyl ethanoate (an ester). While it has some solubility, it is significantly less soluble than the corresponding carboxylic acid because it lacks the hydroxyl (-OH) group necessary for strong hydrogen bonding.

    Comparing these, $CH_3COOH$ is the most soluble due to its ability to form extensive hydrogen bonds.

    Conclusion

    The correct option is A.

    Exam Tip

    When you see questions about solubility in organic chemistry, look for the functional group. If you see -OH (alcohols) or -COOH (carboxylic acids), those molecules are usually water-soluble, especially if the carbon chain is short (1–4 carbons). Hydrocarbons (compounds with only Carbon and Hydrogen) are almost always insoluble in water!

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49. 49.

    Which of the following is used to hasten the ripening of fruit?

    1. A.Ethene
    2. B.Ethanol
    3. C.Ethyne
    4. D.Ethane

    Show answer

    Answer: A. Ethene

    AI explanation

    Analysis

    The question asks for the specific chemical compound used in agriculture and food science to accelerate the ripening process of fruits.

    Concept Explanation

    The process of fruit ripening is a natural physiological change, but it can be artificially induced or accelerated using plant hormones or specific hydrocarbons.

    • Ethene (Ethylene): This is a naturally occurring plant hormone that regulates growth and development. In its gaseous form ($C_2H_4$), it is widely used commercially to trigger the ripening process in climacteric fruits like bananas, tomatoes, and mangoes.
    • Ethane ($C_2H_6$): This is a saturated hydrocarbon (alkane) and has no effect on fruit ripening.
    • Ethyne (Acetylene, $C_2H_2$): While sometimes used illegally or in crude settings to ripen fruit, it is not the standard biological or industrial agent used for this purpose; it is often considered a contaminant.
    • Ethanol ($C_2H_5OH$): This is an alcohol and is not used to ripen fruit.

    Step-by-Step Solution

    1. Identify the chemical nature of ripening agents: Ripening is a biological process triggered by the hormone ethylene.
    2. Match the hormone to the chemical name: The IUPAC name for the gas ethylene is ethene.
    3. Evaluate the options: • A. Ethene: Correct. It is the alkene used to stimulate the ripening process. • B. Ethanol: Incorrect; it is an alcohol. • C. Ethyne: Incorrect; although it can force ripening, it is not the standard biological agent and can be toxic. • D. Ethane: Incorrect; it is an unreactive alkane.

    Conclusion

    Ethene is the correct chemical compound used to hasten the ripening of fruits.

    Exam Tip: Remember the "ene" suffix for plant hormones. If you see "Ethene" (or Ethylene) in a question about plant growth, ripening, or fruit development, that is almost certainly your answer!

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