If there is no change in volume in a gaseous reaction, the pressure will

C. have no effect on equilibrium position

D. have effect on equilibrium position

Which of the following statements is false about hard water?

B. Cannot be supplied in pipes made of leads

A. Helps animals to build strong teeth

B. Cannot be supplied in pipes made of leads

C. Tastes better than soft water

D. Helps animals to build strong bones.

What is the function of concentrated H$_2$SO$_4$?

A. It reacts with metals to give a salt and hydro gas

D. It produces a cryolite precipitate with basic solution

When ΔH is positive and small, and ΔS is positive and large, the reaction will be

B. favour forward reaction only

C. favour backward reaction only

What accounts for the low melting and boiling points of covalent molecules?

A. They have weak intermolecular forces

C. They are three dimensional structures

D. They possess shared electron pairs

NH$_3$ $(_g$) + HCl$(_g$) → NH$_4$Cl $_(g)$ In the reaction above, increase in pressure will

A. lower the equilibrium constant

D. increase the equilibrium constant

A molecule with a polar covalent bond between atoms exhibits

A. unequal sharing of electron pair

B. equal sharing of electron pair

C. equal sharing of electron lone pair

D. unequal sharing of electron lone pair

In the laboratory preparation of oxygen using potassium trioxochlorate(V), tetraoxomanganate (VII), usually not recommended in modern laboratory because

A. it forms explosive mixture with carbonaceous materials

JAMB 2025 · Part 1 of 2

2025 JAMB Chemistry Past Questions and Answers

The 2025 JAMB Chemistry past questions with answers: 107 objective questions from the 2025 paper in the DeeepSolve question bank, each with its answer behind "Show answer". This is part 1 of 2: questions 1 to 54. Try each question before you check, then practise the paper as a timed JAMB-style CBT.

Practise this paper as a timed JAMB-style CBT All JAMB Chemistry years

Questions 1–54

1.
In welding and cutting of metals, the organic gas commonly used in the heating process is
1. A.methane
2. B.propene
3. C.butane
4. D.ethyne
Show answer
Answer: D. ethyne
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2.
The most suitable indicator to be used when reacting ethanedioic acid with potassium hydroxide is
1. A.phenolphthalein
2. B.methyl orange
3. C.methyl red
4. D.methyl purple
Show answer
Answer: A. phenolphthalein
AI explanation
1. Analyze the question
The question asks for the most suitable indicator for a titration between ethanedioic acid (also known as oxalic acid, $H_2C_2O_4$ ) and potassium hydroxide ( $KOH$ ).
2. Explain the concept
To choose the right indicator, we must identify the type of acid-base reaction:
Ethanedioic acid ( $H_2C_2O_4$ ): This is a weak acid.
Potassium hydroxide ( $KOH$ ): This is a strong base.
When a weak acid reacts with a strong base, the resulting salt undergoes hydrolysis, making the solution at the equivalence point alkaline (basic), typically with a pH greater than 7 (usually around 8–9).
Indicators change color over a specific pH range. We need an indicator that changes color within the pH range of the equivalence point:
Phenolphthalein: Changes color in the pH range of 8.2 to 10.0. This perfectly matches the alkaline equivalence point of a weak acid-strong base titration.
Methyl orange/Methyl red: These change color in the acidic range (pH 3.1–4.4 and 4.4–6.2, respectively), making them unsuitable for this specific reaction because they would change color before the equivalence point is reached.
3. Show the solution
Step 1: Identify the reactants. Ethanedioic acid is a weak organic acid, and potassium hydroxide is a strong inorganic base.
Step 2: Determine the nature of the salt formed. The reaction produces potassium ethanedioate and water. Because the base is stronger than the acid, the salt solution is basic.
Step 3: Match the indicator to the pH. Since the equivalence point is in the basic region (pH > 7), we require an indicator that functions in the basic range.
Step 4: Evaluate the options. Phenolphthalein is the only indicator listed that operates effectively in the basic pH range (8.2–10.0).
4. Identify the answer
5. Exam Tip
A simple rule of thumb for choosing indicators:
Strong Acid + Strong Base: Any indicator (Phenolphthalein or Methyl Orange).
Strong Acid + Weak Base: Methyl Orange.
Weak Acid + Strong Base: Phenolphthalein.
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3.
If there is no change in volume in a gaseous reaction, the pressure will
1. A.favour forward reaction
2. B.favour backward reaction
3. C.have no effect on equilibrium position
4. D.have effect on equilibrium position
Show answer
Answer: C. have no effect on equilibrium position
AI explanation
Analysis
This question tests your understanding of Le Chatelier’s Principle as it applies to gaseous equilibrium reactions, specifically focusing on the effect of pressure changes.
Concept Explanation
Le Chatelier’s Principle states that if a dynamic equilibrium is disturbed by changing the conditions (like pressure, temperature, or concentration), the position of equilibrium shifts to counteract the change.
When dealing with pressure in gaseous reactions:
Pressure changes only affect the equilibrium position if there is a change in the number of moles of gas between the reactants and the products.
If the total number of moles of gaseous reactants equals the total number of moles of gaseous products (i.e., $\Delta n = 0$ ), the volume remains constant during the reaction.
In such cases, increasing or decreasing the pressure will affect both sides of the equation equally, meaning there is no net shift in the equilibrium position.
Step-by-Step Solution
Identify the condition: The question specifies "no change in volume" in the reaction. This mathematically means the sum of the stoichiometric coefficients of the gaseous reactants equals the sum of the stoichiometric coefficients of the gaseous products.
Apply the principle: Since the number of moles of gas on both sides is the same, changing the pressure does not favor either the forward or the backward reaction.
Conclusion: Because the pressure change exerts an equal force on both sides of the equilibrium, the system remains balanced. Therefore, pressure has no effect on the equilibrium position.
Exam Tip
Whenever you see a question about pressure and equilibrium, quickly write out the balanced equation and count the moles of gas on both sides.
If moles are different, pressure change shifts the equilibrium.
If moles are the same, pressure change has no effect.
Memory trick: Think of it like a seesaw with equal weights on both sides; moving the fulcrum (pressure) doesn't tip the balance!
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4.
The compound responsible for the pleasant scent of fruits and perfumes belongs to the class of
1. A.alkanamide
2. B.alkanone
3. C.alkanal
4. D.alkanoate
Show answer
Answer: D. alkanoate
AI explanation
Analysis
The question asks for the specific class of organic compounds that are characterized by their sweet, fruity, and pleasant smells, commonly used in the fragrance and flavoring industry.
Concept Explanation
In organic chemistry, different functional groups give compounds distinct physical and chemical properties.
Alkanamide (Amides): These are derivatives of carboxylic acids where the -OH group is replaced by an -NH₂ group. They are generally odorless or have a pungent smell.
Alkanone (Ketones): These contain a carbonyl group (C=O) bonded to two carbon atoms. While some have distinct smells (like acetone), they are not primarily known for "pleasant fruity scents."
Alkanal (Aldehydes): These contain a carbonyl group bonded to at least one hydrogen atom. Many have strong, sometimes sharp or pungent odors.
Alkanoate (Esters): These are formed by the reaction between a carboxylic acid and an alcohol (esterification). Esters are widely known in chemistry for their characteristic sweet, fruity, and floral aromas. They are the primary components in artificial fruit flavorings and perfumes.
Step-by-Step Solution
Identify the functional group: The question focuses on "pleasant scent of fruits and perfumes."
Relate to organic chemistry: Esters are the products of the reaction: • Carboxylic Acid + Alcohol → Ester + Water
Evaluate the options: • A, B, and C do not possess the characteristic sweet, fruity aroma associated with esters. • D (Alkanoate) is the IUPAC name for the salt or ester of an alkanoic acid. Esters are chemically referred to as alkyl alkanoates.
Conclusion: Because esters (alkanoates) are responsible for the natural scents of fruits like bananas, pineapples, and oranges, they are the correct class of compounds.
Correct Answer
The correct option is D.
Exam Tip
To remember this for your JAMB exam, use the mnemonic "Ester = Essence." Whenever you see a question about "fruity smells," "perfumes," or "flavorings," immediately look for Esters or Alkanoates in the options.
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5.
Which of the following statements is false about hard water?
1. A.Helps animals to build strong teeth
2. B.Cannot be supplied in pipes made of leads
3. C.Tastes better than soft water
4. D.Helps animals to build strong bones.
Show answer
Answer: B. Cannot be supplied in pipes made of leads
AI explanation
Analysis of the Question
The question asks us to identify the false statement regarding hard water among the given options. Hard water is water that contains high concentrations of dissolved minerals, specifically calcium ( $Ca^{2+}$ ) and magnesium ( $Mg^{2+}$ ) ions.
Concept Explanation
To solve this, let's evaluate the properties of hard water:
Biological Benefits: Hard water contains calcium and magnesium ions. Calcium is essential for the development and maintenance of strong bones and teeth in animals and humans. Therefore, consuming hard water is generally considered beneficial for skeletal health.
Taste: Many people prefer the taste of hard water over soft water because the dissolved minerals give it a distinct, "refreshing" mineral taste compared to the "flat" taste of soft water.
Interaction with Pipes: Hard water is generally less corrosive than soft water. Soft water (which is often acidic) tends to dissolve lead from lead pipes, leading to lead poisoning. Hard water, however, often forms a protective layer of calcium carbonate ( $CaCO_3$ ) inside pipes, which prevents the leaching of lead. Therefore, hard water is actually safer to transport through lead pipes than soft water.
Step-by-Step Evaluation
Option A: True. The calcium content helps in the formation of strong teeth.
Option B: False. Hard water is not prohibited from being supplied in lead pipes; in fact, it is safer than soft water because the mineral deposits form a protective coating that prevents lead from dissolving into the water.
Option C: True. The mineral content is widely considered to improve the taste of drinking water.
Option D: True. Similar to teeth, the calcium in hard water is vital for bone density and strength.
Conclusion
Since options A, C, and D are scientifically accurate descriptions of the benefits and properties of hard water, option B is the incorrect statement.
Exam Tip
When dealing with "Which of the following is false/not true" questions, always look for the statement that contradicts established scientific facts. Remember the mnemonic "Hard water = Healthy bones/teeth, but bad for soap"—this helps you quickly identify benefits versus industrial disadvantages (like scale formation in boilers).
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6.
Commercial deodorant is an example of a colloid called
1. A.aerosol
2. B.sol
3. C.foam
4. D.emulsion
Show answer
Answer: A. aerosol
AI explanation
Analysis
This question tests your knowledge of colloidal systems, specifically the classification of colloids based on the physical state of the dispersed phase and the dispersion medium.
Concept Explanation
A colloid is a mixture where one substance of microscopically dispersed insoluble particles is suspended throughout another substance. We classify them based on what is being dispersed in what:
Aerosol: A system where a liquid or solid is dispersed in a gas. (Examples: Fog, smoke, hair spray, and commercial deodorants).
Sol: A system where a solid is dispersed in a liquid. (Examples: Paint, blood, ink).
Foam: A system where a gas is dispersed in a liquid. (Examples: Whipped cream, soda water).
Emulsion: A system where a liquid is dispersed in another liquid. (Examples: Milk, mayonnaise).
Step-by-Step Solution
Identify the components of a commercial deodorant: A spray-type deodorant consists of fine liquid droplets (the dispersed phase) suspended in a gas (the dispersion medium, which is the propellant).
Match with definitions: Since we have a liquid dispersed in a gas, we look for the definition that fits this category.
Evaluate the options: • Aerosol: Liquid/Solid in Gas. This matches our deodorant spray. • Sol: Solid in Liquid. Incorrect. • Foam: Gas in Liquid. Incorrect. • Emulsion: Liquid in Liquid. Incorrect.
Conclusion
Commercial deodorants (specifically spray types) are classified as aerosols because they involve the dispersion of liquid particles into a gaseous medium.
Exam Tip
To remember these easily, think of the "medium":
If it ends in "-sol" (Aerosol), think of the air (gas) as the medium.
If it is an Emulsion, think of "emulsified" liquids like oil and water.
If it is Foam, think of the bubbles (gas) trapped in a liquid (like soap suds).
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7.
Calculate the pH of 0.001M KOH solution.
1. A.11
2. B.14
3. C.12
4. D.3
Show answer
Answer: A. 11
AI explanation
Analysis
To find the pH of a KOH solution, we must recognize that KOH (Potassium Hydroxide) is a strong base. It dissociates completely in water to produce hydroxide ions ( $OH^-$ ). We need to calculate the concentration of hydroxide ions, find the pOH, and then convert that to pH.
Concept
Dissociation: $KOH \rightarrow K^+ + OH^-$ . Since it is a strong base, the concentration of $OH^-$ is equal to the concentration of the KOH solution.
pOH formula: $pOH = -\log_{10}[OH^-]$
pH relationship: $pH + pOH = 14$ (at 25°C)
Step-by-Step Solution
Step 1: Determine the concentration of $[OH^-]$ Since $KOH$ is a strong base, $[OH^-] = 0.001 \, M$ . We can write this in scientific notation as: $[OH^-] = 1 \times 10^{-3} \, M$
Step 2: Calculate the pOH Using the formula $pOH = -\log[OH^-]$ : $pOH = -\log(1 \times 10^{-3})$ $pOH = -(-3)$ $pOH = 3$
Step 3: Calculate the pH Using the relationship $pH + pOH = 14$ : $pH + 3 = 14$ $pH = 14 - 3$ $pH = 11$
Conclusion
The pH of the 0.001M KOH solution is 11.
Exam Tip
Always check if the substance is an acid or a base before calculating. If you calculate the $-\log$ of a base concentration, you get the pOH, not the pH. Many students stop at 3 and choose option D, which is a common distractor! Always remember to subtract your result from 14 if you started with a base.
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8.
What is the function of concentrated H $_2$ SO $_4$ ?
1. A.It reacts with metals to give a salt and hydro gas
2. B.It is a dehydrating agent
3. C.It neutralizes alkali
4. D.It produces a cryolite precipitate with basic solution
Show answer
Answer: B. It is a dehydrating agent
AI explanation
Analysis of the Question
The question asks for the primary chemical function of concentrated tetraoxosulphate(VI) acid ( $H_2SO_4$ ). We need to evaluate each option based on the chemical properties of concentrated $H_2SO_4$ .
Concept Explanation
Concentrated $H_2SO_4$ is a versatile chemical reagent in the laboratory. Its most distinct properties include:
Dehydrating Property: It has a very strong affinity for water. It can remove chemically combined water (hydrogen and oxygen in the ratio 2:1) from organic compounds like sugar ( $C_{12}H_{22}O_{11}$ ), ethanol, and oxalic acid.
Oxidizing Property: Concentrated $H_2SO_4$ is a strong oxidizing agent, especially when hot.
Acidic Property: It reacts with bases to form salts and water (neutralization).
Step-by-Step Evaluation
Option A: While dilute $H_2SO_4$ reacts with reactive metals to produce hydrogen gas, concentrated $H_2SO_4$ acts as an oxidizing agent. Instead of hydrogen gas, it typically produces sulphur(IV) oxide ( $SO_2$ ), water, and a salt. Therefore, this is not its primary defining function.
Option B: Concentrated $H_2SO_4$ is widely used in chemistry as a dehydrating agent. For example, when added to sugar, it removes water molecules, leaving behind a black carbon residue. This is one of its most famous laboratory roles.
Option C: While it does neutralize alkalis, this is a property of acids in general (including dilute acids). It is not the unique or primary function that distinguishes concentrated $H_2SO_4$ from other acids.
Option D: This statement is chemically incorrect; concentrated $H_2SO_4$ does not produce a cryolite precipitate.
Conclusion
The most significant and characteristic function of concentrated $H_2SO_4$ among the given choices is its ability to act as a dehydrating agent.
Exam Tip
When you see "Concentrated $H_2SO_4$ " in a JAMB question, immediately think of two keywords: Dehydration (removing water) and Oxidation. If you see a question about removing water from a gas or an organic compound, $H_2SO_4$ is almost always the answer!
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9.
When ΔH is positive and small, and ΔS is positive and large, the reaction will be
1. A.Non-spontaneous
2. B.favour forward reaction only
3. C.favour backward reaction only
4. D.spontaneous
Show answer
Answer: D. spontaneous
AI explanation
Analysis
To solve this, we need to apply the Gibbs Free Energy equation, which determines the spontaneity of a chemical reaction.
Concept Explanation
The spontaneity of a reaction is governed by 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).
The rules for spontaneity are:
If $\Delta G < 0$ (negative), the reaction is spontaneous.
If $\Delta G > 0$ (positive), the reaction is non-spontaneous.
If $\Delta G = 0$ , the reaction is at equilibrium.
Step-by-Step Solution
Identify the given values: • $\Delta H$ is positive ( $+$ ). • $\Delta S$ is positive ( $+$ ). • The term $T\Delta S$ is "large" because $\Delta S$ is large.
Substitute into the equation: $ $\Delta G = (\text{positive small value}) - (T \times \text{positive large value})$ $
Analyze the result: Since the $T\Delta S$ term is large and it is being subtracted from a small positive $\Delta H$ , the resulting value of $\Delta G$ will be negative.
Conclusion: Because $\Delta G$ is negative, the reaction will proceed spontaneously.
Identification of the Answer
Based on the Gibbs Free Energy criteria, when the entropy factor ( $T\Delta S$ ) outweighs the enthalpy factor ( $\Delta H$ ), the reaction becomes spontaneous.
The correct option is D.
Exam Tip
Remember the "Spontaneity Table":
$\Delta H (-)$ and $\Delta S (+)$ : Always spontaneous.
$\Delta H (+)$ and $\Delta S (-)$ : Never spontaneous.
$\Delta H (+)$ and $\Delta S (+)$ : Spontaneous at high temperatures (because $T\Delta S$ becomes large).
$\Delta H (-)$ and $\Delta S (-)$ : Spontaneous at low temperatures.
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10.
Mg + Pb $^{2+}$ → Mg $^{2+}$ + Pb What is the cell notation for the cell reaction above?
1. A.MgIPb $^{2+}$ IIMg $^{2+}$ IPb
2. B.MgIMg $^{2+}$ IIPbIPb $^{2+}$
3. C.MgIMg $^{2+}$ IIPb $^{2+}$ IPb
4. D.MgIMg $^+$ IIPbIPb $^+$
Show answer
Answer: C. MgIMg $^{2+}$ IIPb $^{2+}$ IPb
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11.
Alkenes are represented with the general molecular formula
1. A.C $_n$ H $_{2n}$
2. B.C $_n$ H $_{2n-2}$
3. C.C $_n$ H $_{2n+2}$
4. D.C $_n$ H $_{2n+1}$ OH
Show answer
Answer: A. C $_n$ H $_{2n}$
AI explanation
Analysis
The question asks for the general molecular formula of alkenes, which are a class of hydrocarbons characterized by the presence of at least one carbon-carbon double bond ( $C=C$ ).
Concept Explanation
In organic chemistry, hydrocarbons are classified based on the type of bonds between their carbon atoms:
Alkanes: These are saturated hydrocarbons containing only single bonds. Their general formula is $C_nH_{2n+2}$ .
Alkenes: These are unsaturated hydrocarbons containing at least one double bond. Because a double bond "uses up" two hydrogen atoms that would otherwise be attached to the carbons, the formula is reduced by two hydrogens compared to alkanes. The general formula is $C_nH_{2n}$ .
Alkynes: These are unsaturated hydrocarbons containing at least one triple bond. Their general formula is $C_nH_{2n-2}$ .
Alkanols (Alcohols): These are organic compounds containing a hydroxyl (-OH) group. Their general formula is $C_nH_{2n+1}OH$ .
Step-by-Step Solution
Step 1: Identify the homologous series. We are looking for Alkenes.
Step 2: Recall the degree of unsaturation. Alkenes have one double bond, which removes two hydrogen atoms from the corresponding alkane ( $C_nH_{2n+2} - 2H = C_nH_{2n}$ ).
Step 3: Compare with the given options: • A: $C_nH_{2n}$ (Matches the alkene formula) • B: $C_nH_{2n-2}$ (This is the formula for Alkynes) • C: $C_nH_{2n+2}$ (This is the formula for Alkanes) • D: $C_nH_{2n+1}OH$ (This is the formula for Alkanols)
Conclusion
The general molecular formula for alkenes is $C_nH_{2n}$ .
Exam Tip
To remember these easily, think of the hydrogen count:
Alkanes (Single bond): $2n + 2$ (The "plus" gives you more)
Alkenes (Double bond): $2n$ (The "middle" ground)
Alkynes (Triple bond): $2n - 2$ (The "minus" takes away)
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12.
2Na + Cl $_2$ → 2NaCl In the reaction above, the specie that undergoes reduction is
1. A.Na
2. B.NaCl
3. C.Cl $^-$
4. D.Cl $_2$
Show answer
Answer: D. Cl $_2$
AI explanation
Analysis
To determine which species undergoes reduction, we need to look at the change in the oxidation states of the elements involved in the reaction: 2Na + Cl $_2$ → 2NaCl
Concept: Oxidation and Reduction (Redox)
In chemistry, we use the acronym OIL RIG to remember redox reactions:
Oxidation Is Loss (of electrons, increase in oxidation number).
Reduction Is Gain (of electrons, decrease in oxidation number).
The oxidation state of an element in its free or uncombined state (like Na or Cl $_2$ ) is always 0.
Step-by-Step Solution
Assign oxidation numbers to the reactants: • Na (elemental form) = 0 • Cl $_2$ (elemental form) = 0
Assign oxidation numbers to the products: • In NaCl, Sodium (Na) is in Group 1, so it has an oxidation state of +1. • Chlorine (Cl) in NaCl has an oxidation state of -1.
Track the changes: • Na: Goes from 0 to +1 (Increase in oxidation number = Oxidation) • Cl: Goes from 0 to -1 (Decrease in oxidation number = Reduction)
Identify the species: The chlorine atoms started as part of the Cl $_2$ molecule. Since the oxidation state of the chlorine atoms decreased from 0 to -1, the species that underwent reduction is Cl $_2$ .
Conclusion
The chlorine molecule (Cl $_2$ ) gained electrons to become chloride ions (Cl $^-$ ) in the NaCl compound. Therefore, Cl $_2$ is the species that is reduced.
Exam Tip
Always remember that in a reaction between a metal and a non-metal, the metal is almost always oxidized (loses electrons), and the non-metal is almost always reduced (gains electrons).
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13.
The above structure is
1. A.alkylamine
2. B.alkanone
3. C.alkanamide
4. D.amino acid
Show answer
Answer: A. alkylamine
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14.
If a gold bar and a silver bar are tied together firmly and left for years, some of the gold particles will be found in the silver bar due to
1. A.Brownian movement
2. B.diffusion
3. C.displacement
4. D.osmosis
Show answer
Answer: B. diffusion
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15.
The products of the thermal decomposition of ammonium trioxonitrate(v) are
1. A.N $_2$ O and H $_2$ O
2. B.NO $_2$ and H $_2$ O
3. C.N $_2$ O and O $_2$
4. D.NO $_3$ and H $_2$ O
Show answer
Answer: A. N $_2$ O and H $_2$ O
AI explanation
Analysis
The question asks for the products formed when ammonium trioxonitrate(V), also known as ammonium nitrate ( $NH_4NO_3$ ), undergoes thermal decomposition. This is a classic reaction in inorganic chemistry often tested in JAMB.
Concept
Thermal decomposition involves breaking down a chemical compound into simpler substances by heating it. Ammonium salts of oxidizing acids (like nitric acid) decompose differently than other ammonium salts. When ammonium nitrate is heated, it undergoes a redox reaction where the nitrogen in the ammonium ion ( $NH_4^+$ ) is oxidized and the nitrogen in the nitrate ion ( $NO_3^-$ ) is reduced.
Solution
Write the chemical formula: Ammonium trioxonitrate(V) is $NH_4NO_3$ .
Write the balanced chemical equation: When $NH_4NO_3$ is heated gently, it decomposes into dinitrogen oxide (nitrous oxide) and water vapor.
$NH_4NO_3(s) \xrightarrow{\Delta} N_2O(g) + 2H_2O(g)$
Breakdown of the products: • $N_2O$ is Dinitrogen(I) oxide (also known as laughing gas). • $H_2O$ is water vapor.
Comparing this to your options:
Option A: $N_2O$ and $H_2O$ (Matches our equation)
Option B: $NO_2$ and $H_2O$ (This would be the product for a different nitrate decomposition)
Option C: $N_2O$ and $O_2$ (Incorrect)
Option D: $NO_3$ and $H_2O$ (Incorrect)
Conclusion
The thermal decomposition of ammonium trioxonitrate(V) yields dinitrogen oxide and water.
Exam Tip
Remember the "Ammonium Nitrate Rule":
Gentle heating: $NH_4NO_3 \rightarrow N_2O + 2H_2O$
Explosive/Strong heating: $2NH_4NO_3 \rightarrow 2N_2 + O_2 + 4H_2O$ If the question doesn't specify "explosive," always assume the standard decomposition into $N_2O$ and water.
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16.
The relative molecular mass of an alkanoic acid with the empirical formula of CH $_2$ O is [ H = 1, C = 12, O = 16 ]
1. A.46
2. B.88
3. C.60
4. D.72
Show answer
Answer: C. 60
AI explanation
Analysis
To find the relative molecular mass of an alkanoic acid given its empirical formula, we need to understand the general formula for alkanoic acids (carboxylic acids) and how empirical formulas relate to molecular formulas.
Concept
Empirical Formula: This is the simplest whole-number ratio of atoms in a compound. Here, it is $CH_2O$ .
General Formula of Alkanoic Acids: Alkanoic acids have the general formula $C_nH_{2n}O_2$ .
Relationship: The molecular formula is always a multiple of the empirical formula ( $n \times \text{Empirical Formula}$ ). We need to find the value of $n$ that fits the general formula of an alkanoic acid.
Step-by-Step Solution
Step 1: Determine the empirical mass. Using the atomic masses provided ( $H=1, C=12, O=16$ ): Mass of $CH_2O = 12 + (2 \times 1) + 16$ Mass of $CH_2O = 12 + 2 + 16 = 30\text{ g/mol}$
Step 2: Relate to the general formula. The empirical formula $CH_2O$ has a ratio of $1:2:1$ . The general formula for alkanoic acids is $C_nH_{2n}O_2$ . If we multiply the empirical formula $CH_2O$ by $2$ , we get: $2 \times (CH_2O) = C_2H_4O_2$
Step 3: Check if $C_2H_4O_2$ is an alkanoic acid. $C_2H_4O_2$ is Ethanoic acid ( $CH_3COOH$ ). This fits the definition of an alkanoic acid.
Step 4: Calculate the molecular mass. Molecular Mass $= 2 \times (\text{Empirical Mass})$ Molecular Mass $= 2 \times 30 = 60\text{ g/mol}$
(Note: If we multiplied by 1, we would get $CH_2O$ (Methanal, an aldehyde, not an acid). If we multiplied by 3, we would get $C_3H_6O_3$ , which is not the standard form for a simple alkanoic acid.)
Conclusion
The relative molecular mass of the alkanoic acid (Ethanoic acid) is 60.
Exam Tip: Always remember that for alkanoic acids, the number of Oxygen atoms is always 2. If your empirical formula has only one oxygen, you must multiply the entire formula by 2 to satisfy the requirement for an alkanoic acid ( $C_nH_{2n}O_2$ ).
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17.
On heating 12.5g of saturated solution to dryness at 60 $^0$ C, 2g of anhydrous salt was recovered, calculate its solubility in grams per 100g of water.
1. A.16.0
2. B.4.0
3. C.2.0
4. D.20.0
Show answer
Answer: A. 16.0
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18.
Na $_2$ X ⇌ 2Na $^+$ + X $^{2-}$ The bond between Na and X is likely to be
1. A.dative
2. B.ionic
3. C.metallic
4. D.covalent
Show answer
Answer: B. ionic
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19.
Atom with the electron configuration of 1S $^2$ 2S $^2$ 2P $^6$ 3S $^2$ belongs to
1. A.Group 3 and Period 2
2. B.Group 8 and Period 3
3. C.Group 2 and Period 3
4. D.Group 2 and Period 6
Show answer
Answer: C. Group 2 and Period 3
AI explanation
To determine the position of an element in the Periodic Table based on its electron configuration, we follow these two simple rules:
1. Analyze the Question
We are given the electron configuration: 1s² 2s² 2p⁶ 3s². We need to identify the Group and Period of this atom.
2. Explain the Concept
Period: The period of an element is determined by the highest principal quantum number (n) in its electron configuration. This represents the outermost shell (energy level) that contains electrons.
Group: The group of an element is determined by the number of valence electrons (electrons in the outermost shell). For elements in the s-block, the number of valence electrons directly corresponds to the Group number.
3. Show the Solution
Step 1: Identify the Period. Look at the highest number appearing before the orbitals (s, p, d, f). In 1s² 2s² 2p⁶ 3s², the highest number is 3. Therefore, the element is in Period 3.
Step 2: Identify the Group. Look at the outermost shell (the shell with the highest number, which is 3). The configuration shows 3s². There are 2 electrons in this outermost shell. Therefore, the element is in Group 2.
Step 3: Conclusion. The element has 2 valence electrons in the 3rd energy level, placing it in Group 2, Period 3. (This element is Magnesium, Mg).
4. Identify the Answer
Comparing our result with the options provided: A. Group 3 and Period 2 B. Group 8 and Period 3 C. Group 2 and Period 3 D. Group 2 and Period 6
The correct option is C.
5. Exam Tip
To quickly find the group for s-block elements, just look at the exponent of the last 's' orbital. If it ends in 's¹', it's Group 1; if it ends in 's²', it's Group 2. For the period, just look at the largest number in front of the orbitals!
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20.
The following is not a water pollutant?
1. A.warm water affluent
2. B.inorganic fertilizers
3. C.oxygen gas
4. D.biodegradable waste
Show answer
Answer: C. oxygen gas
AI explanation
Analysis
The question asks us to identify which of the listed substances does not cause water pollution. A water pollutant is any substance that, when added to a water body, makes it harmful to living organisms or unsuitable for its intended use.
Concept Explanation
Warm water affluent (Thermal pollution): When industries discharge hot water into rivers, it reduces the amount of dissolved oxygen available for aquatic life, causing thermal pollution.
Inorganic fertilizers: These contain nitrates and phosphates. When washed into water bodies, they cause eutrophication (excessive growth of algae), which depletes oxygen and kills aquatic life.
Oxygen gas ( $O_2$ ): Oxygen is essential for the survival of aquatic organisms (fish, plants, etc.). It is a natural component of water and is necessary for the aerobic decomposition of organic matter. It is not a pollutant; rather, its depletion is a sign of pollution.
Biodegradable waste: While these can be broken down by bacteria, if they enter water in large quantities, they increase the Biological Oxygen Demand (BOD), which leads to oxygen depletion and pollution.
Step-by-Step Solution
Evaluate A: Warm water alters the temperature of the ecosystem, which is a form of pollution.
Evaluate B: Inorganic fertilizers lead to nutrient enrichment (eutrophication), which is a major water pollutant.
Evaluate C: Oxygen gas is a life-sustaining gas. It is naturally present in water and is required for the health of the aquatic ecosystem. It does not cause harm or contamination.
Evaluate D: Large amounts of biodegradable waste lead to the depletion of dissolved oxygen as microorganisms consume it to break down the waste, thus acting as a pollutant.
Comparing these, oxygen gas is the only substance that is beneficial and natural to the aquatic environment in its standard state.
Conclusion
Exam Tip
When answering "which is not a pollutant" questions, look for substances that are naturally required for life (like Oxygen) versus substances that disrupt the chemical or physical balance of the environment (like heavy metals, nitrates, or heat).
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21.
The gas that is commonly used to demonstrate the fountain experiment is
1. A.hydrogen chloride
2. B.hydrogen sulphide
3. C.dinitrogen(I) oxide
4. D.nitrogen(II) oxide
Show answer
Answer: A. hydrogen chloride
AI explanation
Analysis
The question asks for the specific gas used in the "fountain experiment." This is a classic demonstration in chemistry used to illustrate the high solubility of certain gases in water.
Concept Explanation
The fountain experiment relies on the principle of extreme solubility. For the experiment to work, a gas must be highly soluble in water. When a small amount of water is introduced into a flask filled with such a gas, the gas dissolves rapidly, creating a sudden drop in pressure inside the flask. The external atmospheric pressure then forces water from a reservoir up through a tube and into the flask, creating a "fountain" effect.
Hydrogen chloride (HCl): This gas is extremely soluble in water. One volume of water can dissolve about 450 volumes of HCl gas at room temperature. This rapid dissolution creates the necessary vacuum for the fountain experiment.
Hydrogen sulphide ( $H_2S$ ): This is only moderately soluble in water and is not used for this demonstration.
Dinitrogen(I) oxide ( $N_2O$ ) and Nitrogen(II) oxide ( $NO$ ): These gases have low solubility in water and cannot create the pressure differential required for the experiment.
Step-by-Step Solution
Identify the requirement: The experiment requires a gas that dissolves almost instantaneously in water to create a partial vacuum.
Evaluate the options: • HCl is highly polar and reacts with water to form hydrochloric acid, leading to rapid absorption. • $H_2S$ , $N_2O$ , and $NO$ do not possess the high degree of solubility required to displace the volume of air needed to pull water up a tube effectively.
Conclusion: Hydrogen chloride is the standard gas used in laboratory demonstrations to show this principle.
Exam Tip
Whenever you see "fountain experiment" in a chemistry question, immediately look for highly soluble gases like Hydrogen chloride (HCl) or Ammonia ( $NH_3$ ). These are the two most common gases used for this demonstration because they are both extremely soluble in water.
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22.
Magnesium tetraoxosulphate(VI) salt is commonly used as a
1. A.disinfectant
2. B.plasticizer
3. C.fertilizer
4. D.laxative
Show answer
Answer: D. laxative
AI explanation
Analysis
The question asks for the common application of Magnesium tetraoxosulphate(VI), which is chemically represented as $\text{MgSO}_4$ . This compound is more commonly known as Epsom salt.
Concept Explanation
To identify the correct use, we look at the properties and common household/medical uses of $\text{MgSO}_4$ :
Chemical Nature: It is a hydrated salt containing magnesium, sulfur, and oxygen.
Medical/Biological Application: Magnesium sulfate is widely recognized in pharmacology for its ability to draw water into the intestines through osmosis. This action softens the stool and stimulates bowel movements.
Other uses: While it is sometimes used in agriculture (as a magnesium supplement for plants), its most prominent and direct "common" use in a medical context is as a saline laxative.
Step-by-Step Solution
Identify the compound: Magnesium tetraoxosulphate(VI) is $\text{MgSO}_4 \cdot 7\text{H}_2\text{O}$ (Epsom salt).
Evaluate the options: • A. Disinfectant: Substances like chlorine or hydrogen peroxide are used for this. $\text{MgSO}_4$ does not kill bacteria. • B. Plasticizer: These are additives used to increase the flexibility of plastics (e.g., phthalates). $\text{MgSO}_4$ is an inorganic salt, not a plasticizer. • C. Fertilizer: While magnesium is a nutrient for plants, it is not the primary or "common" definition for this salt in the context of standard chemistry examinations compared to its medical use. • D. Laxative: $\text{MgSO}_4$ is a well-known saline laxative used to treat constipation.
Conclusion
Based on standard pharmaceutical and chemical classifications, Magnesium tetraoxosulphate(VI) is primarily classified as a laxative.
Exam Tip
When you see "tetraoxosulphate(VI)," remember that this is the IUPAC name for "sulfate." Whenever you encounter Epsom salt ( $\text{MgSO}_4$ ) in an exam, immediately associate it with its medical use as a laxative to relieve constipation.
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23.
A by-product of the alkaline hydrolysis of tristearin is a
1. A.dihydric alkanol
2. B.tertiary alkanol
3. C.secondary alkanol
4. D.trihydric alkanol
Show answer
Answer: D. trihydric alkanol
AI explanation
Analysis
The question asks for the identity of the by-product formed during the alkaline hydrolysis of tristearin. Tristearin is a common fat (a triglyceride).
Concept Explanation
Tristearin: This is a triglyceride formed from the reaction of glycerol (a trihydric alcohol) and stearic acid (a fatty acid). Its chemical structure consists of a glycerol backbone attached to three stearate groups.
Alkaline Hydrolysis (Saponification): When a fat or oil is heated with an alkali (like NaOH or KOH), the ester bonds are broken. This process is known as saponification, which is the basis for soap making.
The Reaction: $ $\text{Tristearin} + 3\text{NaOH} \rightarrow 3\text{Sodium Stearate (Soap)} + \text{Glycerol}$ $
Glycerol: The chemical name for glycerol is propane-1,2,3-triol. Because it contains three hydroxyl (-OH) groups attached to different carbon atoms, it is classified as a trihydric alkanol.
Step-by-Step Solution
Identify the reactants: Tristearin (an ester of glycerol) and an alkali (e.g., NaOH).
Identify the products: The main product is soap (sodium stearate), and the by-product is glycerol.
Analyze the structure of the by-product: Glycerol has the formula $CH_2(OH)CH(OH)CH_2(OH)$ .
Classify the by-product: Since it has three -OH groups, it is a trihydric alcohol (trihydric alkanol).
Conclusion
The by-product of the alkaline hydrolysis of fats like tristearin is glycerol, which is a trihydric alkanol. This matches option D.
Exam Tip
Remember the "Saponification Rule": Fat + Alkali $\rightarrow$ Soap + Glycerol. Whenever you see "alkaline hydrolysis of fats/oils," always look for glycerol or trihydric alkanol as the by-product.
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24.
The acid used in making baking soda and soft drink is
1. A.tartaric acid
2. B.fatty acid
3. C.boric acid
4. D.citric acid
Show answer
Answer: A. tartaric acid
AI explanation
1. Analyze the question
The question asks for the specific acid used in the production of baking soda and soft drinks. We need to identify which of the listed acids is commonly used as a food additive for these purposes.
2. Explain the concept
Baking Soda (Sodium Bicarbonate): In baking, an acid is required to react with sodium bicarbonate to release carbon dioxide gas, which makes dough rise. Tartaric acid (often in the form of cream of tartar) is the classic acid used in baking powder formulations.
Soft Drinks: While citric acid is also used in soft drinks, tartaric acid is specifically noted in chemistry curricula for its dual role in both the baking industry (as a leavening agent component) and the beverage industry (as an acidulant to provide a sharp, tart taste).
3. Show the solution
Tartaric acid is a naturally occurring organic acid found in grapes and other fruits. It is widely used in the food industry as an acidulant and a leavening agent.
Fatty acids are components of fats and oils, not typically used as leavening agents or drink acidulants.
Boric acid is toxic and is used as an antiseptic or insecticide; it is definitely not used in food production.
Citric acid is very common in soft drinks, but in the context of standard chemistry examinations (like JAMB), tartaric acid is the specific answer associated with the combination of baking soda applications and soft drink production.
4. Identify the answer
Comparing the options, tartaric acid is the correct chemical compound associated with these specific industrial food applications in the context of your syllabus.
5. Tip
When you see questions about "acids in food," remember:
Tartaric acid: Baking powder and soft drinks.
Citric acid: Citrus fruits and soft drinks.
Acetic acid: Vinegar.
Lactic acid: Sour milk/yoghurt.
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25.
The process by which iron corrodes is
1. A.rusting
2. B.burning
3. C.galvanizing
4. D.alloying
Show answer
Answer: A. rusting
AI explanation
Analysis of the Question
The question asks for the specific term used to describe the chemical process of corrosion when it occurs on iron.
Concept Explanation
Corrosion is the gradual destruction of materials (usually metals) by chemical reaction with their environment.
When this process involves iron reacting with oxygen and moisture (water) to form hydrated iron(III) oxide, it is specifically referred to as rusting.
Burning (Combustion) is a rapid chemical reaction with oxygen that produces heat and light, which is different from the slow oxidation process of corrosion.
Galvanizing is a method used to prevent corrosion by coating iron with a layer of zinc.
Alloying is the process of mixing iron with other elements (like carbon or chromium) to improve its properties, such as making it resistant to corrosion (e.g., stainless steel).
Step-by-Step Solution
Identify the metal: The question specifies "iron."
Identify the reaction: Iron reacts with oxygen ( $O_2$ ) and water ( $H_2O$ ) to form a reddish-brown flaky substance known as rust ( $Fe_2O_3 \cdot xH_2O$ ).
Match the term: In chemistry, the oxidation of iron in the presence of air and water is uniquely termed "rusting."
Evaluate the options: • A (Rusting): Correct, as it is the specific term for iron corrosion. • B (Burning): Incorrect; this is a rapid combustion process. • C (Galvanizing): Incorrect; this is a protective measure against corrosion. • D (Alloying): Incorrect; this is a method of modifying the metal's composition.
Conclusion
The process of iron corrosion is scientifically known as rusting.
Exam Tip: Always remember that while all rusting is corrosion, not all corrosion is rusting. "Rusting" is a term reserved exclusively for iron and its alloys. If the metal were copper, for example, the corrosion product would be called "verdigris."
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26.
The compound CH $_3$ CH(NH $_2$ )CH $_2$ CH $_2$ CH $_3$ is an example of a
1. A.primary amine
2. B.tertiary amine
3. C.quaternary amine
4. D.secondary amine
Show answer
Answer: A. primary amine
AI explanation
Analysis of the Question
The question asks us to classify an organic compound based on its functional group. Specifically, we need to determine the degree of the amine group attached to the carbon chain.
Concept Explanation
In organic chemistry, amines are classified as primary ( $1^\circ$ ), secondary ( $2^\circ$ ), or tertiary ( $3^\circ$ ) based on the number of carbon groups (alkyl or aryl groups) directly attached to the nitrogen atom:
Primary ( $1^\circ$ ) amine: The nitrogen is attached to only one carbon atom (General formula: $R-NH_2$ ).
Secondary ( $2^\circ$ ) amine: The nitrogen is attached to two carbon atoms (General formula: $R_2NH$ ).
Tertiary ( $3^\circ$ ) amine: The nitrogen is attached to three carbon atoms (General formula: $R_3N$ ).
To classify the compound, we look at the nitrogen atom and count how many carbon atoms are directly bonded to it.
Step-by-Step Solution
Examine the structure: The compound is $CH_3-CH(NH_2)-CH_2-CH_2-CH_3$ .
Identify the functional group: The functional group is the amino group ( $-NH_2$ ) attached to the second carbon atom of the pentane chain.
Analyze the Nitrogen atom: • The nitrogen atom is bonded to two hydrogen atoms ( $H_2$ ). • The nitrogen atom is bonded to exactly one carbon atom (the carbon at the second position of the chain).
Conclusion: Since the nitrogen is attached to only one alkyl group (a pentyl group), it is classified as a primary amine.
Note: Do not be confused by the fact that the carbon holding the nitrogen is a secondary carbon (a carbon attached to two other carbons). The classification of an amine depends solely on the number of carbons attached to the nitrogen, not the carbon holding the nitrogen.
Identifying the Answer
Based on the classification rules, the compound is a primary amine. This matches Option A.
Exam Tip
Always remember:
Amines are classified by the number of carbons attached to the Nitrogen.
Alcohols are classified by the number of carbons attached to the Carbon holding the $-OH$ group. Don't mix these two rules up in the exam hall!
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27.
Iron produced directly from a blast furnace is
1. A.wrought iron
2. B.cast iron
3. C.pig iron
4. D.carbon steel
Show answer
Answer: C. pig iron
AI explanation
Analysis
The question asks for the specific form of iron that is obtained immediately after the extraction process in a blast furnace.
Concept Explanation
In the metallurgy of iron, the blast furnace is used to reduce iron ore (usually hematite, $Fe_2O_3$ ) into metallic iron. The raw material fed into the furnace includes iron ore, coke (carbon), and limestone.
As the iron melts, it absorbs a significant amount of carbon from the coke (typically 3% to 4.5%). Because of this high carbon content and the presence of impurities like silicon, phosphorus, and sulfur, the resulting product is brittle and cannot be forged. This raw, impure product is known as pig iron.
To obtain other forms of iron:
Cast iron is produced by remelting pig iron with scrap iron.
Wrought iron is produced by further refining pig iron to remove almost all carbon and impurities.
Steel is produced by controlling the carbon content of iron (usually between 0.1% and 1.5%).
Step-by-Step Solution
Identify the process: The blast furnace process is the primary extraction method for iron.
Analyze the output: The molten iron collected at the base of the blast furnace contains high levels of carbon (approx. 4%) and other impurities.
Classify the product: This specific high-carbon, impure product is commercially and chemically referred to as "pig iron."
Evaluate the options: • A (Wrought iron) is the purest form, not the raw product. • B (Cast iron) is a secondary product made from pig iron. • C (Pig iron) is the direct product of the blast furnace. • D (Carbon steel) is an alloy created through further processing.
Conclusion
The product obtained directly from the blast furnace is pig iron.
Exam Tip: Remember the "P" in Pig iron stands for Primary or Production. It is the "raw" iron that must undergo further processing to become useful materials like steel or wrought iron.
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28.
The process employed in the industrial preparation of tetraoxosulphate(VI) acid is
1. A.Contact
2. B.Bosch
3. C.Frasch
4. D.Haber
Show answer
Answer: A. Contact
AI explanation
1. Analyze the Question
The question asks for the specific industrial process used to manufacture tetraoxosulphate(VI) acid, which is commonly known as Sulphuric acid ( $H_2SO_4$ ).
2. Explain the Concept
Industrial chemistry involves specific named processes for the large-scale production of chemicals. To answer this, we must identify the purpose of each process listed in the options:
Contact Process: This is the standard industrial method for the production of concentrated sulphuric acid. It involves the catalytic oxidation of sulphur(IV) oxide to sulphur(VI) oxide, which is then absorbed in concentrated sulphuric acid to form oleum, followed by dilution.
Bosch Process: This is used for the production of hydrogen gas by the reaction of steam with water gas (a mixture of carbon(II) oxide and hydrogen).
Frasch Process: This is the method used for the extraction of elemental sulphur from underground deposits, not for the production of the acid itself.
Haber Process: This is the industrial process for the manufacture of ammonia ( $NH_3$ ) from nitrogen and hydrogen gases.
3. Show the Solution
Step 1: Identify the chemical: Tetraoxosulphate(VI) acid is $H_2SO_4$ .
Step 2: Evaluate the options based on their industrial application.
Step 3: Since the Contact Process is specifically designed to convert sulphur dioxide into sulphur trioxide and subsequently into sulphuric acid, it is the correct industrial method.
4. Identify the Answer
The correct option is A.
5. Exam Tip
To avoid confusion in the exam hall, remember these "Big Four" industrial processes:
Contact = Sulphuric Acid ( $H_2SO_4$ )
Haber = Ammonia ( $NH_3$ )
Ostwald = Nitric Acid ( $HNO_3$ )
Solvay = Sodium Trioxocarbonate(IV) ( $Na_2CO_3$ )
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29.
A metal that can be found in the free state in nature is
1. A.silver
2. B.Zinc
3. C.Iron
4. D.copper
Show answer
Answer: A. silver
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30.
The molecule with the highest number of lone pair of electrons is
1. A.H $_2$ O
2. B.NH $_3$
3. C.CH $_4$
4. D.C0 $_2$
Show answer
Answer: D. C0 $_2$
AI explanation
To solve this, we need to determine the Lewis structure for each molecule and count the lone pairs of electrons on the central atom and the surrounding atoms.
1. Analyze the concept
A lone pair is a pair of valence electrons that is not shared with another atom in a covalent bond. To find the total number of lone pairs, we look at the valence electrons of each atom and how many are used in bonding.
2. Step-by-step breakdown
A. H $_2$ O (Water): • Oxygen (Group 6) has 6 valence electrons. It uses 2 for bonding with two Hydrogen atoms. • This leaves 4 electrons (2 lone pairs) on the Oxygen atom. • Total lone pairs = 2.
B. NH $_3$ (Ammonia): • Nitrogen (Group 5) has 5 valence electrons. It uses 3 for bonding with three Hydrogen atoms. • This leaves 2 electrons (1 lone pair) on the Nitrogen atom. • Total lone pairs = 1.
C. CH $_4$ (Methane): • Carbon (Group 4) has 4 valence electrons. It uses all 4 for bonding with four Hydrogen atoms. • There are no electrons left on the Carbon atom. • Total lone pairs = 0.
D. CO $_2$ (Carbon Dioxide): • Carbon is in the center, double-bonded to two Oxygen atoms. • Carbon uses all 4 of its valence electrons for bonding (no lone pairs on Carbon). • Each Oxygen atom (Group 6) uses 2 electrons for the double bond, leaving 4 electrons (2 lone pairs) on each Oxygen atom. • Total lone pairs = 2 (on first O) + 2 (on second O) = 4.
3. Conclusion
Comparing the totals:
H $_2$ O: 2 lone pairs
NH $_3$ : 1 lone pair
CH $_4$ : 0 lone pairs
CO $_2$ : 4 lone pairs
The molecule with the highest number of lone pairs is CO $_2$ .
Exam Tip
When counting lone pairs, don't just look at the central atom! Always check the surrounding atoms (like Oxygen or Halogens) as they often carry lone pairs that contribute to the total count of the molecule.
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31.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.
32.
What accounts for the low melting and boiling points of covalent molecules?
1. A.They have weak intermolecular forces
2. B.They have definite shapes
3. C.They are three dimensional structures
4. D.They possess shared electron pairs
Show answer
Answer: A. They have weak intermolecular forces
AI explanation
Analysis
The question asks for the fundamental reason why covalent molecular substances (like water, methane, or carbon dioxide) exhibit low melting and boiling points compared to ionic or metallic substances.
Concept Explanation
To understand this, we must distinguish between intramolecular forces and intermolecular forces:
Intramolecular forces: These are the strong covalent bonds within the molecule (the shared electron pairs holding the atoms together). These are very strong and require a lot of energy to break.
Intermolecular forces: These are the weak forces of attraction between separate molecules (such as Van der Waals forces or dipole-dipole interactions).
When a covalent substance melts or boils, you are not breaking the strong covalent bonds within the molecules; you are only overcoming the weak intermolecular forces that hold the molecules together in a solid or liquid state. Because these forces are weak, very little heat energy is required to pull the molecules apart, resulting in low melting and boiling points.
Step-by-Step Solution
Step 1: Evaluate Option B and C. Having a definite shape or a 3D structure describes the geometry of the molecule, but it does not dictate the energy required to change its state.
Step 2: Evaluate Option D. Shared electron pairs are what form the covalent bond itself. These are strong, so if we had to break these to melt a substance, covalent compounds would have extremely high melting points (like diamond or silica).
Step 3: Evaluate Option A. Since the molecules are held together by weak intermolecular forces, the energy required to overcome these forces is minimal. This directly explains the low melting and boiling points.
Conclusion
Exam Tip: Always remember the "Inter vs. Intra" rule! If a question asks about melting/boiling points, it is almost always about the intermolecular forces (the "glue" between molecules), not the strong bonds inside the molecule.
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33.
The composition of petroleum varies because it is a
1. A.hydrocarbon
2. B.liquid
3. C.mixture
4. D.natural resource
Show answer
Answer: C. mixture
AI explanation
Analysis
The question asks why the composition of petroleum (crude oil) is not constant but varies depending on its source. We need to identify the chemical nature of petroleum that accounts for this variability.
Concept Explanation
Petroleum is a complex substance formed from the remains of ancient marine organisms buried under layers of sediment for millions of years.
Chemical Nature: Petroleum is not a single chemical compound with a fixed molecular formula. Instead, it is a mixture of various hydrocarbons (alkanes, cycloalkanes, and aromatic hydrocarbons) along with small amounts of impurities like sulfur, nitrogen, and oxygen compounds.
Why it varies: Because it is a mixture, the proportions of these different hydrocarbons and impurities depend entirely on the geological conditions, temperature, pressure, and the specific organic matter present in the location where the oil was formed. Since there is no fixed chemical ratio (unlike a pure compound like water, $H_2O$ ), the composition changes from one oil field to another.
Step-by-Step Solution
Evaluate Option A (Hydrocarbon): While petroleum is composed of hydrocarbons, being a hydrocarbon does not explain why the composition varies. Many pure hydrocarbons (like methane) have a fixed composition.
Evaluate Option B (Liquid): Being a liquid is a physical state, not a chemical property that dictates composition.
Evaluate Option C (Mixture): By definition, a mixture consists of two or more substances that are not chemically bonded. The components of a mixture can be present in any proportion. Therefore, the fact that petroleum is a mixture is the direct reason why its composition is inconsistent.
Evaluate Option D (Natural resource): This describes the origin and economic classification of petroleum, not its chemical composition.
Conclusion
Because petroleum is a mixture of various organic compounds, its composition is not fixed and varies depending on the geological location.
Exam Tip: Whenever you see a question about "composition varying" or "components being separated by physical means" (like fractional distillation), the answer is almost always that the substance is a mixture.
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34.
The metal used as a packaging material is
1. A.Al
2. B.Cu
3. C.Fe
4. D.Na
Show answer
Answer: A. Al
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35.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.
36.
The reaction above is
1. A.neutralization
2. B.substitution
3. C.polymerization
4. D.oxidation
Show answer
Answer: C. polymerization
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37.
NH $_3$ $(_g$ ) + HCl $(_g$ ) → NH $_4$ Cl $_(g)$ In the reaction above, increase in pressure will
1. A.lower the equilibrium constant
2. B.favour the product
3. C.favour the reactant
4. D.increase the equilibrium constant
Show answer
Answer: B. favour the product
AI explanation
1. Analyze the question
The question asks how an increase in pressure affects the chemical equilibrium of the reaction: NH $_3(g) + \text{HCl}(g) \rightleftharpoons \text{NH}_4\text{Cl}(s)$ (Note: While the question writes NH $_4$ Cl as a gas, in standard chemical equilibrium problems of this type, NH $_4$ Cl is typically treated as a solid. Let's analyze the moles of gas on both sides.)
2. Explain the concept
According to Le Chatelier’s Principle, if a system at equilibrium is subjected to a change in pressure, the system will shift its equilibrium position to counteract that change.
Increase in pressure: The system shifts toward the side with fewer moles of gas to reduce the pressure.
Decrease in pressure: The system shifts toward the side with more moles of gas to increase the pressure.
Equilibrium Constant ( $K$ ): The equilibrium constant is only affected by temperature. Changes in pressure or concentration do not change the value of $K$ .
3. Show the solution
Step 1: Count the moles of gas on the reactant side. NH $_3(g)$ = 1 mole HCl $(g)$ = 1 mole Total = 2 moles of gas
Step 2: Count the moles of gas on the product side. NH $_4$ Cl $(s)$ = 0 moles of gas (solids are not counted in pressure equilibrium shifts).
Step 3: Apply the pressure change. Since there are 2 moles of gas on the left and 0 moles of gas on the right, an increase in pressure will force the reaction to shift to the right (the product side) to reduce the total number of gas particles.
Step 4: Evaluate the options. • Options A and D are incorrect because pressure does not change the equilibrium constant ( $K$ ). • Option C is incorrect because increasing pressure favors the side with fewer moles of gas (the product). • Option B is correct because the reaction shifts toward the product side.
4. Identify the answer
The increase in pressure shifts the equilibrium to the right, thereby favoring the formation of the product.
5. Tip
Memory Trick: Think of pressure like a crowded room. If you increase the pressure (squeeze the room), the molecules will try to "shrink" by moving to the side of the equation that has fewer moles of gas to save space!
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38.
CH $_3$ - CH $_2$ - COOCH $_2$ - CH $_3$ From the condensed structure above, the reactants are
1. A.ethanol and ethanoic acid
2. B.propanol and propanoic acid
3. C.propanol and ethanoic acid
4. D.propanoic acid and ethanol
Show answer
Answer: D. propanoic acid and ethanol
AI explanation
To solve this, we need to understand the structure of an ester and how it is formed through the process of esterification.
1. Analyze the structure
The given compound is: CH $_3$ - CH $_2$ - COOCH $_2$ - CH $_3$
An ester is formed by the reaction between a carboxylic acid and an alcohol. The general formula for an ester is R-COO-R', where:
R-COO- comes from the carboxylic acid.
-R' comes from the alcohol.
2. Break down the molecule
Let’s split the structure at the ester linkage (-COO-):
Part 1 (Acid side): CH $_3$ - CH $_2$ - COO- • This part contains 3 carbon atoms (Propanoate). The corresponding acid is Propanoic acid (CH $_3$ CH $_2$ COOH).
Part 2 (Alcohol side): -CH $_2$ - CH $_3$ • This part contains 2 carbon atoms. The corresponding alcohol is Ethanol (CH $_3$ CH $_2$ OH).
3. Explain the concept (Esterification)
Esterification is a condensation reaction where an acid and an alcohol react to form an ester and water:
$\text{Acid} + \text{Alcohol} \rightleftharpoons \text{Ester} + \text{Water}$
In this specific case:
$\text{Propanoic acid} + \text{Ethanol} \rightarrow \text{Ethyl propanoate} + \text{Water}$
4. Identify the reactants
Looking at our breakdown:
The acid part is Propanoic acid.
The alcohol part is Ethanol.
Comparing this to your options, we see that Option D matches our findings perfectly.
5. Exam Tip
When naming or identifying esters, remember the "Alcohol-Acid" rule:
The first part of the name (the alkyl group) comes from the alcohol (e.g., Ethyl from Ethanol).
The second part of the name (the alkanoate) comes from the acid (e.g., Propanoate from Propanoic acid).
Always look for the "COO" group; the carbon attached to the two oxygens is the "head" of the acid part!
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39.
CH $_3$ C ≡ CHCH(CH $_3$ ) $_2$ The IUPAC nomenclature of the compound above is
1. A.3-ethyl but-2-yne
2. B.4-methylpent-2-yne
3. C.2-ethyl but-2-yne
4. D.2-methyl pent-2-yne
Show answer
Answer: B. 4-methylpent-2-yne
AI explanation
To determine the IUPAC name of the compound CH $_3$ C ≡ CCH(CH $_3$ ) $_2$ , we must follow the standard IUPAC rules for naming alkynes.
1. Analyze the Structure
First, let's expand the condensed formula to see the carbon chain clearly:
The structure is: CH $_3$ — C ≡ C — CH(CH $_3$ ) $_2$
The group CH(CH $_3$ ) $_2$ is an isopropyl group attached to the main chain.
2. Identify the Longest Carbon Chain
We need to find the longest continuous carbon chain that contains the triple bond.
Let's count the carbons: • CH $_3$ (1) — C (2) ≡ C (3) — CH (4) — CH $_3$ (5) • The branch (the other CH $_3$ group) is attached to carbon 4.
The longest chain has 5 carbons, which corresponds to the parent alkane pentane. Since there is a triple bond, it becomes pentyne.
3. Number the Chain
Numbering must start from the end that gives the triple bond the lowest possible locant (position number):
If we number from left to right: The triple bond starts at carbon 2.
If we number from right to left: The triple bond starts at carbon 3.
Therefore, we number from left to right. The parent name is pent-2-yne.
4. Identify and Locate Substituents
With the chain numbered from left to right, the methyl group (–CH $_3$ ) is attached to carbon 4.
Putting it all together: 4-methylpent-2-yne.
5. Conclusion
Parent chain: Pent-2-yne
Substituent: 4-methyl
Full Name: 4-methylpent-2-yne
Comparing this to your options: A. 3-ethyl but-2-yne (Incorrect) B. 4-methylpent-2-yne (Correct) C. 2-ethyl but-2-yne (Incorrect) D. 2-methyl pent-2-yne (Incorrect)
Exam Tip: When naming hydrocarbons with multiple bonds, always prioritize the position of the functional group (the triple bond) over the position of the alkyl branches. Number the chain so the triple bond gets the smallest number possible, then name the branches alphabetically if there are multiple types.
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40.
Nitrogen, a component of air is used for
1. A.production of cooling agent
2. B.production of margarine
3. C.production of HNO $_3$
4. D.manufacturing of oil
Show answer
Answer: C. production of HNO $_3$
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41.
The disintegration of radioactive phosphorus to silicon follows the first order kinetics with rate constant k $_1$ = 3.85 x 10 $^{-2}$ . Determine the half life of phosphorus.
1. A.8 s
2. B.180 s
3. C.80 s
4. D.100 s
Show answer
Answer: A. 8 s
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42.
The compound in which the oxidation state of nitrogen is + 3 is
1. A.N $_2$ O $_5$
2. B.N $_2$ O $_3$
3. C.N $_2$ O $_4$
4. D.N $_2$ O
Show answer
Answer: B. N $_2$ O $_3$
AI explanation
To find the oxidation state of nitrogen in these compounds, we use the rule that the sum of oxidation states in a neutral compound is zero, and the oxidation state of oxygen is typically -2.
Step-by-Step Analysis
Let the oxidation state of nitrogen be $x$ .
A. N $_2$ O $_5$ $2(x) + 5(-2) = 0$ $2x - 10 = 0$ $2x = 10$ $x = +5$
B. N $_2$ O $_3$ $2(x) + 3(-2) = 0$ $2x - 6 = 0$ $2x = 6$ $x = +3$
C. N $_2$ O $_4$ $2(x) + 4(-2) = 0$ $2x - 8 = 0$ $2x = 8$ $x = +4$
D. N $_2$ O $2(x) + 1(-2) = 0$ $2x - 2 = 0$ $2x = 2$ $x = +1$
Conclusion
Comparing our results, the compound where nitrogen has an oxidation state of +3 is N $_2$ O $_3$ .
Exam Tip
When calculating oxidation states for oxides of nitrogen, remember that nitrogen can exhibit multiple oxidation states (+1 to +5). A quick trick is to look at the subscript of the oxygen: in $N_2O_y$ , the oxidation state of nitrogen is simply $+y$ . For example, in $N_2O_3$ , $y=3$ , so the state is $+3$ . This works perfectly for these specific nitrogen oxides!
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43.
If 5g of Iron filling was reacted with excess dilute H $_2$ SO $_4$ to evolve hydrogen gas, which came to completion after 10 min, calculate the rate of reaction in g/hr.
1. A.5
2. B.30
3. C.20
4. D.10
Show answer
Answer: B. 30
AI explanation
1. Analyze the question
We are given the mass of a reactant (Iron) consumed in a chemical reaction and the time taken for the reaction to complete. We need to calculate the rate of reaction specifically in the unit of grams per hour (g/hr).
2. Explain the concept
The rate of reaction is defined as the change in the amount of a reactant or product per unit time. The formula is:
$\text{Rate} = \frac{\text{Mass of reactant consumed}}{\text{Time taken}}$
To get the answer in g/hr, we must ensure the mass is in grams (which it is: 5g) and the time is converted from minutes to hours.
3. Show the solution
Step 1: Identify the given values. • Mass of Iron = 5g • Time = 10 minutes
Step 2: Convert time from minutes to hours. Since there are 60 minutes in 1 hour: $ $\text{Time in hours} = \frac{10 \text{ minutes}}{60 \text{ minutes/hour}} = \frac{1}{6} \text{ hours}$ $
Step 3: Calculate the rate. $ $\text{Rate} = \frac{\text{Mass}}{\text{Time}}$ $ $ $\text{Rate} = \frac{5 \text{ g}}{1/6 \text{ hr}}$ $ $ $\text{Rate} = 5 \times 6 = 30 \text{ g/hr}$ $
4. Identify the answer
Comparing our result to the given options: A. 5 B. 30 C. 20 D. 10
The calculated value matches option B.
5. Tip
Always check the units requested in the question! JAMB examiners often provide time in minutes or seconds but ask for the rate in hours to test if you remember to perform the unit conversion. Don't let a simple unit oversight cost you marks.
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44.
PCl $_5$ $(_g$ ) → PCl $_3$ $(_s$ ) + Cl $_2$ $(_g$ ) In the equation above, the reaction will be spontaneous if
1. A.ΔH = constant
2. B.ΔH = + ve
3. C.ΔH = zero
4. D.ΔH = - ve
Show answer
Answer: D. ΔH = - ve
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45.
A molecule with a polar covalent bond between atoms exhibits
1. A.unequal sharing of electron pair
2. B.equal sharing of electron pair
3. C.equal sharing of electron lone pair
4. D.unequal sharing of electron lone pair
Show answer
Answer: A. unequal sharing of electron pair
AI explanation
Analysis
The question asks for the defining characteristic of a polar covalent bond. To answer this, we must understand how electrons behave when two atoms with different electronegativities form a bond.
Concept Explanation
Covalent Bond: This is a chemical bond formed by the sharing of electron pairs between atoms.
Non-polar Covalent Bond: This occurs when the atoms involved have similar electronegativities (e.g., $H_2$ or $Cl_2$ ). The electrons are shared equally because neither atom has a stronger "pull" on the electrons.
Polar Covalent Bond: This occurs when there is a significant difference in the electronegativity of the two atoms (e.g., $HCl$ or $H_2O$ ). The more electronegative atom attracts the shared electron pair more strongly toward itself. Because the electrons spend more time near the more electronegative atom, the sharing is unequal, resulting in a partial negative charge ( $\delta-$ ) on one atom and a partial positive charge ( $\delta+$ ) on the other.
Step-by-Step Solution
Identify the nature of the bond: A polar covalent bond is characterized by a dipole moment created by the uneven distribution of electron density.
Evaluate the options: • Option A: "Unequal sharing of electron pair" accurately describes the shift of electron density toward the more electronegative atom. • Option B: "Equal sharing" describes a non-polar covalent bond. • Options C & D: These refer to "lone pairs," which are non-bonding electrons. While lone pairs affect molecular geometry, the definition of a polar covalent bond specifically concerns the shared bonding pair.
Conclusion
The unequal sharing of the bonding electron pair is the fundamental reason why a covalent bond becomes polar.
Exam Tip: Remember the mnemonic "P-U-E": Polar = Unequal sharing of Electrons. If you see a bond between two different non-metals (like $C$ and $O$ ), it is almost always polar because they have different electronegativity values!
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46.
The source of carbon(II)oxide that acts as air pollutant is
1. A.photochemical smog
2. B.respiration
3. C.combustion of fuel
4. D.decomposition of sewage
Show answer
Answer: C. combustion of fuel
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47.
The IUPAC nomenclature of the compound above is
1. A.propyl ethanoate
2. B.ethyl ethanoate
3. C.ethyl propanoate
4. D.propyl methanoate
Show answer
Answer: A. propyl ethanoate
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48.
In the laboratory preparation of oxygen using potassium trioxochlorate(V), tetraoxomanganate (VII), usually not recommended in modern laboratory because
1. A.it forms explosive mixture with carbonaceous materials
2. B.it is explosive.
3. C.it reacts with the catalyst
4. D.it is hazardous
Show answer
Answer: A. it forms explosive mixture with carbonaceous materials
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49.
The nitrogenous compound in dead materials in the soil is converted to
1. A.ammonia
2. B.trioxonitrate(V)
3. C.ammonium salts
4. D.dioxonitrate(III)
Show answer
Answer: A. ammonia
AI explanation
1. Analyze the Question
The question asks about the biological and chemical process that occurs when organic matter (dead plants and animals) containing nitrogenous compounds (like proteins) decomposes in the soil. Specifically, it asks for the primary product formed during the initial stage of this decomposition.
2. Explain the Concept
This process is part of the Nitrogen Cycle, specifically the stage known as Ammonification.
When plants and animals die, their bodies contain complex nitrogenous compounds such as proteins and amino acids.
Decomposers, such as bacteria and fungi in the soil, break down these complex organic nitrogen compounds.
The first major inorganic nitrogen product released into the soil during this microbial breakdown is ammonia ( $NH_3$ ).
In the presence of soil moisture, this ammonia quickly reacts to form ammonium ions ( $NH_4^+$ ), but the primary conversion product from the organic material is ammonia.
3. Show the Solution
Step 1: Identify the source material. Dead materials contain organic nitrogen (proteins).
Step 2: Identify the process. Microorganisms (decomposers) act on these proteins.
Step 3: Determine the product. The enzymatic breakdown of proteins by soil bacteria releases nitrogen in the form of ammonia ( $NH_3$ ).
Step 4: Evaluate the options. While ammonia can later be converted into ammonium salts or nitrates (through nitrification), the immediate conversion product from the nitrogenous compound in dead matter is ammonia.
4. Identify the Answer
The correct option is A.
5. Exam Tip
Remember the sequence of the Nitrogen Cycle: Nitrogen Fixation → Ammonification → Nitrification → Denitrification.
Ammonification = Dead matter to Ammonia.
Nitrification = Ammonia to Nitrites ( $NO_2^-$ ) then to Nitrates ( $NO_3^-$ ).
If you see "dead materials," always look for ammonia as the starting point of the cycle in the soil.
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50.
The fractions of crude oil are best separated by
1. A.steam distillation
2. B.fractional crystallization
3. C.precipitation
4. D.fractional distillation
Show answer
Answer: D. fractional distillation
AI explanation
Analysis
The question asks for the most effective method used to separate the various components (fractions) found in crude oil.
Concept Explanation
Crude oil is a complex mixture of hydrocarbons with different molecular sizes and chain lengths. Because these hydrocarbons have different molecular masses, they possess different boiling points.
To separate these components, we use a process that relies on these varying boiling points. When crude oil is heated in a fractionating column, the components vaporize and rise. As the vapors rise, they cool and condense at different levels of the column based on their specific boiling points. This process is known as fractional distillation.
Step-by-Step Solution
Understand the mixture: Crude oil is a liquid mixture of many different substances (petrol, kerosene, diesel, bitumen, etc.).
Evaluate the options: • Steam distillation: Used for substances that decompose at their boiling points or are immiscible with water (e.g., essential oils). • Fractional crystallization: Used to separate solids based on differences in solubility at different temperatures. • Precipitation: Used to form a solid from a solution, usually through a chemical reaction. • Fractional distillation: Specifically designed to separate liquids that are miscible and have boiling points that are relatively close to each other.
Conclusion: Since crude oil is a mixture of liquids with different boiling points, fractional distillation is the industrial standard for separating it into useful fractions.
Exam Tip
Whenever you see a question about separating crude oil, always look for the word "fractional" and "distillation". Remember that the column used in the refinery is called a "fractionating column."
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51.
The use of CFCs as a blowing agent has application in
1. A.foam industry
2. B.tyre industry
3. C.refrigerant
4. D.propellant
Show answer
Answer: A. foam industry
AI explanation
1. Analyze the question
The question asks for the specific industrial application of Chlorofluorocarbons (CFCs) when used as a "blowing agent."
2. Explain the concept
What are CFCs? CFCs are organic compounds containing carbon, chlorine, and fluorine. They were historically used in many applications due to their stability and low toxicity.
What is a "blowing agent"? A blowing agent is a substance used to produce a cellular structure (like bubbles or pores) in a plastic or rubber matrix. This process is known as "foaming."
The Process: When a blowing agent is added to a polymer (like polyurethane), it vaporizes or releases gas during the manufacturing process. This gas creates tiny bubbles within the material, turning it into a lightweight, insulating foam.
3. Show the solution
Evaluating the options: • A. Foam industry: CFCs were widely used to create polyurethane foams (like those in mattresses, upholstery, and insulation) because they could easily turn into gas and expand the plastic into a foam structure. This is the definition of a blowing agent. • B. Tyre industry: Tyres are made through vulcanization and molding; they do not require blowing agents to create a foam structure. • C. Refrigerant: While CFCs were famously used as refrigerants in fridges and air conditioners, this is a different application. Being a refrigerant relies on the gas's ability to absorb and release heat during phase changes, not its ability to create a foam structure. • D. Propellant: CFCs were used as propellants in aerosol cans (like hairspray), but this is also a different application based on the pressure of the gas to push liquid out of a container.
Since the question specifically asks for the application of CFCs as a blowing agent, the correct context is the production of foam.
4. Identify the answer
The correct option is A.
5. Exam Tip
When you see questions about CFCs, remember they have three major historical uses: Refrigerants (cooling), Propellants (aerosols), and Blowing agents (foam). Always read the question carefully to see which specific function is being described!
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52.
The catalytic hydrogenation of benzene produces
1. A.hexane
2. B.margarine
3. C.hexene
4. D.cyclohexane
Show answer
Answer: D. cyclohexane
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53.
The gas that ammoniacal solution of CuCl $_2$ is used to absorb from producer or water gas is
1. A.H $_2$
2. B.CO
3. C.N $_2$
4. D.CO $_2$
Show answer
Answer: B. CO
AI explanation
Analysis of the Question
The question asks for the specific gas component in producer gas or water gas that is selectively absorbed by an ammoniacal solution of copper(I) chloride ( $CuCl_2$ or more accurately $CuCl$ ).
Concept Explanation
Producer gas and water gas are both mixtures of gases used primarily as industrial fuels.
Producer gas: A mixture of Carbon(II) oxide ( $CO$ ) and Nitrogen ( $N_2$ ).
Water gas: A mixture of Carbon(II) oxide ( $CO$ ) and Hydrogen ( $H_2$ ).
In industrial chemistry, it is often necessary to purify these gases. Ammoniacal copper(I) chloride is a specific chemical reagent used in gas analysis and purification. It reacts with Carbon(II) oxide ( $CO$ ) to form a coordination complex, effectively removing it from the gas mixture. The reaction is:
$CuCl + CO + 2H_2O \rightarrow Cu(CO)Cl \cdot 2H_2O$
Step-by-Step Solution
Identify the components: We know producer gas contains $CO$ and $N_2$ , while water gas contains $CO$ and $H_2$ .
Evaluate the reagent: Ammoniacal copper(I) chloride is a well-known absorbent for Carbon(II) oxide ( $CO$ ).
Eliminate other options: • $H_2$ and $N_2$ do not react with ammoniacal copper(I) chloride. • $CO_2$ is typically absorbed by alkaline solutions like Potassium Hydroxide ( $KOH$ ), not copper(I) chloride.
Conclusion: Since $CO$ is the common component in both gases that reacts with this specific reagent, it is the gas being absorbed.
Identification of the Answer
The correct option is B.
Exam Tip
Remember the "Absorbent Rule" for gas analysis:
$CO_2$ is absorbed by $KOH$ (Potassium hydroxide).
$O_2$ is absorbed by Alkaline pyrogallol.
$CO$ is absorbed by Ammoniacal copper(I) chloride.
$H_2$ is often removed by passing the gas over heated copper(II) oxide ( $CuO$ ).
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54.
After breathing in a test tube that contains acidified K $_2$ Cr $_2$ O $_7$ , a man noticed a change in the colour of the acidified K $_2$ Cr $_2$ O $_7$ from orange to green. This suggests the presence of
1. A.alkanoic acid
2. B.alkanal
3. C.alkanol
4. D.alkanone
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 observing a color change in acidified potassium dichromate(VI) ( $K_2Cr_2O_7$ ) from orange to green.
Concept Explanation
The Reagent: Acidified potassium dichromate ( $K_2Cr_2O_7$ ) 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 Functional Group: This specific color change occurs when the reagent is reduced by a substance that can be oxidized. Alkanols (primary and secondary alcohols) are easily oxidized by acidified $K_2Cr_2O_7$ . • Primary alcohols are oxidized to alkanals (aldehydes) and then to alkanoic acids. • Secondary alcohols are oxidized to alkanones (ketones). • Alkanals can also be oxidized, but in the context of breathalyzer tests (which this question describes), the test is specifically designed to detect the presence of ethanol (an alkanol) in the breath.
Step-by-Step Solution
Identify the reactant: The man is breathing into the test tube. If he has consumed alcohol, his breath will contain ethanol vapor ( $C_2H_5OH$ ).
Analyze the chemical change: Ethanol is a primary alkanol. When it comes into contact with the acidified $K_2Cr_2O_7$ , the ethanol is oxidized to ethanal and then to ethanoic acid.
Observe the color change: During this redox reaction, the chromium in the dichromate is reduced from an oxidation state of +6 (orange) to +3 (green).
Evaluate the options: • Alkanoic acids and alkanones are generally resistant to further oxidation by this reagent under these conditions. • Alkanols are the primary targets for this oxidation test.
Conclusion
The change from orange to green confirms the presence of an alkanol (specifically ethanol in the context of a breath test).
Exam Tip
Remember the mnemonic "Alcohol is Green" for this test. If you see "acidified potassium dichromate" and "orange to green," immediately look for an alcohol (alkanol) in your options. This is a very common question in JAMB Chemistry!
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2025 JAMB Chemistry Past Questions and Answers

Questions 1–54

1. Analyze the question

2. Explain the concept

3. Show the solution

4. Identify the answer

5. Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Correct Answer

Exam Tip

Analysis of the Question

Concept Explanation

Step-by-Step Evaluation

Conclusion

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Exam Tip

Analysis

Concept

Step-by-Step Solution

Conclusion

Exam Tip

Analysis of the Question

Concept Explanation

Step-by-Step Evaluation

Conclusion

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Identification of the Answer

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Exam Tip

Analysis

Concept: Oxidation and Reduction (Redox)

Step-by-Step Solution

Conclusion

Exam Tip

Analysis

Concept

Solution

Conclusion

Exam Tip

Analysis

Concept

Step-by-Step Solution

Conclusion

1. Analyze the Question

2. Explain the Concept

3. Show the Solution

4. Identify the Answer

5. Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Exam Tip

Analysis