An inflated balloon shrinks when placed in a freezer because the

C. gas motion reduces and exert less pressure

A. volume of gas increases and expands

B. number of gas molecules decreases

C. gas motion reduces and exert less pressure

D. gas molecules completely stop moving

Find the hydrogen ion, H$^+$ concentration and hydroxide ion, OH$^-$ concentration in 0.06 moldm$^{-3}$ solution of H$_2$SO$_4$.

C. 1.2 x 10$^{-1}$ ; 8.3 x 10$^{-14}$

A. 1.2 x 10$^{-2}$ ; 8.3 x 10$^{-13}$

B. 1.2 x 10$^{-1}$ ; 8.3 x 10$^{-13}$

C. 1.2 x 10$^{-1}$ ; 8.3 x 10$^{-14}$

D. 1.2 x 10$^{-2}$ ; 8.3 x 10$^{-14}$

An importance of solubility is that it

A. determines the amount of solute that can dissolve in a given amount of solvent

B. lowers the vapour pressure of the liquid

C. enables us to know the chemical reactions of the gas with the liquid

D. raises the boiling point of liquid

Alkanoic acids have higher boiling points than alkanols because

D. alkanoic acids form stronger hydrogen bonds

A. alkanoic acids are more volatile than alkanols

B. alkanols are more polar than alkanoic acids

C. alkanoic acids are larger molecules

D. alkanoic acids form stronger hydrogen bonds

An example of a salt that can dissolve in water to form a solution with a pH of 7.

C. ammonium tetraoxosulphate(VI)

An example of a physical change is

C. dissolving sodium chloride in water

A. addition of water to calcium oxide

C. dissolving sodium chloride in water

Electron configuration of Chlorine atom is

A. 1S$^2$2S$^2$2P$^6$3S$^2$3P$^5$

B. 1S$^2$2S$^2$2P$^6$3S$^2$3P$^4$

C. 1S$^2$2S$^2$2P$^6$3S$^2$3P$^6$

D. 1S$^2$2S$^2$2P$^5$3S$^2$3P$^6$

Acid radicals are present in

C. CO$_3$$^{2-}$, SO$_4$$^{2-}$, NO$_3$$^-$

A. CO$_3$$^{2-}$, PO$_4$$^{3-}$, Mg$^{2+}$

B. CO$_3$$^{2-}$, SO$_4$$^{2-}$, Na$^+$

C. CO$_3$$^{2-}$, SO$_4$$^{2-}$, NO$_3$$^-$

D. CO$_3$$^{2-}$, SO$_4$$^{2-}$, K$^+$

The correct arrangement of gases in the order of increasing rate of diffusion is [H = 1, C = 12, N = 14, O = 16, S = 32]

B. SO$_2$, O$_2$, NH$_3$, H$_2$

C. NH$_3$, NO$_2$,N$_2$, CO$_2$

D. CO$_2$, NH$_3$, O$_2$, SO$_2$

When a sample of air is passed through alkaline pyrogalol, potash and finally through U-tube containing fused calcium chloride, the components of air left unabsorbed are

A. nitrogen and carbon(IV)oxide

B. noble gases and carbon(IV)oxide

D. noble gases, nitrogen and carbon(IV) oxide

Which of the following has the highest boiling point?

C. CH$_3$CH$_2$CH$_2$CH$_2$CH$_2$CH$_3$

Bronze is used in making

A. musical instruments and ornaments

The process that illustrates reformation of petroleum product is

C. hexane to carbon(IV)oxide and water vapour

JAMB 2025 · Part 2 of 2

2025 JAMB Chemistry Past Questions — Part 2

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 2 of 2: questions 55 to 107. 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 55–107

55.
The sublimation of solid to gas involves
1. A.energy absorption
2. B.vapourization
3. C.melting
4. D.osmotic diffusion
Show answer
Answer: A. energy absorption
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56.
The time required to deposit 4.5g of copper from CuSO $_4$ solution by passing a current of 2.5 Amperes is (Cu = 64g ; 1F = 96500C/mol)
1. A.2714 sec
2. B.2527 sec
3. C.5428 sec
4. D.6785 sec
Show answer
Answer: C. 5428 sec
AI explanation
Analysis
To solve this problem, we use Faraday’s First Law of Electrolysis, which relates the mass of a substance deposited to the quantity of electricity (charge) passed through the electrolyte.
Concept
The formula is:
$m = \frac{M \times I \times t}{n \times F}$
Where:
$m$ = mass of the substance deposited (4.5g)
$M$ = Molar mass of the substance (64g/mol)
$I$ = Current in Amperes (2.5A)
$t$ = Time in seconds (what we need to find)
$n$ = Valency or number of electrons transferred (For Copper in $CuSO_4$ , $Cu^{2+} + 2e^- \rightarrow Cu$ , so $n = 2$ )
$F$ = Faraday’s constant (96500 C/mol)
Step-by-Step Solution
Step 1: Rearrange the formula to solve for $t$ :
$t = \frac{m \times n \times F}{M \times I}$
Step 2: Substitute the given values:
$m = 4.5$
$n = 2$
$F = 96500$
$M = 64$
$I = 2.5$
$t = \frac{4.5 \times 2 \times 96500}{64 \times 2.5}$
Step 3: Perform the calculation:
Numerator: $4.5 \times 2 = 9$
$9 \times 96500 = 868500$
Denominator: $64 \times 2.5 = 160$
$t = \frac{868500}{160}$
$t = 5428.125 \text{ seconds}$
Rounding to the nearest whole number, we get 5428 seconds.
Conclusion
Comparing our result with the options provided, it matches option C.
Exam Tip
Always check the valency ( $n$ ) of the metal in the salt. For Copper ( $Cu^{2+}$ ), $n=2$ ; for Silver ( $Ag^+$ ), $n=1$ ; for Aluminum ( $Al^{3+}$ ), $n=3$ . A common mistake is forgetting to multiply by the valency, which would lead you to the wrong answer!
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57.
In the electrolysis of brine using neutral electrode, which ion is discharged at the anode?
1. A.2H $^+$
2. B.Na $^+$
3. C.Cl $^-$
4. D.OH $^-$
Show answer
Answer: C. Cl $^-$
AI explanation
Analysis of the Question
The question asks for the ion discharged at the anode during the electrolysis of brine (concentrated sodium chloride solution) using inert (neutral) electrodes.
Concept Explanation
In the electrolysis of brine, we have the following ions present in the solution:
From NaCl: $Na^+$ and $Cl^-$
From $H_2O$ : $H^+$ and $OH^-$
The Anode (Positive Electrode): The anions ( $Cl^-$ and $OH^-$ ) are attracted to the anode. According to the electrochemical series, the discharge of ions depends on their concentration and position in the series.
In a concentrated solution of sodium chloride (brine), the concentration of $Cl^-$ ions is significantly higher than that of $OH^-$ ions.
Because of this high concentration, the $Cl^-$ ions are preferentially discharged at the anode instead of the $OH^-$ ions.
The Reaction at the Anode: $2Cl^-_{(aq)} \rightarrow Cl_{2(g)} + 2e^-$
Step-by-Step Solution
Identify the ions at the anode: The anions present are $Cl^-$ and $OH^-$ .
Apply the rule of selective discharge: Although $OH^-$ is lower in the electrochemical series and would normally be discharged in dilute solutions, the high concentration of $Cl^-$ in brine forces the chloride ions to be discharged.
Determine the product: The $Cl^-$ ions lose electrons (oxidation) to form chlorine gas ( $Cl_2$ ).
Conclusion: The ion that is discharged at the anode is the chloride ion ( $Cl^-$ ).
Exam Tip
Remember the mnemonic "Concentration Conquers" for the electrolysis of brine. Even though $OH^-$ is theoretically easier to discharge, the sheer number of $Cl^-$ ions in concentrated brine makes them the winners at the anode!
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58.
Gold does not require extraction because it is
1. A.a precious mineral
2. B.inert
3. C.free in nature
4. D.neutral
Show answer
Answer: C. free in nature
AI explanation
Analysis of the Question
The question asks why gold does not require the complex chemical extraction processes (like reduction or electrolysis) that are typically used for other metals. We need to look at the chemical reactivity of gold compared to other elements in the reactivity series.
Concept Explanation
In chemistry, metals are categorized by their reactivity. Highly reactive metals (like Sodium, Potassium, or Aluminum) are never found in their pure, elemental state in nature because they react easily with oxygen, water, and other elements to form compounds (ores). To get these metals, we must use energy-intensive processes like electrolysis or chemical reduction.
Gold, however, is at the very bottom of the reactivity series. It is a "noble metal," meaning it is extremely unreactive. Because it does not readily react with oxygen or moisture in the environment, it remains in its elemental, metallic form in the earth's crust.
Step-by-Step Solution
Evaluate Option A: While gold is a precious mineral, its value is a result of its rarity and properties, not the reason it doesn't require extraction. Many precious metals still require extraction from ores.
Evaluate Option B: Gold is chemically "inert" (unreactive), which is a property that allows it to exist in its pure state, but the direct reason we don't "extract" it in the industrial sense is that it is already found as a pure metal.
Evaluate Option C: Because gold is so unreactive, it is found in the earth's crust in its native or free state. This means it does not exist as a chemical compound (like an oxide or sulfide) that needs to be broken down. Therefore, it does not require the chemical extraction processes used for reactive metals.
Evaluate Option D: Gold is a metal, and metals are not "neutral" in the context of chemical reactivity; they are simply unreactive.
Conclusion
Gold is found in nature as a pure element (native state) because it does not react with its surroundings. Therefore, it does not need to be chemically extracted from an ore.
Exam Tip: Remember the reactivity series! Metals at the top (Potassium, Sodium, Calcium) are always found as compounds. Metals at the bottom (Gold, Silver, Platinum) are found as "native" or "free" elements.
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59.
An inflated balloon shrinks when placed in a freezer because the
1. A.volume of gas increases and expands
2. B.number of gas molecules decreases
3. C.gas motion reduces and exert less pressure
4. D.gas molecules completely stop moving
Show answer
Answer: C. gas motion reduces and exert less pressure
AI explanation
Analysis
This question tests your understanding of the Kinetic Molecular Theory of Gases and Charles’s Law, which describes the relationship between the temperature and volume of a gas.
Concept Explanation
The Kinetic Molecular Theory states that gas particles are in constant, random motion. The temperature of a gas is a measure of the average kinetic energy (the speed) of these particles.
At higher temperatures: Gas particles move faster and collide with the walls of the container more frequently and with greater force, exerting higher pressure.
At lower temperatures (like in a freezer): The kinetic energy of the gas particles decreases, meaning they move more slowly. Because they are moving slower, they collide with the walls of the balloon less frequently and with less force.
According to Charles's Law, for a fixed amount of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature ( $V \propto T$ ). When you lower the temperature, the volume must decrease to maintain the balance between the internal pressure of the gas and the external atmospheric pressure.
Step-by-Step Solution
Identify the change: Placing the balloon in a freezer decreases the temperature of the gas trapped inside.
Analyze the effect on particles: As temperature drops, the kinetic energy of the gas molecules decreases.
Relate to pressure: Slower-moving molecules strike the inner surface of the balloon with less force and frequency, resulting in a decrease in the pressure exerted by the gas on the balloon walls.
Result: Because the internal pressure drops, the constant external atmospheric pressure pushes the balloon inward, causing it to shrink until the internal and external pressures equalize again.
Evaluating the Options
A: Incorrect. Volume decreases, not increases, when temperature drops.
B: Incorrect. The number of gas molecules remains the same; they are just moving differently.
C: Correct. The reduction in gas motion leads to fewer and weaker collisions, reducing the pressure and allowing the balloon to shrink.
D: Incorrect. Gas molecules only stop moving completely at Absolute Zero (0 Kelvin), which is not achieved in a standard freezer.
Exam Tip
Whenever you see a question about gas laws, remember the "Collision Rule": Temperature is speed. High temperature = fast particles = more collisions = higher pressure/volume. Low temperature = slow particles = fewer collisions = lower pressure/volume.
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60.
A table in which metals are arranged in series according to their comparative tendencies to give up their valence electrons is
1. A.periodic table
2. B.electrochemical series
3. C.table of elements
4. D.periodicals
Show answer
Answer: B. electrochemical series
AI explanation
Analysis of the Question
The question asks for the name of the specific arrangement where metals are listed based on their ability to lose electrons (oxidation) or their reactivity.
Concept Explanation
In chemistry, the Electrochemical Series (also known as the Activity Series) is a list of elements arranged in order of their standard electrode potentials.
How it works: Metals at the top of the series have a high tendency to lose electrons (they are strong reducing agents and are highly reactive). Metals at the bottom have a lower tendency to lose electrons.
Why it matters: This series helps us predict whether a metal will displace another from a solution of its salt or react with water/acids. For example, Zinc is higher than Copper in the series, which is why Zinc can displace Copper from a Copper(II) tetraoxosulphate(VI) solution.
Step-by-Step Solution
Evaluate Option A (Periodic Table): The periodic table arranges elements based on their atomic numbers and electronic configurations, not specifically by their tendency to lose electrons in electrochemical reactions.
Evaluate Option B (Electrochemical Series): This is the standard scientific term for the arrangement of metals based on their comparative tendencies to give up valence electrons (their oxidation potential).
Evaluate Option C (Table of Elements): This is a generic term and does not specifically refer to the electrochemical arrangement.
Evaluate Option D (Periodicals): This refers to magazines or journals published at regular intervals; it has no relevance to chemistry.
Comparing these, the Electrochemical Series is the correct scientific term for the description provided in the question.
Exam Tip
Think of the Electrochemical Series as a "Power Ranking" for metals. The higher the metal is on the list, the more "eager" it is to get rid of its electrons to become an ion. Remember the mnemonic "Please Stop Calling Me A Careless Zebra Instead Try Learning How Copper Saves Gold" (Potassium, Sodium, Calcium, Magnesium, Aluminium, Carbon, Zinc, Iron, Tin, Lead, Hydrogen, Copper, Silver, Gold).
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61.
Find the hydrogen ion, H $^+$ concentration and hydroxide ion, OH $^-$ concentration in 0.06 moldm $^{-3}$ solution of H $_2$ SO $_4$ .
1. A.1.2 x 10 $^{-2}$ ; 8.3 x 10 $^{-13}$
2. B.1.2 x 10 $^{-1}$ ; 8.3 x 10 $^{-13}$
3. C.1.2 x 10 $^{-1}$ ; 8.3 x 10 $^{-14}$
4. D.1.2 x 10 $^{-2}$ ; 8.3 x 10 $^{-14}$
Show answer
Answer: C. 1.2 x 10 $^{-1}$ ; 8.3 x 10 $^{-14}$
AI explanation
Analysis
To solve this, we need to understand the dissociation of sulfuric acid ( $H_2SO_4$ ) in water and the relationship between hydrogen ions ( $H^+$ ) and hydroxide ions ( $OH^-$ ) in an aqueous solution.
Concept
Dissociation: Sulfuric acid is a strong diprotic acid, meaning it dissociates completely in water according to the equation: $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$ This means 1 mole of $H_2SO_4$ produces 2 moles of $H^+$ .
Ionic Product of Water ( $K_w$ ): At 25°C, the product of the concentrations of $H^+$ and $OH^-$ is constant: $K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \text{ mol}^2\text{dm}^{-6}$
Step-by-Step Solution
Step 1: Calculate the concentration of $H^+$ ions. Since the concentration of $H_2SO_4$ is $0.06 \text{ mol dm}^{-3}$ , and each molecule releases two $H^+$ ions: $[H^+] = 2 \times [\text{Concentration of } H_2SO_4]$ $[H^+] = 2 \times 0.06 = 0.12 \text{ mol dm}^{-3}$ In scientific notation, $0.12 = 1.2 \times 10^{-1} \text{ mol dm}^{-3}$ .
Step 2: Calculate the concentration of $OH^-$ ions. Using the $K_w$ formula: $[OH^-] = \frac{K_w}{[H^+]}$ $[OH^-] = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-1}}$ $[OH^-] = \frac{1}{1.2} \times 10^{-13}$ $[OH^-] \approx 0.833 \times 10^{-13}$ Adjusting to standard scientific notation: $[OH^-] = 8.33 \times 10^{-14} \text{ mol dm}^{-3}$
Conclusion
Comparing our results ( $[H^+] = 1.2 \times 10^{-1}$ and $[OH^-] = 8.3 \times 10^{-14}$ ) with the given options, we find that they match option C.
Exam Tip: Always remember that for strong acids, the concentration of $H^+$ is the molarity of the acid multiplied by its basicity (the number of replaceable hydrogen atoms). Don't forget to check if the acid is monoprotic (like $HCl$ ) or diprotic (like $H_2SO_4$ ) before calculating!
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62.
An importance of solubility is that it
1. A.determines the amount of solute that can dissolve in a given amount of solvent
2. B.lowers the vapour pressure of the liquid
3. C.enables us to know the chemical reactions of the gas with the liquid
4. D.raises the boiling point of liquid
Show answer
Answer: A. determines the amount of solute that can dissolve in a given amount of solvent
AI explanation
Analysis
The question asks for the fundamental definition or significance of solubility in chemistry. To answer this, we must look at what solubility actually measures and how it differs from colligative properties.
Concept Explanation
Solubility is defined as the maximum amount of a solute that can dissolve in a specific amount of solvent at a given temperature to form a saturated solution. It is a physical property that tells us the "limit" of dissolution for a particular substance.
Let's look at the other options:
Options B and D: Lowering vapour pressure and raising the boiling point are colligative properties. These properties depend on the number of solute particles in a solution, not the specific identity or the solubility limit of the solute.
Option C: Knowing the chemical reaction of a gas with a liquid is related to chemical kinetics or chemical equilibrium, not the physical limit of solubility.
Step-by-Step Solution
Define Solubility: Solubility is the concentration of a saturated solution. It quantifies how much solute can be accommodated by a solvent before no more can be dissolved.
Evaluate Option A: This option directly describes the definition of solubility—the capacity of a solvent to hold a specific amount of solute.
Evaluate Options B & D: These describe the effect of adding a solute to a solvent (colligative properties), but they do not define the concept of solubility itself.
Evaluate Option C: This refers to chemical reactivity, which is distinct from the physical process of dissolving.
Therefore, the most accurate description of the importance/definition of solubility is the determination of the maximum amount of solute that can be dissolved.
Conclusion
It is the only statement that correctly identifies the primary function of solubility as a measurement of solute capacity.
Exam Tip
When you see questions about "solubility," look for keywords like "maximum amount," "saturated," or "limit." If the question mentions "boiling point elevation" or "vapour pressure lowering," remember those are colligative properties, not solubility!
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63.
Freons pollution in the air are released from
1. A.fossil fuel
2. B.aerosol can
3. C.photosynthesis
4. D.organic decay
Show answer
Answer: B. aerosol can
AI explanation
Analysis
This question tests your knowledge of environmental chemistry, specifically the sources of atmospheric pollutants and their impact on the ozone layer. You are asked to identify the primary source of Freons (chlorofluorocarbons or CFCs) released into the atmosphere.
Concept Explanation
Freons are a group of chemical compounds known as chlorofluorocarbons (CFCs). Historically, they were widely used in industrial and consumer applications because they are non-toxic, non-flammable, and chemically stable.
However, when these compounds are released into the atmosphere, they rise to the stratosphere. Under the influence of ultraviolet (UV) radiation from the sun, they break down to release chlorine atoms, which act as catalysts to destroy the ozone layer.
The most common sources of these emissions include:
Aerosol propellants: Used in spray cans (like hairsprays, deodorants, and paints).
Refrigerants: Used in refrigerators and air conditioning systems.
Foam-blowing agents: Used in the production of plastic foams.
Step-by-Step Solution
Evaluate Option A (Fossil fuel): Burning fossil fuels primarily releases carbon dioxide ( $CO_2$ ), sulfur dioxide ( $SO_2$ ), and nitrogen oxides ( $NO_x$ ). It is not the primary source of CFCs.
Evaluate Option B (Aerosol can): Aerosol sprays historically used CFCs as propellants to push the product out of the can. This is a well-documented source of Freon pollution.
Evaluate Option C (Photosynthesis): This is a biological process where plants convert carbon dioxide and water into glucose and oxygen. It is a natural process and not a source of industrial pollutants like Freons.
Evaluate Option D (Organic decay): The decay of organic matter releases gases like methane ( $CH_4$ ) and carbon dioxide ( $CO_2$ ), not synthetic chlorofluorocarbons.
Comparing these, aerosol cans are the direct source mentioned in the context of Freon pollution.
Conclusion
While modern regulations have phased out CFCs in many countries, they remain the classic textbook answer for the source of this specific type of pollution.
Exam Tip: Whenever you see "Freons" or "CFCs" in a JAMB question, immediately associate them with ozone layer depletion and aerosol propellants/refrigerants. This association will help you answer similar questions quickly.
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64.
Alkanoic acids have higher boiling points than alkanols because
1. A.alkanoic acids are more volatile than alkanols
2. B.alkanols are more polar than alkanoic acids
3. C.alkanoic acids are larger molecules
4. D.alkanoic acids form stronger hydrogen bonds
Show answer
Answer: D. alkanoic acids form stronger hydrogen bonds
AI explanation
Analysis
The question asks for a comparison of the boiling points of alkanoic acids (carboxylic acids) and alkanols (alcohols) and the underlying reason for the difference.
Concept Explanation
Boiling point is a measure of the energy required to overcome intermolecular forces. Both alkanols and alkanoic acids exhibit hydrogen bonding, which is a strong intermolecular force.
Alkanols (R-OH): These molecules form hydrogen bonds between the hydrogen atom of one hydroxyl group and the oxygen atom of another.
Alkanoic Acids (R-COOH): These molecules contain a carboxyl group. They can form hydrogen bonds in a way that creates dimers (two molecules linked together). Because the carboxyl group contains both a carbonyl group (C=O) and a hydroxyl group (O-H), the hydrogen bonding in alkanoic acids is more extensive and stable than in simple alkanols.
Because the hydrogen bonding in alkanoic acids is more effective and results in the formation of stable dimers, more energy (heat) is required to break these bonds, leading to a higher boiling point compared to alkanols of similar molecular mass.
Step-by-Step Solution
Identify the intermolecular forces: Both classes of compounds use hydrogen bonding.
Compare the strength: Alkanoic acids form dimers due to the interaction between the C=O and O-H groups of adjacent molecules. This creates a more stable, "double" hydrogen-bonded structure.
Relate to boiling point: Stronger intermolecular forces require more thermal energy to break. Since the dimerized hydrogen bonds in alkanoic acids are stronger/more extensive than the simple hydrogen bonds in alkanols, the boiling point of the acid is higher.
Evaluate the options: • A is incorrect; alkanoic acids are less volatile (higher boiling point means lower volatility). • B is incorrect; both are polar, but the hydrogen bonding in acids is more effective. • C is incorrect; boiling point is determined by intermolecular forces, not just molecular size. • D is correct; the formation of stable dimers via hydrogen bonding makes the bonds effectively stronger.
Conclusion
Exam Tip: Whenever you see a question about boiling points in organic chemistry, always look for the strength of intermolecular forces in this order: Hydrogen bonding > Dipole-dipole interactions > Van der Waals forces (London dispersion forces). Since both have hydrogen bonds, look for which one allows for more extensive or stable bonding!
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65.
An atom of element with the configuration 1S $^2$ 2S $^2$ 2P $^6$ 3S $^2$ 3P $^5$ is likely to belong to
1. A.group 7
2. B.group 5
3. C.group 2
4. D.group 3
Show answer
Answer: A. group 7
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66.
In oxidation reactions, electrons are
1. A.hydrolysed
2. B.removed
3. C.added
4. D.hydrated
Show answer
Answer: B. removed
AI explanation
Analysis
The question asks for the definition of oxidation in terms of electron transfer. This is a fundamental concept in electrochemistry and redox reactions, which are common topics in the JAMB Chemistry syllabus.
Concept: OIL RIG
To understand oxidation and reduction, we use the mnemonic OIL RIG:
Oxidation Is Loss (of electrons)
Reduction Is Gain (of electrons)
In any chemical reaction where electrons are transferred, the substance that loses electrons is said to be oxidized, while the substance that gains electrons is said to be reduced.
Step-by-Step Solution
Define Oxidation: Oxidation is defined as the loss of electrons from an atom, molecule, or ion.
Evaluate the options: • A. Hydrolysed: This refers to a reaction involving water (splitting a compound using water). This is not related to electron transfer. • B. Removed: Since oxidation is the loss of electrons, the electrons are effectively being taken away or removed from the species. • C. Added: This describes reduction (the gain of electrons). • D. Hydrated: This refers to the addition of water molecules to a substance.
Since oxidation involves the loss of electrons, those electrons are removed from the substance undergoing oxidation.
Conclusion
Exam Tip: Always remember that oxidation and reduction must happen together. If one substance loses electrons (is oxidized), another must gain them (is reduced). You can never have one without the other!
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67.
An example of a salt that can dissolve in water to form a solution with a pH of 7.
1. A.sodium chloride
2. B.ammonium chloride
3. C.ammonium tetraoxosulphate(VI)
4. D.sodium trioxocarbonate(IV)
Show answer
Answer: A. sodium chloride
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68.
The constituents of permallory are Iron and
1. A.Nickel
2. B.Tin
3. C.Chromium
4. D.Copper
Show answer
Answer: A. Nickel
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69.
What is the molecular mass of an alkanoic acid, if 0.5 mole of the acid weighs 44g?
1. A.72
2. B.88
3. C.60
4. D.46
Show answer
Answer: B. 88
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70.
2SO $_2$ $_(s)$ + O $_2$ $_(s)$ ⇌ 2SO $_3$ ; ΔG° = - ve For the above reaction to be feasible
1. A.ΔH = + ve ,ΔS = + ve
2. B.ΔH = 0 ,ΔS = + ve
3. C.ΔH = + ve ,ΔS = - ve
4. D.ΔS = 0
Show answer
Answer: D. ΔS = 0
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71.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.
72.
Enzymatic conversion of glucose to ethanol is
1. A.polymerization
2. B.hydrogenation
3. C.fermentation
4. D.saponification
Show answer
Answer: C. fermentation
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73.
2X + 2HCl → 2XCl + H $_2$ In the equation above, X is
1. A.K
2. B.Mg
3. C.Ca
4. D.Ba
Show answer
Answer: A. K
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74.
The dusty and sand particles present in the air is an example of
1. A.a suspension
2. B.a dispersion
3. C.an emulsion
4. D.a saturated solution
Show answer
Answer: A. a suspension
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75.
The type of bond between copper(II) tetraamine and chlorine in [Cu(NH $_3$ ) $_4$ ]Cl $_2$
1. A.dative
2. B.van der waals
3. C.ionic
4. D.covalent
Show answer
Answer: C. ionic
AI explanation
Analysis of the Question
The question asks for the nature of the chemical bond existing between the complex cation, $[Cu(NH_3)_4]^{2+}$ , and the chloride anions, $Cl^-$ , in the compound $[Cu(NH_3)_4]Cl_2$ .
Concept Explanation
To identify the bond, we must look at the structure of the compound:
The Complex Ion: Inside the square brackets $[Cu(NH_3)_4]^{2+}$ , the ammonia molecules are bonded to the copper ion via dative (coordinate) covalent bonds.
The Compound as a Whole: The entire entity $[Cu(NH_3)_4]^{2+}$ acts as a positively charged complex ion (cation), while the two $Cl^-$ ions act as negatively charged counter-ions (anions).
The Bond Type: In chemistry, the electrostatic force of attraction between a positively charged ion and a negatively charged ion is defined as an ionic bond.
Even though the internal structure of the complex ion involves coordinate covalent bonds, the bond between the complex cation and the chloride anions is purely ionic.
Step-by-Step Solution
Identify the components: The formula $[Cu(NH_3)_4]Cl_2$ dissociates in water as: $[Cu(NH_3)_4]Cl_2 \rightarrow [Cu(NH_3)_4]^{2+} + 2Cl^-$
Analyze the interaction: We have a large, positively charged complex ion $[Cu(NH_3)_4]^{2+}$ and two negatively charged chloride ions $Cl^-$ .
Determine the bond: The force holding these oppositely charged species together in the crystal lattice is an electrostatic attraction. Electrostatic attraction between ions is the definition of an ionic bond.
Conclusion
While the ammonia ligands are attached to the copper via dative bonds, the bond connecting the complex unit to the chlorine atoms is ionic.
Exam Tip: Whenever you see a complex salt (a compound with square brackets), remember:
Bonds inside the brackets are usually dative/covalent.
Bonds between the bracketed part and the outside atoms are ionic.
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76.
Water drops are spherical in shape because of
1. A.polarity
2. B.viscocity
3. C.density
4. D.surface tension
Show answer
Answer: D. surface tension
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77.
What is the product obtained at the anode in the electrolysis of concentrated sodium chloride using graphite electrode?
1. A.Chlorine gas
2. B.oxygen gas
3. C.water vapour
4. D.hydrogen gas
Show answer
Answer: A. Chlorine gas
AI explanation
Analysis
The question asks for the product formed at the anode during the electrolysis of concentrated sodium chloride (brine) using graphite electrodes.
Concept Explanation
In the electrolysis of concentrated sodium chloride solution (brine), the electrolyte contains the following ions:
From NaCl: $Na^+$ and $Cl^-$
From $H_2O$ : $H^+$ and $OH^-$
During electrolysis, ions migrate to the electrodes based on their discharge potential:
At the Cathode (Negative Electrode): Both $Na^+$ and $H^+$ ions migrate here. However, $H^+$ ions have a lower discharge potential than $Na^+$ ions. Therefore, $H^+$ ions are preferentially discharged to form Hydrogen gas ( $H_2$ ).
At the Anode (Positive Electrode): Both $Cl^-$ and $OH^-$ ions migrate here. In a concentrated solution, the high concentration of chloride ions ( $Cl^-$ ) overcomes the discharge potential of the hydroxide ions ( $OH^-$ ). Consequently, $Cl^-$ ions are preferentially discharged to form Chlorine gas ( $Cl_2$ ).
Step-by-Step Solution
Identify the ions present: $Na^+$ , $Cl^-$ , $H^+$ , and $OH^-$ .
Determine the migration: The negatively charged ions ( $Cl^-$ and $OH^-$ ) move toward the positive anode.
Apply the concentration rule: Because the solution is concentrated, the chloride ions ( $Cl^-$ ) are in excess.
Write the half-equation at the anode: $2Cl^-_{(aq)} \rightarrow Cl_{2(g)} + 2e^-$
Conclusion: The gas evolved at the anode is Chlorine gas.
Exam Tip
Remember the "Concentration Effect": In dilute NaCl, Oxygen gas is produced at the anode because $OH^-$ is preferentially discharged. But in concentrated NaCl (brine), Chlorine gas is always the product at the anode. Think of it as "Concentrated = Chlorine".
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78.
The gas produced at the cathode during electrolysis of brine is
1. A.steam
2. B.chlorine
3. C.oxygen
4. D.hydrogen
Show answer
Answer: D. hydrogen
AI explanation
Analysis of the Question
The question asks for the product formed at the cathode during the electrolysis of brine. Brine is a concentrated aqueous solution of sodium chloride ( $\text{NaCl}$ ).
Concept Explanation
During the electrolysis of brine (using inert electrodes like carbon or platinum), there are two competing ions at each electrode:
At the Cathode (Negative Electrode): • The ions present are $\text{Na}^+$ (from $\text{NaCl}$ ) and $\text{H}^+$ (from the dissociation of water, $\text{H}_2\text{O}$ ). • According to the electrochemical series, the ion that is lower in the series is discharged more easily. $\text{H}^+$ is lower than $\text{Na}^+$ , so hydrogen ions are preferentially discharged. • The reaction is: $2\text{H}^+_{(aq)} + 2e^- \rightarrow \text{H}_{2(g)}$
At the Anode (Positive Electrode): • The ions present are $\text{Cl}^-$ and $\text{OH}^-$ . • Due to the high concentration of chloride ions in brine, $\text{Cl}^-$ is preferentially discharged over $\text{OH}^-$ . • The reaction is: $2\text{Cl}^-_{(aq)} \rightarrow \text{Cl}_{2(g)} + 2e^-$
Step-by-Step Solution
Step 1: Identify the ions moving to the cathode. These are $\text{Na}^+$ and $\text{H}^+$ .
Step 2: Apply the rule of selective discharge. Hydrogen ions ( $\text{H}^+$ ) have a lower discharge potential than sodium ions ( $\text{Na}^+$ ).
Step 3: Write the reduction half-equation at the cathode: $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$ .
Step 4: Conclude that the gas produced is hydrogen.
Conclusion
The gas produced at the cathode is hydrogen, while the gas produced at the anode is chlorine. Therefore, the correct option is D.
Exam Tip
Remember the mnemonic "PANIC": Positive Anode, Negative Is Cathode. Also, always remember that in the electrolysis of brine, the cathode produces Hydrogen and the anode produces Chlorine—this is a very common JAMB question!
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79.
The stainless steel alloy is used for making
1. A.cutlery
2. B.permanent magnets
3. C.rock crushers
4. D.bridges
Show answer
Answer: A. cutlery
AI explanation
Analysis of the Question
The question asks for the common application of stainless steel, which is a specific type of alloy. To answer this, we need to understand the properties of stainless steel and why it is chosen for certain industrial or domestic purposes.
Concept Explanation
An alloy is a mixture of a metal with other elements (metals or non-metals) to improve its properties, such as strength, hardness, or resistance to corrosion.
Stainless Steel: This is an alloy of iron, carbon, and a significant amount of chromium (usually at least 10.5%).
Key Property: The addition of chromium creates a thin, invisible layer of chromium oxide on the surface of the metal. This layer is "passive," meaning it prevents the iron underneath from reacting with oxygen and moisture, thereby preventing rust (corrosion).
Application: Because stainless steel is resistant to rust, stains, and chemical reactions with food, it is the ideal material for items that come into contact with water and food regularly.
Step-by-Step Solution
Evaluate Option A (Cutlery): Cutlery (knives, forks, spoons) is frequently washed and exposed to acidic foods. Stainless steel is perfect here because it does not rust or impart a metallic taste to food.
Evaluate Option B (Permanent Magnets): Permanent magnets are typically made from alloys like Alnico (Aluminum, Nickel, Cobalt) or Neodymium, not stainless steel.
Evaluate Option C (Rock Crushers): Rock crushers require extreme hardness and wear resistance. They are usually made from Manganese steel (Hadfield steel), which becomes harder under impact.
Evaluate Option D (Bridges): Bridges are typically constructed from structural carbon steel. While stainless steel could be used, it is too expensive for the massive scale of bridge construction; instead, carbon steel is used and protected with paint or galvanization.
Conclusion
Given the properties of stainless steel—specifically its resistance to corrosion and its hygienic nature—it is most widely and characteristically used for manufacturing kitchenware and cutlery.
Exam Tip: When you see questions about alloys, always link the metal to its "special property."
Stainless Steel = Chromium = Corrosion resistance (Cutlery).
Duralumin = Aluminum + Copper = Lightweight (Aircraft parts).
Brass = Copper + Zinc = Decorative/Musical instruments.
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80.
C $_2$ H $_5$ OH + CH $_3$ COOH ⇌ CH $_3$ COOC $_2$ H $_5$ + H $_2$ O The reaction above is
1. A.condensation
2. B.fermentation
3. C.esterification
4. D.hydrolysis
Show answer
Answer: C. esterification
AI explanation
1. Analyze the question
The question asks you to identify the specific type of chemical reaction occurring between ethanol ( $C_2H_5OH$ ) and ethanoic acid ( $CH_3COOH$ ) to produce ethyl ethanoate ( $CH_3COOC_2H_5$ ) and water ( $H_2O$ ).
2. Explain the concept
This reaction is a classic example of a reaction between an alcohol and a carboxylic acid.
The Principle: When an alcohol reacts with a carboxylic acid in the presence of a strong acid catalyst (usually concentrated $H_2SO_4$ ), it forms an ester and water. This process is known as esterification.
The Equation: $R-OH + R'-COOH \rightleftharpoons R'-COOR + H_2O$ In this specific case: • $C_2H_5OH$ (Ethanol) + $CH_3COOH$ (Ethanoic acid) $\rightleftharpoons$ $CH_3COOC_2H_5$ (Ethyl ethanoate) + $H_2O$ (Water).
3. Show the solution
Step 1: Identify the reactants. We have an alcohol ( $C_2H_5OH$ ) and an organic acid ( $CH_3COOH$ ).
Step 2: Identify the products. The product $CH_3COOC_2H_5$ is an ester (specifically ethyl ethanoate), and water is produced as a byproduct.
Step 3: Match the reaction type. Since the reaction produces an ester from an alcohol and an acid, it is defined as esterification.
4. Identify the answer
A (Condensation): While esterification is a type of condensation reaction (because a small molecule like water is removed), "esterification" is the more specific and accurate term for this reaction in chemistry.
B (Fermentation): This is the process of breaking down sugars into alcohol and $CO_2$ using enzymes.
C (Esterification): This is the correct term for the formation of an ester from an alcohol and an acid.
D (Hydrolysis): This is the reverse of esterification (breaking an ester down using water).
5. Tip
Memory Trick: Think of the name "Ester-ification" as "making an Ester." Whenever you see an alcohol and a carboxylic acid on the left side of the arrow, look for the ester functional group ( $-COOC-$ ) on the right side. If you see it, the answer is almost certainly esterification!
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81.
An example of a physical change is
1. A.addition of water to calcium oxide
2. B.corrosion of iron
3. C.dissolving sodium chloride in water
4. D.burning of charcoal
Show answer
Answer: C. dissolving sodium chloride in water
AI explanation
Analysis
To solve this, we must distinguish between a physical change and a chemical change:
Physical Change: A change where no new substance is formed. The chemical composition remains the same, and the change is often reversible (e.g., melting, boiling, dissolving).
Chemical Change: A change that results in the formation of one or more new substances with different chemical properties. These are usually difficult to reverse.
Concept Explanation
Addition of water to calcium oxide: This is a chemical reaction (slaking of lime). Calcium oxide reacts with water to form calcium hydroxide ( $CaO + H_2O \rightarrow Ca(OH)_2$ ). This releases heat, indicating a chemical change.
Corrosion of iron: This is a chemical process (oxidation). Iron reacts with oxygen and moisture to form hydrated iron(III) oxide (rust). A new substance is formed.
Dissolving sodium chloride in water: When salt dissolves in water, it dissociates into its constituent ions ( $Na^+$ and $Cl^-$ ). However, the chemical identity of the salt remains. If you evaporate the water, you recover the original sodium chloride crystals. No new chemical substance is created.
Burning of charcoal: This is a combustion reaction. Carbon reacts with oxygen to form carbon dioxide ( $C + O_2 \rightarrow CO_2$ ). This is a classic chemical change.
Step-by-Step Solution
Evaluate A: Calcium oxide + water $\rightarrow$ Calcium hydroxide. New substance formed = Chemical change.
Evaluate B: Iron + Oxygen $\rightarrow$ Rust. New substance formed = Chemical change.
Evaluate C: Sodium chloride + water $\rightarrow$ Aqueous solution. The salt can be recovered by evaporation. No new substance = Physical change.
Evaluate D: Carbon + Oxygen $\rightarrow$ Carbon dioxide. New substance formed = Chemical change.
Comparing these, only option C represents a process where the original substances can be recovered without a chemical reaction occurring.
Conclusion
The correct option is C.
Exam Tip
A simple way to remember the difference: If you can easily "undo" the process to get your original materials back (like evaporating water from salt), it is likely a physical change. If you see heat, light, gas bubbles, or a permanent color change, it is almost certainly a chemical change.
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82.
Sodium in the above reaction is produced by
1. A.Browning process
2. B.Chlor process
3. C.Downs process
4. D.Bosch process
Show answer
Answer: C. Downs process
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83.
Which of the following pairs of elements will exhibit diagonal relationship?
1. A.Be and Si
2. B.Na and Be
3. C.Li and Al
4. D.B and Si
Show answer
Answer: D. B and Si
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84.
The major product when 2-methylpropene reacts with HCl is
1. A.2-chloro, 2-methylpropane
2. B.1-chloro,2-methylpropane
3. C.3-chloro, 2-methylpropane
4. D.2-chloro,3-methylbutane
Show answer
Answer: A. 2-chloro, 2-methylpropane
AI explanation
Analysis
The question asks for the major product of the addition reaction between 2-methylpropene (an alkene) and hydrogen chloride (HCl). This is a classic electrophilic addition reaction.
Concept: Markovnikov’s Rule
When an unsymmetrical reagent (like HCl) adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon atom that already has the greater number of hydrogen atoms. The halogen (chlorine) attaches to the carbon atom with fewer hydrogen atoms (the more substituted carbon).
This rule works because the reaction proceeds through the formation of a carbocation intermediate. A more substituted carbocation (tertiary > secondary > primary) is more stable and forms faster, leading to the major product.
Step-by-Step Solution
Identify the structure of 2-methylpropene: The structure is $CH_2=C(CH_3)_2$ . • Carbon 1 ( $CH_2$ ) has 2 hydrogens. • Carbon 2 ( $C$ ) has 0 hydrogens and is bonded to two methyl groups.
The Reaction Mechanism: • The double bond breaks. The $H^+$ from HCl attacks the $CH_2$ group (because it has more hydrogens). • This leaves a positive charge on the central carbon (Carbon 2). • This forms a tertiary carbocation $(CH_3)_3C^+$ , which is very stable.
Nucleophilic Attack: • The chloride ion ( $Cl^-$ ) then attacks the positively charged central carbon. • The resulting structure is $CH_3-C(Cl)(CH_3)-CH_3$ .
Naming the Product: • The longest carbon chain has 3 carbons (propane). • There is a chlorine atom on carbon 2 and a methyl group on carbon 2. • The IUPAC name is 2-chloro-2-methylpropane.
Conclusion
Following Markovnikov's rule, the chlorine atom attaches to the more substituted carbon (the central carbon), resulting in 2-chloro-2-methylpropane.
Exam Tip: Whenever you see an addition reaction with an alkene, always look for the "most substituted" carbon. The halogen will almost always end up there!
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85.
Electron configuration of Chlorine atom is
1. A.1S $^2$ 2S $^2$ 2P $^6$ 3S $^2$ 3P $^5$
2. B.1S $^2$ 2S $^2$ 2P $^6$ 3S $^2$ 3P $^4$
3. C.1S $^2$ 2S $^2$ 2P $^6$ 3S $^2$ 3P $^6$
4. D.1S $^2$ 2S $^2$ 2P $^5$ 3S $^2$ 3P $^6$
Show answer
Answer: A. 1S $^2$ 2S $^2$ 2P $^6$ 3S $^2$ 3P $^5$
AI explanation
Analysis
To find the electron configuration of an atom, we must first determine its atomic number ( $Z$ ). Chlorine ( $Cl$ ) is a halogen located in Group 7 of the periodic table, and its atomic number is 17. This means a neutral chlorine atom has 17 electrons that must be arranged in orbitals according to the Aufbau principle.
Concept: The Aufbau Principle
The Aufbau principle states that electrons fill atomic orbitals in order of increasing energy levels. The sequence for filling orbitals is: $1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \dots$
Capacity of orbitals:
s-orbital: holds a maximum of 2 electrons
p-orbital: holds a maximum of 6 electrons
Step-by-Step Solution
We need to distribute 17 electrons into the orbitals following the sequence:
1s orbital: Fill with 2 electrons ( $1s^2$ ). Remaining: $17 - 2 = 15$ .
2s orbital: Fill with 2 electrons ( $2s^2$ ). Remaining: $15 - 2 = 13$ .
2p orbital: Fill with 6 electrons ( $2p^6$ ). Remaining: $13 - 6 = 7$ .
3s orbital: Fill with 2 electrons ( $3s^2$ ). Remaining: $7 - 2 = 5$ .
3p orbital: Place the remaining 5 electrons here ( $3p^5$ ).
Combining these, we get: $1s^2 2s^2 2p^6 3s^2 3p^5$
Verification
Adding the superscripts: $2 + 2 + 6 + 2 + 5 = 17$ . This matches the atomic number of Chlorine.
Comparing this to the provided options:
Option A: $1s^2 2s^2 2p^6 3s^2 3p^5$ (Matches our calculation)
Option B: $1s^2 2s^2 2p^6 3s^2 3p^4$ (Total = 16 electrons)
Option C: $1s^2 2s^2 2p^6 3s^2 3p^6$ (Total = 18 electrons, which is Argon)
Option D: $1s^2 2s^2 2p^5 3s^2 3p^6$ (Total = 17 electrons, but violates the order of filling)
Exam Tip
Always remember the "2-8-7" shell structure for Chlorine. The $1s^2$ is the 1st shell (2 electrons), $2s^2 2p^6$ is the 2nd shell (8 electrons), and $3s^2 3p^5$ is the 3rd shell (7 electrons). If you ever forget the long-form configuration, just remember the shell distribution and fill the sub-shells accordingly!
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86.
The evaluation of petrol from crude oil before it is put on sale is to prevent
1. A.pinking
2. B.reforming
3. C.pyrolysis
4. D.cracking
Show answer
Answer: A. pinking
AI explanation
Analysis
The question asks why petrol (gasoline) obtained from crude oil must be evaluated or treated before it is sold for use in internal combustion engines. We need to identify the specific phenomenon that occurs when low-quality petrol is used in an engine.
Concept Explanation
When petrol is used in an engine, it should burn smoothly. However, if the petrol has a low octane rating (meaning it contains a high proportion of straight-chain alkanes), it tends to ignite prematurely due to high pressure and heat before the spark plug fires.
This premature, explosive combustion causes a metallic knocking or rattling sound in the engine, which can lead to engine damage and loss of power. This phenomenon is known as knocking or pinking. To prevent this, petrol is treated (often by adding anti-knock agents like tetraethyl lead or by reforming the hydrocarbons into branched-chain alkanes) to ensure it burns smoothly.
Step-by-Step Solution
Analyze the options: • Cracking (D): This is the process of breaking down large hydrocarbon molecules into smaller, more useful ones. It is a production method, not a prevention method. • Pyrolysis (C): This is the thermal decomposition of substances at high temperatures in the absence of oxygen. It is a chemical process, not a problem to be prevented. • Reforming (B): This is a process used to improve the quality of petrol by converting straight-chain alkanes into branched-chain alkanes or cyclic hydrocarbons to increase the octane rating. It is a solution to the problem, not the problem itself. • Pinking (A): This is the common term used to describe "knocking" in an engine. It is the undesirable phenomenon that occurs when petrol quality is poor.
Conclusion: The evaluation and treatment of petrol are specifically carried out to prevent the engine from "pinking" (knocking).
Exam Tip
Remember the "Octane Rating" rule: High octane = smooth burning (no pinking). Low octane = "pinking" or "knocking." If you see "pinking" or "knocking" in a JAMB question, it is almost always referring to the quality of petrol in an internal combustion engine.
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87.
In the laboratory preparation of Chlorine, the gas is passed through a wash bottle of water to
1. A.dilute it
2. B.absorb HCl gas
3. C.neutralize it
4. D.absorb the Cl $_2$ gas
Show answer
Answer: B. absorb HCl gas
AI explanation
Analysis of the Question
The question asks for the specific purpose of passing chlorine gas through a wash bottle containing water during its laboratory preparation.
Concept Explanation
In the laboratory, chlorine gas ( $Cl_2$ ) is typically prepared by the reaction of concentrated hydrochloric acid ( $HCl$ ) with an oxidizing agent like manganese(IV) oxide ( $MnO_2$ ). The chemical equation is: $MnO_2(s) + 4HCl(aq) \rightarrow MnCl_2(aq) + 2H_2O(l) + Cl_2(g)$
Because concentrated $HCl$ is volatile, the chlorine gas produced is often contaminated with hydrogen chloride ( $HCl$ ) gas fumes. To obtain pure chlorine gas, these impurities must be removed before the gas is collected.
Step-by-Step Solution
Identify the impurity: The gas produced is a mixture of $Cl_2$ and $HCl$ vapor.
Analyze the solubility: Hydrogen chloride ( $HCl$ ) is extremely soluble in water. Chlorine gas ( $Cl_2$ ), however, is only sparingly soluble in water.
The process: When the mixture is bubbled through water, the $HCl$ gas dissolves readily into the water, effectively "washing" the gas stream.
The result: The $Cl_2$ gas passes through the water relatively unaffected, while the $HCl$ is trapped in the wash bottle. (Note: A second wash bottle containing concentrated sulfuric acid is usually used afterward to dry the gas).
Conclusion
Passing the gas through water is a purification step specifically designed to remove the acidic $HCl$ impurity.
Option A: Incorrect. Dilution does not purify the gas.
Option B: Correct. Water effectively absorbs the highly soluble $HCl$ gas.
Option C: Incorrect. While $HCl$ is an acid, the primary goal is purification by absorption, not neutralization.
Option D: Incorrect. If the water absorbed the $Cl_2$ gas, you would not be able to collect the product.
Exam Tip
Always remember the "purification train" for chlorine gas:
Water: To remove $HCl$ gas.
Conc. $H_2SO_4$ : To remove moisture (dry the gas).
Downward delivery: Because chlorine is denser than air.
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88.
(CH $_3$ ) $_2$ CHCH(OH)CH $_2$ C(CH $_3$ ) $_3$ The IUPAC name of the compound above is
1. A.2,5,5-trimethylhexan-3-ol
2. B.1,1-dimethylisopentanol
3. C.1,1,4,4-pentamethylbutanol
4. D.2,5-dimethylhexan-4-ol
Show answer
Answer: A. 2,5,5-trimethylhexan-3-ol
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89.
The basicity of C $_2$ H $_2$ O $_4$ is
1. A.3
2. B.4
3. C.2
4. D.8
Show answer
Answer: C. 2
AI explanation
1. Analyze the Question
The question asks for the basicity of the compound $\text{C}_2\text{H}_2\text{O}_4$ . Basicity refers to the number of replaceable hydrogen ions ( $\text{H}^+$ ) that one molecule of an acid can release in an aqueous solution.
2. Explain the Concept
Basicity of an acid: This is determined by the number of ionizable hydrogen atoms present in the acid molecule.
Structure of $\text{C}_2\text{H}_2\text{O}_4$ : This chemical formula represents Oxalic acid. Its structural formula is $(\text{COOH})_2$ or $\text{HOOC-COOH}$ .
In this structure, there are two hydroxyl groups ( $-\text{OH}$ ) attached to carbonyl groups. The hydrogen atoms in these $-\text{OH}$ groups are acidic and can be replaced by a base during a neutralization reaction.
3. Show the Solution
Step 1: Write out the dissociation equation for Oxalic acid in water: $ $\text{C}_2\text{H}_2\text{O}_4(aq) \rightleftharpoons 2\text{H}^+(aq) + \text{C}_2\text{O}_4^{2-}(aq)$ $
Step 2: Count the number of hydrogen ions ( $\text{H}^+$ ) produced. As seen in the equation, one molecule of oxalic acid produces two hydrogen ions.
Step 3: Since it releases two $\text{H}^+$ ions, the acid is described as dibasic. Therefore, its basicity is 2.
4. Identify the Answer
Comparing our result to the given options: A. 3 B. 4 C. 2 D. 8
The correct option is C.
5. Exam Tip
To quickly determine the basicity of organic acids (carboxylic acids), look at the number of -COOH groups in the formula.
$\text{CH}_3\text{COOH}$ (Ethanoic acid) has one $-\text{COOH}$ group $\rightarrow$ Basicity = 1 (Monobasic).
$(\text{COOH})_2$ (Oxalic acid) has two $-\text{COOH}$ groups $\rightarrow$ Basicity = 2 (Dibasic).
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90.
How many molecules are there in 4.0 moles of glucose? [Avogadro's number = 6.02 x 10 $^{23}$ ]
1. A.1.51 x 10 $^{23}$
2. B.2.41 x 10 $^{24}$
3. C.24.08 x 10 $^{24}$
4. D.1.51 x 10 $^{24}$
Show answer
Answer: B. 2.41 x 10 $^{24}$
AI explanation
Analysis
To find the number of molecules in a given number of moles, we use the relationship between the mole and Avogadro's constant. The mole is a standard unit in chemistry that represents a specific quantity of particles (atoms, molecules, or ions).
Concept
The relationship is defined by the formula: Number of molecules = Number of moles (n) × Avogadro's constant ( $N_A$ )
Where:
n = 4.0 moles
$N_A$ = $6.02 \times 10^{23}$ molecules/mol
Solution
Set up the calculation: Number of molecules = $4.0 \times (6.02 \times 10^{23})$
Multiply the coefficients: $4.0 \times 6.02 = 24.08$
Combine with the power of 10: $24.08 \times 10^{23}$
Convert to standard form (scientific notation): To express $24.08 \times 10^{23}$ in standard form, we move the decimal point one place to the left, which increases the exponent by 1: $2.408 \times 10^{24}$
Rounding to two decimal places (as seen in the options), we get $2.41 \times 10^{24}$ .
Conclusion
Comparing our result to the given options, the correct value is $2.41 \times 10^{24}$ . This matches option B.
Exam Tip
When dealing with Avogadro's constant, always watch your decimal placement. A common mistake is forgetting to adjust the exponent when converting from a number like $24.08$ to $2.41$ . Remember: "Left adds, Right subtracts" (moving the decimal to the left adds to the exponent; moving it to the right subtracts from it).
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91.
Carbohydrates can generally be represented by the general formula C $_x$ (H $_2$ O) $_y$ , for fructose the value for "X" is
1. A.3
2. B.12
3. C.5
4. D.6
Show answer
Answer: D. 6
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92.
Acid radicals are present in
1. A.CO $_3$ $^{2-}$ , PO $_4$ $^{3-}$ , Mg $^{2+}$
2. B.CO $_3$ $^{2-}$ , SO $_4$ $^{2-}$ , Na $^+$
3. C.CO $_3$ $^{2-}$ , SO $_4$ $^{2-}$ , NO $_3$ $^-$
4. D.CO $_3$ $^{2-}$ , SO $_4$ $^{2-}$ , K $^+$
Show answer
Answer: C. CO $_3$ $^{2-}$ , SO $_4$ $^{2-}$ , NO $_3$ $^-$
AI explanation
1. Analyze the question
The question asks you to identify which group of ions consists entirely of acid radicals. An acid radical is the negative ion (anion) that remains when an acid loses its hydrogen ions ( $H^+$ ).
2. Explain the concept
Acid Radicals: These are negatively charged ions derived from acids. For example, when Sulfuric acid ( $H_2SO_4$ ) loses its hydrogen ions, it forms the sulfate radical ( $SO_4^{2-}$ ).
Basic Radicals: These are positively charged ions (cations), usually metals or the ammonium ion ( $NH_4^+$ ).
Key Rule: To identify the correct option, look for the set that contains only anions (negative ions). Any option containing a metal cation (like $Mg^{2+}$ , $Na^+$ , or $K^+$ ) is incorrect because those are basic radicals.
3. Show the solution
Let's evaluate each option:
Option A: $CO_3^{2-}$ (Anion), $PO_4^{3-}$ (Anion), $Mg^{2+}$ (Cation). Since $Mg^{2+}$ is a basic radical, this is incorrect.
Option B: $CO_3^{2-}$ (Anion), $SO_4^{2-}$ (Anion), $Na^+$ (Cation). Since $Na^+$ is a basic radical, this is incorrect.
Option C: $CO_3^{2-}$ (Carbonate), $SO_4^{2-}$ (Sulfate), $NO_3^-$ (Nitrate). All three are negatively charged anions derived from carbonic acid, sulfuric acid, and nitric acid respectively. This fits the definition perfectly.
Option D: $CO_3^{2-}$ (Anion), $SO_4^{2-}$ (Anion), $K^+$ (Cation). Since $K^+$ is a basic radical, this is incorrect.
4. Identify the answer
The correct option is C.
5. Tip for your exam
Remember the "Cation/Anion" split:
Cations are Connected to the Cathode (Positive ions).
Anions are Attracted to the Anode (Negative ions). In salt chemistry, the metal part is always the basic radical (cation), and the non-metal part (or polyatomic group) is the acid radical (anion). If you see a metal symbol like Na, K, Mg, or Ca, it cannot be an acid radical!
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93.
The metal that will librate H $_2$ gas from dilute HNO $_3$ is
1. A.Cu
2. B.Mg
3. C.Ca
4. D.Zn
Show answer
Answer: B. Mg
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94.
The correct arrangement of gases in the order of increasing rate of diffusion is [H = 1, C = 12, N = 14, O = 16, S = 32]
1. A.CO $_2$ , H $_2$ , SO $_2$ , CO
2. B.SO $_2$ , O $_2$ , NH $_3$ , H $_2$
3. C.NH $_3$ , NO $_2$ ,N $_2$ , CO $_2$
4. D.CO $_2$ , NH $_3$ , O $_2$ , SO $_2$
Show answer
Answer: B. SO $_2$ , O $_2$ , NH $_3$ , H $_2$
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95.
When a sample of air is passed through alkaline pyrogalol, potash and finally through U-tube containing fused calcium chloride, the components of air left unabsorbed are
1. A.nitrogen and carbon(IV)oxide
2. B.noble gases and carbon(IV)oxide
3. C.noble gases and nitrogen
4. D.noble gases, nitrogen and carbon(IV) oxide
Show answer
Answer: C. noble gases and nitrogen
AI explanation
To solve this question, we need to understand the chemical process of removing specific components from a sample of air. Let’s break down the function of each substance used in the experiment:
1. Analyze the components of air
Air is a mixture primarily composed of nitrogen (~78%), oxygen (~21%), noble gases (~0.9%), and carbon(IV) oxide (~0.04%).
2. Explain the chemical processes
Alkaline Pyrogallol: This is a strong reducing agent used specifically to absorb oxygen from the air.
Potash (Potassium Hydroxide, KOH): This is a strong base used to absorb acidic gases, specifically carbon(IV) oxide ( $CO_2$ ).
Fused Calcium Chloride ( $CaCl_2$ ): This is a drying agent (desiccant) used to absorb water vapor (moisture) from the air.
3. Step-by-step solution
Step 1: The air passes through alkaline pyrogallol, which removes the oxygen.
Step 2: The remaining air passes through potash, which removes the carbon(IV) oxide.
Step 3: The air then passes through the U-tube containing fused calcium chloride, which removes the water vapor.
After these three substances have done their work, the components that were not absorbed are the ones that do not react with these chemicals. These are:
Nitrogen: It is chemically inert under these conditions and does not react with pyrogallol, KOH, or $CaCl_2$ .
Noble Gases: These are chemically inert (unreactive) and will pass through all the reagents unchanged.
4. Conclusion
Since oxygen, carbon(IV) oxide, and water vapor have been removed, the remaining components are nitrogen and the noble gases.
Comparing this to your options:
Option A is incorrect because $CO_2$ was removed by the potash.
Option B is incorrect because $CO_2$ was removed by the potash.
Option C is correct because both nitrogen and noble gases remain unabsorbed.
Option D is incorrect because $CO_2$ was removed.
Exam Tip: Remember the "Order of Removal" mnemonic: P-O-W (Pyrogallol for Oxygen, Potash for Water-gas/ $CO_2$ , Water-absorber for moisture). Always check if the question asks what is removed or what is left behind to avoid silly mistakes!
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96.
In a series of solutions with pH of 2.5, 3.5, 7.0 and 8.0, which is likely to turn red moist litmus paper blue?
1. A.3.5
2. B.8.0
3. C.2.5
4. D.7.0
Show answer
Answer: B. 8.0
AI explanation
1. Analyze the question
The question asks you to identify which of the given pH values corresponds to a substance that will turn red moist litmus paper blue. You are provided with four pH values: 2.5, 3.5, 7.0, and 8.0.
2. Explain the concept
To solve this, you need to understand the pH scale and the behavior of litmus paper:
The pH Scale: This scale measures the acidity or alkalinity of a solution. • pH < 7: The solution is acidic. • pH = 7: The solution is neutral (like pure water). • pH > 7: The solution is alkaline (basic).
Litmus Paper Test: • Blue litmus paper turns red in acidic solutions. • Red litmus paper turns blue in alkaline (basic) solutions.
Since the question asks for a substance that turns red litmus paper blue, we are looking for a substance that is alkaline (basic). Therefore, we need to find the pH value that is greater than 7.
3. Show the solution
pH 2.5: Since 2.5 < 7, this solution is acidic. It will turn blue litmus paper red.
pH 3.5: Since 3.5 < 7, this solution is acidic. It will turn blue litmus paper red.
pH 7.0: This is neutral. It will not change the color of either red or blue litmus paper.
pH 8.0: Since 8.0 > 7, this solution is alkaline (basic). Alkaline solutions turn red litmus paper blue.
4. Identify the answer
Comparing our findings with the options provided, the only value that represents an alkaline solution is 8.0. This matches option B.
5. Tip
Remember the mnemonic "B-A-R": Blue to Acid turns Red. If you remember that blue litmus turns red in acid, you can easily deduce that the opposite (red to blue) must happen in a base!
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97.
Which of the following has the highest boiling point?
1. A.CH $_3$ CH $_2$ OH
2. B.CH $_3$ CH $_2$ CH $_2$ Cl
3. C.CH $_3$ CH $_2$ CH $_2$ CH $_2$ CH $_2$ CH $_3$
4. D.CH $_3$ CH $_2$ CH $_2$ OH
Show answer
Answer: D. CH $_3$ CH $_2$ CH $_2$ OH
AI explanation
Analysis
To determine which compound has the highest boiling point, we must look at the intermolecular forces present in each molecule. Boiling point is a measure of the energy required to overcome the forces holding molecules together in the liquid state.
Concept Explanation
Van der Waals forces (London Dispersion Forces): These exist in all molecules. The strength increases with the molecular mass and the length of the carbon chain.
Dipole-Dipole interactions: Present in polar molecules (like the chloroalkane in option B).
Hydrogen Bonding: This is the strongest intermolecular force among these options. It occurs when a hydrogen atom is bonded to a highly electronegative atom like Oxygen, Nitrogen, or Fluorine. Molecules capable of hydrogen bonding generally have significantly higher boiling points than those that are not.
Comparing the options:
A (Ethanol, $CH_3CH_2OH$ ): Contains an -OH group, so it exhibits hydrogen bonding.
B (1-Chloropropane, $CH_3CH_2CH_2Cl$ ): Polar, but lacks hydrogen bonding.
C (Hexane, $CH_3CH_2CH_2CH_2CH_2CH_3$ ): Non-polar, relies only on Van der Waals forces. While it has a long chain, it lacks the strong dipole or hydrogen bonding of the alcohols.
D (Propan-1-ol, $CH_3CH_2CH_2OH$ ): Contains an -OH group, so it exhibits hydrogen bonding.
Step-by-Step Solution
Identify the strongest force: Both A and D have -OH groups, meaning they both undergo hydrogen bonding.
Compare molecular size: When comparing molecules with the same functional group (alcohols), the boiling point increases as the carbon chain length increases because the Van der Waals forces become stronger as the molecule gets larger.
Evaluate the chain length: • Ethanol (A) has 2 carbons. • Propan-1-ol (D) has 3 carbons.
Conclusion: Because Propan-1-ol has a longer carbon chain than Ethanol, it has a larger surface area and higher molecular mass, leading to stronger Van der Waals forces in addition to its hydrogen bonding. Therefore, it requires more energy to boil.
Answer
The correct option is D.
Exam Tip
When comparing boiling points of organic compounds:
Check for Hydrogen Bonding first: Alcohols, carboxylic acids, and amines usually have higher boiling points than alkanes or halides of similar size.
Check the Carbon Chain: If the functional group is the same, the molecule with the longer carbon chain (higher molar mass) will always have the higher boiling point.
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98.
The property of metal that makes it suitable as a catalyst is
1. A.filled f-orbital
2. B.partially filled d-orbitals
3. C.partially filled f-orbital
4. D.filled d-orbital
Show answer
Answer: B. partially filled d-orbitals
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99.
The liquid state of water at room temperature is as a result of
1. A.electrovalent bond
2. B.covalent bond
3. C.van der waals forces
4. D.hydrogen bond
Show answer
Answer: D. hydrogen bond
AI explanation
Analysis
The question asks why water exists as a liquid at room temperature, despite having a relatively low molecular mass compared to other substances that are gases at the same temperature.
Concept Explanation
To understand this, we must look at Intermolecular Forces. These are the forces of attraction that exist between molecules.
Covalent bonds (Option B) exist within the water molecule (between the Oxygen and Hydrogen atoms). These are very strong, but they do not determine the physical state (solid, liquid, or gas) of the substance.
Hydrogen bonding is a special, strong type of dipole-dipole attraction that occurs when a hydrogen atom is covalently bonded to a highly electronegative atom (like Oxygen, Nitrogen, or Fluorine).
In water ( $H_2O$ ), the oxygen atom is highly electronegative and pulls the shared electrons towards itself, creating a partial negative charge ( $\delta-$ ) on the oxygen and a partial positive charge ( $\delta+$ ) on the hydrogen atoms. This causes the hydrogen of one water molecule to be strongly attracted to the oxygen of a neighboring water molecule.
Step-by-Step Solution
Identify the forces: Water molecules are held together by hydrogen bonds.
Evaluate the effect: Because hydrogen bonds are significantly stronger than standard dipole-dipole interactions or Van der Waals forces, they require more energy (heat) to break.
Relate to state: At room temperature, the kinetic energy of water molecules is not high enough to overcome these strong hydrogen bonds. Consequently, the molecules remain "tethered" together in a liquid state rather than escaping into the atmosphere as a gas.
Conclusion: If water did not have hydrogen bonding, it would be a gas at room temperature (similar to $H_2S$ , which is a gas because it lacks significant hydrogen bonding).
Identifying the Answer
A. Electrovalent bond: Incorrect; this involves the transfer of electrons between ions (e.g., in NaCl).
B. Covalent bond: Incorrect; this holds the atoms inside the molecule together, not the molecules to each other.
C. Van der Waals forces: Incorrect; these are weak forces present in all molecules, but they are not the primary reason for water's liquid state.
D. Hydrogen bond: Correct. This is the specific intermolecular force responsible for the high boiling point and liquid state of water at room temperature.
Exam Tip
Whenever you see a question about the physical properties of water (like its high boiling point, surface tension, or liquid state), always look for "Hydrogen bonding" as the keyword. It is a favorite topic for JAMB examiners!
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100.
Bronze is used in making
1. A.musical instruments and ornaments
2. B.coins and medals
3. C.electromagnets and cutlery
4. D.aircrafts and ship
Show answer
Answer: B. coins and medals
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101.
An example of an alkaline gas is
1. A.HCl
2. B.N $_2$
3. C.NH $_3$
4. D.NO $_2$
Show answer
Answer: C. NH $_3$
AI explanation
1. Analyze the question
The question asks us to identify which of the given gases is "alkaline." In chemistry, an alkaline substance is one that acts as a base. Specifically, for a gas to be alkaline, it must be able to react with water or acids to produce hydroxide ions ( $OH^-$ ) or accept protons ( $H^+$ ).
2. Explain the concept
Alkaline/Basic Gases: These are gases that dissolve in water to form an alkaline solution (a base). The most common example is Ammonia ( $NH_3$ ). When $NH_3$ dissolves in water, it reacts to form ammonium hydroxide ( $NH_4OH$ ), which dissociates to release $OH^-$ ions: $ $NH_3(g) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)$ $
Acidic Gases: Gases like $HCl$ (hydrogen chloride) and $NO_2$ (nitrogen dioxide) are acidic. $HCl$ dissolves in water to form hydrochloric acid, and $NO_2$ dissolves to form a mixture of nitrous and nitric acids.
Neutral Gases: $N_2$ (nitrogen) is generally inert and does not exhibit acidic or basic properties under standard conditions.
3. Show the solution
Option A ( $HCl$ ): This is a strong acid gas. It turns blue litmus paper red.
Option B ( $N_2$ ): This is a neutral, diatomic gas that makes up most of our atmosphere. It is neither acidic nor alkaline.
Option C ( $NH_3$ ): Ammonia has a lone pair of electrons on the nitrogen atom, which allows it to act as a Lewis base (proton acceptor). It turns red litmus paper blue, confirming its alkaline nature.
Option D ( $NO_2$ ): This is an acidic oxide of nitrogen.
4. Identify the answer
Based on the chemical properties, Ammonia ( $NH_3$ ) is the only gas listed that exhibits alkaline properties.
5. Tip
Exam Strategy: Remember the "Litmus Test" rule for gases! If you are asked about the nature of a gas, think about how it affects litmus paper. Alkaline gases (like $NH_3$ ) turn red litmus paper blue, while acidic gases (like $HCl$ or $SO_2$ ) turn blue litmus paper red.
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102.
Mg $_{(s)}$ + 2HCl $_{(aq)}$ → MgCl $_2$ $_{(aq)}$ + H $_2$ $_{(g)}$ In the reaction above, magnesium is a reducing agent because
1. A.oxidized
2. B.reduced
3. C.a metal
4. D.a solid
Show answer
Answer: A. oxidized
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103.
Anti-freeze used in a vehicle radiator is a mixture of water and
1. A.propanetriol
2. B.propanol
3. C.propanone
4. D.propanal
Show answer
Answer: A. propanetriol
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104.
The process that illustrates reformation of petroleum product is
1. A.hexene to hexane
2. B.cyclohexane to benzene
3. C.hexane to carbon(IV)oxide and water vapour
4. D.hexane to butane and ethane
Show answer
Answer: B. cyclohexane to benzene
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105.
The expression above represents
1. A.Avogadro's law
2. B.Boyle's law
3. C.General gas law
4. D.Charles' law
Show answer
Answer: C. General gas law
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106.
Calculate the time required to liberate 9g of Aluminium metal, when a current of 18A is passed through it.
1. A.1769.17 minutes
2. B.89.34 minutes
3. C.5360.58 minutes
4. D.29.48 minutes
Show answer
Answer: B. 89.34 minutes
AI explanation
To solve this problem, we use Faraday’s First Law of Electrolysis, which states that the mass of a substance deposited during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
1. Analyze the Question
Mass ( $m$ ): 9g
Current ( $I$ ): 18A
Element: Aluminium ( $Al$ )
Molar Mass of Al: 27 g/mol
Valency of Al: 3 (Aluminium forms $Al^{3+}$ ions, so $n = 3$ )
Faraday’s Constant ( $F$ ): 96,500 C/mol
2. Explain the Concept
The formula for Faraday’s Law is:
$m = \frac{M \times I \times t}{n \times F}$
Where:
$m$ = mass deposited (g)
$M$ = molar mass (g/mol)
$I$ = current (A)
$t$ = time (seconds)
$n$ = number of electrons transferred (valency)
$F$ = Faraday’s constant (96,500 C/mol)
We need to rearrange this formula to solve for time ( $t$ ):
$t = \frac{m \times n \times F}{M \times I}$
3. Show the Solution
Step 1: Plug the values into the formula
$t = \frac{9 \times 3 \times 96500}{27 \times 18}$
Step 2: Simplify the calculation
$t = \frac{2,605,500}{486}$
$t \approx 5361.11 \text{ seconds}$
Step 3: Convert seconds to minutes Since the options are in minutes, divide by 60:
$t (\text{minutes}) = \frac{5361.11}{60}$
$t \approx 89.35 \text{ minutes}$
4. Identify the Answer
Comparing our result to the options provided, 89.35 minutes is approximately 89.34 minutes. This matches option B.
(Note: The slight difference is due to rounding variations in the value of Faraday's constant used in different textbooks, but 89.34 is the intended answer.)
5. Exam Tip
Always remember the "M-I-T-N-F" mnemonic for Faraday's law: Mass = (Molar mass × Current × Time) / (Number of electrons × Faraday constant). Also, always check if your final answer needs to be in seconds or minutes—many students lose marks by forgetting to divide by 60!
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107.
The empirical mass of C $_6$ H $_{12}$ O $_6$ is [H =1, C = 12, O = 16]
1. A.90
2. B.180
3. C.45
4. D.30
Show answer
Answer: D. 30
AI explanation
1. Analyze the Question
The question asks for the empirical mass of glucose ( $C_6H_{12}O_6$ ). To find this, we must first determine the empirical formula of the compound and then calculate the sum of the atomic masses of the elements in that formula.
2. Explain the Concept
Empirical Formula: This is the simplest whole-number ratio of atoms of each element in a compound.
Molecular Formula: This shows the actual number of atoms of each element in a molecule. For glucose, the molecular formula is $C_6H_{12}O_6$ .
Empirical Mass: This is the molar mass calculated using the empirical formula.
3. Show the Solution
Step 1: Find the empirical formula. The molecular formula is $C_6H_{12}O_6$ . To get the empirical formula, divide the subscripts by their greatest common divisor (which is 6):
$C: 6/6 = 1$
$H: 12/6 = 2$
$O: 6/6 = 1$
The empirical formula is $CH_2O$ .
Step 2: Calculate the empirical mass. Using the provided atomic masses ( $C=12, H=1, O=16$ ):
Mass of $C = 1 \times 12 = 12$
Mass of $H = 2 \times 1 = 2$
Mass of $O = 1 \times 16 = 16$
Total Empirical Mass = $12 + 2 + 16 = 30$
4. Identify the Answer
Comparing our result to the options provided: A. 90 B. 180 C. 45 D. 30
The correct option is D.
5. Tip
When asked for an empirical formula or mass, always look for the simplest ratio first. If you are given a molecular formula like $C_6H_{12}O_6$ , just divide all the subscripts by the smallest number present (if possible) to quickly find the empirical unit. Remember: the molecular mass is always a multiple of the empirical mass ( $Molecular\ Mass = n \times Empirical\ Mass$ ). In this case, $180 = 6 \times 30$ .
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2025 JAMB Chemistry Past Questions — Part 2

Questions 55–107

Analysis

Concept

Step-by-Step Solution

Conclusion

Exam Tip

Analysis of the Question

Concept Explanation

Step-by-Step Solution

Exam Tip

Analysis of the Question

Concept Explanation

Step-by-Step Solution

Conclusion

Analysis

Concept Explanation

Step-by-Step Solution

Evaluating the Options

Exam Tip

Analysis of the Question

Concept Explanation

Step-by-Step Solution

Exam Tip

Analysis

Concept

Step-by-Step Solution

Conclusion

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Analysis

Concept: OIL RIG

Step-by-Step Solution

Conclusion

Analysis of the Question

Concept Explanation

Step-by-Step Solution

Conclusion

Analysis

Concept Explanation

Step-by-Step Solution

Exam Tip

Analysis of the Question

Concept Explanation

Step-by-Step Solution

Conclusion

Exam Tip

Analysis of the Question

Concept Explanation

Step-by-Step Solution

Conclusion

1. Analyze the question

2. Explain the concept

3. Show the solution

4. Identify the answer

5. Tip

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Exam Tip

Analysis

Concept: Markovnikov’s Rule

Step-by-Step Solution

Conclusion

Analysis

Concept: The Aufbau Principle

Step-by-Step Solution

Verification