2017 JAMB Chemistry Past Questions

Questions

1. 1.

   Due to the high reactivity of sodium, it is usually stored under

   1. A.water
   2. B.mercury
   3. C.paraffin
   4. D.phenol

   Show answer

   Answer: C. paraffin

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2. 2.

   The ideal gas laws and equations are true for all gases at

   1. A.low pressures and lower temperatures
   2. B.low temperatures and high pressures
   3. C.high pressures and high temperatures
   4. D.low pressure and high temperatures

   Show answer

   Answer: D. low pressure and high temperatures

   AI explanation

   Analysis

   The question asks for the specific conditions under which real gases behave most like an "ideal gas." In chemistry, we assume ideal gas behavior to simplify calculations using the Ideal Gas Equation ($PV = nRT$).

   Concept Explanation

   Real gases deviate from ideal behavior because:

   1. Molecular Volume: Real gas molecules occupy space, whereas ideal gas particles are considered point masses with negligible volume.
   2. Intermolecular Forces: Real gas molecules exert attractive forces on each other (Van der Waals forces), whereas ideal gas particles are assumed to have no such interactions.

   To make a real gas behave like an ideal gas, we must minimize these two factors:

   • High Temperature: At high temperatures, gas particles have high kinetic energy. This energy allows them to overcome the weak intermolecular attractive forces.
   • Low Pressure: At low pressures, the gas particles are spread far apart. This makes the volume of the individual gas molecules negligible compared to the large empty space between them.

   Step-by-Step Solution

   1. Evaluate the conditions: • If pressure is high, particles are crowded together, making the volume of the molecules significant and increasing the frequency of collisions/interactions. This leads to deviation from ideal behavior. • If temperature is low, particles move slowly, allowing intermolecular forces to pull them together, which also leads to deviation.
   2. Identify the ideal conditions: • To minimize interactions, we need low pressure (to increase distance between particles). • To minimize the effect of intermolecular forces, we need high temperature (to increase kinetic energy).
   3. Match with options: Comparing our findings with the given options, the condition "low pressure and high temperatures" is the correct physical state for ideal gas behavior.

   Conclusion

   Exam Tip: Remember the mnemonic "High T, Low P" for Ideal Gases. Think of it as "High Temperature gives them the energy to ignore each other, and Low Pressure gives them the space to ignore each other."

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3. 3.

   The constituent of air necessary in the rusting process are

   1. A.O2 and H2O
   2. B.Ar and CO2
   3. C.CO2 and H2O
   4. D.O2 and CO2

   Show answer

   Answer: A. O2 and H2O

   AI explanation

   Analysis

   The question asks for the specific components of air that must be present for the chemical process of rusting to occur on iron surfaces.

   Concept Explanation

   Rusting is a chemical process known as oxidation. It is the corrosion of iron in the presence of oxygen and moisture.

   1. Oxygen ($O_2$): Iron reacts with oxygen to form iron(III) oxide. Without oxygen, the oxidation process cannot begin.
   2. Water ($H_2O$): Water acts as a medium for the reaction. It is essential for the formation of hydrated iron(III) oxide, which is the reddish-brown substance we call "rust." The chemical formula for rust is typically represented as $Fe_2O_3 \cdot xH_2O$.

   While other gases like Nitrogen, Argon, and Carbon Dioxide are present in the air, they do not participate in the fundamental chemical reaction of rusting.

   Step-by-Step Solution

   • Step 1: Identify the chemical requirements for rusting. Rusting is an electrochemical process that requires both an oxidizing agent (Oxygen) and an electrolyte/medium (Water/Moisture).
   • Step 2: Evaluate the options: • A ($O_2$ and $H_2O$): Both are essential for the formation of hydrated iron(III) oxide. • B (Ar and $CO_2$): Argon is an inert gas and does not react; $CO_2$ is not required for rusting. • C ($CO_2$ and $H_2O$): While water is needed, $CO_2$ is not the primary reactant for rusting. • D ($O_2$ and $CO_2$): Oxygen is needed, but without water, the reaction is extremely slow or non-existent in atmospheric conditions.
   • Step 3: Conclude that the combination of Oxygen and Water is the scientifically correct requirement for rusting.

   Exam Tip

   Think of rusting as a "partnership." Iron needs both Oxygen (the fuel) and Water (the bridge) to create rust. If you keep iron in a dry environment (no water) or a vacuum (no oxygen), it will not rust. This is why we paint iron gates or grease car parts—to create a barrier that keeps $O_2$ and $H_2O$ away from the metal!

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4. 4.

   The densities of two gases, X and Y are 0.5g$dm^{-3}$ and 2.0g$dm^{-3}$ respectively. What is the rate of diffusion of X relative to Y?

   1. A.0.1
   2. B.0.5
   3. C.2.0
   4. D.4.0

   Show answer

   Answer: C. 2.0

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5. 5.

   When ΔH is negative, a reaction is said to be

   1. A.endothermic
   2. B.exothermic
   3. C.reversible
   4. D.ionic

   Show answer

   Answer: B. exothermic

   AI explanation

   1. Analyze the question

   The question asks us to identify the type of chemical reaction that occurs when the change in enthalpy ($\Delta H$) is negative.

   2. Explain the concept

   In thermochemistry, enthalpy ($\Delta H$) represents the heat content of a system.

   • Exothermic reactions: These are reactions that release heat energy into the surroundings. Because the system loses energy, the final enthalpy is lower than the initial enthalpy, resulting in a negative $\Delta H$ value ($\Delta H < 0$).
   • Endothermic reactions: These are reactions that absorb heat energy from the surroundings. Because the system gains energy, the final enthalpy is higher than the initial enthalpy, resulting in a positive $\Delta H$ value ($\Delta H > 0$).

   3. Show the solution

   • If $\Delta H$ is negative, the reaction releases heat.
   • By definition, a reaction that releases heat is an exothermic reaction.
   • Option A (Endothermic) corresponds to a positive $\Delta H$.
   • Option C (Reversible) and D (Ionic) describe the nature of the reaction mechanism or bond type, not the heat change.

   4. Identify the answer

   Based on the principles of thermodynamics, a negative $\Delta H$ indicates an exothermic process.

   Correct Option: B

   5. Exam Tip

   Think of the prefix "Exo-" like "Exit". Just as you exit a room to leave it, heat "exits" the reaction system in an exothermic process, leaving the system with less energy (negative). Conversely, "Endo-" sounds like "Into"; heat goes "into" the system, increasing its energy (positive).

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6. 6.

   Temporary hard water is formed when rain water containing dissolved carbon(IV) oxide flows over deposits of

   1. A.CaCO3
   2. B.Na2CO3
   3. C.Na2SO4
   4. D.CaSO4

   Show answer

   Answer: A. CaCO3

   AI explanation

   Analysis of the Question

   The question asks for the specific mineral deposit that, when reacted with rainwater containing dissolved carbon(IV) oxide ($CO_2$), results in the formation of temporary hard water.

   Concept Explanation

   Temporary hardness in water is caused by the presence of dissolved calcium hydrogen trioxocarbonate(IV) or magnesium hydrogen trioxocarbonate(IV).

   When rainwater falls, it absorbs $CO_2$ from the atmosphere to form a weak acid called trioxocarbonate(IV) acid ($H_2CO_3$):

   $H_2O_{(l)} + CO_{2(g)} \rightarrow H_2CO_{3(aq)}$

   When this acidic water flows over limestone deposits (which is calcium trioxocarbonate(IV), $CaCO_3$), the acid reacts with the insoluble limestone to form soluble calcium hydrogen trioxocarbonate(IV):

   $CaCO_{3(s)} + H_2O_{(l)} + CO_{2(g)} \rightarrow Ca(HCO_3)_{2(aq)}$

   The resulting $Ca(HCO_3)_2$ makes the water "temporarily hard" because the hardness can be removed by simply boiling the water, which converts the soluble hydrogen trioxocarbonate(IV) back into insoluble $CaCO_3$.

   Step-by-Step Solution

   1. Identify the cause: Temporary hardness is caused by $Ca(HCO_3)_2$ or $Mg(HCO_3)_2$.
   2. Identify the source: These are formed when rainwater (containing $CO_2$) reacts with insoluble carbonates like $CaCO_3$.
   3. Evaluate the options: • A. $CaCO_3$ (Calcium trioxocarbonate(IV)): This is limestone. It reacts with $CO_2$ and water to form soluble $Ca(HCO_3)_2$. This is the correct source. • B. $Na_2CO_3$: Sodium salts are generally soluble and do not cause hardness in water. • C. $Na_2SO_4$: Sodium salts do not cause hardness. • D. $CaSO_4$: Calcium tetraoxosulphate(VI) causes permanent hardness, not temporary hardness, because it does not decompose upon boiling.

   Conclusion

   The reaction between rainwater containing $CO_2$ and limestone ($CaCO_3$) is the classic process for creating temporary hard water in nature.

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7. 7.

   Water for town supply is chlorinate to make it free from

   1. A.bad colour
   2. B.bacteria
   3. C.temporary hardness
   4. D.permanent hardness

   Show answer

   Answer: B. bacteria

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8. 8.

   Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500C$mol^{-1}$]

   1. A.0.300g
   2. B.0.250g
   3. C.0.2242g
   4. D.0.448g

   Show answer

   Answer: D. 0.448g

   AI explanation

   Analysis

   To solve this problem, we need to use Faraday’s First Law of Electrolysis, which relates the mass of a substance deposited during electrolysis to the quantity of electricity passed through the electrolyte.

   Concept

   The formula for Faraday's law is:

   $m = \frac{M \times I \times t}{n \times F}$

   Where:

   • $m$ = mass of the substance deposited (in grams)
   • $M$ = Molar mass of the element (for Copper, $Cu = 64$)
   • $I$ = Current (in Amperes) = $0.5\text{ A}$
   • $t$ = Time (in seconds) = $45\text{ minutes} \times 60\text{ seconds/minute} = 2700\text{ s}$
   • $n$ = Number of moles of electrons transferred (valency of Copper in $CuCl_2$ is $2$, so $n = 2$)
   • $F$ = Faraday’s constant = $96500\text{ C/mol}$

   Step-by-Step Solution

   1. Convert time to seconds: $t = 45 \times 60 = 2700\text{ s}$

   2. Determine the quantity of electricity ($Q$): $Q = I \times t$ $Q = 0.5 \times 2700 = 1350\text{ Coulombs}$

   3. Determine the moles of electrons: Since $Cu^{2+} + 2e^- \rightarrow Cu$, we need $2$ moles of electrons to deposit $1$ mole of Copper. Total charge required for $1$ mole of $Cu = 2 \times 96500 = 193000\text{ C}$

   4. Calculate the mass ($m$): Using the ratio: $\frac{m}{64} = \frac{1350}{193000}$ $m = \frac{64 \times 1350}{193000}$ $m = \frac{86400}{193000}$ $m \approx 0.44766\text{ g}$

   Rounding to three decimal places, we get approximately 0.448g.

   Conclusion

   Comparing our result to the options provided, the calculated mass matches option D.

   Exam Tip: Always remember to convert time from minutes to seconds! A common mistake in JAMB Chemistry is forgetting to multiply by 60, which will lead you to an incorrect answer. Also, check the valency of the metal (e.g., $Cu^{2+}$ means $n=2$, $Ag^+$ means $n=1$) before plugging values into the formula.

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9. 9.

   The furring of kettles is caused by the presence in water of

   1. A.calcium hydrogentrioxocarbonate (IV)
   2. B.calcium trioxocarbonate (IV)
   3. C.calcium tetraoxosulphate (VI)
   4. D.calcium hydroxide

   Show answer

   Answer: A. calcium hydrogentrioxocarbonate (IV)

   AI explanation

   Analysis

   The question asks for the chemical substance responsible for the "furring" (the formation of white, crusty deposits) inside kettles when water is boiled. This phenomenon is directly linked to the concept of temporary hardness of water.

   Concept Explanation

   • Hardness of water: This is caused by the presence of dissolved calcium and magnesium ions ($Ca^{2+}$ and $Mg^{2+}$) in water.
   • Temporary Hardness: This type of hardness is caused specifically by the presence of calcium hydrogentrioxocarbonate (IV), also known as calcium bicarbonate, $Ca(HCO_3)_2$.
   • The Process: When water containing calcium hydrogentrioxocarbonate (IV) is boiled, the heat causes the salt to decompose into insoluble calcium trioxocarbonate (IV) (calcium carbonate), water, and carbon (IV) oxide. • Equation: $Ca(HCO_3)_2(aq) \rightarrow CaCO_3(s) + H_2O(l) + CO_2(g)$
   • The insoluble calcium trioxocarbonate (IV) ($CaCO_3$) precipitates out and sticks to the inner walls of the kettle, forming the hard, white crust known as "fur."

   Step-by-Step Solution

   1. Identify the cause of furring: Furring is the result of mineral deposits left behind after boiling water.
   2. Evaluate the options: • A. Calcium hydrogentrioxocarbonate (IV): This is soluble in water and is the primary cause of temporary hardness. Upon heating, it decomposes to form the solid deposit. • B. Calcium trioxocarbonate (IV): This is the product formed after boiling, not the substance dissolved in the water that causes the furring in the first place. • C. Calcium tetraoxosulphate (VI): This causes permanent hardness, which cannot be removed by simple boiling. • D. Calcium hydroxide: This is "limewater" and is not typically responsible for the furring of kettles.
   3. Conclusion: Since the question asks what is present in the water before it forms the fur, the correct substance is calcium hydrogentrioxocarbonate (IV).

   Exam Tip

   Remember the "Boiling Rule": If the hardness can be removed by boiling, it is temporary and caused by hydrogentrioxocarbonates. If it requires chemical treatment (like adding washing soda), it is permanent and caused by chlorides or tetraoxosulphates (VI) of calcium or magnesium.

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10. 10.

    Cu$_2$S(g) + O$_2$(g) → 2Cu + SO$_2$(g) What is the change in the oxidation number of copper in the reaction?

    1. A.0 to + 2
    2. B.0 to + 1
    3. C.+ 1 to 0
    4. D.+ 2 to + 1

    Show answer

    Answer: C. + 1 to 0

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11. 11.

    The mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s is – (Ag = 108 F = 96500($mol^{-1}$)

    1. A.54.0g
    2. B.27.0g
    3. C.13.5g
    4. D.108.0g

    Show answer

    Answer: A. 54.0g

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12. 12.

    The constituent common to duralumin and alnico is

    1. A.Co
    2. B.Mn
    3. C.Al
    4. D.Mg

    Show answer

    Answer: C. Al

    AI explanation

    Analysis

    To solve this question, we need to identify the chemical composition of two specific industrial alloys: duralumin and alnico. We are looking for the common element present in both.

    Concept Explanation

    • Alloys are mixtures of two or more metals (or a metal and a non-metal) designed to improve the properties of the base metal, such as strength, corrosion resistance, or magnetism.
    • Duralumin is a lightweight, high-strength alloy primarily used in the aircraft industry. Its composition is roughly 95% Aluminium (Al), 4% Copper (Cu), 0.5% Manganese (Mn), and 0.5% Magnesium (Mg).
    • Alnico is a powerful magnetic alloy. Its name is derived from its main components: Aluminium (Al), Nickel (Ni), and Cobalt (Co).

    Step-by-Step Solution

    1. Break down Duralumin: The primary constituent is Aluminium (Al). It also contains small amounts of Copper, Manganese, and Magnesium.
    2. Break down Alnico: The name itself gives away the components: Aluminium, Nickel, and Cobalt.
    3. Compare the components: • Duralumin: Al, Cu, Mn, Mg • Alnico: Al, Ni, Co
    4. Identify the common element: Comparing the two lists, the only element that appears in both is Aluminium (Al).

    Conclusion

    My analysis confirms this, as Aluminium is the base metal for both alloys.

    Exam Tip: When dealing with alloy questions in JAMB, always look at the name of the alloy. Often, the name is a portmanteau (a blend) of the chemical symbols or names of the metals involved (e.g., Alnico = Aluminium + Nickel + Cobalt). This is a great way to save time during the exam!

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13. 13.

    When few drops of concentrated trioxonitrate(V) acid is added to an unknown sample and wanned an intense yellow colouration is observed. The likely functional group present in the sample is

    1. A.NH2RCHCOOH
    2. B.CHO
    3. C.CO
    4. D.CNH2

    Show answer

    Answer: A. NH2RCHCOOH

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14. 14.

    The enzyme used in the hydrolysis of starch to dextrin and maltose is?

    1. A.lipase
    2. B.amylase
    3. C.invertase
    4. D.zymase

    Show answer

    Answer: B. amylase

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15. 15.

    According to Charle's law, the volume of a gas becomes zero at

    1. A.−100°c
    2. B.−273°c
    3. C.−373°c
    4. D.0°c

    Show answer

    Answer: B. −273°c

    AI explanation

    Analysis of the Question

    This question tests your understanding of Charles's Law and the concept of Absolute Zero. You are being asked to identify the specific temperature at which the theoretical volume of an ideal gas drops to zero.

    Concept Explanation

    Charles's Law states that the volume ($V$) of a fixed mass of gas is directly proportional to its absolute temperature ($T$) in Kelvin, provided the pressure remains constant. Mathematically, this is expressed as:

    $V \propto T \quad \text{or} \quad V = kT$

    Where:

    • $V$ = Volume
    • $T$ = Temperature in Kelvin (K)
    • $k$ = Constant

    In the Celsius scale, the relationship is $V = k(t + 273)$, where $t$ is the temperature in degrees Celsius. If we want the volume ($V$) to be zero, the temperature in Kelvin must be $0\text{ K}$.

    To convert this to Celsius:

    $T(\text{K}) = t(^\circ\text{C}) + 273.15$

    $0 = t + 273$

    $t = -273^\circ\text{C}$

    This temperature ($-273^\circ\text{C}$) is known as Absolute Zero. At this point, the kinetic energy of the gas particles is theoretically at its minimum, and the volume of an ideal gas would vanish.

    Step-by-Step Solution

    1. Recall the formula: $V \propto T$ (where $T$ is in Kelvin).
    2. Identify the condition: We are looking for the temperature where $V = 0$.
    3. Apply the condition: For $V$ to be $0$, $T$ must be $0\text{ K}$.
    4. Convert to Celsius: Since $T(\text{K}) = t(^\circ\text{C}) + 273$, setting $T = 0$ gives: $0 = t + 273$ $t = -273^\circ\text{C}$

    Conclusion

    The temperature at which the volume of a gas theoretically becomes zero is $-273^\circ\text{C}$.

    Exam Tip

    Always remember that in gas law calculations (Boyle's Law, Charles's Law, Pressure Law, and the Ideal Gas Equation), you must always convert your temperature from Celsius to Kelvin by adding $273$. Never use Celsius directly in these formulas!

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16. 16.

    The oxidation number of iodine in KIO3 is

    1. A.+7
    2. B.+3
    3. C.+5
    4. D.+6

    Show answer

    Answer: C. +5

    AI explanation

    Analysis

    To find the oxidation number of iodine (I) in potassium iodate ($KIO_3$), we use the rules of oxidation states. The sum of the oxidation numbers of all atoms in a neutral compound must equal zero.

    Concept

    • Potassium (K): As a Group 1 element, it always has an oxidation number of +1.
    • Oxygen (O): In most compounds (like this one), it has an oxidation number of -2.
    • Iodine (I): This is the unknown element we need to solve for. Let's represent it as $x$.

    Step-by-Step Solution

    1. Set up the equation: The sum of the oxidation numbers in $KIO_3$ is: $(K) + (I) + 3(O) = 0$

    2. Substitute the known values: $(+1) + (x) + 3(-2) = 0$

    3. Simplify the equation: $1 + x - 6 = 0$ $x - 5 = 0$

    4. Solve for $x$: $x = +5$

    The oxidation number of iodine in $KIO_3$ is +5.

    Conclusion

    Comparing our result with the given options: A. +7 B. +3 C. +5 D. +6

    The correct option is C.

    Exam Tip

    When calculating oxidation numbers, always remember that the sum of oxidation states for a neutral molecule is 0, but if you are dealing with a polyatomic ion (like $SO_4^{2-}$), the sum must equal the charge of the ion (in this case, -2). This is a common trap in JAMB Chemistry!

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17. 17.

    In the laboratory preparation of trioxonitrate (V) acid the nitrogen(iv) oxide formed as a by-product is removed by

    1. A.further heating
    2. B.adding concentrated H2SO4
    3. C.cooling the acid solution with cold water
    4. D.bubbling air through the acid solution

    Show answer

    Answer: D. bubbling air through the acid solution

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18. 18.

    Calculate the amount in moles of silver deposited when 9650C of electricty is passed through a solution of silver salt [= 96500 C$mol^{-1}$]

    1. A.0.05
    2. B.10.80
    3. C.10.00
    4. D.0.10

    Show answer

    Answer: D. 0.10

    AI explanation

    1. Analyze the question

    We are asked to calculate the number of moles of silver ($Ag$) deposited when a specific quantity of electricity (charge) is passed through a silver salt solution.

    Given:

    • Quantity of electricity ($Q$) = $9650\text{ C}$
    • Faraday’s constant ($F$) = $96500\text{ C mol}^{-1}$

    2. Explain the concept

    To solve this, we use Faraday’s First Law of Electrolysis, which states that the mass of a substance deposited is proportional to the quantity of electricity passed. The relationship is:

    $Q = n \times z \times F$

    Where:

    • $Q$ is the quantity of electricity in Coulombs ($C$).
    • $n$ is the number of moles of the substance.
    • $z$ is the valency (or number of electrons involved in the reaction). For silver ($Ag^+$), the reaction is $Ag^+ + e^- \rightarrow Ag(s)$, so $z = 1$.
    • $F$ is Faraday’s constant ($96500\text{ C mol}^{-1}$).

    3. Show the solution

    Step 1: Identify the formula Since we need to find the amount in moles ($n$), we rearrange the formula:

    $n = \frac{Q}{z \times F}$

    Step 2: Substitute the values

    • $Q = 9650\text{ C}$
    • $z = 1$
    • $F = 96500\text{ C mol}^{-1}$

    $n = \frac{9650}{1 \times 96500}$

    Step 3: Calculate

    $n = \frac{9650}{96500}$

    $n = 0.1\text{ mol}$

    4. Identify the answer

    Comparing our result to the given options: A. 0.05 B. 10.80 C. 10.00 D. 0.10

    The calculated value matches option D.

    5. Tip

    When solving electrolysis problems, always check the charge of the metal ion. For Group 1 metals (like $Ag^+$, $Na^+$, $K^+$), $z=1$. For Group 2 metals (like $Cu^{2+}$, $Mg^{2+}$, $Zn^{2+}$), $z=2$. Remembering this saves you time during the exam!

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19. 19.

    Tin is unaffected by air at ordinary temperature due to its?

    1. A.Low melting point
    2. B.Weak electropositive character
    3. C.High boiling point
    4. D.White lustrous appearance

    Show answer

    Answer: B. Weak electropositive character

    AI explanation

    1. Analyze the question

    The question asks why Tin (Sn) remains stable and does not react with air (oxygen) at ordinary room temperatures, despite being a metal.

    2. Explain the concept

    In chemistry, the reactivity of a metal is often linked to its electropositive character (its tendency to lose electrons to form positive ions).

    • Electropositive character: Highly electropositive metals (like Sodium or Potassium) react vigorously with air and water because they lose electrons very easily.
    • Passivation: Some metals, like Tin and Aluminum, form a very thin, invisible, and non-porous layer of oxide on their surface when exposed to air. This layer acts as a protective "shield" that prevents further reaction with oxygen or moisture.
    • Tin's Position: Tin is located in the middle of the reactivity series. It is not as reactive as the alkali metals. Because it has a relatively weak electropositive character compared to highly reactive metals, it does not readily lose electrons to oxygen at room temperature, and the protective oxide film keeps it stable.

    3. Show the solution

    • Option A (Low melting point): Tin has a relatively low melting point (231.9°C), but this property relates to its physical state when heated, not its chemical resistance to air at room temperature.
    • Option B (Weak electropositive character): Because Tin is not strongly electropositive, it does not have a high affinity for oxygen at room temperature. Combined with the formation of a protective oxide layer, this makes it resistant to atmospheric corrosion.
    • Option C (High boiling point): This is a physical property and does not explain chemical reactivity with air.
    • Option D (White lustrous appearance): This describes the physical look of the metal (its luster), which is a result of its metallic bonding, but it is not the reason for its chemical stability.

    4. Identify the answer

    The correct option is B. The weak electropositive character of Tin means it does not readily react with oxygen under ordinary conditions.

    5. Exam Tip

    When dealing with "reactivity" questions in JAMB, always look for keywords like "electropositive," "electronegative," or "oxide layer." If a metal is described as "unaffected by air," it is almost always due to the formation of a protective oxide film or a low tendency to lose electrons (weak electropositive character).

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20. 20.

    The carbon atoms on ethane are

    1. A.sp2 hybridized
    2. B.sp3 hybridized
    3. C.sp4 hybridized
    4. D.sp hybridized

    Show answer

    Answer: B. sp3 hybridized

    AI explanation

    Analysis of the Question

    The question asks for the hybridization state of the carbon atoms in an ethane molecule. To solve this, we need to look at the bonding structure of ethane ($C_2H_6$).

    Concept Explanation

    Hybridization is the process of mixing atomic orbitals to form new hybrid orbitals suitable for the pairing of electrons in chemical bonds.

    In organic chemistry, we determine the hybridization of a carbon atom by counting the number of "sigma bonds" and "lone pairs" attached to it (often called the steric number):

    • sp³ hybridization: Occurs when a carbon atom is bonded to 4 other atoms (single bonds only). The geometry is tetrahedral.
    • sp² hybridization: Occurs when a carbon atom is bonded to 3 other atoms (usually involves one double bond). The geometry is trigonal planar.
    • sp hybridization: Occurs when a carbon atom is bonded to 2 other atoms (usually involves a triple bond or two double bonds). The geometry is linear.

    Step-by-Step Solution

    1. Write the structure of ethane: Ethane has the formula $C_2H_6$. Each carbon atom is bonded to three hydrogen atoms and one other carbon atom. • Structure: $CH_3 - CH_3$
    2. Count the bonds on one carbon atom: • Each carbon atom is bonded to 4 atoms (3 hydrogens and 1 carbon). • All these bonds are single bonds (sigma bonds).
    3. Apply the rule: Since each carbon atom is involved in 4 single sigma bonds, the carbon atom must undergo $sp^3$ hybridization (mixing one $s$ orbital and three $p$ orbitals).

    Conclusion

    Because ethane is a saturated hydrocarbon (an alkane) where every carbon atom is bonded to four other atoms via single bonds, the carbon atoms are $sp^3$ hybridized.

    Exam Tip: Remember the "Alkane Rule": All carbons in alkanes are always $sp^3$ hybridized. Carbons in alkenes (with a double bond) are $sp^2$ hybridized, and carbons in alkynes (with a triple bond) are $sp$ hybridized. This shortcut will save you time in the exam hall!

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21. 21.

    The reddish–brown rust on iron roofing sheets consists of

    1. A.Fe3 + (H2O)6
    2. B.FeO.H2O
    3. C.Fe2O3.XH2O
    4. D.Fe3O4.2H22O

    Show answer

    Answer: C. Fe2O3.XH2O

    AI explanation

    Analysis of the Question

    This question tests your knowledge of corrosion, specifically the chemical composition of rust. You are asked to identify the chemical formula for the reddish-brown substance that forms on iron when it is exposed to oxygen and moisture.

    Concept Explanation

    Rusting is an electrochemical process that occurs when iron reacts with oxygen in the presence of water. The chemical name for rust is hydrated iron(III) oxide.

    • Iron(III) oxide ($Fe_2O_3$): This is the base compound formed when iron oxidizes.
    • Hydrated ($XH_2O$): The "X" represents a variable number of water molecules that are chemically attached to the iron oxide structure. This is why the formula is written with $XH_2O$ rather than a fixed number.

    Step-by-Step Solution

    1. Identify the components: Rust is formed from Iron ($Fe$), Oxygen ($O_2$), and Water ($H_2O$).
    2. Determine the oxidation state: In rust, iron exists in the +3 oxidation state, forming $Fe_2O_3$.
    3. Account for hydration: Because rust is always hydrated (contains water), we add the $XH_2O$ suffix.
    4. Evaluate the options: • Option A represents a complex ion, not rust. • Option B is an incorrect representation of iron oxide. • Option C ($Fe_2O_3.XH_2O$) correctly represents hydrated iron(III) oxide. • Option D is chemically incorrect and contains a typo in the formula.

    Exam Strategy Tip

    When you see questions about "rusting" in Chemistry, always look for the combination of Iron(III) oxide ($Fe_2O_3$) and water ($H_2O$). Remember that the "X" is used because the amount of water trapped in the rust can vary depending on the humidity of the environment.

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22. 22.

    The shape of ammonia molecules is

    1. A.trigonal pyramidal
    2. B.octahedral
    3. C.square planar
    4. D.tetrahedral

    Show answer

    Answer: A. trigonal pyramidal

    AI explanation

    Analysis

    The question asks for the molecular geometry (shape) of an ammonia molecule ($NH_3$). To determine this, we must look at the electron distribution around the central nitrogen atom.

    Concept: VSEPR Theory

    The Valence Shell Electron Pair Repulsion (VSEPR) theory helps us predict the shape of molecules based on the number of bonding pairs and lone pairs of electrons around the central atom.

    1. Electronic Configuration: Nitrogen ($N$) has an atomic number of 7. Its electron configuration is $1s^2 2s^2 2p^3$. It has 5 valence electrons.
    2. Bonding: In $NH_3$, the nitrogen atom forms three covalent bonds with three hydrogen atoms. This uses 3 of its valence electrons.
    3. Lone Pairs: After bonding with the three hydrogens, nitrogen has 2 electrons remaining, which form one lone pair.
    4. Steric Number: The total number of electron domains (bonding pairs + lone pairs) is $3 + 1 = 4$. • A molecule with 4 electron domains typically adopts a tetrahedral electron geometry. • However, because one of these domains is a lone pair, it exerts a stronger repulsive force on the bonding pairs, pushing the hydrogen atoms closer together.

    Solution

    • Electron Geometry: Tetrahedral (due to 4 electron domains).
    • Molecular Geometry: Because one of the four positions is occupied by a lone pair rather than an atom, the shape formed by the atoms is trigonal pyramidal. The nitrogen sits at the apex, and the three hydrogen atoms form the base of a pyramid.

    Conclusion

    While the electron arrangement is tetrahedral, the actual shape of the molecule (the positions of the atoms) is trigonal pyramidal.

    Exam Tip: Always remember the difference between electron geometry and molecular geometry. If a central atom has lone pairs, the molecular geometry will be different from the electron geometry. For $NH_3$, the lone pair "squashes" the bond angles from the ideal $109.5^\circ$ (tetrahedral) down to approximately $107^\circ$.

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23. 23.

    $^{226}_{88}Ra$ → $^x_{86}Rn$ + alpha particle

    1. A.226
    2. B.220
    3. C.227
    4. D.222

    Show answer

    Answer: D. 222

    AI explanation

    Analysis

    This is a nuclear chemistry problem involving radioactive decay. We are given the decay of Radium ($Ra$) into Radon ($Rn$) and an alpha particle. We need to find the value of $x$, which represents the mass number of the Radon isotope produced.

    Concept

    In any nuclear reaction, two fundamental laws must be satisfied:

    1. Conservation of Mass Number: The sum of the mass numbers (the top numbers) on the reactant side must equal the sum of the mass numbers on the product side.
    2. Conservation of Atomic Number: The sum of the atomic numbers (the bottom numbers) on the reactant side must equal the sum of the atomic numbers on the product side.

    An alpha particle is a helium nucleus, represented as $^4_2He$.

    Solution

    The equation is: $^{226}_{88}Ra \rightarrow ^x_{86}Rn + ^4_2He$

    Step 1: Set up the equation for the mass numbers (top numbers). The total mass number on the left is 226. The total mass number on the right is $x + 4$.

    Equating them: $226 = x + 4$

    Step 2: Solve for $x$. $x = 226 - 4$ $x = 222$

    Step 3: Verify with atomic numbers (bottom numbers). The total atomic number on the left is 88. The total atomic number on the right is $86 + 2 = 88$. Since $88 = 88$, the equation is balanced.

    The value of $x$ is 222.

    Answer

    Comparing our result to the given options: A. 226 B. 220 C. 227 D. 222

    The correct option is D.

    Exam Tip

    Whenever you see "alpha decay," always remember to subtract 4 from the mass number and 2 from the atomic number of the parent nucleus. If it were "beta decay," the mass number remains the same, but the atomic number increases by 1!

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24. 24.

    An isomer of C5H12 is

    1. A.2–ethyl butane
    2. B.butane
    3. C.2–methyl butane
    4. D.2–methyl propane

    Show answer

    Answer: C. 2–methyl butane

    AI explanation

    Analysis

    The question asks for an isomer of $C_5H_{12}$. To solve this, we must understand what isomers are and how to name alkanes according to IUPAC rules.

    Concept: Isomerism

    Isomers are compounds that have the same molecular formula but different structural arrangements.

    • The molecular formula $C_5H_{12}$ represents an alkane with 5 carbon atoms (pentane).
    • For a compound to be an isomer of $C_5H_{12}$, it must also have exactly 5 carbon atoms and 12 hydrogen atoms, but arranged in a different structure than the straight-chain pentane.

    Step-by-Step Solution

    1. Analyze the options by counting the total number of carbon atoms: • A. 2–ethyl butane: "Butane" has 4 carbons in the parent chain. An "ethyl" group adds 2 more carbons. Total = $4 + 2 = 6$ carbons. This is an isomer of hexane ($C_6H_{14}$), not pentane. • B. Butane: This has 4 carbon atoms. It is a different compound ($C_4H_{10}$). • C. 2–methyl butane: "Butane" has 4 carbons in the parent chain. A "methyl" group adds 1 more carbon. Total = $4 + 1 = 5$ carbons. This matches the molecular formula $C_5H_{12}$. • D. 2–methyl propane: "Propane" has 3 carbons in the parent chain. A "methyl" group adds 1 more carbon. Total = $3 + 1 = 4$ carbons. This is an isomer of butane ($C_4H_{10}$).

    2. Verify the structure of 2-methyl butane: • The parent chain is butane ($C-C-C-C$). • A methyl group ($CH_3$) is attached to the second carbon. • Total carbons = 5. Total hydrogens = 12. This is a branched isomer of pentane (also known as isopentane).

    Conclusion

    Since 2-methyl butane contains exactly 5 carbon atoms and follows the alkane general formula $C_nH_{2n+2}$, it is a valid isomer of $C_5H_{12}$.

    Exam Tip

    When dealing with isomer questions for alkanes, always count the total number of carbons in the name provided. Remember:

    • Meth- = 1, Eth- = 2, Prop- = 3, But- = 4, Pent- = 5.
    • If the total count doesn't match the molecular formula given in the question, it cannot be an isomer!

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25. 25.

    Ca(OH)2(s) + 2NH4Cl(s) → CaCl2(s) + 2H2O(l) + X. In the reaction above X is

    1. A.NO2
    2. B.NH3
    3. C.N2O
    4. D.NO2

    Show answer

    Answer: B. NH3

    AI explanation

    Analysis of the Question

    The question asks us to identify the missing product X in the chemical equation: Ca(OH)₂(s) + 2NH₄Cl(s) → CaCl₂(s) + 2H₂O(l) + X

    This is a classic laboratory preparation reaction for a specific gas. To find X, we must apply the Law of Conservation of Mass, which states that the number of atoms of each element on the reactant side must equal the number of atoms of each element on the product side.

    Concept: Balancing Chemical Equations

    Let's count the atoms on both sides of the equation:

    Reactant Side:

    • Ca: 1
    • O: 2 (from Ca(OH)₂)
    • H: 2 (from Ca(OH)₂) + 8 (from 2NH₄Cl) = 10
    • N: 2 (from 2NH₄Cl)
    • Cl: 2 (from 2NH₄Cl)

    Product Side (excluding X):

    • Ca: 1
    • Cl: 2
    • H: 4 (from 2H₂O)
    • O: 2 (from 2H₂O)

    Comparing the two sides:

    • Ca, Cl, and O are already balanced.
    • H: We have 10 on the left and 4 on the right. We need 6 more hydrogen atoms.
    • N: We have 2 on the left and 0 on the right. We need 2 nitrogen atoms.

    Step-by-Step Solution

    1. Subtract the atoms already accounted for in the products from the reactants: • Remaining atoms = (Reactants) - (Products excluding X) • Remaining atoms = (10H - 4H) and (2N - 0N) • Remaining atoms = 6H and 2N
    2. The formula for X must be composed of these remaining atoms: N₂H₆.
    3. Since the coefficient for the product must balance the equation, we look for the simplest molecule containing Nitrogen and Hydrogen that fits the stoichiometry.
    4. Dividing N₂H₆ by the coefficient 2 (to match the pattern of the other products), we get NH₃.
    5. Therefore, X = 2NH₃.

    This reaction is the standard method for preparing Ammonia gas (NH₃) in the laboratory by heating an ammonium salt with an alkali.

    Conclusion

    The missing product X is NH₃.

    Note: Looking at your options, option B is NH₃. (Note: Option A and D are identical in your list, but B is the chemically correct product).

    Exam Tip: Whenever you see a reaction involving an ammonium salt (NH₄Cl) and a base (Ca(OH)₂ or NaOH), always expect Ammonia (NH₃) to be one of the products. This is a common pattern in inorganic chemistry questions!

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26. 26.

    If 100cm3 of oxygen pass through a porous plug is 50 seconds, the time taken for the same volume of hydrogen to pass through the same porous plug is? [O = 16, H = 1]

    1. A.10.0s
    2. B.12.5s
    3. C.17.7s
    4. D.32.0s

    Show answer

    Answer: B. 12.5s

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27. 27.

    The alkyl group is represented by the general formula

    1. A.CnH2n
    2. B.CnH2n − 2
    3. C.CnH2n + 1
    4. D.CnH2n + 2

    Show answer

    Answer: C. CnH2n + 1

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28. 28.

    Ethene is prepared industrially by

    1. A.Reforming
    2. B.Polymerization
    3. C.Distillation
    4. D.Cracking

    Show answer

    Answer: D. Cracking

    AI explanation

    Analysis

    The question asks for the industrial method used to produce ethene ($C_2H_4$), which is a simple alkene. We need to identify which of the listed processes is responsible for breaking down larger hydrocarbon molecules into smaller, more useful ones like ethene.

    Concept Explanation

    In the petrochemical industry, crude oil is processed into various fractions. However, the demand for smaller, highly reactive molecules like ethene (used to make plastics like polythene) often exceeds the amount naturally present in crude oil.

    Cracking is the process of breaking down large, long-chain hydrocarbon molecules (found in heavy oil fractions) into smaller, more useful molecules (like ethene and petrol) by heating them to high temperatures, often in the presence of a catalyst.

    • Reforming: This is used to convert low-octane naphtha into high-octane gasoline (petrol) by changing the structure of the molecules (e.g., making them branched or cyclic).
    • Polymerization: This is the opposite of cracking; it is the process of joining many small monomer units (like ethene) together to form a long-chain polymer (like polyethene).
    • Distillation: This is a physical separation process used to separate components of crude oil based on their different boiling points. It does not create new chemical substances.

    Step-by-Step Solution

    1. Identify the goal: We need to produce ethene ($C_2H_4$), a small molecule.
    2. Evaluate the options: • Distillation only separates existing substances; it doesn't create ethene. • Polymerization consumes ethene to make plastics. • Reforming changes the shape of molecules to improve fuel quality. • Cracking breaks large, heavy molecules into smaller, lighter ones.
    3. Conclusion: Since ethene is a small molecule derived from larger hydrocarbons, cracking is the correct industrial process.

    Correct Answer

    The correct option is D.

    Exam Tip

    Think of Cracking as "breaking." Whenever you see a question about converting long-chain hydrocarbons into shorter, more useful ones (like ethene or petrol), the answer is almost always Cracking. Remember: Cracking = Breaking down.

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29. 29.

    In the laboratory preparation of ethyl ethanoate, the water present in the mixture is removed using ?

    1. A.Anhydrous CaCl2
    2. B.concentrated NaCO4
    3. C.dilute NaOH
    4. D.concentrated H2SO4

    Show answer

    Answer: D. concentrated H2SO4

    AI explanation

    Analysis

    The question asks for the chemical agent used to remove water (a drying agent) during the laboratory preparation of ethyl ethanoate (an ester).

    Concept

    The preparation of ethyl ethanoate is an esterification reaction, which involves the reaction between ethanoic acid and ethanol in the presence of a concentrated acid catalyst (usually concentrated $H_2SO_4$).

    The reaction is reversible:

    $\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$

    To shift the equilibrium to the right (to produce more ester according to Le Chatelier’s Principle), the water produced must be removed. Concentrated sulfuric acid ($H_2SO_4$) serves two purposes here:

    1. It acts as a catalyst to speed up the reaction.
    2. It acts as a dehydrating agent to remove the water produced, thereby driving the reaction forward.

    Step-by-Step Solution

    1. Identify the reaction: Esterification is a condensation reaction that produces water as a byproduct.
    2. Understand the role of the reagent: In the laboratory setup, we need a substance that is highly hygroscopic (water-attracting) to absorb the water formed.
    3. Evaluate the options: • A (Anhydrous $CaCl_2$): While it is a drying agent, it is not the primary reagent used during the reaction mixture preparation to drive the equilibrium. • B (Concentrated $NaCO_4$): This is not a standard chemical reagent used for this purpose. • C (Dilute $NaOH$): This would cause the hydrolysis of the ester (saponification), which is the opposite of what we want. • D (Concentrated $H_2SO_4$): This is the standard reagent added to the mixture to catalyze the reaction and dehydrate the system by absorbing the water produced.

    Conclusion

    Concentrated $H_2SO_4$ is the correct agent used in the reaction mixture to remove water and shift the equilibrium toward the formation of the ester.

    Exam Tip: Always remember that in esterification, concentrated $H_2SO_4$ is a "double-duty" worker: it speeds up the reaction (catalyst) and pulls the water out of the way (dehydrating agent) to ensure you get a higher yield of the ester.

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30. 30.

    Incomplete oxidation of ethanol yields

    1. A.CH3COOH
    2. B.CH3COCH3
    3. C.CH3CH2OCH2CH3
    4. D.CH3CHO

    Show answer

    Answer: D. CH3CHO

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

32. 32.

    The gas that can be collected by downward displacement of air is

    1. A.chlorine
    2. B.sulphur (IV) oxide
    3. C.carbon (IV) oxide
    4. D.ammonia

    Show answer

    Answer: D. ammonia

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33. 33.

    The general formula of alkanones is

    1. A.RCHO
    2. B.RCOR'
    3. C.RCOOH
    4. D.RCOOR

    Show answer

    Answer: B. RCOR'

    AI explanation

    Analysis of the Question

    The question asks for the general formula of alkanones. In organic chemistry, alkanones are more commonly known as ketones. To identify the correct formula, we must look at the functional group characteristic of this homologous series.

    Explanation of the Concept

    • Alkanones (Ketones): These are organic compounds characterized by a carbonyl group ($C=O$) bonded to two carbon atoms. The general structure is $R-C(=O)-R'$, where $R$ and $R'$ are alkyl groups (which can be the same or different).
    • Let's evaluate the options provided: • A. RCHO: This is the general formula for alkanals (aldehydes). The carbonyl group is at the end of the carbon chain. • B. RCOR': This represents a carbonyl group bonded to two alkyl groups ($R$ and $R'$). This is the defining structure of ketones (alkanones). • C. RCOOH: This is the general formula for alkanoic acids (carboxylic acids). • D. RCOOR: This is the general formula for alkanoates (esters).

    Step-by-Step Solution

    1. Identify the functional group of alkanones: The carbonyl group ($C=O$) must be attached to two carbon atoms.
    2. Represent the alkyl groups attached to the carbonyl carbon as $R$ and $R'$.
    3. Combine these to form the general representation: $R-CO-R'$ or $RCOR'$.
    4. Compare this with the given options. Option B matches this structure perfectly.

    Conclusion

    The general formula for alkanones (ketones) is $RCOR'$.

    Exam Tip

    To avoid confusion in the exam hall, remember these "functional group suffixes":

    • -al (Aldehyde) ends in -CHO.
    • -one (Ketone) has the -CO- group in the middle.
    • -oic acid (Carboxylic acid) ends in -COOH.
    • -oate (Ester) contains the -COO- linkage.

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34. 34.

    A sample of orange juice is found to have a PH of 3.80. What is the concentration of the hydroxide ion in the juice?

    1. A.1.6 $\times$ 10-4
    2. B.6.3 $\times$ 10-11
    3. C.6.3 $\times$ 10-4
    4. D.1.6 $\times$ 10-11

    Show answer

    Answer: B. 6.3 $\times$ 10-11

    AI explanation

    Analysis

    To find the concentration of the hydroxide ion $[OH^-]$ in a solution when given the pH, we must use the relationship between pH, pOH, and the ion product constant of water ($K_w$).

    Concept

    1. pH and pOH relationship: At $25^\circ C$, the sum of pH and pOH is always 14. • $pH + pOH = 14$
    2. Calculating pOH: Once you have the pH, you find the pOH by subtracting the pH from 14. • $pOH = 14 - pH$
    3. Calculating $[OH^-]$: The concentration of hydroxide ions is calculated using the formula: • $[OH^-] = 10^{-pOH}$

    Step-by-Step Solution

    Step 1: Calculate the pOH. Given $pH = 3.80$ $pOH = 14 - 3.80$ $pOH = 10.20$

    Step 2: Calculate the hydroxide ion concentration $[OH^-]$. $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{-10.20}$

    To solve $10^{-10.20}$ without a calculator: $10^{-10.20} = 10^{0.80} \times 10^{-11}$ Using the value $10^{0.80} \approx 6.31$ (since $\log_{10}(6.31) \approx 0.80$)

    Therefore: $[OH^-] \approx 6.3 \times 10^{-11} \, mol/dm^3$

    Conclusion

    Comparing our result to the given options, the calculated value matches option B.

    Exam Tip

    In JAMB Chemistry, always remember the "14 rule." If you are given pH and asked for $[OH^-]$, you must convert to pOH first. If you are given pOH and asked for $[H^+]$, convert to pH first. Also, keep in mind that $10^{-10.2}$ is the same as $10^{0.8} \times 10^{-11}$; knowing your basic log values (like $\log 2 \approx 0.3$, $\log 3 \approx 0.47$, $\log 6 \approx 0.78$) helps you estimate these answers quickly without a calculator.

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35. 35.

    The process that occurs when two equivalent forms of a compound are in equilibrium is

    1. A.Isotopy
    2. B.Resonance
    3. C.Isomerism
    4. D.Reforming

    Show answer

    Answer: B. Resonance

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36. 36.

    For a general equation of the nature xP + yQ → mR + nS, the expression for the equilibrium constant is

    1. A.k [P]x [Q]y
    2. B.$\frac{[P]^x [Q]^y}{[R]^m [S]^n}$
    3. C.$\frac{[R]^m [S]^n}{[P]^x [Q]^y}$
    4. D.$\frac{m[R] n[S]}{x[P] y[Q] }$

    Show answer

    Answer: C. $\frac{[R]^m [S]^n}{[P]^x [Q]^y}$

    AI explanation

    Analysis

    The question asks for the mathematical expression of the equilibrium constant ($K_c$) for a reversible chemical reaction. In chemistry, the equilibrium constant is defined by the Law of Mass Action.

    Concept: The Law of Mass Action

    The Law of Mass Action states that for a reversible reaction at equilibrium, the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants—each raised to the power of their respective stoichiometric coefficients—is constant at a given temperature.

    For the general reaction:

    $xP + yQ \rightleftharpoons mR + nS$

    The equilibrium constant ($K_c$) is expressed as:

    $K_c = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}}$

    Step-by-Step Solution

    1. Identify the products: In the equation $xP + yQ \rightarrow mR + nS$, the products are R and S.
    2. Identify the reactants: The reactants are P and Q.
    3. Apply the coefficients as powers: • The concentration of R is raised to the power of its coefficient, $m$ $\rightarrow [R]^m$ • The concentration of S is raised to the power of its coefficient, $n$ $\rightarrow [S]^n$ • The concentration of P is raised to the power of its coefficient, $x$ $\rightarrow [P]^x$ • The concentration of Q is raised to the power of its coefficient, $y$ $\rightarrow [Q]^y$
    4. Construct the fraction: Place the product of the products in the numerator and the product of the reactants in the denominator: $$K_c = \frac{[R]^m [S]^n}{[P]^x [Q]^y}$$

    Conclusion

    Comparing this result to the given options, we see that it matches option C.

    Exam Tip: Always remember the phrase "Products over Reactants" (P/R). A common mistake students make is flipping the fraction or multiplying the coefficients instead of using them as exponents. Ensure your coefficients are written as superscripts, not as multipliers in front of the brackets!

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37. 37.

    An oxide XO2 has a vapour density of 32. What is the atomic mass of X?

    1. A.20
    2. B.32
    3. C.14
    4. D.12

    Show answer

    Answer: B. 32

    AI explanation

    1. Analyze the Question

    We are given an oxide with the formula $XO_2$ and its vapour density ($VD = 32$). We need to find the relative atomic mass of the element $X$.

    2. Explain the Concept

    • Vapour Density and Molar Mass: There is a fundamental relationship between the vapour density of a gas and its molar mass: $$\text{Molar Mass} = 2 \times \text{Vapour Density}$$
    • Molar Mass of a Compound: The molar mass of a compound is the sum of the atomic masses of all the atoms in its formula. For $XO_2$, the molar mass is: $$\text{Molar Mass} = (\text{Atomic mass of } X) + 2 \times (\text{Atomic mass of } O)$$(Note: The atomic mass of Oxygen ($O$) is$16.00\text{ g/mol}$).

    3. Show the Solution

    Step 1: Calculate the molar mass of $XO_2$

    $\text{Molar Mass} = 2 \times 32$

    $\text{Molar Mass} = 64\text{ g/mol}$

    Step 2: Set up the equation for the atomic mass of $X$ Let $A_x$ be the atomic mass of $X$.

    $A_x + 2(16) = 64$

    $A_x + 32 = 64$

    Step 3: Solve for $A_x$

    $A_x = 64 - 32$

    $A_x = 32$

    The atomic mass of $X$ is 32. Looking at the periodic table, this corresponds to the element Sulfur ($S$).

    4. Identify the Answer

    Comparing our result to the given options: A. 20 B. 32 C. 14 D. 12

    The correct option is B.

    5. Exam Tip

    Always remember the "Double Vapour Density" rule: Whenever you see "Vapour Density" in a chemistry problem, immediately multiply it by 2 to get the Molar Mass. It is a common trap in JAMB questions to give you the vapour density when you actually need the molar mass to proceed with your calculations.

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38. 38.

    The salt formed from a weak acid and a strong base hydrolyzes in water to form

    1. A.A saturated solution
    2. B.an acidic solution
    3. C.a buffer solution
    4. D.an alkaline solution

    Show answer

    Answer: D. an alkaline solution

    AI explanation

    Analysis

    The question asks about the nature of the solution formed when a salt derived from a weak acid and a strong base undergoes hydrolysis in water.

    Concept Explanation

    Hydrolysis is the reaction of an ion (from a salt) with water to produce either an acidic or alkaline solution.

    • Strong Acid + Strong Base: The resulting salt does not hydrolyze; the solution remains neutral (pH = 7).
    • Weak Acid + Strong Base: The anion of the weak acid reacts with water (hydrolysis) to produce hydroxide ions ($OH^-$). Because the base is strong, it "wins" the character of the solution, making it basic (alkaline).
    • Strong Acid + Weak Base: The cation of the weak base reacts with water to produce hydrogen ions ($H^+$), making the solution acidic.

    Step-by-Step Solution

    1. Identify the components: We have a weak acid (e.g., $CH_3COOH$) and a strong base (e.g., $NaOH$).
    2. Form the salt: The reaction is $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$. The salt is Sodium Ethanoate ($CH_3COONa$).
    3. Hydrolysis reaction: When the salt dissolves in water, it dissociates into $CH_3COO^-$ and $Na^+$. • The $Na^+$ ion comes from a strong base and does not react with water. • The $CH_3COO^-$ ion comes from a weak acid and reacts with water: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$
    4. Conclusion: The production of $OH^-$ ions increases the concentration of hydroxide ions in the solution, which makes the solution alkaline (basic).

    Identification of Answer

    The correct option is D.

    Exam Tip

    Think of it as a "tug-of-war." In a salt, the "stronger" parent (acid or base) determines the pH of the solution. If the base is strong, the solution is alkaline. If the acid is strong, the solution is acidic. If both are strong, it is neutral.

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39. 39.

    The shape of the S-orbital is

    1. A.elliptical
    2. B.spiral
    3. C.circular
    4. D.spherical

    Show answer

    Answer: D. spherical

    AI explanation

    1. Analyze the question

    The question asks for the geometric shape of the s-orbital in an atom. This is a fundamental concept in atomic structure and quantum chemistry, which frequently appears in JAMB Chemistry exams.

    2. Explain the concept

    In quantum mechanics, orbitals are regions in space where there is a high probability of finding an electron. These orbitals have specific shapes determined by the angular momentum quantum number ($l$):

    • s-orbitals ($l=0$): These are spherically symmetrical around the nucleus.
    • p-orbitals ($l=1$): These are dumbbell-shaped.
    • d-orbitals ($l=2$): These are mostly double-dumbbell shaped.

    Because the s-orbital is spherically symmetrical, the probability of finding an electron at a given distance from the nucleus is the same in all directions.

    3. Show the solution

    • The s-orbital is the simplest type of orbital.
    • It is defined by the principal quantum number ($n$) and has no angular dependence.
    • Geometrically, a shape that is uniform in all three dimensions ($x, y, z$ axes) is a sphere.
    • Therefore, the electron cloud of an s-orbital is distributed in a spherical volume around the nucleus.

    4. Identify the answer

    Comparing our explanation to the given options:

    • A. Elliptical (Incorrect - this relates to older models like Bohr-Sommerfeld)
    • B. Spiral (Incorrect - this is not a shape of an orbital)
    • C. Circular (Incorrect - a circle is 2D; an orbital is 3D)
    • D. Spherical (Correct - this accurately describes the 3D shape)

    5. Exam Tip

    To remember orbital shapes, use the mnemonic "S-P-D":

    • S = Spherical
    • P = Propeller (or dumbbell)
    • D = Double-dumbbell

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40. 40.

    X(g) + 3Y(g) ---- 2z(g) H = +ve. if the reaction above takes place at room temperature, the G will be

    1. A.negative
    2. B.zero
    3. C.positive
    4. D.indeterminate

    Show answer

    Answer: C. positive

    AI explanation

    Analysis of the Question

    This question tests your understanding of Gibbs Free Energy ($\Delta G$) and its relationship with Enthalpy ($\Delta H$), Entropy ($\Delta S$), and Temperature ($T$).

    Concept Explanation

    The relationship between these thermodynamic quantities is defined by the Gibbs-Helmholtz equation:

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

    • $\Delta H$ (Enthalpy): The question states $\Delta H$ is positive ($+ve$), meaning the reaction is endothermic.
    • $\Delta S$ (Entropy): We determine the sign of $\Delta S$ by looking at the change in the number of moles of gas. • Reactants: $1X + 3Y = 4$ moles of gas. • Products: $2Z = 2$ moles of gas. • Since the number of moles of gas decreases (from 4 to 2), the system becomes more ordered. Therefore, $\Delta S$ is negative ($-ve$).
    • Temperature ($T$): The reaction occurs at room temperature.

    Step-by-Step Solution

    1. Substitute the signs into the Gibbs equation: $$\Delta G = (+ve) - T(-ve)$$
    2. Simplifying the signs: $$\Delta G = (+ve) + (T \times +ve)$$
    3. Since $T$ (absolute temperature in Kelvin) is always positive, the entire expression becomes: $$\Delta G = (+ve) + (+ve)$$
    4. Adding two positive values always results in a positive value.
    5. For a reaction to be spontaneous, $\Delta G$ must be negative. Since $\Delta G$ is positive, the reaction is non-spontaneous at room temperature.

    Conclusion

    Because the enthalpy change is positive (endothermic) and the entropy change is negative (decreasing disorder), the Gibbs free energy ($\Delta G$) will be positive at all temperatures, including room temperature.

    Exam Tip

    Remember the "Sign Rule" for spontaneity:

    • If $\Delta H$ is negative and $\Delta S$ is positive, $\Delta G$ is always negative (Spontaneous).
    • If $\Delta H$ is positive and $\Delta S$ is negative, $\Delta G$ is always positive (Non-spontaneous).
    • If both are positive or both are negative, the spontaneity depends on the temperature!

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41. 41.

    The tincture of iodine means iodine dissolved in

    1. A.ethanol
    2. B.bromine chloride
    3. C.chlorine water
    4. D.water

    Show answer

    Answer: A. ethanol

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42. 42.

    A particle that contains 9 protons, 10 neutrons and 10 electrons is

    1. A.positive ion
    2. B.neutral atom of a metal
    3. C.neutral atom of a non-metal
    4. D.negative ion

    Show answer

    Answer: D. negative ion

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43. 43.

    Aluminium hydroxide is used in the dyeing industry as a

    1. A.dye
    2. B.dispersant
    3. C.salt
    4. D.mordant

    Show answer

    Answer: D. mordant

    AI explanation

    Analysis

    The question asks for the specific function of aluminium hydroxide in the dyeing industry. To answer this, we need to understand how dyes interact with fabrics and the role of chemical additives in that process.

    Concept Explanation

    In the dyeing industry, some dyes do not have a natural affinity for certain types of fabrics (like cotton). If you simply dip the fabric in the dye, the color will wash out easily. To solve this, a substance called a mordant is used.

    A mordant is a chemical substance that acts as a "bridge" or a fixing agent. It binds to both the fabric and the dye molecule, forming an insoluble complex (often called a "lake") within the fabric fibers. This ensures that the color becomes permanent and resistant to washing. Aluminium hydroxide is a classic example of a metallic hydroxide used for this purpose.

    Step-by-Step Solution

    1. Identify the role of the substance: Aluminium hydroxide is a gelatinous precipitate. When applied to fabric, it adheres to the fibers.
    2. Examine the interaction: When the dye is added, the aluminium hydroxide reacts with the dye to form an insoluble compound that stays trapped in the fabric.
    3. Match with terminology: The specific term for a substance that fixes a dye to a fabric is a mordant.
    4. Evaluate the options: • A (Dye): Incorrect. Aluminium hydroxide provides the base for the color, but it is not the dye itself. • B (Dispersant): Incorrect. A dispersant is used to keep particles suspended in a liquid, not to fix color to fabric. • C (Salt): Incorrect. While aluminium compounds can be salts, "salt" is a broad chemical classification and does not describe its functional role in dyeing. • D (Mordant): Correct. This is the technical term for a dye-fixing agent.

    Conclusion

    Aluminium hydroxide is specifically utilized in the dyeing industry to fix dyes onto fabrics, making it a mordant.

    Exam Tip: Remember the "M" in Mordant stands for "Make it stick." Whenever you see questions about fixing dyes to fabrics in Chemistry, think of mordants like aluminium hydroxide or alum.

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44. 44.

    The IUPAC nomenclature of the structure above is

    1. A.3-methybut-3-ene
    2. B.2-methylbut-1-ene
    3. C.2-ethylprop-1-ene
    4. D.2-methylbut-2-ene

    Show answer

    Answer: B. 2-methylbut-1-ene

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45. 45.

    The acid anhydride that will produce weak acid in water is

    1. A.SO3
    2. B.NO2
    3. C.SO2
    4. D.CO2

    Show answer

    Answer: D. CO2

    AI explanation

    Analysis

    The question asks us to identify which of the given acid anhydrides produces a weak acid when dissolved in water. An acid anhydride is a compound that reacts with water to form an acid.

    Concept Explanation

    • Acid Anhydride: These are typically non-metal oxides that react with water to form acidic solutions.
    • Strong vs. Weak Acids: • Strong acids (like $H_2SO_4$ or $HNO_3$) dissociate completely in water. • Weak acids (like $H_2CO_3$ or $H_2SO_3$) dissociate only partially, meaning they exist in equilibrium with their ions.

    Step-by-Step Solution

    Let's look at the reaction of each option with water:

    1. Option A ($SO_3$): $SO_3 + H_2O \rightarrow H_2SO_4$ (Sulfuric acid). This is a strong acid.
    2. Option B ($NO_2$): $2NO_2 + H_2O \rightarrow HNO_3 + HNO_2$ (Nitric acid and Nitrous acid). Nitric acid is a strong acid.
    3. Option C ($SO_2$): $SO_2 + H_2O \rightleftharpoons H_2SO_3$ (Sulfurous acid). While this is an acid, it is relatively weak compared to sulfuric acid, but let's compare it with $CO_2$.
    4. Option D ($CO_2$): $CO_2 + H_2O \rightleftharpoons H_2CO_3$ (Carbonic acid). Carbonic acid is a classic example of a weak acid used in carbonated drinks. It is very unstable and dissociates only slightly in water.

    Comparing the options, $CO_2$ produces carbonic acid, which is significantly weaker than the mineral acids produced by the other oxides listed. In the context of standard chemistry examinations, $CO_2$ is the standard answer for a weak acid anhydride.

    Conclusion

    Carbonic acid ($H_2CO_3$) is a very weak acid compared to the others listed, which form strong mineral acids.

    Exam Tip

    Remember the "Strong Acid" rule: Most oxides of non-metals in high oxidation states (like $S$ in $+6$ or $N$ in $+5$) form strong acids. Carbon, being in a lower oxidation state ($+4$) and forming a weak, unstable acid, is your go-to for "weak acid" questions.

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

47. 47.

    The reaction of halogens with alkanes in the presence of sunlight is an example of

    1. A.oxidation reaction
    2. B.addition reaction
    3. C.hydrogenation reaction
    4. D.substitution reaction

    Show answer

    Answer: D. substitution reaction

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48. 48.

    A given mass of gas occupies 2dm3 at 300k. At what temperature will its volume be doubled, keeping the pressure constant?

    1. A.400k
    2. B.480k
    3. C.550k
    4. D.600k

    Show answer

    Answer: D. 600k

    AI explanation

    1. Analyze the question

    We are given the initial volume ($V_1$) and initial temperature ($T_1$) of a gas. We need to find the new temperature ($T_2$) when the volume is doubled ($V_2 = 2 \times V_1$), while the pressure remains constant.

    2. Explain the concept

    This problem is governed by Charles's Law. Charles's Law states that for a fixed mass of gas at constant pressure, the volume is directly proportional to its absolute temperature (measured in Kelvin).

    The formula is:

    $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

    3. Show the solution

    • Step 1: Identify the given values • $V_1 = 2 \text{ dm}^3$ • $T_1 = 300 \text{ K}$ • $V_2 = 2 \times V_1 = 2 \times 2 = 4 \text{ dm}^3$ • $T_2 = ?$

    • Step 2: Substitute the values into the formula $$\frac{2}{300} = \frac{4}{T_2}$$

    • Step 3: Solve for $T_2$ Cross-multiply to isolate $T_2$: $$2 \times T_2 = 4 \times 300$$ $$2 \times T_2 = 1200$$ $$T_2 = \frac{1200}{2}$$ $$T_2 = 600 \text{ K}$$

    4. Identify the answer

    Comparing our result ($600 \text{ K}$) with the given options, the correct choice is D.

    5. Tip

    Whenever you see "constant pressure" in a gas law problem, immediately think of Charles's Law. A quick mental shortcut: if the pressure is constant, doubling the volume will always require doubling the absolute temperature. Since $300 \text{ K} \times 2 = 600 \text{ K}$, you can solve this in seconds!

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49. 49.

    Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC [R = 0.082 atm dm3 K-1 $mol^{-1}$]

    1. A.2.536
    2. B.1.623
    3. C.4.736
    4. D.0.394

    Show answer

    Answer: A. 2.536

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