All 49 questions from the Joint Admissions and Matriculation Board (JAMB) Chemistry 2011 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.
Which of the following properties is NOT peculiar to matter?
A. A. Random motion of particles increases from gas to solidCorrect
B. B. Kinetic energy of particles increases from solid to gas
C. C. Random motion of particles increases from liquid to gas
D. D. Orderliness of particles increases from gas to liquid
Explanation
The highest entropy is found in gases while solids have the lowest entropy. Random motion of particles increases from solids to gases, not the other way round, so option A is the false statement.
[Diagram: number of moles of solute vs temperature graph] From the diagram above, find the amount of solute deposited when 200cm³ of the solution is cooled from 55°C to 40°C.
A. A. 0.02 mole
B. B. 0.10 mole
C. C. 0.20 moleCorrect
D. D. 0.01 mole
Explanation
From the graph, solubility at 55°C = 6 moldm⁻³ and at 40°C = 5 moldm⁻³. Mole present in 200cm³ at 55°C = (200/1000)x6 = 1.2 mole. Mole present at 40°C = (200/1000)x5 = 1.0 mole. Mole crystallized out = 1.2 - 1.0 = 0.2 mole.
What type of bond exists between an element X with atomic number 12 and Y with atomic number 17?
A. A. Dative
B. B. ElectrovalentCorrect
C. C. Metallic
D. D. Covalent
Explanation
X=12X (2,8,2); Y=17Y (2,8,7). To attain octet, X loses two electrons while Y gains an electron, so two atoms of Y combine with one atom of X. The bonding type involving transfer of electrons is electrovalent bond.
Ca(HCO3)2 and Mg(HCO3)2 are responsible for temporary hardness of water, while CaSO4 and MgSO4 are responsible for permanent hardness; calcium trioxocarbonate (IV), CaCO3, is the underlying compound mainly responsible for hardness.
N2O4(g) ⇌ 2NO2(g); ΔH=+ve. In the reaction above, an increase in temperature will
A. A. shift the equilibrium to the left
B. B. increase the value of the equilibrium constantCorrect
C. C. decrease the value of the equilibrium constant
D. D. increase the reactant production
Explanation
Since the forward reaction is endothermic (ΔH=+ve), an increase in temperature will shift the equilibrium to the right and increase the value of the equilibrium constant.
CH3COOH(aq)+OH(aq) ⇌ CH3COO(aq)+H2O(l). In the reaction above, CH3COO(aq) is the
A. A. conjugate acid
B. B. conjugate baseCorrect
C. C. acid
D. D. base
Explanation
A Bronsted-Lowry acid donates a proton, while the base accepts a proton. A conjugate acid-base pair differ by a proton. CH3COOH and CH3COO constitute a conjugate acid-base pair, with CH3COO as the conjugate base.
How many cations will be produced from a solution of potassium tetraoxosulphate (VI)?
A. A. 2
B. B. 3
C. C. 4
D. D. 1Correct
Explanation
Potassium tetraoxosulphate (VI), K2SO4, having the formula with '2' does not mean it produces two different cations - it produces just one type of cation, the potassium ion (K+). It is only a double salt that can have two different cations.
An effect of thermal pollution on water bodies is that the
A. A. level of oxygen reducesCorrect
B. B. volume of water reduces
C. C. chemical waste increases
D. D. level of oxides of nitrogen increases
Explanation
The solubility of gases decreases as temperature increases. Thus, thermal pollution (raising water temperature) reduces the level of dissolved oxygen in water bodies.
A chemical reaction in which the hydration energy is greater than the lattice energy is referred to as
A. A. a reversible reaction
B. B. a spontaneous reaction
C. C. an endothermic reactionCorrect
D. D. an exothermic reaction
Explanation
For an exothermic reaction, the lattice energy is greater than the hydration energy, while the reverse (hydration energy greater than lattice energy) is the case for an endothermic reaction.
[Diagram: rate of reaction decreasing over reaction time] The diagram above best illustrates the effect of decrease in
A. A. pressure
B. B. surface area
C. C. temperature
D. D. concentrationCorrect
Explanation
The graph represents the effect of a decrease in the concentration of reactants on reaction rate. As the concentration of reactant decreases, the reaction rate decreases as well.
Given that M is the mass of a substance deposited during electrolysis and Q is the quantity of electricity consumed, then Faraday's first law can be written as [E=Electrochemical equivalent]
A. A. M=E/Q
B. B. M=E/Q
C. C. M=EQCorrect
D. D. M=Q/E
Explanation
Faraday's first law states that the mass deposited is directly proportional to the quantity of electricity: M=EQ, where E is the electrochemical equivalent.
The impurities formed during the laboratory preparation of chlorine gas are removed by
A. A. HCl
B. B. H2OCorrect
C. C. NH3
D. D. H2SO4
Explanation
In the laboratory preparation of chlorine gas, hydrogen chloride gas is produced as an impurity. It is removed by passing the gas into water, since HCl is highly soluble in water.
The effect of the presence of impurities such as carbon and sulphur on iron is that they
A. A. lower its melting pointCorrect
B. B. give it high tensile strength
C. C. make it malleable and ductile
D. D. increase its melting point
Explanation
Presence of impurities in iron lowers its melting point (and also increases the boiling point of a liquid it may be dissolved in, per the general effect of impurities).
A few drops of concentrated HNO3 is added to an unknown solution and boiled for a while. If this produces a brown solution, the cation present is likely to be
A. A. Fe2+Correct
B. B. Pb2+
C. C. Cu2+
D. D. Fe3+
Explanation
The unknown solution originally contains Fe2+, whose heating with HNO3 will oxidize it to Fe3+, which manifests as brown coloration.
The bleaching action of chlorine gas is effective due to the presence of
A. A. oxygen
B. B. hydrogen chloride
C. C. waterCorrect
D. D. air
Explanation
The bleaching action of chlorine gas is effective due to the presence of water: Cl2+H2O→HCl+HOCl, then HOCl + coloured material → HCl+[O]+bleached material.
The property of concentrated H2SO4 that makes it suitable for preparing HNO3 is its
A. A. dehydrating property
B. B. boiling pointCorrect
C. C. density
D. D. oxidizing property
Explanation
The property of H2SO4 that makes it suitable for preparing HNO3 is its boiling point (high boiling point, low volatility) - it is common practice to prepare a more volatile acid from a less volatile one.
The constituent of baking powder that makes the dough to rise is
A. A. NaCl
B. B. NaHCO3Correct
C. C. NaOH
D. D. Na2CO3
Explanation
The constituent that makes dough rise is NaHCO3. The rising is due to CO2 gas filling the spaces within the flour, released upon decomposition: 2NaHCO3 → Na2CO3+H2O+CO2.
A primary amine has only one alkyl group attached to the nitrogen atom, e.g. CH3NH2. A secondary amine has two alkyl groups, e.g. (CH3)2NH, while a tertiary amine has three, e.g. (CH3)3N.
Two organic compounds K and L were treated with a few drops of Fehling's solution respectively. K formed a brick-red precipitate while L remains unaffected. The compound K is an
A. A. alkanone
B. B. alkanol
C. C. alkane
D. D. alkanalCorrect
Explanation
Fehling's solution distinguishes alkanals (aldehydes) from alkanones (ketones). Alkanals produce a red precipitate with Fehling's solution (they have an oxidizable hydrogen atom, R-CHO) while alkanones do not (R-CO-R).
Which of the following statements is true about 2-methylpropane and butane?
A. A. They have the same chemical propertiesCorrect
B. B. They are members of the same homologous series
C. C. They have the same boiling points
D. D. They have different number of carbon atoms
Explanation
2-methylpropane and butane are isomers - they have the same molecular formula and chemical properties, but different physical properties such as boiling point.
CH3COOH + C2H5OH -Conc.H2SO4→ CH3COOC2H5 + H2O. The reaction above is best described as
A. A. neutralization
B. B. esterificationCorrect
C. C. condensation
D. D. saponification
Explanation
This equation demonstrates esterification - the reaction between an alkanoic acid and an alkanol (in the presence of H+ catalyst) to produce an ester and water.
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