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JAMB Chemistry 2012 past questions and answers

45 questions. Try each one, then open the answer to check yourself and read the explanation.

Question 2

Separation of Mixtures

Which of the following methods can be used to obtain pure water from a mixture of sand, water and methanoic acid?

  1. A.Neutralization with NaOH followed by filtration
  2. B.Neutralization with NaOH followed by distillation
  3. C.Fractional distillation
  4. D.Filtration followed by distillation.
Show the answer and explanation

Answer: B (Neutralization with NaOH followed by distillation)

Explanation

Methanoic acid boils at almost the same temperature as water, so it would distil over with it. Neutralising turns it into non-volatile sodium methanoate; distillation then leaves the salt and sand behind and gives pure water.

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Question 3

Mole Concept and Stoichiometry

How many atoms are present in 6.0g of magnesium? [Mg = 24, NA = 6.02 × 1023 mol−1]

  1. A.1.20 × 1022
  2. B.2.41 × 1022
  3. C.1.51 × 1023
  4. D.3.02 × 1023
Show the answer and explanation

Answer: C (1.51 × 1023)

Explanation

6.0/24 = 0.25 mol; 0.25 × 6.02 × 1023 = 1.51 × 1023 atoms.

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Question 4

Gas Laws

50 cm3 of a gas was collected over water at 10 °C and 765 mm Hg. Calculate the volume of the gas at s.t.p. if the saturated vapour pressure of water at 10 °C is 5mm Hg.

  1. A.49.19 cm3
  2. B.48.87 cm3
  3. C.48.55 cm3
  4. D.48.23 cm3
Show the answer and explanation

Answer: D (48.23 cm3)

Explanation

Dry gas pressure = 765 − 5 = 760 mmHg. V2 = 50 × (760/760) × (273/283) = 48.23 cm3.

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Question 5

Gas Laws

An increase in the pressure exerted on a gas at a constant temperature results in

  1. A.a decrease in the number of effective collisions
  2. B.a decrease in volume
  3. C.an increase in the average inter-molecular distance
  4. D.an increase in volume.
Show the answer and explanation

Answer: B (a decrease in volume)

Explanation

Boyle's law: at constant temperature, raising the pressure makes the volume smaller.

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Question 6

Gas Laws

2H2(g) + O2(g) → 2H2O(g)
In the reaction above, what volume of hydrogen would be left over when 300 cm3 of oxygen and 1000 cm3 of hydrogen are exploded in a sealed tube?

  1. A.200 cm3
  2. B.400 cm3
  3. C.600 cm3
  4. D.700 cm3
Show the answer and explanation

Answer: B (400 cm3)

Explanation

Hydrogen and oxygen react 2 : 1 by volume, so 300 cm3 O2 uses 600 cm3 H2; 1000 − 600 = 400 cm3 is left.

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Question 7

Kinetic Theory and States of Matter

I. Evaporation. II. Sublimation. III. Diffusion. IV. Brownian motion.
Which of the above can correctly be listed as evidences for the particulate nature of matter?

  1. A.I and III only
  2. B.II and IV only
  3. C.I, II and III only
  4. D.I, II, III and IV
Show the answer and explanation

Answer: D (I, II, III and IV)

Explanation

All four can only be explained if matter is made of tiny moving particles that can escape from a surface, spread out, or jostle larger particles.

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Question 8

Chemical Bonding

If the elements X and Y have atomic numbers 11 and 17 respectively, what type of bond can they form?

  1. A.Dative
  2. B.Covalent
  3. C.Ionic
  4. D.Metallic.
Show the answer and explanation

Answer: C (Ionic)

Explanation

X is sodium (loses 1 electron) and Y is chlorine (gains 1), so they form an ionic bond in NaCl.

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Question 9

Atomic Structure

A hydrogen atom which has lost an electron contains

  1. A.one proton only
  2. B.one neutron only
  3. C.one proton and one neutron
  4. D.one proton, one electron and one neutron.
Show the answer and explanation

Answer: A (one proton only)

Explanation

An ordinary hydrogen atom has one proton, one electron and no neutron; when it loses the electron, only the proton (H+) remains.

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Question 10

Atomic Structure

The electronic configuration of Mg2+ is

  1. A.1s22s22p63s23p2
  2. B.1s22s22p63s2
  3. C.1s22s22p6
  4. D.1s22s22p4
Show the answer and explanation

Answer: C (1s22s22p6)

Explanation

Mg (12 electrons) loses its two 3s electrons, leaving 10: 1s22s22p6.

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Question 11

Halogens

Group VII elements are

  1. A.monoatomic
  2. B.good oxidizing agents
  3. C.highly electropositive
  4. D.electron donors.
Show the answer and explanation

Answer: B (good oxidizing agents)

Explanation

Halogens need one electron to complete their octet, so they readily accept electrons: they are good oxidizing agents.

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Question 12

Chemical Bonding

Which of the following is used to study the arrangement of particles in crystal lattices?

  1. A.Alpha-particles
  2. B.Beta-particles
  3. C.Gamma-rays
  4. D.X-rays.
Show the answer and explanation

Answer: D (X-rays.)

Explanation

X-ray diffraction is used to work out how particles are arranged in crystals.

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Question 13

Air

I. It has a varied composition from one place to another. II. Its constituents can be separated by physical means III. It contains unreactive noble gases.
Which of the above shows that air is a mixture?

  1. A.I and II only
  2. B.II and III only
  3. C.I and III only
  4. D.I, II and III
Show the answer and explanation

Answer: A (I and II only)

Explanation

A mixture has no fixed composition and can be separated physically. Containing noble gases says nothing about whether air is a mixture.

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Question 14

Water

The chemicals used to soften hard water involves the addition of

  1. A.insoluble sodium compounds which form soluble solutions of calcium and magnesium
  2. B.soluble sodium compounds which form soluble solutions of calcium and magnesium ions
  3. C.soluble sodium compounds which form insoluble precipitates of calcium and magnesium ions
  4. D.insoluble precipitates of calcium and magnesium ions.
Show the answer and explanation

Answer: C (soluble sodium compounds which form insoluble precipitates of calcium and magnesium ions)

Explanation

Washing soda (soluble Na2CO3) removes Ca2+ and Mg2+ as insoluble trioxocarbonates(IV).

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Question 15

Water

Chlorination of water for town supply is carried out to

  1. A.make the water colourless
  2. B.remove germs from the water
  3. C.make the water tasteful
  4. D.remove odour from the water.
Show the answer and explanation

Answer: B (remove germs from the water)

Explanation

Chlorine kills disease-causing bacteria in the water.

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Question 16

Solubility and Solutions

The solubilities of different solutes in a given solvent can be compared by

  1. A.plotting their solubility curves on separate axes
  2. B.plotting their solubility curves on the same axes
  3. C.plotting some of the solubility curves on the x-axis and others on the y-axis
  4. D.plotting their solubility curves on the x-axis only.
Show the answer and explanation

Answer: B (plotting their solubility curves on the same axes)

Explanation

Drawing all the curves on one set of axes lets you compare the solubilities directly at any temperature.

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Question 17

Solubility and Solutions

Potassium trioxochlorate(V) has a solubility of 1.5 mol dm−3 at 45 °C. On cooling this solution to a temperature of 20 °C, the solubility was found to be 0.5 mol dm−3. What mass of KClO3 was crystalized out? [K = 39, Cl = 35.5, O = 16]

  1. A.1.00 g
  2. B.10.00 g
  3. C.12.25 g
  4. D.122.50 g
Show the answer and explanation

Answer: D (122.50 g)

Explanation

1.5 − 0.5 = 1.0 mol crystallises from each dm3. KClO3 = 122.5 g/mol, so 122.5 g crystallises (taking 1 dm3 of solution).

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Question 19

Acids, Bases and Salts

Which of the following will produce a solution with pH less than 7 at equivalent point?

  1. A.HNO3 + NaOH
  2. B.H2SO4 + KOH
  3. C.HCl + Mg(OH)2
  4. D.HNO3 + KOH
Show the answer and explanation

Answer: C (HCl + Mg(OH)2)

Explanation

A strong acid with a weak base gives a slightly acidic salt solution; Mg(OH)2 is the only weak base here.

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Question 20

Acids, Bases and Salts

The number of hydroxonium ions produced by one molecule of an acid in aqueous solution is its

  1. A.basicity
  2. B.acid strength
  3. C.pH
  4. D.concentration.
Show the answer and explanation

Answer: A (basicity)

Explanation

Basicity is the number of replaceable hydrogen ions (H3O+) one molecule of an acid can produce, e.g. 2 for H2SO4.

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Question 21

Mole Concept and Stoichiometry

During a titration experiment, 0.05 moles of carbon(IV) oxide is liberated. What is the volume of gas liberated? [Molar volume of a gas at s.t.p. = 22.4 dm3]

  1. A.22.40 dm3
  2. B.11.20 dm3
  3. C.2.24 dm3
  4. D.1.12 dm3
Show the answer and explanation

Answer: D (1.12 dm3)

Explanation

Volume = 0.05 × 22.4 = 1.12 dm3.

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Question 22

Acids, Bases and Salts

A major factor considered in selecting a suitable method for preparing a simple salt is its

  1. A.crystalline form
  2. B.melting point
  3. C.reactivity with dilute acids
  4. D.solubility in water.
Show the answer and explanation

Answer: D (solubility in water.)

Explanation

Soluble salts are made by neutralisation or acid reactions and crystallisation; insoluble salts by precipitation. So solubility decides the method.

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Question 23

Redox Reactions

The oxidation number of boron in NaBH4 is

  1. A.−3
  2. B.−1
  3. C.+1
  4. D.+3
Show the answer and explanation

Answer: D (+3)

Explanation

Na is +1 and each H (a hydride) is −1: 1 + x − 4 = 0, so x = +3.

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Question 24

Redox Reactions

2Na2O2(s) + 2H2O(l) → 4NaOH(s) + O2(s)
The substance that is oxidized in the reaction above is

  1. A.Na2O2(s)
  2. B.NaOH(aq)
  3. C.H2O(l)
  4. D.O2(g)
Show the answer and explanation

Answer: A (Na2O2(s))

Explanation

Oxygen in the peroxide is −1; some of it rises to 0 in O2, so Na2O2 is oxidized (and also reduced, to −2 in NaOH).

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Question 25

Electrolysis

What number of moles of Cu2+ will be deposited by 360 coulombs of electricity? [F = 96500 C mol−1]

  1. A.5.36 × 10−4 mole
  2. B.1.87 × 10−3 mole
  3. C.9.35 × 10−4 mole
  4. D.3.73 × 10−3 mole
Show the answer and explanation

Answer: B (1.87 × 10−3 mole)

Explanation

Cu2+ needs 2 F per mole: 360/(2 × 96 500) = 1.87 × 10−3 mole.

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Question 26

Electrolysis

A metal M displaces zinc from ZnCl2 solution. This shows that

  1. A.electrons flow from zinc to M
  2. B.M is more electropositive than zinc
  3. C.M is more electonegative than zinc
  4. D.zinc is more electropositive than M.
Show the answer and explanation

Answer: B (M is more electropositive than zinc)

Explanation

A metal can only displace another from its salt if it is higher in the activity series, i.e. more electropositive.

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Question 28

Energy Changes

An increase in entropy can best be illustrated by

  1. A.mixing of gases
  2. B.freezing of water
  3. C.the condensation of vapour
  4. D.solidifying candle wax.
Show the answer and explanation

Answer: A (mixing of gases)

Explanation

When gases mix, the particles become more spread out and disordered; the other changes make particles more ordered.

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Question 29

Rates of Reaction

The highest rate of production of carbon(IV) oxide can be achieved using

  1. A.0.05 mol dm−3 HCl and 5g powdered CaCO3
  2. B.0.05 mol dm−3 HCl and 5g lump CaCO3
  3. C.0.10 mol dm−3 HCl and 5g powdered CaCO3
  4. D.0.025 mol dm−3 HCl and 5g powdered CaCO3
Show the answer and explanation

Answer: C (0.10 mol dm−3 HCl and 5g powdered CaCO3)

Explanation

The fastest rate needs the highest acid concentration and the largest surface area (powder).

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Question 31

Chemical Equilibrium

2CO(g) + O2(g) ⇌ 2CO2(g)
In the reaction above, high pressure will favour the foward reaction because

  1. A.high pressure favours gas formation
  2. B.the reaction is in dynamic equilibrium
  3. C.the reaction is exothermic
  4. D.the process occurs with a decrease in volume.
Show the answer and explanation

Answer: D (the process occurs with a decrease in volume.)

Explanation

3 volumes of gas become 2, so the forward reaction reduces the volume, which high pressure favours.

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Question 32

Sulphur and its Compounds

A piece of filter paper moistened with lead(II) ethanoate solution turns black when the paper is dropped into a gas jar containing an unknown gas. The gas is likely to be

  1. A.sulphur(IV) oxide
  2. B.hydrogen chloride
  3. C.sulphur(VI) oxide
  4. D.hydrogen sulphide.
Show the answer and explanation

Answer: D (hydrogen sulphide.)

Explanation

H2S reacts with lead(II) ethanoate to form black lead(II) sulphide.

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Question 33

Nitrogen and its Compounds

Which of the following gases has a characteristic pungent smell, turns red litmus paper blue and forms dense white fumes with hydrogen chloride gas?

  1. A.N2
  2. B.N2O
  3. C.Cl2
  4. D.NH3
Show the answer and explanation

Answer: D (NH3)

Explanation

Ammonia is the only common alkaline gas; with HCl it forms white fumes of NH4Cl.

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Question 34

Sulphur and its Compounds

Commercial bleaching can be carried out using

  1. A.sulphur(IV) oxide and ammonia
  2. B.hydrogen sulphide and chlorine
  3. C.chlorine and sulphur(IV) oxide
  4. D.ammonia and chlorine.
Show the answer and explanation

Answer: C (chlorine and sulphur(IV) oxide)

Explanation

Chlorine bleaches by oxidation and sulphur(IV) oxide by reduction; both are used as industrial bleaches.

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Question 35

Hydrogen, Oxygen and Oxides

Mineral acids are usually added to commercial hydrogen peroxide to

  1. A.oxidize it
  2. B.decompose it
  3. C.minimize its decomposition
  4. D.reduce it to water and oxygen.
Show the answer and explanation

Answer: C (minimize its decomposition)

Explanation

Hydrogen peroxide decomposes faster in alkaline conditions; a little acid stabilises it.

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Question 37

Metals and their Compounds

The purest form of iron which contains only about 0.1% carbon is

  1. A.pig iron
  2. B.wrought iron
  3. C.cast iron
  4. D.iron pyrite.
Show the answer and explanation

Answer: B (wrought iron)

Explanation

Wrought iron has the least carbon (about 0.1%); pig and cast iron contain 3–5%.

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Question 38

Metals and their Compounds

A common characteristic between zinc and the other transition elements is the ability to

  1. A.have variable oxidation states
  2. B.form complex ions
  3. C.act as a catalyst
  4. D.form coloured ions.
Show the answer and explanation

Answer: B (form complex ions)

Explanation

Zinc shows only +2 and its ions are colourless, but like transition metals it forms complex ions such as [Zn(NH3)4]2+.

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Question 39

Metals and their Compounds

Which of the following metals is the least reactive?

  1. A.Pb
  2. B.Sn
  3. C.Hg
  4. D.Au
Show the answer and explanation

Answer: D (Au)

Explanation

Gold is at the bottom of the activity series, below mercury, lead and tin.

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Question 40

Introduction to Organic Chemistry

Geometric isomerism can exist in

  1. A.hex-3-ene
  2. B.hexane
  3. C.prop-l-ene
  4. D.3-methyl but-l-ene
Show the answer and explanation

Answer: A (hex-3-ene)

Explanation

Hex-3-ene has two different groups on each carbon of the C=C, so cis and trans forms exist; the others have two identical groups on one double-bond carbon or no double bond.

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Question 41

Alkanals and Alkanones

Alkanals can be distinguished from alkanones by the reaction with

  1. A.sudan III stain
  2. B.starch iodide paper
  3. C.lithium tetrahydrido aluminate(III)
  4. D.Fehling's solution
Show the answer and explanation

Answer: D (Fehling's solution)

Explanation

Alkanals reduce Fehling's solution to a red precipitate; alkanones do not.

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Question 42

Introduction to Organic Chemistry

The isomers of C3H8O are

  1. A.1 - propanol and 2 - propanol
  2. B.1 - propanol and 1 - propanal
  3. C.2 - propanol and 2 - propanone
  4. D.2 - propanol and 1 - propanal.
Show the answer and explanation

Answer: A (1 - propanol and 2 - propanol)

Explanation

Propan-1-ol and propan-2-ol both have the formula C3H8O; propanal and propanone are C3H6O.

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Question 43

Carbohydrates, Proteins and Amines

Carbohydrates are large molecules with the molecular formula Cx(H2O)y. In which of the following pairs is x not equal to y?

  1. A.Glucose and starch
  2. B.Maltose and fructose
  3. C.Sucrose and fructose
  4. D.Maltose and starch.
Show the answer and explanation

Answer: D (Maltose and starch.)

Explanation

Maltose is C12(H2O)11 and starch (C6H10O5)n is C6n(H2O)5n, so x ≠ y for both. Glucose and fructose are C6(H2O)6 (x = y).

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Question 44

Introduction to Organic Chemistry

A compound contains 40.0% C, 6.7% H 53.3% O. If the molecular mass of the compound is 180, its molecular formula is [C = 12, H = 1, O = 16]

  1. A.CH2O
  2. B.C3H6O3
  3. C.C6H6O3
  4. D.C6H12O6
Show the answer and explanation

Answer: D (C6H12O6)

Explanation

C : H : O = 40/12 : 6.7/1 : 53.3/16 = 3.33 : 6.7 : 3.33 = 1 : 2 : 1, so CH2O (30). 180/30 = 6, giving C6H12O6.

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Question 45

Hydrocarbons

The alkyne that will give a white precipitate silver trioxonitrate(V) is

  1. A.CH3CH2C ≡ CCH2CH3
  2. B.CH3C ≡ CCH2CH2CH3
  3. C.CH3CH2CH2CH2C ≡ CH
  4. D.CH3CH2CH2C ≡ CCH2CH3
Show the answer and explanation

Answer: C (CH3CH2CH2CH2C ≡ CH)

Explanation

Only a terminal alkyne (−C≡CH) has the acidic hydrogen that is replaced by silver to give a white silver alkynide precipitate.

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Question 46

Fats, Oils and Soaps

The saponification of an alkanoate to produce soap and alkanol involves

  1. A.dehydration
  2. B.esterification
  3. C.hydrolysis
  4. D.oxidation.
Show the answer and explanation

Answer: C (hydrolysis)

Explanation

Saponification is alkaline hydrolysis of an ester (fat or oil) into soap and an alkanol (glycerol).

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Question 47

Alkanols

2-methyl propan -2- ol is an example of a

  1. A.primary alkanol
  2. B.secondary alkanol
  3. C.tertiary alkanol
  4. D.quartenary alkanol.
Show the answer and explanation

Answer: C (tertiary alkanol)

Explanation

The carbon carrying −OH is bonded to three other carbons, so it is a tertiary alkanol.

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Question 48

Alkanols

The final oxidation product of alkanol, alkanal and alkanones is

  1. A.alkanoic acid
  2. B.alkanoyl halide
  3. C.alkanoate
  4. D.alkanamide
Show the answer and explanation

Answer: A (alkanoic acid)

Explanation

Alkanols oxidise to alkanals and then to alkanoic acids; strong oxidation of alkanones also ends in alkanoic acids.

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Question 49

Alkanols

Ethanol reacts with concentrated tetraoxosulphate(V) acid at a temperature above170 °C to form

  1. A.ethanone
  2. B.ethene
  3. C.ethyne
  4. D.ethanal.
Show the answer and explanation

Answer: B (ethene)

Explanation

Hot concentrated H2SO4 removes water from ethanol (dehydration), giving ethene.

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Question 50

Carbohydrates, Proteins and Amines

An example of oxidation - reduction enzyme is

  1. A.amylase
  2. B.protease
  3. C.lipase
  4. D.dehydrogenase.
Show the answer and explanation

Answer: D (dehydrogenase.)

Explanation

Dehydrogenases remove hydrogen from molecules, which is oxidation; amylase, protease and lipase are hydrolysing enzymes.

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