JAMB Chemistry 1990 past questions and answers

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

Question 1

Separation of Mixtures

Which of the following is a physical change?

  1. A.The bubbling of chlorine into water.
  2. B.The bubbling of chlorine into a jar containing hydrogen
  3. C.The dissolution of sodium chloride in water.
  4. D.The passing of steam over heated iron
Show the answer and explanation

Answer: C (The dissolution of sodium chloride in water.)

Explanation

Dissolving salt in water makes no new substance; evaporating the water gives the salt back. Chlorine reacts with water and with hydrogen, and steam reacts with hot iron, so those are chemical changes.

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Question 2

Kinetic Theory and States of Matter

Changes in the physical states of a chemical substance T are shown in the scheme below.
The letters X, Y and Z respectively represent

Diagram for this question
  1. A.sublimation, condensation and freezing
  2. B.sublimation, vapourization and solidification
  3. C.freezing, condensation and sublimation
  4. D.evaporation, liquefaction and solidification
Show the answer and explanation

Answer: A (sublimation, condensation and freezing)

Explanation

X goes straight from solid to gas (sublimation), Y goes from gas to liquid (condensation) and Z goes from liquid to solid (freezing).

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Question 3

Mole Concept and Stoichiometry

In the reaction: SnO2 + 2C → Sn + 2CO, the mass of coke containing 80% carbon required to reduce 0.302 kg of pure tin oxide is [Sn = 119, O = 16, C = 12]

  1. A.0.40 kg
  2. B.0.20 kg
  3. C.0.06 kg
  4. D.0.04 kg
Show the answer and explanation

Answer: C (0.06 kg)

Explanation

151 g of SnO2 needs 24 g of carbon, so 0.302 kg needs 0.048 kg of carbon. Coke is only 80% carbon, so mass of coke = 0.048 ÷ 0.80 = 0.06 kg.

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Question 4

Mole Concept and Stoichiometry

The Avogadro number of 24 g of magnesium is the same as that of

  1. A.1g of hydrogen molecules
  2. B.16 g of oxygen molecules
  3. C.32 g of oxygen molecules
  4. D.35.5 g of chlorine molecules.
Show the answer and explanation

Answer: C (32 g of oxygen molecules)

Explanation

24 g of magnesium is 1 mole. 32 g of O2 is also 1 mole, so it has the same number of particles; the other masses are only half a mole of molecules.

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Question 5

Gas Laws

If a gas occupies a container of volume 146 cm3 at 18°C and 0.971 atm, its volume in cm3 at s.t.p is

  1. A.133
  2. B.146
  3. C.266
  4. D.292
Show the answer and explanation

Answer: A (133)

Explanation

Use P1V1/T1 = P2V2/T2: V2 = 0.971 × 146 × 273 ÷ (1 × 291) ≈ 133 cm3.

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Question 6

Mole Concept and Stoichiometry

The volume occupied by 1.58 g of a gas at s.t.p is 500 cm3. What is the relative molecular mass of the gas? [G.M.V at s.t.p = 22.40 dm3].

  1. A.28
  2. B.32
  3. C.44
  4. D.71
Show the answer and explanation

Answer: D (71)

Explanation

One mole occupies 22 400 cm3, so molar mass = 1.58 × 22 400 ÷ 500 ≈ 71.

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Question 7

Gas Laws

Equal volumes of CO, SO2, NO2 and H2S were released into a room at the same point and time. Which of the following gives the order of diffusion of the gases to the opposite corner of the room? [S = 32, C = 12, O = 16, N = 14, H = 1].

  1. A.CO, SO2, NO2, H2S
  2. B.SO2, NO2, H2S, CO
  3. C.CO, H2S, SO2, NO2
  4. D.CO, H2S, NO2, SO2
Show the answer and explanation

Answer: D (CO, H2S, NO2, SO2)

Explanation

By Graham's law lighter gases diffuse faster. The molar masses are CO 28, H2S 34, NO2 46 and SO2 64, so they arrive in that order.

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Question 8

Gas Laws

A basic postulate of the kinetic theory of gases is that the molecules of a gas move in straight lines between collisions. This implies that

  1. A.collisions are perfectly elastic
  2. B.forces of repulsion exist
  3. C.forces of repulsion and attraction are in equilibrium
  4. D.collisions are inelastic.
Show the answer and explanation

Answer: C (forces of repulsion and attraction are in equilibrium)

Explanation

An object moves in a straight line when there is no net force on it. So between collisions the attractive and repulsive forces on a gas molecule must cancel out.

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Question 10

Chemical Bonding

Which of the following terms indicates the number of bonds that can be formed by an atom?

  1. A.Oxidation number
  2. B.Valence
  3. C.Atomic number
  4. D.Electronegativity
Show the answer and explanation

Answer: B (Valence)

Explanation

Valence (valency) is the combining power of an atom, i.e. the number of bonds it can form.

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Question 12

Atomic Structure

Chlorine, consisting of two isotopes of mass numbers 35 and 37, has an atomic mass of 35.5. The relative abundance of the isotope of mass number 37 is

  1. A.20
  2. B.25
  3. C.50
  4. D.75
Show the answer and explanation

Answer: B (25)

Explanation

Let x% be chlorine-37: 37x + 35(100 − x) = 3550, so 2x = 50 and x = 25%.

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Question 13

Mole Concept and Stoichiometry

10.0 dm3 of air containing H2S as an impurity was passed through a solution of Pb(NO3)2 until all the H2S had reacted. The precipitate of PbS was found to weigh 5.02 g. According to the equation: Pb(NO3)2 + H2S → PbS + 2HNO3 the percentage by volume of hydrogen sulphide in the air is [Pb = 207, S = 32, GMV at s.t.p = 22.4 dm3]

  1. A.50.2
  2. B.47.0
  3. C.4.70
  4. D.0.47
Show the answer and explanation

Answer: C (4.70)

Explanation

Moles of PbS = 5.02 ÷ 239 = 0.021, which equals moles of H2S. Volume = 0.021 × 22.4 = 0.470 dm3, which is 4.70% of 10.0 dm3.

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Question 14

Water

A blue solid, T, which weighed 5.0 g was placed on a table. After 8 hours, the resulting pink solid was found to weigh 5.5 g. It can be inferred that substance T

  1. A.is deliquescent
  2. B.is hydroscopic
  3. C.has some molecules of water of crystallization
  4. D.is efflorescent
Show the answer and explanation

Answer: B (is hydroscopic)

Explanation

T gained mass by absorbing water vapour from the air, yet it was still a solid. Absorbing moisture without forming a solution is what a hygroscopic substance does.

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Question 15

Electrolysis

The effluent of an industrial plant used in the electrolysis of concentrated brine, with a flowing mercury cathode may obtain impurities like

  1. A.oxygen
  2. B.hydrogen
  3. C.mercury (II) chloride
  4. D.hydrogen chloride.
Show the answer and explanation

Answer: C (mercury (II) chloride)

Explanation

In the mercury-cathode cell some mercury escapes and reacts with chlorine, so the waste water can contain poisonous mercury(II) chloride.

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Question 16

Solubility and Solutions

The solubility in moles per dm3 of 20 g of CuSO4 dissolved in 100 g of water at 180°C is [Cu = 63.5, S = 32, O = 16].

  1. A.0.13
  2. B.0.25
  3. C.1.25
  4. D.2.00
Show the answer and explanation

Answer: C (1.25)

Explanation

20 g ÷ 159.5 g/mol = 0.125 mol in 100 g of water. Scaling to 1000 g (about 1 dm3) gives 1.25 mol/dm3.

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Question 17

Solubility and Solutions

Smoke consists of

  1. A.solid particles dispersed in liquid
  2. B.solid or liquid particles dispersed in gas
  3. C.gas or liquid particles dispersed in liquid
  4. D.liquid particles dispersed in liquid.
Show the answer and explanation

Answer: B (solid or liquid particles dispersed in gas)

Explanation

Smoke is a colloid of tiny solid (and some liquid) particles spread through air, a gas.

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Question 18

Mole Concept and Stoichiometry

Na2C2O4 + CaCl2 → CaC2O4 + 2NaCl. Given a solution of 1.9 g of sodium oxalate in 50 g of water at room temperature, calculate the minimum volume of 0.1 M calcium chloride required to produce maximum calcium oxalate using the above equation.

  1. A.1.40 × 102 dm3
  2. B.1.40 × 102 cm3
  3. C.1.40 × 10−2 dm−2
  4. D.1.40 × 10−2 cm3.
Show the answer and explanation

Answer: B (1.40 × 102 cm3)

Explanation

Moles of Na2C2O4 = 1.9 ÷ 134 = 0.0142, and the same number of moles of CaCl2 is needed. Volume = 0.0142 ÷ 0.1 = 0.142 dm3 = 1.40 × 102 cm3.

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Question 19

Volumetric Analysis

2.0 g of a monobasic acid was made up to 250 cm3 with distilled water. 25.00 cm3 of this solution required 20.00 cm3 of 0.1 M NaOH solution for complete neutralization. The molar mass of the acid is

  1. A.200 g
  2. B.160 g
  3. C.100 g
  4. D.50 g
Show the answer and explanation

Answer: C (100 g)

Explanation

Moles of NaOH = 0.1 × 20/1000 = 0.002, so 25 cm3 holds 0.002 mol of acid and 250 cm3 holds 0.02 mol. Molar mass = 2.0 ÷ 0.02 = 100 g/mol.

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Question 20

Acids, Bases and Salts

What is the concentration of H+ ions in moles per dm3 of a solution of pH 4.398?

  1. A.4.0 × 10−5
  2. B.0.4 × 10−5
  3. C.4.0 × 10−3
  4. D.0.4 × 10−3
Show the answer and explanation

Answer: A (4.0 × 10−5)

Explanation

[H+] = 10−pH = 10−4.398 ≈ 4.0 × 10−5 mol/dm3.

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Question 22

Electrolysis

If 10.8 g of silver is deposited in a silver coulometer connected in series with a copper coulometer, the volume of oxygen liberated is [Ag = 108, Cu = 64, GMV at s.t.p = 22.40 dm3]

  1. A.0.56 dm3
  2. B.5.60 dm3
  3. C.11.20 dm3
  4. D.22.40 dm3
Show the answer and explanation

Answer: A (0.56 dm3)

Explanation

10.8 g of silver is 0.1 mol, so 0.1 faraday passed. Oxygen needs 4 faradays per mole (4OH− → O2 + 2H2O + 4e−), giving 0.025 mol × 22.4 = 0.56 dm3.

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Question 23

Electrolysis

0.1 faraday of electricity deposited 2.95 g of nickel during electrolysis of an aqueous solution. Calculate the number of moles of nickel that will be deposited by 0.4 faraday. [Ni = 58.7]

  1. A.0.20
  2. B.0.30
  3. C.0.04
  4. D.5.87
Show the answer and explanation

Answer: A (0.20)

Explanation

0.1 F deposits 2.95 ÷ 58.7 = 0.0503 mol of nickel. Four times the charge (0.4 F) deposits 4 × 0.0503 ≈ 0.20 mol.

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Question 24

Redox Reactions

Cr2O72− + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O. In the above equation, the oxidation number of chromium changed from

  1. A.+7 to +3
  2. B.+6 to +3
  3. C.+5 to +3
  4. D.−2 to +3.
Show the answer and explanation

Answer: B (+6 to +3)

Explanation

In Cr2O72−: 2x + 7(−2) = −2, so x = +6. It ends as Cr3+, so it goes from +6 to +3.

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Question 25

Redox Reactions

In the reaction IO3− + 5I− + 6H+ → 3I2 + 3H2O, the oxidizing agent is

  1. A.H+
  2. B.I−
  3. C.IO3−
  4. D.I2.
Show the answer and explanation

Answer: C (IO3−)

Explanation

Iodine in IO3− goes from +5 to 0, so IO3− is reduced; the substance that is reduced is the oxidizing agent.

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Question 26

Energy Changes

Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(s). If the heats of formation of Al2O3 and Fe2O3 are −1670 kJ mol−1 and −822 kJ mol−1 respectively, the enthalpy change in kJ for the reaction is

  1. A.+2492
  2. B.+848
  3. C.−848
  4. D.−2492.
Show the answer and explanation

Answer: C (−848)

Explanation

ΔH = sum of heats of formation of products − reactants = (−1670) − (−822) = −848 kJ. Elements (Al, Fe) have zero heat of formation.

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Question 27

Electrolysis

Iron galvanized with zinc is cathodically protected from corrosion. This is because

  1. A.zinc has a more positive oxidation potential than iron
  2. B.zinc has a less positive oxidation potential than iron
  3. C.both have the same oxidation potential
  4. D.zinc is harder than iron.
Show the answer and explanation

Answer: A (zinc has a more positive oxidation potential than iron)

Explanation

Zinc is higher in the activity series, so it loses electrons more easily than iron (more positive oxidation potential). Zinc corrodes instead of the iron.

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Question 28

Rates of Reaction

Which of the following samples will react fastest with dilute trioxonitrate (V) acid?

  1. A.5 g of lumps of CaCO3 at 25°C
  2. B.5 g of powdered CaCO3 at 25°C
  3. C.5 g of lumps of CaCO3 at 50°C
  4. D.5g of powdered CaCO3 at 50°C.
Show the answer and explanation

Answer: D (5g of powdered CaCO3 at 50°C.)

Explanation

A powder has a larger surface area and a higher temperature gives faster particles, so powdered CaCO3 at 50°C reacts fastest.

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Question 29

Chemical Equilibrium

In the reaction 2HI(g) ⇌ H2(g) + I2(g), ΔH = 10 kJ; the concentration of iodine in the equilibrium mixture can be increased by

  1. A.raising the pressure
  2. B.raising the temperature
  3. C.adding a catalyst
  4. D.lowering the pressure
Show the answer and explanation

Answer: B (raising the temperature)

Explanation

The forward reaction is endothermic (ΔH positive), so heating shifts the equilibrium to the right, making more I2. Pressure has no effect because both sides have 2 moles of gas.

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Question 30

Halogens

Which of the following gases can be collected by the upward displacement of air?

  1. A.NO
  2. B.H2
  3. C.NH3
  4. D.Cl2.
Show the answer and explanation

Answer: D (Cl2.)

Explanation

Upward displacement of air (downward delivery) is used for gases denser than air: the gas sinks and pushes the air up. Chlorine is much denser than air.

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Question 31

Nitrogen and its Compounds

The brown fumes given off when trioxonitrate (V) acid is heated consist of

  1. A.NO2 and O2
  2. B.H2O and NO2
  3. C.NO2, O2 and H2O
  4. D.NO, NO2 and H2O.
Show the answer and explanation

Answer: C (NO2, O2 and H2O)

Explanation

Heated nitric acid decomposes: 4HNO3 → 4NO2 + O2 + 2H2O. The brown colour is NO2, mixed with oxygen and steam.

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Question 32

Qualitative Analysis

Which of the following tests will completely identify any one of sulphur (IV) oxide, hydrogen, carbon (IV) oxide and nitrogen (II) oxide?

  1. A.Pass each gas into water and test with blue litmus paper.
  2. B.Pass each gas into lime water
  3. C.Expose each gas to atmospheric air
  4. D.Pass each gas to concentrated tetraoxosulphate (VI) acid.
Show the answer and explanation

Answer: C (Expose each gas to atmospheric air)

Explanation

Nitrogen(II) oxide is the only one of these gases that changes on meeting air: it turns brown as it forms NO2. Lime water and litmus each give the same result for both SO2 and CO2.

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Question 33

Nitrogen and its Compounds

In the Haber process for the manufacture of ammonia, the catalyst commonly used is finely divided

  1. A.vanadium
  2. B.platinum
  3. C.iron
  4. D.copper
Show the answer and explanation

Answer: C (iron)

Explanation

The Haber process uses finely divided iron as the catalyst for N2 + 3H2 ⇌ 2NH3.

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Question 34

Hydrogen, Oxygen and Oxides

A metallic oxide which reacts with both HCl and NaOH to give salt and water only can be classified as

  1. A.an acidic oxide
  2. B.a basic oxide
  3. C.a neutral oxide
  4. D.an amphoteric oxide
Show the answer and explanation

Answer: D (an amphoteric oxide)

Explanation

An oxide that reacts with both acids and alkalis to form a salt and water is amphoteric, e.g. ZnO or Al2O3.

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Question 35

Metals and their Compounds

Which of the following metals will liberate hydrogen from steam or dilute acid?

  1. A.Copper
  2. B.Iron
  3. C.Lead
  4. D.Mercury
Show the answer and explanation

Answer: B (Iron)

Explanation

Iron is above hydrogen in the activity series, so it reacts with steam and with dilute acids to give hydrogen. Copper and mercury do neither, and lead barely reacts.

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Question 36

Carbon and its Compounds

Coal fire should not be used in poorly ventilated rooms because

  1. A.of the accumulation of CO2 which causes deep sleep
  2. B.it is usually too hot
  3. C.of the accumulation of CO which causes suffocation
  4. D.it removes most of the gases in the rooms.
Show the answer and explanation

Answer: C (of the accumulation of CO which causes suffocation)

Explanation

With little air, coal burns incompletely to carbon(II) oxide, CO. CO binds to haemoglobin and stops the blood carrying oxygen, causing suffocation.

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Question 37

Metals and their Compounds

The major component of the slag from the production of iron is

  1. A.an alloy of calcium and iron
  2. B.coke
  3. C.impure iron
  4. D.calcium trioxosilicate (IV).
Show the answer and explanation

Answer: D (calcium trioxosilicate (IV).)

Explanation

In the blast furnace, calcium oxide from limestone reacts with silica impurities to form calcium silicate, CaSiO3 (calcium trioxosilicate(IV)), which is the slag.

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Question 38

Metals and their Compounds

Sodium hydroxide should be stored in properly closed containers because it

  1. A.readily absorbs water vapour from the air
  2. B.is easily oxidized by atmospheric oxygen
  3. C.turns golden yellow when exposed to light
  4. D.melts at a low temperature
Show the answer and explanation

Answer: A (readily absorbs water vapour from the air)

Explanation

Sodium hydroxide is deliquescent: it absorbs water vapour (and carbon(IV) oxide) from the air and turns into a solution if left open.

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Question 39

Metals and their Compounds

To make coloured glasses, small quantities of oxides of metals which form coloured silicates are often added to the reaction mixture consisting of Na2CO3 and SiO2. Such a metal is

  1. A.potassium
  2. B.barium
  3. C.zinc
  4. D.copper
Show the answer and explanation

Answer: D (copper)

Explanation

Transition metals such as copper form coloured compounds; copper oxide gives glass a blue-green colour. Potassium, barium and zinc compounds are colourless.

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Question 40

Qualitative Analysis

Which of the following compounds gives a yellow residue when heated and also reacts with aqueous sodium hydroxide to give a white gelatinous precipitate soluble in excess sodium hydroxide solution?

  1. A.(NH4)2CO3
  2. B.ZnCO3
  3. C.Al2(SO4)3
  4. D.PbCO3.
Show the answer and explanation

Answer: B (ZnCO3)

Explanation

ZnCO3 decomposes on heating to ZnO, which is yellow when hot. Zn2+ gives a white gelatinous Zn(OH)2 precipitate that dissolves in excess NaOH.

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Question 42

Introduction to Organic Chemistry

The structure of cis-2-butene is (see the structures in the figure)

Diagram for this question
  1. A.CH3—CH = CH—CH3.
  2. B.Structure B: CH3 and H on each carbon of C=C, with both CH3 groups on the same side
  3. C.Structure C: CH3 and H on each carbon of C=C, with the CH3 groups on opposite sides
  4. D.Structure D: CH3 and H on one carbon of C=C, two CH3 groups on the other
Show the answer and explanation

Answer: B (Structure B: CH3 and H on each carbon of C=C, with both CH3 groups on the same side)

Explanation

In a cis isomer the two like groups are on the same side of the double bond. Structure B has both CH3 groups on the same side; C is the trans isomer.

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Question 43

Introduction to Organic Chemistry

What is the IUPAC name for the hydrocarbon CH3—C(CH2CH3) = CH—CH(CH3)—CH3? (structure in the figure)

Diagram for this question
  1. A.2-ethyl-4-methylpent-2-ene
  2. B.3,5-dimethylhex-3-ene
  3. C.2, 4-dimethylhex-3-ene
  4. D.2-methyl-4-ethylpent-3-ene.
Show the answer and explanation

Answer: C (2, 4-dimethylhex-3-ene)

Explanation

The longest chain with the double bond has six carbons (it runs through the ethyl group). Numbered to give the lowest locants, the double bond is at C3 and the methyl groups at C2 and C4: 2,4-dimethylhex-3-ene.

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Question 44

Hydrocarbons

CH3—C ≡ CH (Na, liq NH3) → P. Compound P, in the above reaction, is

  1. A.CH3—C(Na) = CH NH2
  2. B.CH3—C(NH2) = CH Na
  3. C.CH3—C ≡ C—Na
  4. D.CH3—C ≡ C—NH2.
Show the answer and explanation

Answer: C (CH3—C ≡ C—Na)

Explanation

The hydrogen on a C≡C carbon at the end of a chain is slightly acidic. Sodium (in liquid ammonia) replaces it to form sodium propynide, CH3—C≡C—Na, and the triple bond stays.

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Question 45

Alkanols

The label on a reagent bottle containing a clear organic liquid dropped off. The liquid was neutral to litmus and gave a colourless gas with metallic sodium. The liquid must be an

  1. A.alkanoate
  2. B.alkene
  3. C.alkanol
  4. D.alkane.
Show the answer and explanation

Answer: C (alkanol)

Explanation

Alkanols are neutral to litmus but their –OH group reacts with sodium to give hydrogen gas. Alkanoic acids would turn litmus red; alkanes and alkenes don't react with sodium.

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Question 46

Acids, Bases and Salts

HOOC—COOH + NaOH → HOOC—COO−Na+ + H2O
The above reaction is an example of

  1. A.a displacement reaction
  2. B.a neutralization reaction
  3. C.an elimination reaction
  4. D.saponification
Show the answer and explanation

Answer: B (a neutralization reaction)

Explanation

An acid (ethanedioic acid) reacts with a base (NaOH) to give a salt and water, which is neutralization.

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Question 47

Alkanoic Acids and Esters

Alkanoic acids have low volatility compared with alkanols because they

  1. A.are more polar than alkanols
  2. B.have two oxygen atoms while alkanols have one
  3. C.form hydrogen bonds while alkanols do not
  4. D.form two hydrogen bonds while alkanols form one.
Show the answer and explanation

Answer: D (form two hydrogen bonds while alkanols form one.)

Explanation

Two acid molecules join by two hydrogen bonds to form a dimer, while alkanol molecules link by single hydrogen bonds. The stronger attraction makes acids less volatile.

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Question 48

Petroleum and Coal

The octane number of a fuel whose performance is the same as that of a mixture of 55 g of 2,2,4-trimethyl pentane and 45 g of n-heptane is

  1. A.45
  2. B.55
  3. C.80
  4. D.100.
Show the answer and explanation

Answer: B (55)

Explanation

Octane number is the percentage of 2,2,4-trimethylpentane (iso-octane) in a mixture with heptane that performs the same: 55 out of 100 g is 55%.

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Question 49

Carbohydrates, Proteins and Amines

Which of the following is formed when maltose reacts with concentrated tetraoxosulphate (VI) acid?

  1. A.Carbon (IV) oxide
  2. B.Coal tar
  3. C.Charcoal
  4. D.Toxic fumes.
Show the answer and explanation

Answer: C (Charcoal)

Explanation

Concentrated sulphuric acid is a strong dehydrating agent. It removes the hydrogen and oxygen of the sugar as water, leaving black carbon (charcoal).

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Question 50

Hydrocarbons

Which of the following compounds represents the polymerization product of ethyne? (see the structures in the figure)

Diagram for this question
  1. A.Ring A: six-membered ring with three alternating double bonds
  2. B.Ring B: six-membered ring with two double bonds
  3. C.Ring C: two fused six-membered rings
  4. D.Ring D: six-membered ring with one double bond
Show the answer and explanation

Answer: A (Ring A: six-membered ring with three alternating double bonds)

Explanation

Three ethyne molecules join (passed over a hot catalyst) to form benzene, C6H6, the ring with three alternating double bonds.

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