JAMB Chemistry 2013 Objective Past Questions

All 50 questions from the Joint Admissions and Matriculation Board (JAMB) Chemistry 2013 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

Chemistry 2013 Objective — Question 1

Which Question Paper Type of Chemistry is given to you?

• A. Type D
• B. Type I
• C. Type B
• D. Type UCorrect

Explanation

This is a Type U Chemistry Question Paper.

Chemistry 2013 Objective — Question 2

The presence of an impurity in a substance will cause the melting point to

• A. increase
• B. be stable
• C. be zero
• D. reduceCorrect

Explanation

The presence of impurity causes the boiling point of a liquid to increase and the melting point of a solid to decrease.

Chemistry 2013 Objective — Question 3

What volume of carbon(II) oxide is produced by reacting excess carbon with 10dm³ of oxygen?

• A. 15dm³
• B. 10dm³
• C. 5dm³
• D. 20dm³Correct

Explanation

2C(s) + O2(g) → 2CO(g). Since the mole ratio of oxygen to carbon(II) oxide (from the balanced equation) is 1:2, 10dm³ of oxygen will produce 20dm³ of carbon(II) oxide.

Chemistry 2013 Objective — Question 4

From the diagram above, at P a real gas is represented by

• A. X
• B. Y
• C. ZCorrect
• D. W

Explanation

Graph Z portrays a typical ideal gas. For an ideal gas, the value of PV is constant regardless of the value of pressure. A real gas (at low pressure P) deviates from this and is represented by curve Z's neighbouring/curving line — per the printed key the answer is C.

Chemistry 2013 Objective — Question 5

The rate of diffusion of a gas Y is twice that of Z. If the relative molecular mass of Y is 64 and the two gases diffuse under the same conditions, find the relative molecular mass of Z.

• A. 8
• B. 16
• C. 32
• D. 4

Explanation

Using Graham's law: R₁/R₂ = √(M₂/M₁). Since R(Y)=2R(Z): 2 = √(M(Z)/64), so M(Z)/64 = 4, giving M(Z) = 256. This value does not match any of the given options — the printed key states 'No Correct Option'.

Chemistry 2013 Objective — Question 6

The radioisotope used in industrial radiography for the rapid checking of faults in welds and casting is

• A. cobalt-60Correct
• B. iodine-131
• C. carbon-14
• D. phosphorus-32

Explanation

The radioisotope used in industrial radiography for the rapid checking of faults in welds and casting is cobalt-60.

Chemistry 2013 Objective — Question 7

How many unpaired electrons are in the p-orbitals of a fluorine atom?

• A. 1Correct
• B. 2
• C. 3
• D. 0

Explanation

F = 1s²2s²2p⁵. As can be seen, there is only one unpaired electron, and it is in the 2p orbital.

Chemistry 2013 Objective — Question 8

The radioactive emission with the least ionization power is

• A. γ-raysCorrect
• B. β-particles
• C. α-particles
• D. X-rays

Explanation

Gamma rays have the lowest ionization power and the highest penetration power. The opposite is true for alpha particles.

Chemistry 2013 Objective — Question 9

The shape of the carbon(IV) oxide molecule is

• A. angular
• B. tetrahedral
• C. pyramidal
• D. linearCorrect

Explanation

Carbon(IV) oxide is a linear molecule. Other linear molecules include BeCl2, H2, O2, N2, F2, Cl2, Br2 and I2.

Chemistry 2013 Objective — Question 10

Which of the following molecules is held together by hydrogen bond?

• A. H2SO4
• B. HFCorrect
• C. CH4
• D. HBr

Explanation

Hydrogen bonding occurs in compounds where hydrogen is linked to very small electronegative atoms such as nitrogen, oxygen and fluorine. The strongest hydrogen bonding occurs in hydrogen fluoride (HF).

Chemistry 2013 Objective — Question 11

The bond formed between two elements with electron configurations 1s²2s²2p⁶3s¹ and 1s²2s²2p⁵ is

• A. dative
• B. ionicCorrect
• C. metallic
• D. covalent

Explanation

The element with configuration 1s²2s²2p⁶3s¹ will lose an electron to the one with 1s²2s²2p⁵; each then attains 1s²2s²2p⁶. This is ionic bonding.

Chemistry 2013 Objective — Question 12

The constituent of air that acts as a diluent is

• A. noble gases
• B. oxygen
• C. nitrogenCorrect
• D. carbon(IV) oxide

Explanation

Nitrogen is an important diluent of air. Without nitrogen, combustion would proceed out of proportion.

Chemistry 2013 Objective — Question 13

Steam changes the colour of anhydrous cobalt(II) chloride from

• A. blue to pinkCorrect
• B. white to blue
• C. white to red
• D. blue to white

Explanation

Water (steam or otherwise) turns anhydrous cobalt(II) chloride from blue to pink. Anhydrous CuSO4, by contrast, changes from white to blue when water is added.

Chemistry 2013 Objective — Question 14

An example of a hygroscopic substance is

• A. CaCl2
• B. NaOH(aq)
• C. CaO(s)Correct
• D. MgCl2(aq)

Explanation

Hygroscopic substances absorb water and become sticky when exposed to the atmosphere. CaO is a good example.

Chemistry 2013 Objective — Question 15

If 24.4g of lead(II) trioxonitrate(V) were dissolved in 42g of distilled water at 20°C, calculate the solubility of the solute in gdm⁻³.

• A. 5.810
• B. 58.100
• C. 581.000Correct
• D. 0.581

Explanation

Volume of water = 42cm³ = 0.042dm³ (taking density of water as 1gcm⁻³). Solubility = 24.4g ÷ 0.042dm³ = 580.95 ≈ 581gdm⁻³.

Chemistry 2013 Objective — Question 16

The solvent used for removing grease stain is

• A. ethanol
• B. solution of borax in water
• C. turpentine
• D. ammonia solutionCorrect

Explanation

In laundries, liquid ammonia is a favourite choice for removing grease and oil stains.

Chemistry 2013 Objective — Question 17

In a water body, too much sewage leads to

• A. an increase in the bacterial population which reduces the level of oxygen in the waterCorrect
• B. a decrease in bacterial population which increases the level of oxygen in the water
• C. a decrease in the temperature of the water which causes death of aquatic animals
• D. an increase in the number of aquatic animals in the water

Explanation

Too much sewage causes the population of (usually aerobic) bacteria to increase, which consequently decreases the level of oxygen in the water body.

Chemistry 2013 Objective — Question 18

10.0dm³ of water added to 2.0dm³ of 2.5dm³ solution of HCl. What is the concentration of the final solution in moldm⁻³?

• A. 2.0
• B. 0.5
• C. 0.4Correct
• D. 8.0

Explanation

C₁=2moldm⁻³, V₁=2.5dm³, V₂=10dm³+2.5dm³=12.5dm³. Using C₁V₁=C₂V₂: C₂ = (2×2.5)/12.5 = 0.4moldm⁻³.

Chemistry 2013 Objective — Question 19

Three drops of a 1.0moldm⁻³ solution of HCl was added to 20cm³ of a solution of pH 6.4. The pH of the resulting solution will

• A. be greater than 6.4
• B. be unaltered
• C. close to that of pure water
• D. be less than 6.4Correct

Explanation

Adding HCl (an acid) to a solution makes it more acidic, so the pH value will decrease (be less than 6.4).

Chemistry 2013 Objective — Question 20

Which of the following compounds is not a salt?

• A. sodium trioxocarbonate (IV)
• B. zinc chloride
• C. aluminium oxideCorrect
• D. sodium hydrogen trioxosulphate (IV)

Explanation

Sodium trioxocarbonate(IV), zinc chloride and sodium hydrogen trioxosulphate(IV) are salts, while aluminium oxide (Al2O3) is an amphoteric oxide, not a salt.

Chemistry 2013 Objective — Question 21

An insoluble salt can be prepared by

• A. the action of dilute acid on an insoluble base
• B. the reaction of metals with an acid
• C. reaction of trioxocarbonate (IV) with an acid
• D. double decompositionCorrect

Explanation

The best way of preparing insoluble salts is double decomposition. For instance, silver chloride (AgCl) can be produced from the double decomposition reaction between AgNO3 and NaCl.

Chemistry 2013 Objective — Question 22

2H2O(l) + 2F2(g) ⇌ 4HF(aq) + O2(g). In the reaction above, the substance that undergoes reduction here is

• A. fluorineCorrect
• B. oxygen
• C. hydrogen
• D. water

Explanation

The substance that undergoes reduction is fluorine. Its oxidation number decreases from 0 in F2 to -1 in HF.

Chemistry 2013 Objective — Question 23

Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s). In the reaction above, the oxidizing agent is

• A. Zn(s)
• B. Cu(s)
• C. CuSO4(aq)Correct
• D. ZnO(s)

Explanation

The oxidation number of zinc increases from zero to +2 in ZnSO4; zinc has undergone oxidation, so it is the reducing agent, while CuSO4 is the oxidizing agent.

Chemistry 2013 Objective — Question 24

In an electrochemical cell, polarization is caused by A. tetraoxosulphate (VI) acid B. hydrogen sulphide C. chlorine D. oxygen

• A. tetraoxosulphate (VI) acid
• B. hydrogen sulphide
• C. chlorine
• D. oxygenCorrect

Explanation

Polarization is caused by the formation of hydrogen (from reduction) coating the electrode; it is reduced by the use of depolarizers such as manganese(IV) oxide (MnO2). The gas responsible for polarization here is oxygen/hydrogen gas evolved at the electrode — per the printed key the answer is D.

Chemistry 2013 Objective — Question 25

Calculate the volume in cm³ of oxygen evolved at s.t.p. when a current of 5A is passed through acidified water for 193s.

• A. 0.224Correct
• B. 56.000
• C. 224.000
• D. 0.056

Explanation

4OH⁻ → 2H2O + O2 + 4e⁻. Q = It = (5×193)C = 965C. 1 mole of O2 (22400cm³) requires 4F (96500C×4) of electricity. Thus 965C produces (22400/(96500×4))×965 = 56cm³ — per the printed key the selected answer is A.

Chemistry 2013 Objective — Question 26

In an endothermic reaction, if there is a loss in entropy, the reaction will

• A. not be spontaneousCorrect
• B. be spontaneous
• C. be at equilibrium
• D. be indeterminate

Explanation

In an endothermic reaction, if there is a loss in entropy, the reaction will not be spontaneous.

Chemistry 2013 Objective — Question 27

2SO2(g) + O2(g) ⇌ 2SO3(g) ΔH = -395.7kJmol⁻¹. In the reaction above, the concentration of SO3(g) can be increased by

• A. increasing temperature
• B. the addition of catalyst
• C. decreasing the pressure
• D. decreasing the temperatureCorrect

Explanation

Since the reaction is exothermic, the forward reaction (production of more SO3) is favoured by a decrease in temperature.

Chemistry 2013 Objective — Question 28

The minimum amount of energy required for a reaction to take place is A. activation energy B. kinetic energy C. lattice energy D. ionization energy

• A. activation energyCorrect
• B. kinetic energy
• C. lattice energy
• D. ionization energy

Explanation

For a reaction to take place, the colliding particles must have energies greater than or equal to the activation energy.

Chemistry 2013 Objective — Question 29

In the graph above, the activation energy of the catalyzed reaction is

• A. 250kJ
• B. 200kJ
• C. 100kJCorrect
• D. 300kJ

Explanation

The lower of the two curves is always the profile diagram of the catalysed reaction. For the catalyzed reaction, activation energy = 200kJ − 100kJ = 100kJ.

Chemistry 2013 Objective — Question 30

3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g). The equilibrium constant K of the reaction above is represented as

• A. [H2]⁴/[H2O]⁴Correct
• B. [Fe]³[H2O]⁴/[Fe3O4][H2]⁴
• C. [Fe3O4][H2]⁴/[Fe][H2O]⁴
• D. [H2O]⁴/[H2]⁴

Explanation

For a heterogeneous equilibrium, species in the solid and liquid phases are not featured in the expression for the equilibrium constant. K = [H2]⁴/[H2O]⁴.

Chemistry 2013 Objective — Question 31

Which of the following compounds is a neutral oxide?

• A. Sulphur (IV) oxide
• B. Carbon (II) oxideCorrect
• C. Carbon (IV) oxide
• D. Sulphur (VI) oxide

Explanation

Sulphur(IV) oxide, carbon(IV) oxide and sulphur(VI) oxide are acidic oxides, while carbon(II) oxide is a neutral oxide.

Chemistry 2013 Objective — Question 32

In the laboratory preparation of ammonia, the flask is placed in a slanting position so as to

• A. enhance the speed of the reaction
• B. prevent formation of precipitate
• C. prevent condensed water from breaking the reaction flaskCorrect
• D. enable the proper mixing of the reactants in the flask

Explanation

The flask is placed in a slanting position to prevent condensed water from flowing back and breaking the hot reaction flask.

Chemistry 2013 Objective — Question 33

Which of the following gases is employed as an anaesthesia?

• A. NH3
• B. NO
• C. N2OCorrect
• D. NO2

Explanation

Dinitrogen oxide, N2O, also known as laughing gas, is used as a mild anaesthetic for minor surgical operations.

Chemistry 2013 Objective — Question 34

Sulphur(IV) oxide is a strong reducing agent in the presence of water due to the formation of

• A. hydrogen sulphide
• B. trioxosulphate (IV) ionCorrect
• C. hydroxide ion
• D. sulphur (IV) oxide

Explanation

Sulphur(IV) oxide is a good reducing agent in water, as it produces trioxosulphate(IV) ion in solution — a species that readily donates electrons to potential oxidizing agents.

Chemistry 2013 Objective — Question 35

A metal that forms soluble trioxosulphate (IV) salt is

• A. manganese
• B. aluminium
• C. barium
• D. potassiumCorrect

Explanation

All salts of sodium, potassium and ammonium are soluble, regardless of the accompanying negative ion.

Chemistry 2013 Objective — Question 36

Copper is displaced from the solution of its salts by most metals because it A. is a transition element B. has completely filled 3d-orbitals C. is at the bottom of the activity series D. is very reactive

• A. is a transition element
• B. has completely filled 3d-orbitals
• C. is at the bottom of the activity seriesCorrect
• D. is very reactive

Explanation

Copper is commonly displaced from its solutions by most metals because it is almost at the bottom of the electrochemical (activity) series. Only mercury, silver, gold and platinum are less reactive than copper.

Chemistry 2013 Objective — Question 37

The coloured nature of transition metal ions is associated with their partially filled

• A. p-orbital
• B. d-orbitalCorrect
• C. f-orbital
• D. s-orbital

Explanation

The coloured nature of transition metal ions is associated with their partially filled d-orbital.

Chemistry 2013 Objective — Question 38

Aluminium containers are frequently used to transport trioxonitrate (V) acid because aluminium

• A. does not corrode
• B. has a silvery-white appearance
• C. has a low density
• D. does not react with the acidCorrect

Explanation

No matter how concentrated the acid is, aluminium does not react with trioxonitrate(V) acid — this is attributable to the formation of the protective aluminium oxide layer.

Chemistry 2013 Objective — Question 39

2-methylbutan-2-ol is an example of a

• A. secondary alkanol
• B. tertiary alkanolCorrect
• C. dihydric alkanol
• D. primary alkanol

Explanation

2-methylbutan-2-ol is a tertiary alkanol — a tertiary alkanol is one in which the carbon bearing the hydroxyl group is attached to three alkyl groups.

Chemistry 2013 Objective — Question 40

The reaction between ammonia and ethylethanoate produces

• A. ethanol and propanamide
• B. ethanol and ethanamideCorrect
• C. propanol and ethanamide
• D. propanol and propanamide

Explanation

Ammonia reacts with ethylethanoate to produce ethanol and ethanamide: CH3COOC2H5 + NH3 → CH3CONH2 + C2H5OH.

Chemistry 2013 Objective — Question 41

The decarboxylation of ethanoic acid will produce carbon(IV) oxide and

• A. propane
• B. butane
• C. methaneCorrect
• D. ethane

Explanation

The decarboxylation of ethanoic acid produces carbon(IV) oxide and methane: CH3COOH → CH4 + CO2.

Chemistry 2013 Objective — Question 42

The compound H–C(OH)(H)–C(H)(H)–H (structure showing an -OH on a carbon chain) above is an

• A. alkanal
• B. alkanolCorrect
• C. alkanone
• D. alkanoate

Explanation

The compound shown has a hydroxyl (–OH) group attached to a carbon chain, making it an alkanol.

Chemistry 2013 Objective — Question 43

The compound that will react with sodium hydroxide to form salt and water only is

• A. CH3CH=CH2
• B. CH3CH2COOHCorrect
• C. C6H12O6
• D. (CH3)3COH

Explanation

A compound that reacts with sodium hydroxide (a base) to give salt and water only must be an acid. CH3CH2COOH (propanoic acid) is an acid: CH3CH2COOH + NaOH → CH3CH2COONa + H2O.

Chemistry 2013 Objective — Question 44

Which of the following compounds in solution turn red litmus paper blue?

• A. RNH2Correct
• B. R-C(=O)-R
• C. R'OR"
• D. R-C(=O)-N(R)R

Explanation

Just like ammonia, amines (RNH2) can turn damp red litmus paper blue.

Chemistry 2013 Objective — Question 45

The dehydration of ammonium salt of alkanoic acids produces a compound with the general formula

• A. R-NH2
• B. R-C(=O)-R
• C. R-C(=O)-OR
• D. R-C(=O)-NH2Correct

Explanation

The dehydration of the ammonium salt of alkanoic acids produces alkanamides. Amides/alkanamides have the general formula RCONH2.

Chemistry 2013 Objective — Question 46

Which of the following fractions is used as raw material for the cracking process?

• A. Bitumen
• B. Diesel oilsCorrect
• C. Kerosine
• D. Lubricating oils

Explanation

In cracking, less volatile crude oil fractions (such as diesel oils) are passed through a chamber at very high temperature to produce more volatile fractions.

Chemistry 2013 Objective — Question 47

An organic compound with a pleasant smell is likely to have a general formula

• A. CnH2n+1COOCnH2n+1Correct
• B. CnH2n+1COCnH2n+1
• C. CnH2n+1CHO
• D. CnH2n+1COOH

Explanation

Alkanoates (esters) are sweet-smelling organic compounds. Their general molecular formula is CnH2n+1COOCnH2n+1, or RCOOR', where R and R' may be the same or different alkyl/aryl groups.

Chemistry 2013 Objective — Question 48

A primary amide is generally represented by the formula

• A. RCONHR
• B. RCONR2
• C. RCOOR
• D. RCONH2Correct

Explanation

A primary amide has the general formula RCONH2.

Chemistry 2013 Objective — Question 49

The IUPAC nomenclature for the compound CH3-C(CH3)(H)-CH2-CH=CH2 above is

• A. 2-methylpent-1-ene
• B. 2-methylpent-4-ene
• C. 4-methylpent-1-eneCorrect
• D. 4-methylpent-2-ene

Explanation

The longest continuous chain has five carbon atoms with a double bond starting at the first carbon atom. The parent name is pent-1-ene, with a methyl attachment on carbon-4. Hence the complete name is 4-methylpent-1-ene.

Chemistry 2013 Objective — Question 50

An organic compound contains 60% carbon, 13.3% hydrogen and 26.7% oxygen. Calculate the empirical formula.

• A. C2H5O2
• B. C2H6O
• C. C3H12O
• D. C3H8OCorrect

Explanation

Moles: C = 60/12 = 5, H = 13.3/1 = 13.3, O = 26.7/16 = 1.67. Dividing by the smallest (1.67): C = 3, H = 8, O = 1. Empirical formula = C3H8O.