JAMB Chemistry 1999

33 reviewed questions with answers and explanations.

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Question 1

200 cm³ each of 0.1 mol dm⁻³ lead(II) nitrate and hydrochloric acid are mixed. Assuming lead(II) chloride is completely insoluble, what mass precipitates? [Pb = 207, Cl = 35.5]

  1. 2.78 g
  2. 5.56 g
  3. 8.34 g
  4. 11.12 g
Answer and explanation

A: 2.78 g

Each solution contains 0.200 × 0.100 = 0.0200 mol solute. Pb²⁺ + 2Cl⁻ → PbCl₂ requires two chloride ions per lead ion, so chloride limits the precipitate to 0.0100 mol. M(PbCl₂) = 207 + 2(35.5) = 278 g mol⁻¹, giving 2.78 g.

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Question 2

56.00 cm³ of a gas at STP weighs 0.11 g. What is its vapour density relative to hydrogen? [Molar gas volume at STP = 22.4 dm³ mol⁻¹]

  1. 11.00
  2. 22.00
  3. 33.00
  4. 44.00
Answer and explanation

B: 22.00

The amount of gas is 0.05600/22.4 = 0.00250 mol. Its molar mass is 0.11/0.00250 = 44 g mol⁻¹. Vapour density relative to hydrogen is half the relative molecular mass, so it is 22.00.

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Question 3

Which gas diffuses fastest through a porous plug under the same conditions? [H = 1, C = 12, N = 14, O = 16]

  1. Propane
  2. Oxygen
  3. Methane
  4. Ammonia
Answer and explanation

C: Methane

At the same conditions, diffusion through a porous plug is faster for lower molar mass. The masses are propane 44, oxygen 32, methane 16 and ammonia 17 g mol⁻¹. Methane is the lightest and diffuses fastest.

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Question 4

Which listed substance gains mass by combining with oxygen when heated in air?

  1. Helium
  2. Magnesium
  3. Copper pyrites
  4. Glass
Answer and explanation

B: Magnesium

Magnesium combines with oxygen when heated in air: 2Mg + O₂ → 2MgO. The oxygen added to the solid increases its mass. This is different from roasting a sulfide, where sulfur can leave as a gas.

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Question 5

A fixed amount of ideal gas is initially at 0 °C and 9 atm. What is its final temperature if the pressure falls to 3 atm at constant volume? [0 °C = 273 K]

  1. 91 K
  2. 182 K
  3. 273 K
  4. 819 K
Answer and explanation

A: 91 K

At constant volume for a fixed amount of ideal gas, P/T is constant. The initial temperature is 273 K. Thus T₂ = 273 × (3/9) = 91 K.

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Question 6

Solubility curves for two solids P and Q both rise as temperature increases. P is more soluble than Q at each shown temperature. Which method uses this difference to separate their mixture into successive crystal fractions?

  1. Distillation
  2. Fractional distillation
  3. Crystallisation
  4. Fractional crystallisation
Answer and explanation

D: Fractional crystallisation

The solids have different solubilities at the same temperature. Controlled cooling and successive collection of crystals can separate them into fractions. This process is fractional crystallisation.

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Question 7

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). What mass of magnesium reacts completely with 250 cm³ of 0.5 mol dm⁻³ HCl? [Mg = 24]

  1. 0.3 g
  2. 1.5 g
  3. 2.4 g
  4. 3.0 g
Answer and explanation

B: 1.5 g

The amount of HCl is 0.250 × 0.5 = 0.125 mol. Mg + 2HCl → MgCl₂ + H₂ requires half as many moles of Mg: 0.0625 mol. With Mg = 24, the mass is 0.0625 × 24 = 1.5 g.

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Question 9

In which sample do water molecules have the most disordered spatial arrangement?

  1. Ice at −10 °C
  2. Ice at 0 °C
  3. Liquid water at 100 °C
  4. Steam at 100 °C
Answer and explanation

D: Steam at 100 °C

The gaseous state has far more possible spatial arrangements than the solid or liquid states. Steam at 100 °C therefore has the greatest positional disorder among the choices. Equal temperature does not imply liquid and vapour have equal entropy.

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Question 10

Removing one 3s electron from each atom in one mole of gaseous sodium atoms requires about 496 kJ. What is this energy called?

  1. Electron affinity
  2. Ionisation energy
  3. Activation energy
  4. Electronegativity
Answer and explanation

B: Ionisation energy

First ionisation energy is the energy required to remove one electron from each gaseous neutral atom in one mole, forming gaseous singly charged ions. For sodium, that electron is in the 3s orbital.

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Question 12

Which method can soften hard water?

  1. Chlorination
  2. Passing over activated charcoal
  3. Using an ion-exchange resin
  4. Aeration
Answer and explanation

C: Using an ion-exchange resin

A suitable ion-exchange resin removes hardness-causing Ca²⁺ and Mg²⁺ ions by replacing them with ions such as Na⁺. Chlorination and aeration do not perform this exchange.

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Question 14

Four positions in a periodic-table section are labelled: W is in Group I, period 3; X in Group IV, period 2; Y in Group VII, period 3; and Z in Group 0, period 2. Which element is least reactive?

  1. W
  2. X
  3. Y
  4. Z
Answer and explanation

D: Z

Z occupies Group 0, the noble-gas group, and has a complete outer electron shell. It is therefore the least reactive of the labelled elements. W is an alkali metal, X is carbon and Y is a halogen.

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Question 16

Which listed compound has the strongest intermolecular hydrogen bonding, considering atomic size and electronegativity?

  1. HF
  2. NH₃
  3. CH₄
  4. HCl
Answer and explanation

A: HF

Fluorine is very small and highly electronegative, making the H–F bond strongly polar and allowing strong intermolecular hydrogen bonding. Hydrogen bonding in NH₃ is weaker; CH₄ and HCl lack the strong conventional N–H, O–H or F–H pattern.

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Question 17

0.25 mol HCl is dissolved and made up to 0.50 dm³. A 15.00 cm³ portion neutralises 12.50 cm³ of aqueous sodium carbonate completely. What is the carbonate concentration?

  1. 0.30 mol dm⁻³
  2. 0.40 mol dm⁻³
  3. 0.50 mol dm⁻³
  4. 0.60 mol dm⁻³
Answer and explanation

A: 0.30 mol dm⁻³

The HCl concentration is 0.25/0.50 = 0.50 mol dm⁻³. The 15.00 cm³ portion contains 0.00750 mol HCl. Since Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O, the carbonate amount is 0.00375 mol. Dividing by 0.01250 dm³ gives 0.30 mol dm⁻³.

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Question 18

Which order gives increasing oxidation number of the transition elements in these compounds?

  1. V₂O₅ < K₂Cr₂O₇ < KMnO₄
  2. K₂Cr₂O₇ < KMnO₄ < V₂O₅
  3. KMnO₄ < K₂Cr₂O₇ < V₂O₅
  4. KMnO₄ < V₂O₅ < K₂Cr₂O₇
Answer and explanation

A: V₂O₅ < K₂Cr₂O₇ < KMnO₄

Oxygen is −2 and potassium +1. In V₂O₅, vanadium is +5; in K₂Cr₂O₇, chromium is +6; and in KMnO₄, manganese is +7. The increasing order is therefore V₂O₅, K₂Cr₂O₇, KMnO₄.

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Question 20

What is observed when aqueous sulfuric acid, potassium iodate(V) and potassium iodide are mixed?

  1. A white precipitate forms
  2. A green precipitate forms
  3. The mixture remains colourless
  4. The mixture becomes reddish-brown
Answer and explanation

D: The mixture becomes reddish-brown

Acidified iodate oxidises iodide to iodine: IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O. Dissolved iodine, including triiodide in excess iodide, gives the reddish-brown colour.

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Question 23

Cu(IO₃)₂ has Ksp = 1.08 × 10⁻⁷. Assuming negligible reaction of its ions with water, what is its molar solubility in pure water?

  1. 2.7 × 10⁻⁸ mol dm⁻³
  2. 9.0 × 10⁻⁸ mol dm⁻³
  3. 3.0 × 10⁻³ mol dm⁻³
  4. 9.0 × 10⁻³ mol dm⁻³
Answer and explanation

C: 3.0 × 10⁻³ mol dm⁻³

Let the molar solubility be s. Dissociation gives [Cu²⁺] = s and [IO₃⁻] = 2s. Hence Ksp = s(2s)² = 4s³ = 1.08 × 10⁻⁷. Therefore s = ∛(2.70 × 10⁻⁸) = 3.0 × 10⁻³ mol dm⁻³.

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Question 24

In introductory thermodynamic terminology, entropy and enthalpy are associated respectively with

  1. Degree of disorder and heat content
  2. Heat content and degree of disorder
  3. Heat content only
  4. Degree of disorder only
Answer and explanation

A: Degree of disorder and heat content

Entropy describes the number of accessible microscopic arrangements, often introduced as disorder. Enthalpy is H = U + pV; its change equals heat transferred at constant pressure when only pressure–volume work occurs. The introductory pairing is disorder and heat content, respectively.

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Question 25

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), which listed catalyst increases the rate of sulfur trioxide production in the Contact process?

  1. Manganese(IV) oxide
  2. Finely divided iron
  3. Vanadium(V) oxide
  4. Nickel
Answer and explanation

C: Vanadium(V) oxide

Vanadium(V) oxide is the catalyst used in the Contact process for converting SO₂ to SO₃. It provides a faster reaction pathway without changing the equilibrium composition.

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Question 27

What charge is required to liberate 0.125 mol O₂ during electrolysis of dilute sodium chloride solution, assuming full current efficiency for oxygen evolution? [F = 96,500 C mol⁻¹]

  1. 24,125 C
  2. 48,250 C
  3. 72,375 C
  4. 96,500 C
Answer and explanation

B: 48,250 C

Forming one mole of O₂ requires transfer of four moles of electrons. For 0.125 mol O₂, the charge is 0.125 × 4 × 96,500 = 48,250 C.

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Question 29

Current I deposits X g of a univalent metal in 40 minutes. What mass is deposited by current 2I in 10 minutes under the same conditions?

  1. X/4 g
  2. X/2 g
  3. 2X g
  4. 4X g
Answer and explanation

B: X/2 g

For the same metal and current efficiency, deposited mass is proportional to charge It. The second charge is (2I × 10)/(I × 40) = 1/2 of the first, so the deposited mass is X/2 g.

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Question 30

RS(aq) + HF(aq) → RF(s) + HS(aq), ΔH = −65.7 kJ mol⁻¹. What can be deduced?

  1. The reactants have lower enthalpy than the products
  2. The reactants have higher enthalpy than the products
  3. The reaction is slow
  4. A large amount of heat is absorbed
Answer and explanation

B: The reactants have higher enthalpy than the products

The enthalpy change is H(products) − H(reactants). A negative value of −65.7 kJ mol⁻¹ means that the products have lower enthalpy and heat is released. It does not determine the reaction speed.

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Question 32

Which gas forms a white precipitate when passed into aqueous silver nitrate acidified with nitric acid?

  1. NH₃
  2. SO₂
  3. CO₂
  4. HCl
Answer and explanation

D: HCl

Hydrogen chloride dissolves to supply chloride ions. These react with silver ions to form white silver chloride: Ag⁺ + Cl⁻ → AgCl(s). Nitric-acid acidification suppresses interference from carbonate and sulfite precipitates.

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Question 33

Chlorine, bromine and iodine resemble one another in that they

  1. Dissolve in alkalis
  2. React violently with hydrogen without heating
  3. Are liquids
  4. Displace one another from solutions of their salts
Answer and explanation

A: Dissolve in alkalis

All three halogens react with aqueous alkalis. For example, chlorine disproportionates in cold dilute alkali to chloride and hypochlorite. They are not all liquids, and halogen displacement works in the direction of decreasing oxidising strength rather than mutually in both directions.

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Question 34

Which salt reacts with dilute hydrochloric acid to produce a pungent-smelling gas that decolourises acidified purple potassium permanganate solution?

  1. Na₂SO₄
  2. Na₂SO₃
  3. Na₂S
  4. Na₂CO₃
Answer and explanation

B: Na₂SO₃

Sodium sulfite reacts with acid to release sulfur dioxide: Na₂SO₃ + 2HCl → 2NaCl + H₂O + SO₂. Its pungent gas reduces purple permanganate to Mn²⁺. Sodium sulfide instead gives hydrogen sulfide, characterised by a rotten-egg smell.

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Question 36

Hydrogen is used in oxyhydrogen flames for melting metals because it

  1. Evolves a lot of heat when burnt
  2. Combines explosively with oxygen
  3. Is a very light gas
  4. Is a rocket fuel
Answer and explanation

A: Evolves a lot of heat when burnt

Hydrogen combustion in oxygen releases substantial energy: 2H₂ + O₂ → 2H₂O. The resulting high-temperature flame provides heat for melting metals; low gas density is not the reason.

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Question 37

A heated flask contains mixture Y. Gas passes from the flask through a drying vessel containing calcium oxide, then into an inverted gas jar labelled ammonia. What is mixture Y?

  1. Calcium hydroxide and ammonium chloride
  2. Calcium hydroxide and sodium chloride
  3. Sodium chloride and ammonium nitrate
  4. Sodium nitrite and ammonium chloride
Answer and explanation

A: Calcium hydroxide and ammonium chloride

Heating ammonium chloride with calcium hydroxide releases ammonia: 2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O. Calcium oxide dries the ammonia without reacting with it as an acidic drying agent would.

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Question 38

Which properties make duralumin especially useful compared with its constituent metals?

  1. Heavy with a high melting point
  2. Malleable with high density
  3. Strong and light
  4. Hard and ductile
Answer and explanation

C: Strong and light

Duralumin is an aluminium alloy that combines low density with greater strength than pure aluminium. That strength-to-weight advantage makes it useful for structures where mass matters.

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Question 39

Which listed pair of reactive metals is extracted by electrolysis of suitable molten compounds?

  1. Magnesium and zinc
  2. Magnesium and calcium
  3. Copper and zinc
  4. Lead and calcium
Answer and explanation

B: Magnesium and calcium

Magnesium and calcium are highly reactive metals for which electrolysis of molten compounds is an established extraction route. Zinc, copper and lead can be obtained by chemical reduction routes instead.

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Question 41

Which listed metal becomes passive in concentrated nitric acid?

  1. Iron
  2. Tin
  3. Copper
  4. Zinc
Answer and explanation

A: Iron

Concentrated nitric acid can make iron passive by producing a protective surface film that greatly slows further attack. This is the standard iron passivation example.

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Question 45

When excess ethanol is heated to 145 °C in the presence of concentrated H₂SO₄, the main organic product is

  1. Ethyne
  2. Diethyl sulfate
  3. Diethyl ether
  4. Acetone
Answer and explanation

C: Diethyl ether

At about 140–145 °C with excess ethanol and concentrated sulfuric acid, intermolecular dehydration produces diethyl ether: 2C₂H₅OH → C₂H₅OC₂H₅ + H₂O. Higher temperatures favour ethene formation.

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Question 46

How many grams of bromine are required for complete addition to 5.2 g of but-1-en-3-yne? [C = 12, H = 1, Br = 80]

  1. 64.0 g
  2. 48.0 g
  3. 32.0 g
  4. 16.0 g
Answer and explanation

B: 48.0 g

But-1-en-3-yne is CH₂=CH–C≡CH, with formula C₄H₄ and molar mass 52 g mol⁻¹. One C=C bond consumes one Br₂ and one C≡C consumes two Br₂ on full addition. Thus 5.2/52 × 3 × 160 = 48.0 g bromine is required.

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Question 47

Polyvinyl chloride is commonly used to produce

  1. Bread
  2. Pencils
  3. Ink
  4. Pipes
Answer and explanation

D: Pipes

Polyvinyl chloride is a durable thermoplastic widely used for pipes. Its long polymer chains and material properties suit this use.

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Question 49

When the two terminal alkyl groups of ethyl ethanoate are interchanged, the compound formed is

  1. Methyl ethanoate
  2. Ethyl propanoate
  3. Methyl propanoate
  4. Propyl ethanoate
Answer and explanation

C: Methyl propanoate

Ethyl ethanoate is CH₃–C(=O)–O–C₂H₅. Swapping the terminal methyl and ethyl groups gives C₂H₅–C(=O)–O–CH₃. The alcohol-derived group is methyl and the acid-derived part has three carbons, so the ester is methyl propanoate.

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