JAMB Chemistry 1991

25 reviewed questions with answers and explanations.

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Question 1

Which separation is carried out by fractional distillation?

  1. Nitrogen from liquid air
  2. Sodium chloride from seawater
  3. Iodine from a solution in carbon tetrachloride
  4. Sulfur from a solution in carbon disulfide
Answer and explanation

A: Nitrogen from liquid air

Liquid air contains substances with different boiling points. Fractional distillation separates its components, allowing nitrogen to be collected separately from oxygen and other gases. The other listed tasks chiefly involve separating dissolved solids from their solvents.

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Question 3

An iron ore contains 70.0% Fe₂O₃ by mass. What mass of iron can theoretically be obtained from 80 kg of this ore? [Fe = 56, O = 16]

  1. 35.0 kg
  2. 39.2 kg
  3. 70.0 kg
  4. 78.4 kg
Answer and explanation

B: 39.2 kg

The ore contains 80×0.700 = 56.0 kg of Fe₂O₃. Iron contributes 2×56 = 112 of its relative formula mass 160. The theoretical iron mass is therefore 56.0×112/160 = 39.2 kg.

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Question 4

In two experiments, 0.36 g and 0.71 g of chlorine combine with 0.20 g and 0.40 g of a metal X, respectively. Which law is supported by these mass data, allowing for their reported precision?

  1. Multiple proportions
  2. Conservation of mass
  3. Constant composition
  4. Reciprocal proportions
Answer and explanation

C: Constant composition

The chlorine-to-metal mass ratios are 0.36/0.20 = 1.80 and 0.71/0.40 = 1.775. These are approximately the same at the precision of the reported masses. The data therefore support constant composition: the elements combine in essentially the same mass proportion.

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Question 7

What volume of CO₂ at STP is produced by complete thermal decomposition of 1 kg of KHCO₃? [K = 39, H = 1, C = 12, O = 16; molar gas volume at STP = 22.4 dm³ mol⁻¹]

  1. 28 dm³
  2. 56 dm³
  3. 112 dm³
  4. 196 dm³
Answer and explanation

C: 112 dm³

The balanced decomposition is 2KHCO₃ → K₂CO₃ + H₂O + CO₂. KHCO₃ has molar mass 100 g mol⁻¹, so 1 kg is 10 mol. Every two moles give one mole of CO₂; 5 mol therefore occupy 5×22.4 = 112 dm³ at STP.

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Question 9

Atoms of X have two outer-shell electrons and atoms of Y have seven. For their expected ionic compound, which statement is false?

  1. It has formula XY
  2. It is likely to be ionic
  3. It contains X²⁺ ions
  4. It contains Y⁻ ions
Answer and explanation

A: It has formula XY

In the expected electron-transfer model, X loses two electrons to form X²⁺ and each Y gains one to form Y⁻. Charge balance requires two Y⁻ ions for each X²⁺, giving XY₂. A formula of XY would not be electrically neutral.

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Question 10

X⁻ and Y⁺ each contain 10 electrons. How many protons are in the nuclei of X and Y, respectively?

  1. 10 and 10
  2. 9 and 9
  3. 11 and 9
  4. 9 and 11
Answer and explanation

D: 9 and 11

An X⁻ ion has gained one electron, so its 10 electrons correspond to 9 protons. A Y⁺ ion has lost one electron, so its 10 electrons correspond to 11 protons. Ion formation changes electron count, not the number of protons in the nucleus.

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Question 12

Which listed combination describes bonding in ammonium chloride, including the electron-pair donation used to form NH₄⁺?

  1. Ionic only
  2. Covalent only
  3. Ionic and dative covalent
  4. Dative covalent only
Answer and explanation

C: Ionic and dative covalent

Ammonium chloride is an ionic solid of NH₄⁺ and Cl⁻ ions. Within NH₄⁺, the N–H bonds are covalent; formation from NH₃ and H⁺ can be described as donation of a nitrogen lone pair, or dative covalent bonding. Once NH₄⁺ has formed, its four N–H bonds are equivalent.

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Question 13

Which sequence is in increasing order of electronegativity?

  1. Chlorine, aluminium, magnesium, phosphorus, sodium
  2. Sodium, magnesium, aluminium, phosphorus, chlorine
  3. Chlorine, phosphorus, aluminium, magnesium, sodium
  4. Sodium, chlorine, phosphorus, magnesium, aluminium
Answer and explanation

B: Sodium, magnesium, aluminium, phosphorus, chlorine

All five elements lie in period 3. Electronegativity generally increases from left to right across this period, so the order is Na < Mg < Al < P < Cl. Chlorine attracts shared bonding electrons most strongly in this list.

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Question 16

Which listed ion is a particularly toxic drinking-water contaminant at low concentrations?

  1. Ca²⁺
  2. Hg²⁺
  3. Mg²⁺
  4. Fe²⁺
Answer and explanation

B: Hg²⁺

Mercury ions are toxic contaminants and are a concern in drinking water at low concentrations. Calcium and magnesium commonly contribute to water hardness, while iron can affect colour and taste. Mercury is the toxic-metal contaminant identified by this comparison.

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Question 22

What volume of CO₂ at STP results when 10 cm³ of 0.1 mol dm⁻³ Na₂CO₃ solution reacts with excess acid? [Molar gas volume at STP = 22.4 dm³ mol⁻¹]

  1. 2.240 cm³
  2. 22.40 cm³
  3. 224.0 cm³
  4. 2240 cm³
Answer and explanation

B: 22.40 cm³

The carbonate amount is 0.1×(10/1000) = 0.001 mol. Acid converts each CO₃²⁻ into one CO₂ molecule, so the CO₂ amount is also 0.001 mol. Its STP volume is 0.001×22.4 = 0.0224 dm³, or 22.4 cm³.

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Question 26

2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l). Which statement describes this reaction?

  1. It is redox: H₂S is the oxidant and SO₂ is the reductant
  2. It is redox: SO₂ is the oxidant and H₂S is the reductant
  3. It is not redox because there is no oxidant
  4. It is not redox because there is no reductant
Answer and explanation

B: It is redox: SO₂ is the oxidant and H₂S is the reductant

Sulfur in H₂S changes from −2 to 0, so H₂S is oxidised and acts as the reductant. Sulfur in SO₂ changes from +4 to 0, so SO₂ is reduced and acts as the oxidant. Two sulfide sulfur atoms each lose two electrons, balancing the four electrons gained by the SO₂ sulfur atom.

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Question 27

MnO₂ hastens hydrogen peroxide decomposition mainly by

  1. Increasing reactant surface area
  2. Increasing reactant concentration
  3. Lowering the activation energy
  4. Lowering the reaction enthalpy
Answer and explanation

C: Lowering the activation energy

MnO₂ acts as a catalyst. It provides a reaction pathway with a lower activation energy, so more reacting particles can overcome the energy barrier. A catalyst does not change the overall enthalpy change of the reaction.

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Question 29

NO(g) + CO(g) ⇌ ½N₂(g) + CO₂(g), ΔH = −89.3 kJ. Which conditions favour the greatest equilibrium conversion of NO and CO to products?

  1. Low temperature and high pressure
  2. High temperature and low pressure
  3. High temperature and high pressure
  4. Low temperature and low pressure
Answer and explanation

A: Low temperature and high pressure

The negative enthalpy change means the forward reaction releases heat, so lowering temperature favours products. There are 2 moles of reactant gases and 1.5 moles of product gases per equation. Higher pressure therefore also favours products, which occupy the smaller gas volume.

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Question 31

At 25 °C, doubling the initial NO concentration from the first measurement to the second changes the initial rate of its reaction with chlorine from 3.0×10⁻⁵ to 1.2×10⁻⁴ mol s⁻¹. By what factor does the rate increase?

  1. Two
  2. Three
  3. Four
  4. Five
Answer and explanation

C: Four

The rate ratio is (1.2×10⁻⁴)/(3.0×10⁻⁵) = 4. Thus, when the listed initial NO concentration doubles, the measured initial rate increases fourfold.

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Question 32

Which listed gas can rekindle a brightly glowing splint?

  1. NO₂
  2. NO
  3. N₂O
  4. Cl₂
Answer and explanation

C: N₂O

Nitrous oxide, N₂O, can support combustion and relight a glowing splint. At the hot splint it can decompose to nitrogen and oxygen. The glowing-splint observation is therefore not exclusive to a sample of pure oxygen.

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Question 33

Which listed salt can be melted without decomposition under ordinary laboratory conditions?

  1. Na₂CO₃
  2. CaCO₃
  3. MgCO₃
  4. ZnCO₃
Answer and explanation

A: Na₂CO₃

Sodium carbonate is thermally stable enough to melt during ordinary heating. Calcium, magnesium and zinc carbonates instead undergo thermal decomposition to their oxides and carbon dioxide before they can simply be melted under ordinary laboratory conditions.

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Question 36

Why does the white precipitate produced by adding aqueous ammonia to Zn²⁺ solution dissolve in excess ammonia?

  1. Zinc is amphoteric
  2. Zinc hydroxide is readily soluble
  3. Zinc forms a complex soluble in excess ammonia
  4. Ammonia solution is a strong base
Answer and explanation

C: Zinc forms a complex soluble in excess ammonia

Aqueous ammonia first supplies hydroxide ions, forming white Zn(OH)₂. In excess ammonia, zinc forms the soluble complex ion [Zn(NH₃)₄]²⁺, so the precipitate dissolves. This ligand-complex formation explains the observation.

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Question 37

Which substance in clear aqueous solution forms a white precipitate when CO₂ is bubbled through it for a short time?

  1. KOH
  2. NaOH
  3. Ca(OH)₂
  4. Al(OH)₃
Answer and explanation

C: Ca(OH)₂

A small amount of CO₂ reacts with limewater, Ca(OH)₂(aq), to form insoluble calcium carbonate: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O. The CaCO₃ is the white precipitate. With prolonged excess CO₂, that precipitate can dissolve again.

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Question 39

Which pair of metals can be obtained by thermal decomposition of their nitrate salts?

  1. Copper and mercury
  2. Silver and copper
  3. Mercury and silver
  4. Magnesium and mercury
Answer and explanation

C: Mercury and silver

Silver and mercury nitrates can ultimately yield their metals on strong heating because the corresponding oxides are thermally unstable. Copper and magnesium nitrates instead leave stable metal oxides under the usual comparison conditions. The pair is therefore mercury and silver.

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Question 40

Which compound can exist as geometric isomers?

  1. 2-Methylbut-2-ene
  2. But-2-ene
  3. But-1-ene
  4. CH₂BrCl
Answer and explanation

B: But-2-ene

Each carbon of the double bond in but-2-ene has one H and one CH₃ group. Restricted rotation allows distinct cis and trans arrangements. But-1-ene and 2-methylbut-2-ene have identical groups on one double-bond carbon, while CH₂BrCl has no double bond.

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Question 43

What mass of bromine is required for complete addition to 10 g of propyne? [C = 12, H = 1, Br = 80]

  1. 20 g
  2. 40 g
  3. 60 g
  4. 80 g
Answer and explanation

D: 80 g

Propyne has molar mass 3×12 + 4×1 = 40 g mol⁻¹, so 10 g is 0.25 mol. Complete addition across its triple bond uses 2 mol of Br₂ per mole of propyne. The bromine mass is 0.25×2×160 = 80 g.

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Question 44

Ethene is absorbed in concentrated H₂SO₄. The product is diluted with water and warmed. Which organic product is formed?

  1. Ethanol
  2. Diethyl ether
  3. Ethanal
  4. Diethyl sulfate
Answer and explanation

A: Ethanol

Ethene first adds sulfuric acid to form ethyl hydrogen sulfate. Adding water and warming hydrolyses this intermediate to ethanol and regenerates sulfuric acid. The overall change adds H and OH across the ethene double bond.

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Question 45

What is an advantage of common soapless detergents over soap in hard water?

  1. They are always easier to manufacture
  2. They always foam more than soap
  3. Their salts with hard-water ions remain soluble
  4. They always kill more germs than soap
Answer and explanation

C: Their salts with hard-water ions remain soluble

Soap forms insoluble calcium and magnesium salts, seen as scum in hard water. Common soapless detergents avoid this precipitation because their corresponding salts remain sufficiently soluble. They therefore retain cleaning action better in hard water.

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Question 48

What is the IUPAC name of CH₃CH(CH₃)CH₂Cl?

  1. 1-Chloro-2-methylbutane
  2. 1-Chloro-2-methylpropane
  3. 2-Chloromethylethane
  4. 1-Chloro-2,2-dimethylethane
Answer and explanation

B: 1-Chloro-2-methylpropane

The longest carbon chain has three atoms. Numbering from CH₂Cl places chlorine at carbon 1 and the methyl branch at carbon 2. The compound is 1-chloro-2-methylpropane.

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Question 50

Which listed pair is produced by the chlorination of CH₃COOH with Cl₂ in sunlight?

  1. ClCH₂COOH + HCl
  2. CH₃COCl + HOCl
  3. CH₃COOCl + HCl
  4. CH₃COCl + H₂O
Answer and explanation

A: ClCH₂COOH + HCl

In the first chlorination step, one hydrogen on the methyl group of ethanoic acid is replaced by chlorine. The products are chloroethanoic acid, ClCH₂COOH, and HCl. The carboxyl group remains intact.

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