JAMB / UTME · Chemistry · 2005

JAMB Chemistry 2005 Past Questions & Answers

47 JAMB Chemistry 2005 questions with verified answers and worked explanations. Here are the first 10 — sign up free to practise the full paper.

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Q1Acids, Bases and Salts · Acids, bases and salts

The gas that can be dried using concentrated Tetraoxosulphate (VI) acid is

  • A.Ammonia
  • B.Hydrogen bromide
  • C.Sulphur (IV) oxide
  • D.Hydrogen sulphide
Answer: C
Beta explains

Concentrated tetraoxosulphate (VI) acid is an acidic drying agent, which means it should only be used to dry gases that it does not react with. It cannot be used to dry a basic gas like ammonia (NH₃) because they will neutralize each other. It also cannot dry reducing gases like hydrogen sulphide (H₂S) because it will oxidize them. Since sulphur (IV) oxide (SO₂) is also an acidic gas, it does not react with the acid, making it the correct gas to be dried by this method.

Common mistake: Students often choose Ammonia, forgetting that as a basic gas, it will react with the acidic concentrated H₂SO₄ in a neutralization reaction.

Q2Separation and Purification · Separation of mixtures

The property used in the industrial preparation of nitrogen and oxygen from air is

  • A.Rate of diffusion
  • B.Solubility
  • C.Density
  • D.Boiling point
Answer: D
Beta explains

The industrial preparation of nitrogen and oxygen from air is done through the fractional distillation of liquid air. This process works by separating components of a mixture based on their different boiling points. Since liquid nitrogen has a lower boiling point (-196°C) than liquid oxygen (-183°C), it vaporizes first and can be collected separately.

Common mistake: Students might mistakenly choose 'Rate of diffusion' because it is also a method for separating gases, but fractional distillation based on boiling points is the correct industrial method for air.

Q3Acids, Bases and Salts · Acids, bases and salts

A dense white fume is formed when ammonia gas reacts with

  • A.H2(g)
  • B.O2(g)
  • C.HCI(g)
  • D.Cl2(g)
Answer: C
Beta explains

Ammonia gas (NH₃) is a base, while hydrogen chloride gas (HCl) is an acid. When these two gases meet, they undergo an acid-base neutralization reaction to form the salt, ammonium chloride (NH₄Cl). This ammonium chloride is a solid substance that appears as dense white fumes or smoke.

Common mistake: Students might mistakenly choose chlorine gas (Cl₂) because it is a reactive gas, but its reaction with ammonia is a redox reaction and does not produce the specific dense white fumes of ammonium chloride.

Q4Inorganic Chemistry · Metals and their compounds

In a flame test for calcium, the flame appears

  • A.Blue when viewed through a blue glass
  • B.Green when viewed through a blue glass
  • C.Orange-red when viewed through a blue glass
  • D.Brick-red when viewed through a blue glass
Answer: C
Beta explains

The flame test is a qualitative test used to identify certain metals. When a calcium salt is introduced into a Bunsen burner flame, it imparts a characteristic brick-red colour. However, when this flame is viewed through a blue cobalt glass, the glass filters out other interfering colours, and the calcium flame appears orange-red.

Common mistake: Many students choose "brick-red" because they correctly remember the standard flame colour for calcium but forget that viewing it through a blue glass changes the observed colour to orange-red.

Q5Inorganic Chemistry · Metals and their compounds

Which of the following alloys contain iron

  • A.Steel and permalloy
  • B.Duralumlin and steel
  • C.nd Soft Solder a Duralumlin
  • D.Brass and bronze
Answer: A
Beta explains

An alloy is a mixture containing at least one metal. Steel is a very common alloy made primarily from iron and carbon. Permalloy is another important alloy known for its magnetic properties, and it is made from a mixture of iron and nickel. Since both steel and permalloy are made with iron, this is the correct answer.

Common mistake: Students may choose option B because they know steel contains iron, but they are incorrect about duralumin, which is an alloy of aluminum, not iron.

Q6Separation and Purification · Separation of mixtures

To obtain pure carbon(II) oxide from its mixture with carbon (IV)oxide, the mixture should be

  • A.Bubbled through concentrated Tetraoxosulphate(VI) acid
  • B.Bubbled through sodium hydroxide
  • C.Passed over heated copper(II) oxide
  • D.Bubbled through water
Answer: B
Beta explains

The task is to remove carbon(IV) oxide (CO₂) from a mixture with carbon(II) oxide (CO). Carbon(IV) oxide is an acidic oxide, which means it reacts with bases, while carbon(II) oxide is a neutral oxide. When the gas mixture is bubbled through sodium hydroxide solution, a base, the acidic CO₂ reacts and is removed. The neutral CO does not react and passes through, allowing pure carbon(II) oxide to be collected.

Common mistake: Students may mistakenly choose option A, thinking a strong acid is needed for purification, but this is incorrect as an acid will not react with the acidic carbon(IV) oxide.

Q7Physical Chemistry · Chemical equilibrium

N2(g) + 3H2(g) ⇌ 2NH2(g) Δ H = - 90kJ In the equation above, the yield of ammonia can be decreased by

  • A.Adding a catalyst
  • B.Removing ammonia
  • C.Increasing the pressure
  • D.Increasing the temperature
Answer: D
Beta explains

The forward reaction to produce ammonia is exothermic, as shown by the negative ΔH value, meaning it releases heat. According to Le Chatelier's principle, if you increase the temperature, the system will try to counteract this change by favouring the reaction that absorbs heat, which is the backward (endothermic) reaction. This shift to the left breaks down ammonia into nitrogen and hydrogen, thus decreasing the yield of ammonia.

Common mistake: Students often mistakenly choose 'Increasing the pressure' because they know it affects the equilibrium, but they forget that it favours the side with fewer moles of gas, which in this case would increase the yield of ammonia, not decrease it.

Q8Atoms and Bonding · Atomic structure and periodic table

Metalloids are also referred to as

  • A.Metals
  • B.Semi-metals
  • C.Colloids
  • D.Non-metals
Answer: B
Beta explains

Metalloids are elements on the periodic table that have properties that are intermediate between those of metals and non-metals. Because they are not fully metallic but also not completely non-metallic, they are given the alternative name 'semi-metals'. The prefix 'semi-' means 'partly', so 'semi-metals' is a fitting description for elements that are partly like metals.

Common mistake: Students may mistakenly choose 'Metals' or 'Non-metals' because metalloids share properties with both, but 'semi-metals' is the specific alternative name for this distinct group.

Q9Inorganic Chemistry · Metals and their compounds

The ores that can be concentrated by flotation are

  • A.Nitride ores
  • B.Chloride ores
  • C.Oxide ores
  • D.Sulphide ores
Answer: D
Beta explains

Froth flotation is a method used to concentrate ores by separating minerals based on their ability to repel water. Sulphide ores are naturally hydrophobic (water-repelling), so when mixed with water and a frothing agent, they attach to air bubbles and float to the surface. This allows them to be separated from the gangue (waste rock), which is hydrophilic and sinks.

Common mistake: Students might incorrectly choose oxide ores because they are a very common type of ore, but these are usually concentrated by gravity or magnetic methods, not flotation.

Q10Inorganic Chemistry · Metals and their compounds

The most suitable metal that can be used as a lightening conductor is

  • A.Aluminium
  • B.Iron
  • C.Copper
  • D.Silver
Answer: D
Beta explains

A lightning conductor's purpose is to provide a path of least resistance for the massive electrical current from a lightning strike to safely reach the ground. This requires a material with the highest possible electrical conductivity. Among the given options, and indeed among all metals, silver has the best electrical conductivity, making it the most suitable choice.

Common mistake: Students often mistakenly choose copper because it is the most common material used for electrical wiring and practical lightning conductors due to its excellent conductivity and much lower cost compared to silver.

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