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JAMB Chemistry Past Questions 2018

10 questions from the JAMB 2018 Chemistry paper, with answers and explanations. Tap "Show answer" under each one when you are ready.

JAMB 2018Question 1 · Acids, Bases, and Salts
What is the PH of 0.00 solution of the sodium hydroxide
  1. A.14
  2. B.13
  3. C.12
  4. D.11
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Answer: D. 11

[] = 0.001 = mol/dm3, pOH = 3, so pH = 14-3 = 11.

JAMB 2018Question 2 · Atomic Structure and Chemical Bonding
Which of the following are mixtures? I. Petroleum II. Rubber latex III. Vulcanizer's solution IV. Carbon sulphide
  1. A.I, II and III
  2. B.I, II and IV
  3. C.I and II only
  4. D.I and IV
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Answer: A. I, II and III

Petroleum, rubber latex, and vulcanizer's solution are all mixtures; carbon(IV) sulphide is a pure compound.

JAMB 2018Question 3 · Chemistry of Metals
The choice of method for extracting a metal from its ores depends on the
  1. A.strength of the core
  2. B.position of the metal in the electrochemical series
  3. C.source of the core
  4. D.position of the metal in the periodic table
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Answer: B. position of the metal in the electrochemical series

The extraction method for a metal is chosen based on its position (reactivity) in the electrochemical/reactivity series.

JAMB 2018Question 4 · Chemistry of Non-metals and Gases
An element used in the production of matches is
  1. A.nitrogen
  2. B.aluminium
  3. C.copper
  4. D.sulphur
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Answer: D. sulphur

Sulphur is used in the manufacture of matches (in the match head composition).

JAMB 2018Question 5 · Chemistry of Non-metals and Gases
When air which contains the gases Oxygen, nitrogen, carbondioxide, water vapour and the rare gases, is passed through alkaline pyrogallol and then over quicklime, the only gases left are;
  1. A.nitrogen and carbondioxide
  2. B.the rare gases
  3. C.nitrogen and oxygen
  4. D.nitrogen and the rare gases
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Answer: D. nitrogen and the rare gases

Alkaline pyrogallol absorbs oxygen, and quicklime absorbs carbon(IV) oxide and water vapour, leaving nitrogen and the rare (noble) gases.

JAMB 2018Question 6 · Electrochemistry and Redox Reactions
H S + Cl → 2HCl + S In the reaction above, the substance that is reduced is
  1. A.HS
  2. B.S
  3. C.HCl
  4. D.Cl
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Answer: D. Cl

Chlorine (0 in ) is reduced to (-1) in HCl, while sulphur in is oxidized to S.

JAMB 2018Question 7 · Environmental and Industrial Chemistry
In the upper atmosphere, the ultra-violet light breaks off a free chlorine atom from chlorofluorocarbon molecule. The effect of this is that the free chlorine atom will
  1. A.be very reactive and will attack ozone
  2. B.be non-reactive and will not attack ozone
  3. C.not change the level of ozone in the atmosphere
  4. D.increase the level of ozone in the atmosphere
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Answer: A. be very reactive and will attack ozone

The free chlorine atom released by UV light from a CFC molecule is highly reactive and catalytically attacks/depletes ozone molecules.

JAMB 2018Question 8 · Organic Chemistry
When large hydrocarbon molecules are heated at high temperature in the presence of a catalyst to give smaller molecules, the process is known as
  1. A.disintegration
  2. B.polymerization
  3. C.cracking
  4. D.degradation
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Answer: C. cracking

Breaking large hydrocarbon molecules into smaller ones by heating with a catalyst is called cracking.

JAMB 2018Question 9 · Physical Chemistry
2KClO(s) → + 3O(g) The importance of the catalyst in the reaction above is that
  1. A.heating may not be required before the reaction takes place
  2. B.the reaction is controllable even at a high temperature
  3. C.the reaction produces large quantity of oxygen
  4. D.the reaction takes place more rapidly at a lower temperature
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Answer: D. the reaction takes place more rapidly at a lower temperature

The catalyst allows the decomposition of to occur more rapidly at a lower temperature than would otherwise be required.

JAMB 2018Question 10 · Quantitative Chemistry
If 1 litre of 2.2M sulphuric acid is poured into a bucket containing 10 litres of water and the resulting solution mixed thoroughly, the resulting sulphuric acid concentration will be
  1. A.2.2M
  2. B.1.1M
  3. C.0.2M
  4. D.0.11M
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Answer: C. 0.2M

Moles = 1 x 2.2 = . New total volume = 1+10 = . New concentration = = 0.2M.

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