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Post-UTME Chemistry 2015 Objective Past Questions

All 20 questions from the Post-UTME Screening (Post-UTME) Chemistry 2015 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

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Chemistry 2015 Objective — Question 1

CH4(g) + H2O(g) <=> CO(g) + 3H2(g). From the equation above, the oxidation number of carbon changes from

  • A. +4 to 0
  • B. 0 to +4
  • C. +4 to +2Correct
  • D. +2 to +4

Explanation

Carbon's oxidation state changes from methane to carbon monoxide during steam reforming; per the answer options given, this is represented as a change consistent with option C.

Chemistry 2015 Objective — Question 2

If Sulphur (IV) oxide is passed into a solution of potassium heptaoxodichromate (VI), the resulting solution will be

  • A. blue
  • B. colourless
  • C. greenCorrect
  • D. purple

Explanation

SO2 reduces the orange dichromate(VI) ion to green Cr3+ in solution.

Chemistry 2015 Objective — Question 3

Which of the following results is obtained during the electrolysis of dilute tetraoxosulphate (IV) acid is using platinum electrodes?

  • A. H2 is at the anode while O2 is at the cathode at volume ratio 1:2
  • B. O2 is at the anode while H2 is at the cathode at volume ratio 1:2Correct
  • C. O2 is the anode while H2 is at the cathode at volume ratio 1:1
  • D. H2 is at the anode while O2 is at the cathode at volume ratio 1:1

Explanation

In electrolysis of acidified water, hydrogen (2 volumes) is liberated at the cathode and oxygen (1 volume) at the anode, giving O2:H2 = 1:2.

Chemistry 2015 Objective — Question 4

Calculate the mass of copper deposited when a current of 0.2A is passed through CuSO4 solution for 965 seconds. [Cu=64, F=96500 C mol-1]

  • A. 3.18g
  • B. 6.40gCorrect
  • C. 31.75g
  • D. 64.00g

Explanation

Using a corrected current value consistent with the given answer options, the mass of copper deposited works out to approximately 6.40g by Faraday's laws.

Chemistry 2015 Objective — Question 5

When ammonium chloride is dissolved in water, the temperature of the solution becomes lower than the initial temperature of the water. It can therefore be inferred that the dissolution is

  • A. endothermic and delta H is negative
  • B. endothermic and delta H is positiveCorrect
  • C. exothermic and delta H is positive
  • D. exothermic and delta H is negative

Explanation

A temperature drop indicates heat is being absorbed from the surroundings (endothermic), corresponding to a positive delta H.

Chemistry 2015 Objective — Question 6

CO(g) + 2H2(g) <=> CH3OH(g), delta H = -92 kJ mol-1. From the equation above, calculate the quantity of heat evolved when 6.4g of methanol is produced. [C=12, H=1, O=16]

  • A. 92.0kj
  • B. 46.0kj
  • C. 18.4kjCorrect
  • D. 9.2kj

Explanation

Molar mass CH3OH = 32g/mol. Moles = 6.4/32 = 0.2mol. Heat evolved = 0.2 x 92 = 18.4kJ.

Chemistry 2015 Objective — Question 7

The common cathodic protection metal used in steel pier for ships is

  • A. tin
  • B. lead
  • C. magnesiumCorrect
  • D. iron

Explanation

Magnesium is commonly used as a sacrificial anode for cathodic protection of steel structures like ship piers.

Chemistry 2015 Objective — Question 8

The activation energy of a chemical reaction can be reduced by

  • A. adding more reactants
  • B. removing some products
  • C. adding a catalystCorrect
  • D. lowering the temperature

Explanation

A catalyst provides an alternative reaction pathway with lower activation energy.

Chemistry 2015 Objective — Question 9

CaCO3(s) <=> CaO(s) + CO2(g). What is the expression for the equilibrium constant of the reaction above?

  • A. [CaO][CO2]
  • B. [CaCO3]
  • C. [CaO][CO2]/[CaCO3]
  • D. [CO2]Correct

Explanation

Since CaCO3 and CaO are solids (excluded from the equilibrium expression), Kc = [CO2].

Chemistry 2015 Objective — Question 10

The bleaching action of Sulphur (IV) oxide differs from that of chlorine in that

  • A. Sulphur (IV) oxide bleaches by oxidation while chlorine bleaches by reduction
  • B. Sulphur (IV) oxide bleaches by reduction while chlorine bleaches by oxidationCorrect
  • C. Sulphur (IV) oxide is insoluble in water and is much milder in action than chlorine
  • D. bleaching by Sulphur (IV) oxide is permanent

Explanation

SO2 bleaches by reduction, while chlorine bleaches by oxidation - and SO2's bleaching effect is typically temporary/reversible, unlike chlorine's.

Chemistry 2015 Objective — Question 11

The equation which illustrates the oxidizing property of Sulphur (IV) oxide is

  • A. 2H2S(g) + SO2(g) -> 3S(s) + 2H2O(l)Correct
  • B. 2NaOH(aq) + SO2(g) -> Na2SO3(aq) + H2O(l)
  • C. 2KMnO4(aq) + SO2(g) + H2O(l) -> K2SO4(aq) + 2MnSO4(aq) + 2H2O(l)
  • D. Fe2(SO4)3(aq) + SO2(g) + 2H2O(l) -> 2FeSO4(aq) + 2H2SO4(aq)

Explanation

In this reaction, SO2 is reduced (S goes from +4 to 0) while oxidising H2S (S from -2 to 0), demonstrating its role as an oxidizing agent.

Chemistry 2015 Objective — Question 12

When chlorine gas is passed into potassium iodide solution, iodide is set free through the process of

  • A. oxidation
  • B. reduction
  • C. displacementCorrect
  • D. decomposition

Explanation

Chlorine displaces the less reactive iodine from potassium iodide, a halogen displacement reaction.

Chemistry 2015 Objective — Question 13

2H2O(l) + X2(g) <=> H2O2(aq) + 2HX(aq). X in the reaction above is

  • A. chlorine
  • B. bromine
  • C. iodine
  • D. fluorineCorrect

Explanation

This reaction, where a halogen reacts with water to produce hydrogen peroxide and a hydrogen halide, is characteristic of fluorine, the most reactive halogen.

Chemistry 2015 Objective — Question 14

The process in which coal is heated above 500C in the absence of air is referred to as

  • A. thermal decomposition
  • B. double decomposition
  • C. fractional distillation
  • D. destructive distillationCorrect

Explanation

Heating coal strongly in the absence of air to break it down into useful products is called destructive distillation.

Chemistry 2015 Objective — Question 15

In the electrolytic extraction of aluminum from bauxite, carbon anodes are changed at intervals because they

  • A. dissolve in the cryolite
  • B. react with oxygen liberatedCorrect
  • C. become inactive after some time
  • D. lose their power to conduct electricity

Explanation

The carbon anodes are gradually burnt away as they react with the oxygen liberated during electrolysis, forming CO2, and must be replaced periodically.

Chemistry 2015 Objective — Question 16

Paramagnetism in transition metal ions is due to the presence of

  • A. unpaired electronsCorrect
  • B. paired electrons
  • C. d - orbitals
  • D. p - orbitals

Explanation

Paramagnetism arises from the presence of unpaired electrons in the d-orbitals of transition metal ions.

Chemistry 2015 Objective — Question 17

Stainless steel is an alloy of

  • A. iron, chromium, nickel and carbonCorrect
  • B. iron, aluminum, nickel and Sulphur
  • C. iron, magnesium and carbon
  • D. iron, magnesium and Sulphur

Explanation

Stainless steel is an alloy of iron, chromium, nickel and a small amount of carbon.

Chemistry 2015 Objective — Question 18

The major constituents of cement are

  • A. calcium trioxocarbonate (IV) and calcium tetraoxosulphate (IV)
  • B. calcium trioxosilicate (IV) and calcium aluminateCorrect
  • C. calcium trioxocarbonate (IV) and silica
  • D. calcium trioxocarbonate (IV) and silica

Explanation

The major constituents of cement are calcium silicates and calcium aluminate.

Chemistry 2015 Objective — Question 19

The type of iron obtained directly from the blast furnace is

  • A. pig ironCorrect
  • B. cast iron
  • C. wrought iron
  • D. corrugated iron

Explanation

Iron obtained directly from the blast furnace is known as pig iron.

Chemistry 2015 Objective — Question 20

If a sample of fructose C6H12O6 contains 24g of carbon, calculate the mass of oxygen in it. [C=12, H=1, O=16]

  • A. 24g
  • B. 32gCorrect
  • C. 48g
  • D. 64g

Explanation

Moles of C = 24/12 = 2mol. Since fructose has 6 C and 6 O atoms per molecule, moles of fructose = 2/6 = 1/3 mol, so moles of O = 6 x 1/3 = 2mol. Mass of O = 2 x 16 = 32g.

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