WAEC Chemistry Questions and Answers 2022 | Theory and OBJ- Download PDF

Filed in WAEC News by on March 8, 2022 0 Comments

WAEC Chemistry Questions and Answers 2022 | Theory and OBJ- Download PDF

WAEC Chemistry Questions and Answers 2022If you are about to write the West African Senior Certificate Exams (WASSCE), you will realize that you need the WAEC Past Questions and Answers.

WAEC Chemistry Questions and Answers 2021

WAEC Chemistry Questions and Answers 2022

This article provides you with WAEC Chemistry Questions and Answers for the WASSCE 2022. It contains information regarding the WAEC Chemistry Theory and Objective Questions and Answers.

WAEC Chemistry Objective Questions

1. Which of the statements below is correct about Isotopes of the same element?

  1. Have the Same number of Protons, neutrons, and electrons.
  2. Same Protons and neutrons but a different number of electrons.
  3. Same number Protons and electrons but a different number of neutrons.
  4. The number of Neutrons and electrons is the same but the number of protons differs.

2. The chemical bond that is formed by the transfer of electrons is known as?

  1. Covalent Bond
  2. Dative Bond
  3. Ionic bond
  4. Metallic bond

3. Two electrons can fill the same orbital if only?

  1. They have different Angular momentum quantum numbers.
  2. Magnetic quantum numbers differ
  3. Principal quantum numbers are different.
  4. Spin quantum numbers not the same

4. Which of the substances listed below is not a hydrocarbon?

  1. Benzene
  2. Butane
  3. Ethyne
  4. Ethanamide

5. The complete ionization of A substance into hydroxonium ions indicates:

  1. Strong acid.
  2. Strong base.
  3. Weak acid.
  4. Weak base.

6. One of the solutions below that can resist changes in pH when a small quantity of a base or acid is added is?

  1. Buffer solution
  2. Neutral solution
  3. Saturated solution
  4. Supersaturated solution

7. Write a balanced chemical equation for a 2.47g dry pure copper (II) oxide which is completely reduced copper using a laboratory gas. If the mass of the residue left is found to be 1.97g.

    1. Cu + O → CuO.
    2. 2Cu+O2 → 2CuO
    3. Cu → O
    4. 2Cu →O2

The correct answer is B. 2Cu+O2 → 2CuO.
See how this question is solved in the number one (No.1) question in the theory section.

Use this question to answer question 2 & 3

8. Calculate the mass of anhydrous sodium-trioxocarbonate (IV) present in a 300cm3 of 0.1M; (Na =23, C=12, O=16).

  1. 18g
  2. 18kg
  3. 16g
  4. 56g
  5. 16kg

The correct answer is A. 3.18g. See solving in No.2 ques. in the theory section.

9. The number of Na2CO3 particles present in the solution

  1. 6.02 X 1023
  2. 0.1 X 6.02 1023
  3. 1.81 X 1022
  4. 1.81 X 1023
  5. 6.02 X 1022

The correct answer is C. 1.81 X 1022.

WAEC Chemistry Theory Questions

These are possible WAEC Chemistry theory questions and answers 2022 that may appear in this year’s exam.

There are not actual questions and answers but, sample questions for practices.

Correct WAEC Chemistry Theory answers to OBJ paper 1 and 2 solved solution

2.47g of dry pure copper (II) oxide was completely reduced to a copper using laboratory gas. The mass of the residue left was found to be 1.97g. Write a chemical equation for this reaction.


Molar mass of copper atoms =63.5

No. of moles of copper atoms =1.97

=63.5 = 0.03 mole


Molar mass of oxygen atoms = 16

No. of moles Oxygen atom = (2.47 – 1.97) = 0.5

=0.03 mole

Mole ratio of Cu to O = 0.03  :   0.03

=    1   :    1

Therefore, the equation is Cu + O → CuO.

But Oxygen is diatomic, so we have 2Cu+O2 → 2CuO

2. Calculate the;

  • mass of anhydrous sodiumtrioxocarbonate (IV) present in a 300cm3 of 0.1M;
  • Number of Na2CO3 particles present in the solution (Na =23, C=12, O=16)


a. Molar concentration of Na2CO3 =0.1M

Molar mass of Na2CO= 106gmol-1

Mass concentration = Molar concentration X Molar mass

=0.1 X 106


i.e 1000cm3 of 0.1M solution contain 10.6g of Na2CO3

Therfore,300cm3 of 0.1M solution will contain

= 300 X 10.6/1000

=3.18g of Na2CO3

b. Number of Na2CO3 particles

=Molar concentration X 6.02 X 1023

=0.1 X 6.02 1023

=6.02 X 1022

Now, 100cm3 of 0.1M solution contains 6.02X1022 Na2COparticles

Therefore, 300cm3 of 0.1M solution will contain 300X6.02 X1022/1000

= 1.81 X 1022 particular of Na2CO3.

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