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WAEC WASSCE Chemistry 2006 past questions and answers

This 2006 WAEC WASSCE Chemistry set has 40 practice questions. Try these 5 free sample questions, tap a question to reveal the answer and explanation. Then practise the whole 2006 set with instant marking.

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Sample WAEC WASSCE Chemistry 2006 questions

1. A 2.50 g sample of CuSO4·xH2O is heated to constant mass and leaves 1.60 g of anhydrous CuSO4. Given CuSO4 = 160 and H2O = 18, what is x?
  1. 2
  2. 5
  3. 4
  4. 10

Answer: B. 5

The sample contains 0.90/18 = 0.050 mol of water and 1.60/160 = 0.010 mol of CuSO4, giving a ratio of 5:1.

2. When aqueous silver nitrate is mixed with aqueous sodium chloride, which net ionic equation represents the formation of the precipitate?
  1. Na+ + NO3- -> NaNO3(s)
  2. Ag+ + NO3- -> AgNO3(s)
  3. Na+ + Cl- -> NaCl(s)
  4. Ag+ + Cl- -> AgCl(s)

Answer: D. Ag+ + Cl- -> AgCl(s)

Silver ions combine with chloride ions to form insoluble silver chloride.

3. Which method is suitable for preparing pure crystals of copper(II) sulfate from dilute sulfuric acid?
  1. Add excess sodium chloride, filter, then distil
  2. Mix the acid with copper metal and evaporate immediately
  3. Warm the acid with excess copper(II) oxide, filter, then crystallize the filtrate
  4. Neutralize the acid with excess calcium carbonate and collect the gas

Answer: C. Warm the acid with excess copper(II) oxide, filter, then crystallize the filtrate

Excess insoluble copper(II) oxide neutralizes the acid, and filtration removes the unreacted solid before crystallization.

4. What volume of 0.200 mol dm-3 NaOH completely neutralizes 25.0 cm3 of 0.100 mol dm-3 H2SO4? The equation is H2SO4 + 2NaOH -> Na2SO4 + 2H2O.
  1. 25.0 cm3
  2. 12.5 cm3
  3. 50.0 cm3
  4. 6.25 cm3

Answer: A. 25.0 cm3

The acid amount is 0.00250 mol, requiring 0.00500 mol NaOH, which occupies 0.0250 dm3 at 0.200 mol dm-3.

5. Assuming complete ionization, what is the pH of 0.010 mol dm-3 hydrochloric acid?
  1. 1
  2. 2
  3. 12
  4. 13

Answer: B. 2

For this strong acid, [H+] = 0.010 mol dm-3 and pH = -log10(0.010) = 2.

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