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JAMB UTME Chemistry 2020 past questions and answers

This 2020 JAMB UTME 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 2020 set with instant marking.

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Sample JAMB UTME Chemistry 2020 questions

1. A sample of water contains dissolved calcium hydrogencarbonate. Which method will remove its temporary hardness?
  1. Boil the water
  2. Add sodium chloride
  3. Pass the water through a sand filter
  4. Add more calcium hydrogencarbonate

Answer: A. Boil the water

Boiling decomposes dissolved calcium hydrogencarbonate to insoluble calcium carbonate, which can be removed.

2. A solution is prepared by dissolving 150 g of potassium nitrate in 100 g of water at 60°C and then cooled to 20°C. If the solubility at 20°C is 32 g per 100 g of water, what mass of potassium nitrate crystallizes out?
  1. 32 g
  2. 118 g
  3. 150 g
  4. 182 g

Answer: B. 118 g

At 20°C, 32 g remains dissolved in the 100 g of water, so 150 - 32 = 118 g crystallizes.

3. Which observation best distinguishes sulfur dioxide from carbon dioxide when the gases are tested separately?
  1. Both turn damp blue litmus red
  2. Both produce a white precipitate with limewater
  3. Sulfur dioxide decolourizes acidified potassium manganate(VII), but carbon dioxide does not
  4. Sulfur dioxide relights a glowing splint

Answer: C. Sulfur dioxide decolourizes acidified potassium manganate(VII), but carbon dioxide does not

Sulfur dioxide is a reducing agent and decolourizes acidified manganate(VII), whereas carbon dioxide does not.

4. What is the hydrogen ion concentration of a strong monoprotic acid solution with pH 2.0?
  1. 1.0 mol dm^-3
  2. 2.0 mol dm^-3
  3. 0.10 mol dm^-3
  4. 0.010 mol dm^-3

Answer: D. 0.010 mol dm^-3

Since pH = -log10[H+], a pH of 2.0 corresponds to [H+] = 10^-2 mol dm^-3.

5. What is the concentration of sodium hydroxide if 50.0 cm^3 of it is exactly neutralized by 25.0 cm^3 of 0.10 mol dm^-3 sulfuric acid?
  1. 0.10 mol dm^-3
  2. 0.050 mol dm^-3
  3. 0.20 mol dm^-3
  4. 0.40 mol dm^-3

Answer: A. 0.10 mol dm^-3

The acid provides 0.0025 mol H2SO4, or 0.0050 mol NaOH equivalents, giving 0.0050/0.0500 = 0.10 mol dm^-3.

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