Post-UTME Chemistry 2017 Objective — Question 12
Question 12 of 20 from the Post-UTME Screening (Post-UTME) Chemistry 2017 Objective paper, with the correct answer and a full explanation.
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If the solubility of KHCO3 is 0.40 mol/dm3 at room temperature, calculate the mass of KHCO3 in 100cm3 of the solution at that temperature.
- A. 100.0g
- B. 40.0g
- C. 10.0g
- D. 4.0gCorrect
Explanation
Moles = 0.40 x 0.100 = 0.040mol. Molar mass KHCO3 = 100g/mol. Mass = 0.040 x 100 = 4.0g.
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