ChemistryChemical combinationJAMB
Find the mass of 0.25 mol of calcium trioxocarbonate(IV), CaCO₃. (Ca = 40, C = 12, O = 16)
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Answer: D. 25 g
M(CaCO₃) = 40 + 12 + 48 = 100 g/mol; 0.25 × 100 = 25 g.
An AdmitMe practice question, written for the JAMB syllabus. Not an official past question.