Answer:
Solution — Molar mass (Molecular mass in gram) of CaCO3 = 40+12+3×16 = 100 g No. of moles of CaCO3 = No. of molecules/Avogadro constant = 6.022 × 1023/ 6.022 × 1023 = 1 mole Mass of CaCO3 = No. of moles × molar mass = 1 × 100 g = 100 g.
Explanation:
