1. How many molecules of CH, are present in 3.48 moles?

Calculate the mass of 6.022 × 1023 molecule of Calcium carbonate (CaCO3). 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.

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