In the compound CaCO3 at the end of the compound represents the number of

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. of moles × molar mass

= 1 × 100 g = 100 g.

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