Answer:
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.
