Find the number of molecules present in 1miligram of calcium carbonate

1.	Find the number of molecules present in 1miligram of calcium carbonate-
Answer» 6.022 x 1023 x 0.01 molecule is the answer Molecular mass of CaCO3= 40 + 12 + (3 x 16) = 100 g [Molecular weight of Ca = 40, C = 12, O =16] 1 mole of Oxygen atoms = 6.022 x10^23 Oxygen atoms (Avogadro’s number) 1 mole of Calcium Carbonate = 100g of Calcium Carbonate 100 g of Calcium Carbonate (100,000mg of Calcium Carbonate) has 3 moles of Oxygen atoms = 3 x 6.022 x 10^23Oxygen atoms. Thus, 1 mg of Calcium Carbonatehas (3 x 6.022 x 10^23)/100000 Oxygen atoms

Find the number of molecules present in 1miligram of calcium carbonate-

Answer»

6.022 x 1023 x 0.01 molecule is the answer

Molecular mass of CaCO3= 40 + 12 + (3 x 16) = 100 g [Molecular weight of Ca = 40, C = 12, O =16]

1 mole of Oxygen atoms = 6.022 x10^23 Oxygen atoms (Avogadro’s number)

1 mole of Calcium Carbonate = 100g of Calcium Carbonate

100 g of Calcium Carbonate (100,000mg of Calcium Carbonate) has 3 moles of Oxygen atoms = 3 x 6.022 x 10^23Oxygen atoms.

Thus, 1 mg of Calcium Carbonatehas (3 x 6.022 x 10^23)/100000 Oxygen atoms