Question

find number of oxygen atoms present in 100mg of CaCO3

Answer 1

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 x 10^23 Oxygen atoms (Avogadro’s number)

1 mole of Calcium Carbonate = 100 g of Calcium Carbonate

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

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

Therefore, 100 mg of Calcium Carbonate has (3 x 6.022 x 10^23 x 100)/100000 Oxygen atoms

=> 100 mg of Calcium Carbonate has (3 x 6.022 x 10^23)/1000 Oxygen atoms

=> 100 mg of Calcium Carbonate has (3 x 6.022 x 10^23)/10^3 Oxygen atoms

=> 100 mg of Calcium Carbonate has 3 x 6.022 x 10^23 x 10^-3 Oxygen atoms

=> 100 mg of Calcium Carbonate has 3 x 6.022 x 10^20 Oxygen atoms

Answer 2

Answer:

Moles of CaCO 3 = 10power3 ×100/100 =1 millimoles

Number of oxygen atoms in CaCO3 =3

Total number of oxygen atoms in 1 milimoles CaCO 3

​ =10power−3 ×3×6.02×10power23 =1.806×10powre21

Explanation: