Question

calculate no. of atoms present in 15gramsof calcium carbonate​

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

CaCo3 wt =15 g

G.M.wt = 1 Ca + 1 C +3 O

= 40 + 12 + 48

= 100

No. atoms = Wt/G.M.Wt * 6 * 10^23 *atomicity

. =15/100*6*10^23*5

= 15*6*5*10^21

= 450 *10^21