Question

The number of atoms of oxygen present in 10.6 g of na2co3 will be

Answer 1

Mol wt of Na2Co3=46+12+48=106

Moles of Na2CO3= 10.6/106=0.1

3 moles of oxygen = 0.1

1 mole of oxygen = 0.1/3 = 0.033

No of atoms of oxygen

= 0.033*6.023*10^23

= 2 *10^22