Question

Calculate the number of oxygen atoms in 0.22 mole of Na2CO3.10H2O

Answer 1

Weight of oxygen in 1 mole of Na2CO3.10H2O = 208 g

1 mole sodium carbonate decahydrate contains oxygen = 208 g

0.22 mole contains = 208 * 0.22

                                 = 45.76 g

16 g oxygen contains 6.23*10^23 oxygen atoms

Then

45.76 g oxygen contains = 6.23/16 * 45.76 * 10^23

                                          = 17.81 * 10^23 oxygen atoms

                                           = 1.781 * 10^22 oxygen atoms

Hope my answer helps your first question.

Answer 2

it may be ur ans.....