Question

what mass of Na2so4.7H2o contains exactly 6.022×10*22 atoms of oxygen ?

Answer 1

1 molecule of Na₂SO₄.7H₂O contains 11 atoms of oxygen.

Similarly,

1mole of Na₂SO₄.7H₂O contains 11moles of oxygen.

We know 1mole contains 6.022 x 10²³ particles,
so,       0.1 moles contain 6.022 x 10²² particles.

Given that there are 6.022 x 10²² atoms of oxygen.
The above statement indicates there are 0.1 moles of oxygen.

We know,
       11 moles of oxygen atoms are present in 1mole of Na₂SO₄.7H₂O
so,0.1moles of oxygen atoms are present in 0.009moles of Na₂SO₄.7H₂O

Mass of 0.009 moles of Na₂SO₄.7H₂O = 0.009 x 268 = 2.412 grams

Hence 2.412 grams of Na₂SO₄.7H₂O contains 6.022 x 10²² oxygen atoms! 

Answer 2

Mole = number of atom / avagadro no
Mole=6.022×10*22/6.023x 10*23
Mole=6.023×6.022×10*-1
Mole=3.6
Mole= given wt/ molecular weight
3.6=x/266
wt=3.6×266