Question

calculate mass of na2so4.7h2o contains exactly 6.022*10^22 atoms of oxygen

Answer 1

mole = no of atom /avagadro no
mole= 6.022×10^22/6.022× 10^23
= 10^-1 = 1/10 = 0.1
mole= wt/ molecular wt
0.1 = wt / 266
wt = 0.1 × 266
= 26.6

Answer 2

Your answer is ---

Firstly, we find the mol of oxygen

= 6.022×10^22/6.022×10^23

= 0.1mol

Now, in Na2SO4.7H2O

1 mol Na2SO4.7H2O contains 7 mol O

or, 7 mol O = 1 mol Na2SO4.7H2O

So, 0.1 mol O = 0.1/7 mol Na2SO4.7H2O

Now,

molecular mass of Na2SO4.7H2O = 46 + 32 + 64 + 14 + 112

= 268 u

Therefore, mass of Na2SO4.7H2O is

0.1/7 × 268 = 3.8 gram

【 Hope it helps you 】