What proportion of mole are 71 g of sodium sulfate Na2SO4?

given:
m(Na2SO4) = 71 g
find:
ν(Na2SO4) - ?
Solution:
1- way:
М(Na2SO4) = 142 g/mol
71 g Na2SO4  х mol Na2SO4
142 g Na2SO4 1 mol Na2SO4
х= (71 g  • 1 mol)/(142 g)   =0,5 mol
2- way:
М(Na2SO4) = 2 g/mol
ν=m/М  ;ν(Na₂SO₄) =(m(Na₂SO₄))/(М(Na₂SO₄))  = (71 g)/(142 g/mol)  = 0,5 mol.
Answer:ν(Na2SO4) =  0,5 mol.