FeSO4 + 7H2O = FeSO4 x 7H2O
m. sub.diz.(FeSO4)= υ x M =0,1mol x 152 g/mol = 15,2 g
m(H2O) = υ x M = 10 mol x 18 g/mol = 180 g
m.sol. = m(H2O) + m. sub.diz. =180g + 15,2g =195,2g
ω = [tex] \frac{m. sub.diz.}{m.sol} [/tex] x100%= [tex] \frac{15,2 g}{195,2 g} [/tex] x 100% = 7,79%