ch MgSO4*xH2O
Mch=120+18x
in 49,2 g ch exista 25,2g H2O
(120+18x) g ch............18x g H2O
49,2 g ch.......................25,2 g H2O x=7 ==> MgSO4*7H2O
mch=120+126=246g/mol
m sare anhidra din ch = 49,2g - 25,2g = 24g (md)
ms=49,2+174,8= 224 (g)
c=md*100/ms=24*100/224=........
mH2O(din sol.) = 224 - 24 = 200 g
18 g (1mol) H2O.....................6,022*10²³ molecule H2O
200 g H2O.....................................y = 66,91*10²³ molecule H2O