a)
M piatra acra=258+216=474
ms=9,18+200=209,18
474g Cristalohidrat........258g sare anhidra.........216g H2O
9,18g ch...........................x g ...............................y
x=4,99 g sare anhidra =aprox. 5g (md)
y=4,18 g H2O
c=md*100/ms=5*100/209,18=2,39%
b)
474 g ch..........216g H20
100g ch...............z =45,56 % H20
c) apa din solutia finala este:
apa din 9,18 g cristalohidrat (y) +apa adaugata (200g)=4,18+200=204,18g
nH2O=204,18/18=11,34moli in care vor exista 11,34*6,023*10²³ molecule de apa
Pentru siguranta trebuie sa refaci toate calculele !!!