IV.
1.
md = 75g , m.apa = 175g
stim ca ms = md+md = 75+175 = 250g
stim ca c% = mdx100/ms = 75x100/250 = 30%
2.
c2% = c%/2 = 30/2 = 15%
notam cu a = masa de apa adugata
=> ms.final = ms+a = 250+a
=> 15 = 75x100/(250+a)
=> a = 3750/15 = 250 g apa adaugata
V.
1.
ms1 = 200g , c1% = 45%
ms2 = 300g , c2% = 25%
din c% claculam md1 si md2
=> md1 = ms1xc1/100 = 200x45/100 = 90g
=> md2 = ms2xc2/100 = 300x25/100 = 75g
=> md.tot = md1+md2 = 90+75 = 165g
ms.tot = ms1+ms2 = 200+300 = 500g
din ms.tot scoatem m.apa
=> m.apa = ms.tot - md.tot = 500-165 = 335g
2.
c%final = md.totx100/ms.tot = 165x100/500 = 33%
3.
aplicand metoda de separare = cristalizare , apa se evapora si ramane sarea
