ro = p = 1 g/ml
ms = pxVs = 1 x 10 = 10 g sol
c% = mdx100/ms => md = msxc%/100
= 10x33/100 = 3,3 g
ms1 = ms + 50 = 10 + 50 = 60 g sol
c1 = mdx100/ms1 = 3,3x100/60 = 5,5%
Cata apa trebuie adaugata pentru a fi concentratia initiala 5%?
ms2 = ms + a , a = masa de apa adaugata
md = 3,3 g
=> 5 = 3,3x100/(10+a)
=> 50+5a = 330
=> a = 56 g
facem proba:
c=? , ms2 = 10+56 = 66 g , md = 3,3 g
=> c = 3,3x100/66 = 5%