ms initial = 20g c = 20% ⇒ mdi = 4g ni = 0,1 moli NaOH
sol. A : ms = 800g md = 4g
msA/mH2O = 1/0,25 = 800/ 200
sol B : 800 + 200 = 1000g md = 4g
1g sol. B contine : 4/1000 = 0,004g NaOH
sol.C : Vs = 0,1l n = 0,004/40 moli c = n/Vs = 0,004/4 = 0,001 moli/l
c = 10⁻³ mol/l