SE DA
V(SOLUTIEI) =o,5 l
w%(NaOH)=20%
p(sol)=1,28 g/ml
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m(NaOH)-?
W%=m(sub)*100%/m(sol)
m(sub)=m(sol)*w%/100%
P=m/v
m=p*v
0,5l=500ml
m(sol de NaOH)=p(sol)*v(sol)=1,28(g/ml)*500(ml)=4000 g
m(NaOH)=m(sol de Naoh)*w%(NaOH)/100%=4000 g *20% /100%=800 G