Probleme cu cresterea concentratiei.
1)In 2 kg sol HCl 36,5% se adauga 100 g HCl.Calculati concentratia obtinuta
md1=36,5*2/100=0,71 kg HCl
md2=0,71+0,1=0,81 kg HCl
ms2=0,1+2=2,1 kg solutie
c final = md2/ms2*100=0,81/2,1*100=38,57%
2)Ce cantitate de zahar trebuie adaugata in 75 g solutie 10% pt a se obtine o solutie cu concentratia de 20%
md1=10*75/100=7.5 g zahar
md2=7,5+x
ms2=75+x
c final = (7,5+x)/(75+x) * 100 => 750+100x=20*75+20x=>80x=750=>x=9,375 g zahar
3)Sa se calculeze cantitatea de NaOH adaugata unei solutii de 200 g NaOH 20% pentru a obtine o concentratie de 40%.
md1=200*20/100=40 g
md2=40+x
ms2=200+x
c final = (40+x)/(200+x) * 100 =>4000+100x=8000+40x=>60x=4000=>x=66,66 g NaOH
Probleme cu scaderea concentratiei
4)Cate grame de apa sunt necesare pentru a dilua 50 g solutie 40% la o concentratie de 15%?
md1=md2=>md=50*40/100=20 g
ms2=md2/c2*100=20/15*100=133,33 g solutie
ms2=ms1 + m apa => m apa = 133,33 - 50=83,33 g apa
5)Calculati concentratia obtinuta in urma diluarii a 100 g HCl 36,5% cu 200 g apa.
md1=md2=>md=36,5*100/100=36,5 g
ms2=ms 1 + m apa = 100+200=300 g
c%=md2/ms2*100=36.5/300*100=>c%=12,166%
6)Cate grame de apa s-au folosit la diluarea unei solutii cu masa de 200 g NaOH 40% pentru a obtine o concentratie de 20%
md1=md2=>md=200*40/100=80 g
ms2=md2/c2*100=80/20*100=400 g
ms2=ms1 + m apa => m apa = 400 - 200 = 200 g apa
