5. a) C = md/ms ×100 => md = C×ms/100 = 5×30/100 = 1,5g KOH
Cf = md+x/ms+x×100
10 = (1,5+x)/(30+x)×100
10(30+x) = 100(1,5+x)
300+10x = 150+100x
150=90x
x = 1,666g KOH
V: md final = 1,5+1,666 = 3,16g KOH
md final = 30+1,666 = 31,66g sol. KOH
Cf = 3,16/31,66×100 = 10%
b) ms = md+mH2O => mH2O = ms-md = 30-1,5 = 28,5g H2O
C×mH2O = md(100-C)
mH2O = md(100-C)/C = 1,5(100-10)/10 = 13,5g
mH2O evaporata = 28,5-13,5 = 15g
