1.
solutia de 50% contine 50 g alcool la 100 g sol. la care se adauga a g apa si rezulta o sol. de 40%
stim ca c% = mdx100/ms
=> ms2 = ms1 + a = mdx100/c2
=> 100 + a = 50x100/40 <=> 100+a = 125
=> a = 125-100 = 25 g apa adaugata
2.
ms1 = 50 g , c1 = 40%
ms2 = 150 g , c2 = 50%
ms3 = 200 g , c3 = 60%
md4 = 400 g , c%.final = ?
ms.fin = ms1 + ms2 + ms3 + md4
= 50 + 150 + 200 + 400 = 800 g sol finala
stim ca c% = mdx100/ms
md.fin = md1 + md2 + md3 + md4
= ms1xc1/100 + ms2xc2/100 + ms3xc3/100 + md4
= (ms1c1+ms2c2+ms3c3+40000) / 100
=> c%.fin = [(ms1c1+ms2c2+ms3c3+40000) / 100] x100/800
=> c%.fin = [(ms1c1+ms2c2+ms3c3+40000)] / 800
= (50x40+150x50+200x60+40000)/800
= 76,875%