Mesanje rastvora
Submitted by Dušan on Sat, 03/01/2014 - 23:45

Forums:
Koju zapreminu 20% nog rastvora NaOH gustine 1,21 g/cm3 treba odmeriti da bi se mesanjem sa 100 cm3 5%-nog rastvora NaOH dobio rastvor masenog udela 7,5%?
Resenje je 20 cm3
ro2(NaOH, 5 %)=1.055 g/cm3
ro2(NaOH, 5 %)=1.055 g/cm3
mr2=ro2*V2=100*1.055=105.5 g
w=msu/mru
msu=ms1+ms2
ms2=5*105.5/100=5.275 g
mru=mr1+mr2
mr2=105.5 g
ms1=0.2*mr1
0.075=(0.2*mr1+5.275)/(mr1+105.5)
0.2*mr1+5.275=0.075*(mr1+105.5)
0.2*mr1+5.275=0.075*mr1+7.9125
0.2*mr1-0.075*mr1=7.9125-5.275
0.125 mr1=2.6375
mr1=2.6375/0.125=21.1 g
V1=mr1/ro1=21.1/1.21=17.44 cm3