Mesanje rastvora

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

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milicaradenkovic - Sun, 03/02/2014 - 09:03

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