maseni udeo rastvora HNO3

Submitted by mika on Fri, 02/14/2014 - 08:36
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Koje zapremine 70%rastvora gustine p=1.4 g/cm3 i 20% rastvora nitratne kiseline (HNO3) gustine p=1.1g/cm3 treba pomešati da se dobije 500 cm3 40% rastvora te kiseline?

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milicaradenkovic - Fri, 02/14/2014 - 16:29

ro(40 %)=1.2 g/cm3

mru=Vu*ro=500*1.2=600 g

mru=mr1+mr2

600=mr1+mr2

msu=ms1+ms2

w1*mr1+w2*mr2=wu*mru

mr1=600-mr2

0.7*(600-mr2)+0.2*mr2=600*0.4

420-0.5*mr2=240

0.5*mr2=180

mr2=180/0.5=360g

mr1=600-360=240 g

V1=mr1/ro1=240/1.4=171.43 cm3 70 % HNO3

V2=mr2/ro2=360/1.1=327.3 cm3 20 % HNO3