Brom

Submitted by Hemicarka5 on Sun, 03/23/2014 - 13:57
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Na 0 stepeni i atmosferskom pritisku u 1 zapremini vode rastvara se 600 zapremina bromovodonika.Izracunajte maseni udio HBr u dobijenom rastvoru.

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milicaradenkovic - Sun, 03/23/2014 - 14:39

V(HBr)=600 dm3

p*V(HBr)=n(HBr)*R*T

n(HBr)=101.325*10(3)*600*10(-3)/8.314*273.15=26.77 mol

m(HBr)=n(HBr)*Mr=81*26.77=2168.37 g

m(vode)=V(vode)*ro(vode)

ro(vode)=1 g/cm3

m(vode)=1000*1=1000 g

mr=m(vode)+m(HBr)=1000+2168.37=3168.37 g

w(HBr)=m(HBr)/mr

w(HBr)=2168.37/3168.37=0.684=68.4 %