maseni udeo

Jecy's picture

Na O stepena celzijusa i atmosferskom pritisku u 1 zapremini vode rastvara se 600 zapremina bromovodonika .Izracunajte maseni udeo HBr u dobijenom rastvoru.

milicaradenkovic's picture

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

V(HBr)=600 dm3

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

n(HBr)=60795/2270.97=26.77 mol

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

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

ro(vode)=1 g/cm3

m(vode)=1000*1=1000 g

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

w(HBr)=m(HBr)/mr=2168.97/3168.37=0.6843=68.43 %