maseni udeo
Submitted by Jecy on Sat, 03/15/2014 - 22:48
Forums:
Na O stepena celzijusa i atmosferskom pritisku u 1 zapremini vode rastvara se 600 zapremina bromovodonika .Izracunajte maseni udeo HBr u dobijenom rastvoru.
Na O stepena celzijusa i atmosferskom pritisku u 1 zapremini vode rastvara se 600 zapremina bromovodonika .Izracunajte maseni udeo HBr u dobijenom rastvoru.
p*V(HBr)=n(HBr)*R*T
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 %