Maseni udeo-zadatak

Submitted by Lanmy on Thu, 05/15/2014 - 19:30
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Da li neko moze da mi pomogne oko zadatka: Koliko iznosi maseni udeo etanola u %, u rastvoru koji je nastao mešanjem 60g etanola i 100 g vode? Unapred hvala :)

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Dunja - Thu, 05/15/2014 - 20:46

w=mrs/mr x100
mrs=60g
mr=m(etanola)+m(H2O)= 60+120=180

w=60/180 x100 = 0,3333x100=33,33%

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milicaradenkovic - Thu, 05/15/2014 - 21:09

w=ms/mr

mr=m(etanola)+m(vode)=60+100=160 g

w=60/160=0.375=37.5 %

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Dunja - Thu, 05/15/2014 - 21:10

da, moja greska

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Lanmy - Thu, 05/15/2014 - 21:32

Hvala puno :)