Molarna koncentracija

Submitted by gxxrjj on Sun, 06/28/2015 - 14:22
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Izracunati molarnu koncentraciju 42% rastvora azotne kiseline (HNO3) cija je gustina 1,25 mol/dm3

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JOVAN - Sun, 06/28/2015 - 20:34

jel gustina u g/dm3 ili mol?

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buduci_student - Sun, 06/28/2015 - 21:06

w=c*Mr/ro

c=ro*w/Mr

c=8,33(3)*10-3

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gxxrjj - Sun, 06/28/2015 - 21:19

Gustina je 1,25 g/cm3, a resenje je 8,33 mol/dm3 

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milicaradenkovic - Mon, 06/29/2015 - 08:14

c=1.25*1000*0.42/63=8.33 mol/l