Rastvori

Submitted by banejeca943 on Sun, 11/17/2013 - 21:32
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Koliko je grama kalijum-nitrata i vode potrebno za pripremanje 250g rastvora molalnosti 2mol/kg?

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milicaradenkovic - Sun, 11/17/2013 - 22:00

b=n/m(rastvaraca)

n(KNO3)=2 mol

m(KNO3)=n(KNO3)*Mr=2*101==202 g

mr=m(vode)+m(KNO3)

m(vode)=250-202=48 g

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banejeca943 - Sun, 11/17/2013 - 22:15

U resenjima pise 42g i 208g..?
Inace, hvala puno!

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milicaradenkovic - Sun, 11/17/2013 - 22:37

Ja mislim da je greska u resenju.