Rastvor

Submitted by Mimi on Tue, 01/21/2014 - 21:04
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Opšta diskusija

1.) Pripremi 100g 0,1% rastvora KMnO4!

2.) Odrediti masu rastvora koji mozemo pripremiti od 6,4g NaCl ako je W(NaCl)=0,09

milicaradenkovic's picture

milicaradenkovic - Wed, 01/22/2014 - 08:56

1) mr=100 g

ms=mr*w=100*0.001=0.1 g KMnO4 i dodas vode

m(vode)=mr-ms=100-0.1=99.9 g

2) m(NaCl)=6.4 g a w(NaCl)=0.09

w=m(NaCl)/mr

mr=m(NaCl)/w=6.4/0.09=71.11 g

Mimi's picture

Mimi - Wed, 01/22/2014 - 13:07

Hvala :)