Procenat

Submitted by Ana. on Fri, 02/14/2014 - 11:53
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Koliko procenata kiseonika ima rastvor koji sadrži 0.1 mol NaOH u 1000 g vode?

Rješenje: 88.7

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milicaradenkovic - Fri, 02/14/2014 - 16:38

18 g vode----------------------16 g O

1000 g vode--------------------- x g O

x=1000*16/18=888.9 g O

m(NaOH)=n*Mr=40*0.1=4 g

mr=m(NaOH)+m(vode)=10004=1004 g

w(O)=m(O)/mr=888.9/1004=0.885=88.5 %