2.grupa PSE

Submitted by NikolaM on Sun, 12/15/2013 - 19:48
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koliko g gorka so je potrebno rastvoriti u 1dm3 vode da bi se dobio rastvor MgSO4 molalnosti 1 mol/kg

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milicaradenkovic - Sun, 12/15/2013 - 19:55

ro(vode)=1 g/cm3

mr=ro*V=1*1000=1000 g=1 kg

b=n(supstance)/mr

1=n(MgSO4)/1

n(MgSO4)=1 mol

n(MgSO4*7H2O)=n(MgSO4)=1 mol

m(MgSO4*7H2O)=n(MgSO4*7H2O)*Mr=1*246=246 g

 

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NikolaM - Sun, 12/15/2013 - 19:58

U resenju pise 281,46g

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milicaradenkovic - Fri, 01/10/2014 - 19:24

Da li neko zna kako se radi zadatak?

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milicaradenkovic - Fri, 01/24/2014 - 21:21

Konacno sam reslia zadatak

ro(vode)=1 g/cm3 pa je m(vode)=ro(vode)*V=1000*1=1000 g

120:246=x %:100 %

x=120*100/246=48.8 % MgSO4

w(vode)=100-m(vode)=100-48.8=51.2 %

0.488*R:(1000+0.512*R)=120:1000

120*(1000+0.512*R)=1000*0.488*R

488*R-61.44*R=120000

R=120000/426.56=281.3 g MgSO4*7H2O