Koligativne osobine rastvora

Submitted by Micas on Thu, 06/07/2018 - 15:03
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Koliko grama uree treba rastvoriti u 250ml rastvora da bi osmotski pritisak bio 0,15atm na 25°C? 

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zlaja99 - Thu, 06/07/2018 - 16:58

m(CO(NH2)2)=?

p=0.15 atm

V rastvora=250 mL=0.25L=0.00025 m3

T=25 C=298.15 K

R=8.314 J/Kmol

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1 atm=1.01325 *10^5

0.15 atm=0.15*1.01325*10^5=0.1519875=10^5=1.519875*10^4

p=cRT=>c=p/RT

c=1.519875 *10^4/8.314 J/Kmol*298.15 K=6.13147 mol/L

n=c*V=6.13147*0.25=1.5328 mol

m=n*M=60*1.5328=91.97 g