Da li moze objasnjenje postupka za rad ovog zadatka. Rastvor dobijen rastvaranjem 51,3 g saharoze u 300g H2O kljuca na temperaturi :(Ke=0,51 saharoza-342) Resenje je 100,255 C :)
dT=Ke*b b=n/m(H2O)=51.3g/342g/mol / 0.3kg=0.5mol/kg dT=0.51Kkg/mol*0.5mol/kg=0.255K=0.255 stepeni C T=(100+0.255) stepeni C=100.255 stepeni C
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dT=Ke*b b=n/m(H2O)=51.3g/342g
dT=Ke*b
b=n/m(H2O)=51.3g/342g/mol / 0.3kg=0.5mol/kg
dT=0.51Kkg/mol*0.5mol/kg=0.255K=0.255 stepeni C
T=(100+0.255) stepeni C=100.255 stepeni C