1. oblicz masy czasteczkowe propanolu i gliceryny
2. Oblicz stosunki masowe mc:mH:mo w cząsteczce etanolu i metanolu
3. Jaka ilosc etanolu znajduje sie w 1 litrze 35% roztworu gestosc roztworu 0,9 g/cm3
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1)Propanol: C3H7OH, masa cząsteczkowa = 3*12u + 8*1u + 16u = 36u + 8u + 16u = 60u
Glicerol: C3H8O3, masa cząsteczkowa = 3*12u + 8*1u + 3*16u = 36u + 8u + 48u = 92u
2)Etanol: C2H5OH; mC:mH:mO = 24:6:16 = 12:3:8
Metanol: CH3OH; mC:mH:mO = 12:4:16 = 6:2:8 = 3:1:4
3)V = 1l = 1dm^3
Cp = 35%
d = 0.9g/cm^3 = 900g/dm^3
M = 46g/mol
Cp = (Cm*M*100%)/d stąd:
Cm = Cp*d / M*100%
Cm = (35%*900g/dm^3) / (46g/mol*100%)
Cm = 6,85mol/dm^3
Cm=n/V stąd
n = Cm*V
n = 6,85mol/dm^3 * 1dm^3 = 6,85moli
n=m/M stąd
m = n*M
m = 6,85mol*46g/mol = 315,1g