Ile moli gliceryny potrzeba do przygotowania 400g jej 5-procentowego roztworu?
cp=5%
ms=?
mr=400g
400 g - 100%
x g - 5 %
x=20 g
-------------
m=92 g/mol
1 mol - 92 g
x mola - 20g
x= 0,22 mola --->Wynik.
Cp=5%
Cp=(ms*100%)/mr /*mr
Cp*mr=ms*100% /:100%
ms=(Cp*mr)/100%
ms=(5%*400g)/100%=20g
mC₃H₈O₃=3*12u+8*u+3*16u=36u+8u+48u=92u
1 mol C₃H₈O₃ - 92g
x mol C₃H₈O₃ - 20g
x=(1 mol C₃H₈O₃*20g)/92g=0,22 mola C₃H₈O₃
Odp. Potrzeba 0,22 mola C₃H₈O₃.
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cp=5%
ms=?
mr=400g
400 g - 100%
x g - 5 %
x=20 g
-------------
m=92 g/mol
1 mol - 92 g
x mola - 20g
x= 0,22 mola --->Wynik.
Cp=5%
mr=400g
Cp=(ms*100%)/mr /*mr
Cp*mr=ms*100% /:100%
ms=(Cp*mr)/100%
ms=(5%*400g)/100%=20g
mC₃H₈O₃=3*12u+8*u+3*16u=36u+8u+48u=92u
1 mol C₃H₈O₃ - 92g
x mol C₃H₈O₃ - 20g
x=(1 mol C₃H₈O₃*20g)/92g=0,22 mola C₃H₈O₃
Odp. Potrzeba 0,22 mola C₃H₈O₃.