stezenie molowe wodnego roztworu aldehydu mrowkowego wynosi 4,85 mol*dm-3,a gestosc 1,04g*cm-3.oblicz stezenie procentowe tego roztworu.
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Cm=4,85 mol*dm-3
d=1,04 g*cm-3=1040 g*dm-3
M (HCHO)=30 g*mol-1
Cp=Cm*M*100/d
Cp=4,85*30*100/1040=14550/1040=14%
odp:Stezzenie % tego r-ru wynosi ok.14%.