obliczyć stężenie procentowe 10-molowego roztworu ZnCl2, którego gęstość wynosi 1,946 g/cm3
Cm=10mol/dm3 d=1,946g/cm3=1946g/dm3
Cp=Cm*M*100%/d
MZnCl2=65+71=136g/mol
Cp=10mol/dm3*136g/mol*100%//1946g/dm3
Cp=69,9%
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
Cm=10mol/dm3 d=1,946g/cm3=1946g/dm3
Cp=Cm*M*100%/d
MZnCl2=65+71=136g/mol
Cp=10mol/dm3*136g/mol*100%//1946g/dm3
Cp=69,9%