z 320 cm3 26% roztworu o gestosci 1,16 g/cm3 odparowano 100g wody. jakie stezenie procentowe uzyskal roztwor

Dane: mr1=320cm3*1,16g/cm3=371,2g Cp1=26% ms1=? Cp=ms/mr * 100% /*mr Cp*mr=ms*100% /:100% ms=Cp*mr/100% ms1=26%*371,2g/100% ms1=96,5g ms2=ms1=96,5g mr2=mr1-100g=271,2g Cp2=? Cp=ms/mr * 100% Cp=96,5g/271,2g * 100% Cp=35,6%