oblicz stezenie molowe 30% roztworu perhydrolu (H2O2) o gestosci 1.11g.cm3

M(H2O2)=34g/mol zakł. że V=100cm3 mr=V*d mr=100*1,11 mr=111g ms=(30*111)/100 ms=33,3g n=33,3/34 n=0,979mol C=n/V C=0,979/0,1 C=9,79mol/dm3

Cp=30% d=1,11g/cm³ M(H2O2)=2*1g/mol+2*16g/mol=34g/mol Vr=1dm³=1000cm³   d=mr/Vr   /*Vr d*Vr=mr   mr=1000cm³*1,11g/cm³ mr=1110g   ms=? Cp=ms/mr * 100%   /*mr Cp*mr=ms*100%   /:100% ms=Cp*mr/100%   ms=30%*1110g/100% ms=333g   n=? n=ms/M n=333g / 34g/mol n=9,8mol   Cm=? Cm=n/Vr Cm=9,8mol / 1dm³ Cm=9,8mol/dm³