Question

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

Answer 1

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

Answer 2

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