Question

Oblicz stężenie molowe 30% roztworu perhydrolu (H2O2) o gęstości 1.11 g/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%
M H2O2 = 34g/mol
d = 1,11g/cm3
 
Cm = Cp*d*1000/100*M
Cm = 30*1,11*1000/100*34 = 33330/3400 = 9,8 mol/ dm^3