Question

在标况下,将224LHCL气体溶于635ML水中,所得盐酸的密度为1.18g/cm3,求此溶液的质量分数?

Answer 1

m(溶液)=m(HCl)+m(H2O)

ρ(H2O)=1g/cm3

m(HCl)=[V(HCl)/Vm]*M(HCl)

=224L/(22.4L/mol) * 36.5g/mol

=365g

m(H2O0=V(H2O)ρ(H2O)

=635mL*1g/mL

=635g

所以

m(溶液)=365g+635g=1000g

ω=m(HCl)/m(溶液）

=365/1000 * 100%

=36.5%