标准状况下,aLHCl溶于1L水中,得到盐酸的密度为bg*mL-1,则c(HCl)为
2个回答

n(HCl)=a/22.4 1L=1000ml

盐酸溶液的质量m=m(HCl)+m(H2O)

=n(HCl)*36.5+1*1000

=36.5a/22.4+1000 (g)

盐酸溶液的体积V=m/b=(36.5a/22.4+1000)/b

则c(HCl)= n(HCl)/V

=(a/22.4)/[(36.5a/22.4+1000)/b]

=ab/(36.5a+22400)