Question

标准状况下：将67.2LHCl气体溶于440mL水中.所得盐酸的密度为1.1g/㎝3,求盐酸的物质的量浓度?

Answer 1

n(HCl) = 67.2L/(22.4L·mol-1) = 3mol

m(HCl) =3mol*36.5g/mol = 109.5g

m(H2O)= 440ml*1g/ml = 440g

m(溶液)= m(HCl)+m(H2O)= 109.5g+440g=549.5g

V(溶液)= m(溶液)/密度 = 549.5g/(1.1g/㎝3) = 500 mL = 0.5L

c(HCl)= n(HCl)/V(溶液)= 3mol/0.5L =6mol/L