Question

将448L标准状况下的NH3气体通入1LH2O中,所得溶液的密度0.9g/ml,计算溶液的物质的量浓和溶质的质量分数

Answer 1

n=448/22.4=20mol

m(NH3)=20mol*17.0g/mol=340g

m(solution)=m(NH3)+m(water)=1340g

V=m(solution)/ρ=1340g/(0.9g/ml)=1489ml=1.489L

C=n/V=20/1.489=13.(4)mol/L 注：因有效数字原因,括号中的可以忽略.

C%=m(NH3)/m(solution)*100%=340g/1489g*100%=22.8%