Question

在标准状况下，350体积氨气溶解于1体积水中，得到密度是0.92g/cm3的氨水。求此氨水的质量分数和物质量浓度

Answer 1

设1体积为1L，则
n(NH3)=V(NH3)/Vm=350L/(22.4L/mol)=15.625mol
m(NH3)=n(NH3)·M(NH3)=15.625mol*17g/mol=265.625g
m(H2O)=1000ml*1g/cm3 = 1000
m[NH3(aq)]= m(NH3) + m(H2O) = 265.625g + 1000g =1265.625g
w=m(NH3)/m[NH3(aq)]=265.625g/1265.625g≈21.0%
c(NH3)= 1000*ρ[NH3(aq)]*w/M(NH3)=1000cm3/L*0.924g/cm3*0.21 / 17g/mol=11.41mol/L
所以NH3的质量分数为28.7%，物质的量浓度为11.41mol/L

Answer 2

ω=350*22.4*17/（350*22.4*17+1000）≈99.26％
c=350*22.4*920/（350*22.4*17+1000)≈53.7摩尔每升
感觉好假

Answer 3

额 我也纠结中