求所得氨水(密度为0.924g/cm3）中NH3的

求所得氨水(密度为0.924g/cm3）中NH3的质量分数和物质的量浓度

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设1体积为1L,则 n(NH3)=V(NH3)/Vm=350L/(22.4L/mol)=15.625molm(NH3)=n(NH3)·M(NH3)=15.625mol*17g/mol=265.625g m(H2O)=1000ml*1g/cm3 = 1000m[NH3(aq)]= m(NH3) + m(H2O) = 265.625g + 1000g =1265.625gw=m(NH3)/...