要配制0.10mol/L的HCl溶液1000ml,需要密度为1.19g/ cm³、质量分数0.37 要配制0.10摩尔每升的HCL溶液1000mL需密度为1.1...

\u8981\u914d\u52360.1mol/L\u7684HCl\u6eb6\u6db21000ml\uff0c\u9700\u8981\u5bc6\u5ea6\u4e3a1.19g/cm\uff0c\u8d28\u91cf\u5206\u6570\u4e3a0.37\u7684\u6d53

84ml\uff0c\u6c42\u91c7\u7eb3



解:   1000ml=1L

HCl的物质的量为:0.1mol/L×1L=0.1mol

HCl的质量为:       0.1mol×36.5g/mol=3.65g

需要质量分数为0.37的浓盐酸的质量为:3.65g÷0.37=9.86g(取小数点后两位小数)

需要浓盐酸的体积为:9.86g÷1.19g/cm3=8.29cm3

所以配制0.10mol/L的HCl溶液1000ml,需要密度为1.19g/ cm³、质量分数0.37的浓盐酸8.29ml



HCL分子量36.5
36.5*0.1*1=1.19*V*0.37
V=8.3

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