Question

要配制c（HCL）＝0.1000mol/L的溶液300mL,需要含量为37.2%密度的1.19g/mL盐酸多少毫升?

Answer 1

氯化氢的摩尔质量是:36.5克;

所以需要的氯化氢质量是:(36.5g/mol)*(0.1000mol/1000ml)*300ml

=36.5g*0.1000*3/10

=3.65g*3/10

=1.095g

需要的盐酸体积是:1.095g/(1.19g/ml*0.372)

=1.095g/(0.443g/ml)

= 2.47ml