Question

配制0.2mol/L的稀盐酸500ml，需用w=0.36、密度为1.18kg/L的浓盐酸多少毫升

Answer 1

计算：已知w=0.36、密度为1.18kg/LD的浓盐酸的浓度c = 12 mol/L，依据稀释定律：cv = CV,

则 需要浓盐酸体积  V= cv/C = 0.2mol/Lx500ml/12mol/L = 8.33 ml

Answer 2

配置0.2mol／L的稀盐酸500ml，需要HCl的物质的量为0.2*0.5=0.1mol，质量为0.1*36.5=3.65g
需要W=0.36的盐酸为3.65/0.36=10.1g
P＝1.18㎏／L=1.18g/ml
体积为10.1/1.18=8.6ml

Answer 3

V×1.18×0.36/36.5=0.2×0.5
V=0.2×0.5×36.5/1.18/0.36
=0.1×36.5=3.65/1.18
=3.09/0.36
=8.6毫升