设溶液的体积为1L,则n(OH-)=0.01mol,所以n[Ca(OH)2]=0.005mol,
则1L溶液中m[Ca(OH)2]=nM=74g/mol×0.005mol=0.37g,1L该溶液的质量为1000dg,
设该温度下Ca(OH)2的溶解度为Sg,
则:
0.37g |
1000dg |
S |
100+S |
解得:S=
37 |
1000d−0.37 |
故选A.
37 |
1000d−0.37 |
0.37 |
1000d−0.27 |
0.74 |
1000d−0.74 |
7.4 |
1000d−7.4 |
0.37g |
1000dg |
S |
100+S |
37 |
1000d−0.37 |