VmL水的质量为VmL×1g/mL=Vg,则溶液的质量为Wg+Vg=(W+V)g,
溶液的体积为
(W+V)g
ρg/ cm 3 =
(W+V)
ρ cm 3=
(W+V)
ρ ×10 -3L,
n(CuSO 4)=n(CuSO 4•5H 2O)=
Wg
250g/mol =
W
250 mol,
则溶液的物质的量浓度为
W
250 mol
(W+V)
ρ × 10 -3 L =
1000ρW
250(W+V) mol/L=
4ρW
(W+V) mol/L,
溶质的质量为
W
250 mol×160g/mol=
160W
250 g,设该温度下CuSO 4的溶解度为s,则
160W
250 g
(W+V)g =
s
100+s ,解得s=
1600W
(9W+25V) g,
故答案为:
4ρW
(W+V) mol/L;
1600W
(9W+25V) g.