Gọi $n_{CuSO_4.5H_2O} = a(mol) ; m_{dd\ CuSO_4\ 8\%} = b(gam)$
Ta có :
$m_{dung\ dịch} = m_{CuSO_4.5H_2O} + m_{dd\ CuSO_4\ 8\%}$
$\Rightarrow 280 = 250a + b(1)$
$m_{CuSO_4} = 160a + b.8\% = 280.16\% = 44,8(2)$
Từ (1)(2) suy ra : a = 0,16(mol) ; b = 240(gam)
$m_{CuSO_4.5H_2O} = 0,16.250 = 40(gam)$
\(Đặt:m_{CuSO_4.5H_2O}=a\left(g\right);m_{ddCuSO_4\left(8\%\right)}=b\left(g\right)\left(a,b>0\right)\\ m_{CuSO_4}=0,64a+0,08b=280.16\%=44,8\left(1\right)\\ Mà:a+b=280\left(2\right)\\ \left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}0,64a+0,08b=44,8\\a+b=280\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=40\\b=240\end{matrix}\right.\)
Cần 40 gam tinh thể CuSO4.5H2O và 240 gam dd CuSO4 8%