Coi CuSO4.5H2O là dd có \(C\%=\dfrac{160}{250}.100\%=64\%\)
Áp dụng sơ đồ đường chéo, ta có:
\(\dfrac{m_{CuSO_4.5H_2O}}{m_{ddCuSO_4\left(8\%\right)}}=\dfrac{16-8}{64-16}=\dfrac{1}{6}\)
`=>` \(\left\{{}\begin{matrix}m_{CuSO_4.5H_2O}=\dfrac{1}{1+6}.280=40\left(g\right)\\m_{ddCuSO_4\left(8\%\right)}=280-40=240\left(g\right)\end{matrix}\right.\)