Ta có :
$m_{dd} = x + y = 500(gam)(1)$
$n_{CuSO_4} = n_{CuSO_4.5H_2O} + n_{CuSO_4\ trong\ dd\ CuSO_4\ 4\%}$
$\Rightarrow \dfrac{x}{250} + \dfrac{y.4\%}{160} = \dfrac{500.8\%}{160}(2)$
Từ (1)(2) suy ra x = 33,33(gam) ; y = 466,67(gam)
Coi CuSO4.5H2O là dd có \(C\%=\dfrac{160}{250}.100\%=64\%\)
Áp dụng sơ đồ đường chéo, ta có:
\(\dfrac{y}{x}=\dfrac{m_{dd.CuSO_4}}{m_{CuSO_4.5H_2O}}=\dfrac{64-8}{8-4}=\dfrac{14}{1}\)
=> \(\left\{{}\begin{matrix}x=\dfrac{1}{1+14}.500=\dfrac{100}{3}\left(g\right)\\y=500-\dfrac{100}{3}=\dfrac{1400}{3}\left(g\right)\end{matrix}\right.\)