Ở $80^oC$:
$C\%_{\text{bão hoà}}=\dfrac{50.100}{100+50}=\dfrac{100}{3}\%$
$\Rightarrow m_{CuSO_4}=600.\dfrac{100}{3}\%=200g$
Gọi x là mol $CuSO_4.5H_2O$ tách ra.
Sau khi tách:
$m_{CuSO_4}=200-160x$
$m_{dd}=600-250x$
$C\%_{\text{bão hoà}}=\dfrac{15.100}{100+15}=\dfrac{300}{23}\%$
$\Rightarrow 200-160x=\dfrac{3}{23}(600-250x)$
$\Leftrightarrow x=0,955$
$\to m_{CuSO_4.5H_2O}=250x=238,75g$