- Ở 85oC: \(C\%_{CuSO_4\left(bão.hoà\right)}=\dfrac{87,7}{100+87,7}.100\%=46,7234\%\)
=> \(m_{CuSO_4}=1877.46,7234\%=877\left(g\right)\)
=> mH2O = 1877 - 877 = 1000 (g)
- Ở 12oC: Gọi \(n_{CuSO_4.5H_2O}=a\left(mol\right)\)
=> \(\left\{{}\begin{matrix}m_{CuSO_4\left(còn\right)}=877-160a\left(g\right)\\m_{H_2O\left(còn\right)}=1000-90a\left(g\right)\end{matrix}\right.\)
Mà \(S=35,5\left(g\right)\)
=> \(\dfrac{877-160a}{1000-90a}.100=35,5\)
=> a = 4,0765 (mol)
=> mtinh thể = 4,0765.250 = 1019,125 (g)