\(n_{CuSO_4.5H_2O}=\frac{100}{250}=0,4\left(mol\right)\) => \(n_{CuSO_4\left(thêm\right)}=0,4\left(mol\right)\)
\(m_{CuSO_4\left(bđ\right)}=\frac{400.4}{100}=16\left(g\right)=>n_{CuSO_4\left(bđ\right)}=\frac{16}{160}=0,1\left(mol\right)\)
=> nCuSO4 ( dd sau khi hòa tan) = 0,4 + 0,1 = 0,5 (mol)
=> mCuSO4 (dd sau khi hòa tan) = 0,5.160 = 80 (g)
=> C% (dd thu được) = \(\frac{80}{100+400}.100\%=16\%\)