\(a.n_{CuSO_4}=\dfrac{7,5}{250}=0,03\left(mol\right)\\ C_{MddCuSO_4}=\dfrac{0,03}{0,2}=0,15\left(M\right)\\ b.C_{MddBa\left(OH\right)_2}=\dfrac{C\%_{ddBa\left(OH\right)_2}.10.D_{ddBa\left(OH\right)_2}}{M_{Ba\left(OH\right)_2}}\\ =\dfrac{20,684.10.1,25}{171}\approx1,512\left(M\right)\)