How many grams of cuso4 * 5h2o is required to prepare 200 ml of 0.5 m solution?

Given:
V solution (CuSO4) = 200 ml = 0.2 l
Cm (CuSO4) = 0.5 M

Find:
m (CuSO4 * 5H2O) -?

Solution:
1) n (CuSO4) = Cm (CuSO4) * V solution (CuSO4) = 0.5 * 0.2 = 0.1 mol;
2) n (CuSO4 * 5H2O) = n (CuSO4) = 0.1 mol;
3) M (CuSO4 * 5H2O) = Mr (CuSO4 * 5H2O) = Ar (Cu) * N (Cu) + Ar (S) * N (S) + Ar (O) * N (O) + Ar (H) * N (H) + Ar (O) * N (O) = 64 * 1 + 32 * 1 + 16 * 4 + 1 * 2 * 5 + 16 * 1 * 5 = 250 g / mol;
4) m (CuSO4 * 5H2O) = n (CuSO4 * 5H2O) * M (CuSO4 * 5H2O) = 0.1 * 250 = 25 g.

Answer: The mass of CuSO4 * 5H2O is 25 g.