5 g of copper sulfate was dissolved in water and the volume of the solution was brought to 500 cm.

5 g of copper sulfate was dissolved in water and the volume of the solution was brought to 500 cm. What amount of copper sulfate is contained in the resulting solution?

Given:
m (CuSO4 * 5H2O) = 5 g
V solution (CuSO4) = 500 cm3 = 500 ml

To find:
n (CuSO4) -?
m (CuSO4) -?
ω (CuSO4) -?

Solution:
1) M (CuSO4 * 5H2O) = Mr (CuSO4 * 5H2O) = Ar (Cu) + Ar (S) + Ar (O) * 4 + Ar (H) * 2 * 5 + Ar (O) * 5 = 64 + 32 + 16 * 4 + 1 * 2 * 5 + 16 * 5 = 250 g / mol;
M (CuSO4) = Mr (CuSO4) = Ar (Cu) + Ar (S) + Ar (O) * 4 = 64 + 32 + 16 * 4 = 160 g / mol;
2) n (CuSO4 * 5H2O) = m (CuSO4 * 5H2O) / M (CuSO4 * 5H2O) = 5/250 = 0.02 mol;
3) n (CuSO4) = n (CuSO4 * 5H2O) = 0.02 mol;
4) m (CuSO4) = n (CuSO4) * M (CuSO4) = 0.02 * 160 = 3.2 g;
5) Since the density of the CuSO4 solution is not indicated, we will take it as the density of water;
6) m solution (CuSO4) = ρ (H2O) * V solution (CuSO4) = 1 * 500 = 500 g;
7) ω (CuSO4) = m (CuSO4) * 100% / m solution (CuSO4) = 3.2 * 100% / 500 = 0.64%.

Answer: The amount of substance CuSO4 is 0.02 mol; weight – 3.2 g; mass fraction – 0.64%.