On one pan of the beam balance there is 0.5 mol of sodium hydroxide NaOH.

On one pan of the beam balance there is 0.5 mol of sodium hydroxide NaOH. How much copper sulfate CuSO4 should be put on the other pan to balance the balance?

Given:
n (NaOH) = 0.5 mol

To find:
m (CuSO4) -?

1) Using the periodic system of D.I.Mendeleev, calculate the molar mass of sodium hydroxide:
M (NaOH) = Mr (NaOH) = Ar (Na) * N (Na) + Ar (O) * N (O) + Ar (H) * N (H) = 23 * 1 + 16 * 1 + 1 * 1 = 40 g / mol;
2) Calculate the mass of sodium hydroxide:
m (NaOH) = n (NaOH) * M (NaOH) = 0.5 * 40 = 20 g;
3) To balance the scales, the masses of the substances must be equal. Means:
m (CuSO4) = m (NaOH) = 20 g.

Answer: The mass of copper sulfate is 20 g.