how many grams of beryllium chloride are needed to make 125 mL of a 0.050 mol/L solution?

Question

how do you answer this? please explain it, thanks.

Answer 1

BeCl2 ==> 9.012 + 2*35.45 = 79.912 g/mol.
x / (79.912 g/mol.) = (0.125 L)(0.050 mol./L)
x = (79.912 g/mol.)(0.125 L)(0.050 mol./L)
x ≈ 0.49945 g ≈ 499 mg