If I am creating a solution of CuSO4 with 0.500 Molarity (concentration) using 50.0 ml but the solid is copper sulfate PENTAHYDRATE, should I measure out extra to account for the extra mass of the pentahydrate?

Basically should i do this:

(0.500 M)(0.0500 L)= 0.025 moles

(0.025 moles)(CuSO4-5H2O molar mass 249.54 g/mole)= 6.2385 grams of CuSO4-5H2O

and leave it that?

Or should I add (0.0250 moles)(5 moles of the H20 / 1 mole of copper sulfate pentahydrate)(18.01 g/mol of H2O) = 2.25 grams, on top, in order to achieve the real molarity of CuSO4 when in solution?