If I am creating a solution of CuSO4 with 0.500 Molarity (concentration) using 5

Question

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?

Please help.

Explanation / Answer

CuSO4.5H2O(s) => CuSO4(aq) + 5 H2O(l)

1 mole of solid CuSO4.5H2O gives 1 mole of CuSO4 in solution


Moles of CuSO4.5H2O required = moles of CuSO4

= volume x concentration of CuSO4

= 0.050 L x 0.500 mol/L = 0.025 mol


Mass of CuSO4.5H2O required = moles x molar mass of CuSO4.5H2O

= 0.025 mol x 249.54 g/mol

= 6.2385 g = 6.24 g (3 significant figures)


(Note that the mass of the 5H2O is already taken into account when you use the molar mass of 249.54 g/mol for CuSO4.5H2O. There is no need to top up any more.)

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