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For problem G.16 Part C, "Determine the mass of CuSO4 * 5H20 That must be used to prepare 250 mL of 0.20M CuSO4", does the water in the solute not show up as part of the final volume added to the 250 mL? The solution to the problem simply changes the molar mass of the solution when converting from moles of solute to grams of solute, however, if there is water in the solute, wouldn't that effect the concentration of CuSO4?
I think CuSO4 * 5H2O means that for every CuSO4 molecule, there are 5 H2O molecules attached. The 5 H2O molecules are seen as already in the solution, and all the * 5H2O represents is that the CuSO4 is hydrated. Therefore, the * 5H2O only affects the molar mass of the hydrated CuSO4 molecule, and not the volume of the solvent that the CuSO4 occupies.
The key assumption in these questions concerning solutions is that the addition of salt does not affect the volume of the original solution. So, please assume the salt is not affecting the volume when added to water.
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