# Chemistry Lab

posted by Jake

So I have a lab exam tomorrow and our instructor said that one of the questions would be as follows:

Prepare 100mL of 0.03 M Copper Sulfate, beginning from copper sulfate pentahydrate. Calculate the exact molarity to 3 sig figs.

Alright so I understand how to prepare 100mL of 0.003 CuSO4-5H2O by taking 0.03M X 0.1L to get .003 mol.

.003 Mol X the Molar Mass of CuSO4-5H2O which is 250G = .75G

Does this mean I add .75G to 99.25mL of water since the density of water is 1g/mL.

Is he asking us to maybe determine how much of copper sulfate pentahyrdate is actually copper sulfate. Then way out enough pentahyrdate to contain the amount of copper sulfate required to make 100mL of .03M copper sulfate?

So confused this should be basic Gen Chem 1 lab 101 stuff, but I can't seems to wrap my head around it can someone help me? I'd appreciate it thanks!

1. Dr Rebel

gms of CuSO4-5HOH = [Molarity needed][Volume needed (Liters)][formula Wt CuSO4] / [Purity fraction CuSO4 in pentahydrate]
=[(0.03M)(0.100L)(160g/mol)]/[0.64]=0.75gms CuSO4-5HOH + Solvent up to but not to exceed 100 ml total volume.

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