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### G.17

Posted: Tue Jul 11, 2017 7:02 pm
(b) Determine the mass of CuSO45H2O that must be used to prepare 250 mL of 0.20 m CuSO4(aq). Can someone tell me the steps they took to solve this problem?

### Re: G.17  [ENDORSED]

Posted: Tue Jul 11, 2017 7:20 pm
You were given volume of 250ml and Molarity of .20 mol/L.

Use the Molarity equation where Moles/Volume= Molarity and solve for Moles. Which is Moles= Molarity X Volume

Plug in the numbers: Moles= .20mol/L X .25L (250ml X 10-3) =5 X 10-2 mol

Convert moles to Grams using molar mass of CuSO4 5H20 which is 249.68: 5 X 10-2 mol X 249.68 gmol= 12 g

### Re: G.17

Posted: Tue Oct 03, 2017 3:37 pm
Why is the molar mass of CuSO4 5H20 used in part b?

### Re: G.17

Posted: Tue Oct 03, 2017 6:46 pm
You have to use the molar mass of CuSO4 5H20 because the question is asking you to find the mass of the whole compound, including the 5H20.

### Re: G.17

Posted: Sat Oct 07, 2017 2:55 pm
Does the CuSO4 bond with the H2O to form the new compound or do they become more connected due to being in an aqueous solution?
(Claire Woolson 3J)

### Re: G.17

Posted: Sat Oct 07, 2017 3:03 pm
I think the CuSO4*5H2O is a hydrated version of the salt, so while it isn't just a solution, it does definitely have water in it. I kind of imagine it as the Cu and SO4 molecules breaking up slightly enough to allow the H2O to bind with it. Once you've poured in 250 mL though, it's definitely a solution in the sense you'd normally think of it.

### Re: G.17

Posted: Sat Oct 07, 2017 3:24 pm
ClaireHW wrote:Does the CuSO4 bond with the H2O to form the new compound or do they become more connected due to being in an aqueous solution?
(Claire Woolson 3J)

The CuSO4 become a hydrate due to the presence of the H2O molecules, bonded with the CuSO4.