Homework Problem G17 Part B

[Andrew F 2L]

I am confused as to why the process is nearly identical even though there is the 5H2O in part b compared to part a. Is it because it is a new solute so we are simply calculating its molar mass and changing only that?

[Sebastian Lee 1L]

Yeah, the only difference is that in part a, you have anhydrous copper(ii) sulfate whereas in part b, you have copper(ii) sulfate pentahydrate. The 5H2O on the end means that there are water molecules in the crystalline structure of CuSO4 but are not totally bonded to the molecule. The only difference in part b is that, since it's a different molecule, you have a different molar mass to calculate the moles necessary for the solution.

[nshahwan 1L]

Example G2 on page F55 is a good example of this problem.