Q G17 part b. (7th edition)

[Ryan Danis 1J]

Question G 17 asks:
a. Determine the mass of anhydrous copper(II) sulfate that must be used to prepare 250 mL of 0.20 M CuS04 (aq).
b. Determine the mass of CuSO4 x 5(H20) that must be used to prepare 250mL of 0.20 M CuSO4 (aq0.

I was able to solve part a by finding the amount of moles of CuSO4 and multiplying that by its molar mass. However, I am confused about what to do with the 5 H20 in part b. Do I just add the molar mass of 5 H20 to the molar mass of CuSO4 when I multiple moles of CuSO4 by molar mass?

[klarratt2]

The H2O just means that it is copper (ii) sulfate hydrate, so you do add the molar mass of the 5H2O to the copper (ii) sulfate to solve.

[Melody Zhang 3L]

Yes, you add the molar mass of 5H2O to the molar mass of CuSO4 to find the amount in grams of CuSO4*5H2O needed and multiply that with the amount in moles of CuSO4 (0.20mol/L)(.250L).

[Rosha Mamita 2H]

yep, you'd multiply your resulting moles for part b with the molar mass of both the CuSO4 combined with the 5H20, which would be a combined molar mass of 324.65g/mol.

[Michelle Wang 4I]

For some reason when I first saw this problem I thought it would be one where we use the formula Minitial x Vinitial = Mfinal x Vfinal since it had to do with preparing a solution. Can someone explain why we don't solve it that way?

[Lina Petrossian 1D]

I believe the combined molar mass of CuSO4 and 5H2O is 249.68 g/mol. Which when multiplied by 0.05 mol gives 12.484 g CuSO4 5H2O.