## Q G17 part b. (7th edition)

Ryan Danis 1J
Posts: 66
Joined: Fri Sep 28, 2018 12:18 am

### Q G17 part b. (7th edition)

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
Posts: 60
Joined: Fri Sep 28, 2018 12:16 am

### Re: Q G17 part b. (7th edition)

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
Posts: 35
Joined: Fri Sep 28, 2018 12:21 am

### Re: Q G17 part b. (7th edition)

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
Posts: 63
Joined: Fri Sep 28, 2018 12:19 am

### Re: Q G17 part b. (7th edition)

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
Posts: 63
Joined: Fri Sep 28, 2018 12:27 am

### Re: Q G17 part b. (7th edition)

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
Posts: 77
Joined: Fri Oct 05, 2018 12:16 am

### Re: Q G17 part b. (7th edition)

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.