DR.BOB 22 PLS HELP

Question

A chemist dissolves 0.099 g of CuSO4 · 5 H2O
in water and dilutes the solution to the mark
in a 500-mL volumetric flask. A 28.9-mL
sample of this solution is then transferred to a
second 500-mL volumetric flask and diluted.
What is the molarity of CuSO4 in the second
solution?
Answer in units of M.
**answer: .000046**

SECOND PART:
To prepare the second 500 mL of solution
directly, what mass of CuSO4 · 5 H2O would
need to be weighed out?
Answer in units of mg.

Answer 1

impatient much?
Give the guy a chance to see your post!

Answer 2

mols = grams/molar mass = 0.099 g/249.5 = 0.000397 in the first flask so
M = mols/L = 0.000397/0.500 = 0.000793 M.
You can use the dilution formula of
mL1 x M1 = mL2 x M2 but I prefer to to
0.000397 x (28.9/500) = 0.0000229 M CuSO4.5H2O. In reading the question, technically, you asked for molarity of CuSO4 (and not CuSO4.5H2O). I don't know if this was a typo or not; however, if you want M CuSO4, that is
0.0000229 x (159.5/249.5) = 0.0000146
part 2 new post.

Answer 3

Scratch that. I'm using a new calculator. Maybe punched the wrong buttons. Let me check those numbers.

Answer 4

Nope: not the calculator. I copied the wrong number. That line of
0.000397 x (28.9/500) = 0.0000229 M CuSO4.5H2O should be replaced with
0.000793 M x (28.9/500) = 4.59E-5 or 0.0000459 M
In case you wanted CuSO4 instead of CuSO4.5H2O that will be
4.59E-5 x (159.5/249.5) = ? M
I'll do part 2 new post.

Answer 5

i don't see the other new post, could you put the link

Answer 6

So you want to weigh out CuSO4.5H2O directly into a 500 mL volumetric flask to make 4.59E-5 M. You will want how many moles. That's
mols = M x L = 4.59E-5 M x 0.500 L = 2.29E-5 mols.
grams = mols x molar mass = 2.29E-5 mols x 249.5 g/mol = 5.73E-3 g = 0.00573 g = 5.73 mg CuSO4.5H2O
Check my numbers closely: I have poor eyesight and I make some typos>