Question

A chemist dissolves 0.096 g of CuSO4 · 5 H2O in water and dilutes the solution to the mark in a 500-mL volumetric flask. A 10.8-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?


My Work:
Mass of CuSO4.5H2O per mL = 0.096/500g
mass in 43.6mL = 0.096/500 x 43.6 g = 0.0083712

That is put into 500mL for second flask, so mass per liter = 0.096/500 x 43.6 x 2g

To find molarity, divide by molar mass = 0.096/500 x 43.6 x 2/249.7 M

Answer 1

see other post

Answer 2

I don't think so. Where did you get 43.6 mL? That isn't anywhere in the problem.
mols CuSO4.5H2O = 0.096/249.7 = approx 3.8E-4
M = mols/L = 3.8E-4 mols/0.5L = approx 7.7E-4 M for the solution in the first 500 mL flask. Then you dilute 10.8 mL to 500 so
7.7E-4M x (10.8/500) = about 1.7E-5 M