First, calculate the M of the initial solution.
mols CuSO4.5H2O = grams/molar mass = ?
Then mols/L = M
So the diluted solution is M of first x (13.6/500) = ?
For the last part.
How many mols do you have in the second solution? That's mols = M x L = ?
Then mols = grams/molar msss. You know molar mass and mols, solve for grams.
A chemist dissolves 0.097 g of CuSO4 · 5 H2O
in water and dilutes the solution to the mark
in a 500-mL volumetric flask. A 13.6-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?
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
3 answers
y=mx+b
-b+or- square root of b^2 -4ac/2a