Achemist dissolves 0.096 g of cuso4 · 5 h2o in water and dilutes the solution to the mark in a 500-ml volumetric flask. a 6-ml sample of this solution is then transferred to a second 500-ml volumetric flask and diluted.
a) what is the molarity of cuso4 in the second solution? answer in units of m.
b) 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

(A) 4.616 * 10⁻⁶ M

(B) 0.576 mg CuSO₄·5H₂O

Explanation:

63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol

(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M

The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.

(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M

To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8