How would you correctly prepare 125 mL of a 0.30M solution of copper(II) sulfate (CuSO4) from a 2.00M solution of CuSO4?

Question

asked Mar 22, 2023 139k views

1 Answer

Sooobus · Answer 1 · 2023-03-26T08:56:36+0000

Answer:

see below

Step-by-step explanation:

.3 M/L ( .125 L) = .0375 mole required

.0375 mole / 2 mole / l = .01875 l required

take 18.75 ml of the 2.0 solution then dilute it to 125 ml