Respuesta :
Answer:
See answer below
Explanation:
First, you are not putting the whole data, however, I found an exercise very similar to this, so I'm gonna use this data, as an example, and you can use it as guide to solve yours. The exercise is the following:
A chemist must prepare 275. mL of 1967 uM aqueous copper(II) fluoride (Cur) working solution. He'll do this by pouring out some 2.63 mmol/L aqueous copper(II) fluoride stock solution into a graduated cylinder and diluting it with distilled water. Calculate the volume in mL of the copper(II) fluoride stock solution that the chemist should pour out.
According to this, we want a more dilluted solution fo copper Fluoride solution. To do this, we need to know the moles that are required in the desired solution.
1967 uM in just mol/L is:
1967 uM * 1 M / 1x10⁶ uM = 1.967x10⁻³ M
Now that we have the concentration, we can calculate the moles required to prepare this solution:
moles = 1.967x10⁻³ mol/L * 0.275 L = 5.41x10⁻⁴ moles
These are the moles that we need to have to prepare this solution. Now, with the concentration of the stock solution, we just solve for the volume required:
V = moles / M
V = 5.41x10⁻⁴ / 2.63x10⁻³
V = 0.2057 L ----> 205.7 mL or 206 mL
