How would you prepare 2.25L of a 3.00M solution from a 15.0M stock solution?
you want to dilute it five times: which means one part stock, 4 parts water.
1 part= 2.25/5=450 ml
4parts= the rest.
To prepare 2.25L of a 3.00M solution from a 15.0M stock solution, you can use the formula:
M1V1 = M2V2
where M1 and V1 are the initial concentration and volume, and M2 and V2 are the final concentration and volume.
Given:
M1 (initial concentration) = 15.0M
V1 (volume of the stock solution) = ?
M2 (final concentration) = 3.00M
V2 (final volume of the solution) = 2.25L
Using the formula, we can rearrange it to solve for V1:
V1 = (M2 * V2) / M1
Substituting the given values:
V1 = (3.00M * 2.25L) / 15.0M
Lets calculate that:
V1 = (6.75) / (15.0)
V1 ≈ 0.45L
Therefore, to prepare 2.25L of a 3.00M solution from a 15.0M stock solution, you will need to measure 0.45L of the stock solution and then add it to a container. Finally, fill the container with enough solvent (such as water) to reach a total volume of 2.25L.