I can't figure out how to set up this problem...You have a large amount of 8.00 M stock solution. You need 1.80L of 3.00 M solution for an experiment. How would you prepare the desired solution without wasting any stock solution?
The amount of solute in 1L of 1.80M solution would be the same as that in 1.80/8.00L of the 8.00M solution.
So, take 0.225L of the 8M, and fill the flask to 1L.
Sorry. Got my numbers mixed up.
The ratio is 3/8.
The solute in 1L of 3M solution is the same as in 3/8L of 8M.
So, take 3/8*1.8 = 0.675L of the 8M solution, and fill the flask to 1.80L
To prepare the desired solution without wasting any stock solution, we can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the stock solution
V1 = volume of the stock solution used
C2 = final concentration of the desired solution
V2 = final volume of the desired solution
First, let's calculate the volume of stock solution needed:
C1 = 8.00 M
V1 = ?
C2 = 3.00 M
V2 = 1.80 L
Rearranging the formula, we have:
V1 = (C2 * V2) / C1
Substituting the given values:
V1 = (3.00 M * 1.80 L) / 8.00 M
V1 = 0.675 L
Therefore, to prepare the desired solution, you would need to measure out 0.675 L of the 8.00 M stock solution and then dilute it to a final volume of 1.80 L by adding the appropriate amount of solvent (usually water).