How many milliliters of water must be added to 153.2 mL of 2.73 M KOH to give a 1.68 M solution?
Lets see: you want to dilute it by 2.73/1.68 or 1.625 times.
Which means, you need one part KOH, and .625 parts diluting water.
So what is one part? Easy: 153.2 is one part, and then water of 153.2*.625 ml water. Stir, label, and store properly.
using M1*V1=M2*V2
why does this not work?
It does and the answer is the same either way. However, I'm curious, if you knew the answer why did you ask? Or why not show your work and ask if that's the right way?
To solve this problem, we need to use the concept of dilution. Dilution involves adding a solvent (in this case, water) to decrease the concentration of a solution. The equation for dilution is:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
In this case, we have:
C1 = 2.73 M (initial concentration of KOH)
V1 = 153.2 mL (initial volume of KOH)
C2 = 1.68 M (final concentration of the solution)
V2 = ?
We want to find V2, the volume of water to be added.
Rearranging the equation, we have:
V2 = (C1 * V1) / C2
Substituting the given values, we get:
V2 = (2.73 M * 153.2 mL) / 1.68 M
Simplifying, we find:
V2 = 249.94 mL
Therefore, 249.94 mL of water must be added to 153.2 mL of 2.73 M KOH to give a 1.68 M solution.