What volume of a 6.0 M NaOH solution would you need to prepare 250.0 milliliters of a 3.0 M NaOH solution?

How many moles NaOH do you need? THat is M x L = ?

M NaOH = moles NaOH/L NaOH.
YOu know M NaOH (6.0M) and you know moles from the first part. Solve for L and convert to mL.

130 milliliters

Well, let's break this down, shall we? So you want to prepare a 3.0 M NaOH solution, but you have a 6.0 M NaOH solution. Looks like you're going to have to do some diluting!

Now, let's use a little math here. We have the formula for dilution, which is C1V1 = C2V2.

So we know that C1 (the initial concentration) is 6.0 M, V1 (the initial volume) is what we're trying to find, C2 (the final concentration) is 3.0 M, and V2 (the final volume) is 250.0 mL.

So, plugging the numbers in, we get:

(6.0 M)(V1) = (3.0 M)(250.0 mL)

Now, if we divide both sides by 6.0 M, we get:

V1 = (3.0 M)(250.0 mL) / 6.0 M

And if we do the math, we find out that the volume you need of the 6.0 M NaOH solution is approximately 125.0 milliliters.

So, to summarize, you'll need 125.0 mL of the 6.0 M NaOH solution to prepare 250.0 mL of the 3.0 M NaOH solution. And remember, diluting solutions is like watering down your jokes - you can never go wrong with a little extra NaOH, I mean, humor!

To find the volume of a 6.0 M NaOH solution needed to prepare 250.0 milliliters of a 3.0 M NaOH solution, we can use the formula:

M1 * V1 = M2 * V2

where:
M1 = initial molarity of the solution (6.0 M)
V1 = initial volume of the solution (unknown)
M2 = final molarity of the solution (3.0 M)
V2 = final volume of the solution (250.0 mL)

Rearranging the formula to solve for the unknown initial volume (V1):

V1 = (M2 * V2) / M1

Now, let's plug in the values:

V1 = (3.0 M * 250.0 mL) / 6.0 M

V1 = 750.0 mL / 6.0 M

V1 ≈ 125.0 mL

Therefore, you would need to measure approximately 125.0 milliliters of the 6.0 M NaOH solution to prepare 250.0 milliliters of a 3.0 M NaOH solution.