You have 67.0 mL of a 0.400 M stock solution that must be diluted to 0.100 M. Assuming the volumes are additive, how much water should you add?

mL1 x M1 = mL2 x M2

67.0 x 0.4M = mL2 x 0.1M
Solve for mL2 which will be the TOTAL volume. Since you had 67.0 mL initially, subtract from the total to find how much must be added.

26.7

To calculate how much water should be added to dilute the 0.400 M stock solution to 0.100 M, we can use the formula for dilution:

C1V1 = C2V2

where:
C1 = initial concentration (0.400 M)
V1 = initial volume (67.0 mL)
C2 = final concentration (0.100 M)
V2 = final volume (which is the sum of the initial volume and the volume of water added)

Rearranging the equation to solve for V2, we have:

V2 = (C1V1) / C2

Substituting the known values:

V2 = (0.400 M * 67.0 mL) / 0.100 M

Calculating:

V2 = 26.8 mL

Therefore, you should add 26.8 mL of water to the 67.0 mL of the 0.400 M stock solution to obtain a final concentration of 0.100 M.

To calculate how much water should be added, we need to use the dilution equation:

M1V1 = M2V2

Where:
M1 is the initial concentration of the solution (0.400 M),
V1 is the initial volume of the solution (67.0 mL),
M2 is the final concentration you want to achieve (0.100 M), and
V2 is the final volume of the solution after dilution (V1 + volume of water).

We can rearrange the equation to solve for V2:

V2 = (M1 * V1) / M2

Let's plug in the values:
M1 = 0.400 M,
V1 = 67.0 mL,
M2 = 0.100 M.

V2 = (0.400 M * 67.0 mL) / 0.100 M
= 26.8 mL

Therefore, you need to add 26.8 mL of water.