59.0 mL of a 1.60 M solution is diluted to a total volume of 268 mL. A 134-mL portion of that solution is diluted by adding 111 mL of water. What is the final concentration? Assume the volumes are additive.

1.60M x (59.9/208) x [134/(134+111)] = ?

To find the final concentration of the solution after dilution, we need to use the formula:

C1V1 = C2V2

Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution
V2 = final volume of the solution

Let's break down the given information:

Initial concentration (C1) = 1.60 M
Initial volume (V1) = 59.0 mL

Final volume (V2) = 268 mL
Volume of solution used for dilution (V3) = 134 mL
Volume of water added (V4) = 111 mL

We can find the volume of the diluted solution (V5) using the equation:

V5 = V1 + V3

V5 = 59.0 mL + 134 mL
V5 = 193 mL

Now, we can calculate the final concentration (C2) using the formula:

C1V1 = C2V2

(1.60 M)(59.0 mL) = C2(193 mL)

C2 = (1.60 M)(59.0 mL) / (193 mL)

C2 ≈ 0.492 M

Therefore, the final concentration of the solution after dilution is approximately 0.492 M.