59.0 mL of a 1.80 M solution is diluted to a total volume of 268 mL. A 134-mL portion of that solution is diluted by adding 171 mL of water. What is the final concentration? Assume the volumes are additive.

1.80*(59/268)=concentration

concentration*(134/171)= concentration 2

Read it halfway; I thought it was going to say making a solution to 171mL using 134mL of the previous. Dr. Bob222 is correct.

To find the final concentration of the solution, we need to calculate the moles of solute before and after the dilution, and then determine the new concentration.

First, let's find the moles of solute in the original solution using the given volume and concentration:

moles of solute = volume of solution (in liters) * concentration of solution

Volume of solution = 59.0 mL = 0.0590 L
Concentration of solution = 1.80 M

moles of solute in original solution = 0.0590 L * 1.80 M
= 0.1062 moles

Next, let's determine the new concentration after diluting the solution.
When we take a 134 mL portion of the original solution, we remove 0.134 L of the solution. At the same time, we add 171 mL (0.171 L) of water.

Therefore, the final volume of the solution is the sum of the removed solution and added water: 134 mL + 171 mL = 305 mL = 0.305 L

To find the final concentration, we divide the moles of solute we calculated earlier by the new volume of the solution:

final concentration = moles of solute / volume of solution

moles of solute = 0.1062 moles
volume of solution = 0.305 L

final concentration = 0.1062 moles / 0.305 L
≈ 0.3483 M

Therefore, the final concentration of the solution is approximately 0.3483 M.

almost

1.80M x (59.0 mL/268 mL) = about 0.4 but that is not exact.

Then 0.4 x [134 mL/(134 mL + 171 mL)]