A chemist makes 620 mL of barium chloride working solution by adding distilled water to180mL of a 1.39mol/L stock solution of barium chloride in water. Calculate the concentration of the chemist's working solution. Be sure your answer has the correct number of significant digits.

The correct answer is 0.40 mol/L.

Explanation:

In order to solve the problem, we use the following expression, where V1 and C1 are the volume and concentration of stock solution and V2 and C2 are the volume and concentration of working solution:

V1 x C1 = V2 x C2

⇒C2 = V1 x C1 / v2

C2 = (180 ml x 1.39 mol/L) / 620 ml

C2 = 0.40 mol/L

0.404 mol / L

Explanation:

The total moles of barium chloride can be calculated from the volume and concentration of the stock solution:

1.39 mol/L * 0.180 L = 0.2502 moles barium chloride

Now we calculate the concentration of the working solution, by dividing the total moles by the final volume:

0.2502 mol / 0.620 L = 0.404 mol / L