72.0 mL of a 1.60 M solution is diluted to a total volume of 278 mL. A 139-mL portion of that solution is diluted by adding 155 mL of water. What is the final concentration? Assume the volumes are additive.
Let us say the solution is X.
Amount of X at the start `= 1.6/1000xx72 = 0.1152mol`
This 0.1152 mol will be diluted until the volume become 278ml.
So now we have 0.1152 moles of X in 278ml of solution. Assume the the moles of X are equally distributed over the solution.
Then we separate 139ml of the above solution.
Amount of X in 139ml `= 0.1152/278xx139 = 0.0576mol`
So we have 0.0576 moles of X of X in 139ml of the solution. Then we add 155ml of water.
Final volume of the solution `= 155+139 = 294ml`
So final mixture we have 0.0576moles of X diluted in 294ml solution.
Concentration of final mix `= 0.0576/294xx1000 = 0.1959M`
So the concentration of the final mixture is 0.1959M.