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.

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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.*

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