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Two beakers are placed in a small closed container at 25 °C. One contains 416 mL of a 0.112 M aqueous solution of C6H12O6; the second contains 242 mL of a 0.266 M aqueous solution of C6H12O6. Small amounts of water evaporate from both solutions. As time passes, the volume of solution in the second beaker gradually (Increases or decreases) and that in the first gradually (increases or decreases). If we wait long enough, what will the final volumes and concentrations be?

Question

Two beakers are placed in a small closed container at 25 °C. One contains 416 mL of a 0.112 M aqueous solution of C6H12O6; the second contains 242 mL of a 0.266 M aqueous solution of C6H12O6. Small amounts of water evaporate from both solutions. As time passes, the volume of solution in the second beaker gradually (Increases or decreases) and that in the first gradually (increases or decreases). If we wait long enough, what will the final volumes and concentrations be?

check_circleAnswer
Step 1

Concentration of the solute in dilute solution and that of concentrated solution are affected by vapor pressure of the solvent which is governed by Raoult’s law.

Step 2

The concentration of the first beaker solution is less than the concentration of the second beaker. Hence, the vapor pressure of the first beaker is higher.

The partial pressure of vapor in the closed container is an intermediate value. It is less than the vapor pressure of the first beaker, but more than the second beaker.

As a result, vapor leaves the solution from the first beaker (which becomes more concentrated) and condenses on the second beaker (which becomes less concentrated).

Step 3

After we wait a long enough time, the vapor pressure and the concentration of the two beakers become equal. As the concentration increases in the first beaker, t...

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