Two beakers are placed in a small closed container at 25 °C. One contains 284 mL of a 0.296 M aqueous solution of C6H12O6; the second contains 446 mL of a 0.103 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 and that in the first gradually. If we wait long enough, what will the final volumes and concentrations be?

Question

Camden Santos

Chemistry

9 December, 20:27

Two beakers are placed in a small closed container at 25 °C. One contains 284 mL of a 0.296 M aqueous solution of C6H12O6; the second contains 446 mL of a 0.103 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 and that in the first gradually. If we wait long enough, what will the final volumes and concentrations be?

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Answers (2)

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Eleanor Savage · Answer 1

The volume of solutions in the second beaker decreases while the volume in the first beaker increases.

Therefore,

In the first beaker, using the equation,

C1*V1 = C2*V2

Where,

C1 = initial concentration

V1 = initial volume

C2 = final concentration

V2 = final volume

final volume = (0.296 * 284) / 0.184

= 457 ml.

In the second beaker,

final concentration, C2 = 0.184 M

final volume = (0.103 * 446) / 0.184

= 250 ml.

Tara · Answer 2

Final volume = 0.103M x446ml/0.184m = 250ml

Explanation:

As time passes, the volume of solutions in the second beaker decreases and that in the first beaker increases. If we wait long enough, the final volumes and concentration in the beakers would be,

First beaker

Final concentration = 0.184M

Final volume = 0.296M x 284ml/0.184 = 457ml

Second beaker

Final concentration = 0.184M

Final volume = 0.103M x446ml/0.184m = 250ml