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?

Explanation:

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.

the dependant variable is spilled water

the independent is the heat

c (diamond),

sio2 (quartz)