Question

Two beakers are placed in a small closed container at 25 °C. One contains 459 mL of a 0.110 M aqueous solution of NH2CONH2 the second contains 263 mL of a 0.258 M aqueous solution of NH2CONH2. Small amounts of water evaporate from both solutions. As time passes, the volume of solution in the second beaker gradually volumes and concentrations be? and that in the first gradually . If we wait long enough, what will the final First beaker Second beaker Initial 459 mL 0.110 M 263 mL 0.258 M Final mL mL Submit Answer 2 question attempts remaining

Answer 1

As given, water will diffuse until the osmotic pressures of the two solutions are the same.

hence, we can set up equation

Osmotic pressure (P) = MRT and M = moles of solute/ L solution

In the 1st beaker we have ,

0.110 mol/L * 0.459 L = 0.05049 moles of solute

In the 2nd beaker we have ,

0.258mol/L * 0.263 L = 0.067854 moles of solute

Assume that we lose V Liters of water from the 1st beaker and gain V Liters of water in the 2nd beaker.

Then:

Equating osmotic pressure for two beakers

we have,

(0.05049)/(0.459 - V) RT = (0.067854 )/(0.263+V ) RT
(0.05049)/(0.459 - V) = (0.067854 )/(0.263+V )

Solving for V,

V = 0.150 L or 150 mL

so,

Volume in 1st beaker gradually decreases and volume in second beaker gradually increases ... Answer

Final volume in 1st beaker = 459 - 150 mL = 309 mL ................... Answer

and final concentration in 1st beaker = 0.05049 / 0.309 = 0.164 M ............. Answer

Final volume in 2nd beaker = 263 +  150 mL = 413 mL ................ Answer

and final concentration in 1st beaker = 0.067854 / 0.413 = 0.164 M ..............Answer