Two beakers one containing 20 ml of 0.05 molar aquous solution of non-volatile non electrolyte and the other the same volume of 0.03 molar aqueous solution of NaCl our placed side by side in a closed enclosure what are the volumes in the two beakers when the equilibrium is attained? volume of the solution in the first and the second beaker respectively are

Dear Student,

Let the beaker containing the non-volatile non-electrolyte solution be beaker A and the other beaker be beaker B.
Now, since NaCl dissociates into ions, the volume of beaker B will increase because its vapour pressure will be lower than that of beaker A, leading to condensation of the water vapour from beaker A into beaker B.

Water will diffuse till the osmotic pressures of the two solutions become the same.

The solution is as follows:


$$\begin{gathered} In beaker A, \\ Number of moles of solute = 0.02 L x 0.05 M = 1 x {10}^{-3} mol \\ In beaker B, \\ Number of moles of dissociated NaCl=0.03 x 0.02 x 2 = 1.2 x {10}^{-3} mol \\ Now, Osmotic pressure = CRT \\ Let the volume of water which is transferred be x L. \\ At equilibrium, the osmotic pressures for the two beakers are the same. \\ Thus, \\ \frac{1\times {10}^{-3}}{0.02-x}RT=\frac{1.2\times {10}^{-3}}{0.02+x}RT \\ (0.020+x) =(0.020-x)1.2 \\ Solvinf for x, we get x = 0.0018L = 1.8 mL \\ Thus, the final volumes are \\ Beaker A=18.2 mL \\ Beaker B = 21.8 mL \end{gathered}$$

Hope it helps.

Regards