The diagrammatic expression for the solutions in the two arms of the U-tube separated by a membrane that is permeable to water and glucose but not to sucrose is shown in the image attached below.
When the system illustrated above reaches equilibrium, the sugar concentrations on both sides of the U-tube will be 1.5 M sucrose, 1.5 M glucose
From the image attached, if the system approaches equilibrium;
The movement of the solute(i.e. one molecule of sucrose) have the potential to proceed and move to the other arm of the U-shaped tube but with that, the total overall concentration cannot change in the whole U-tube structure due to the fact that the volume of the solvent(water) is constant.
Similarly, both arms of the U-tube shape have the same concentration as well as the same net Kinetic energy of solutes giving them the isotonic condition property.
Now, at equilibrium, the concentration of the sugar on both sides is:
= [tex](\dfrac{1M +2M}{2})[/tex]
[tex]\mathsf{=\dfrac{3M}{2}}[/tex]
= 1.5 M sucrose, 1.5 M glucose
Therefore, we can conclude that the concentration of the sugar on both sides at equilibrium is 1.5 M sucrose, 1.5 M glucose.
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