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Question

The solutions in the two arms of this U-tube are separated by a membrane that is permeable to water and glucose but not sucrose. Side A is half-filled with a solution of 2 M sucrose and 1 M glucose. Side b is half-filled with 1 M sucrose and 2 M glucose. Initially, the liquid levels on both sides are equal.

When it reaches equilibrium, the sugar concentrations on both sides of the U-tube will be 1.5 M sucrose and 1.5 M glucose.

Can somebody explain the answer (D) to me? My understanding is that the membrane is permeable to glucose, therefore glucose will reach equilibrium, with 1.5M on each side. Since the membrane is not permeable to sucrose, the concentration of sucrose will not change. Water will move from B to A. Does the movement of water causes the sucrose solution on both sides to be equal? But wouldn't that alter the glucose concentrations too? I'm sorry I just know what my bio teacher told me but I don't quite get the reasoning. Thanks.

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  • $\begingroup$ Please do not post questions as graphics. They do not index and discriminate against people with visual impairments. $\endgroup$
    – David
    Commented Nov 17, 2023 at 10:06
  • $\begingroup$ OK I'll modify it $\endgroup$ Commented Nov 18, 2023 at 2:33

1 Answer 1

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The membrane allows movement of particles which will cause a balancing of the concentrations. There is one hint in the text: "Initially, the liquid levels on both sides are equal" which gives away how this will happen.

Water and glucose can freely move through the membrane and this is what will happen. Water will move through the membrane, the level in side A will rise until both concentrations of sucrose are equivalent. Glucose will also cross the membrane from B to A until the concentration is equal on both sides.

You are right, the movement of water will lower the concentration of glucose in side A but since the glucose can move across the membrane as well, this will not affect the equilibrium.

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