The solution in the arms of a U-tube are separated at the bottom of the tube by

Question

The solution in the arms of a U-tube are separated at the bottom of the tube by a selectively permeable membrane. The membrane is permeable to sodium chloride but not to glucose. Side A is filled with a solution of 0.4 mols of glucose and 0.5 mols of sodium chloride (NaCl), and side B is filled with the a solution containing 0.8 mols of glucose and 0.4 mols of sodium chloride. Initially, the volume in both arms is the same. A. Initially at the sun of the experiment, which of the following is true? A. side A is hypertonic to side B B. side A is hypotonic to side B C. side A is hypertonic to side B with respect to glucose D. side A is hypotonic to side B with respect to NaCl B. After diffusion and equilibrium: How many mols of glucose are on side A? _____ side B? _____ How many mols of NaCl are on side A? _____ side B? _____ C. If you examine side A after three days you should find l _____. A. a decrease in the number of mols of NaCl and glucose and an increase in the water level B. a decrease in the number of mols of NaCl and a decrease in the water level C. a decrease in the number of mols of NaCl, an increase in water level, and no change in the number of mols of glucose

Explanation / Answer

Part B:

Colligative properties of the dilute solutions depend on the number of solute particles.

The given example shows the development of osmotic pressure.

The concentrations are different in both the com[partments.

The movement of solute particles takes place until both sides concentrations become equal.

The membrane is permeable to only NaCl and not for glucose.

Hence on both sides of the compartments glucose concentration does not change.It remains as it is.

So the concentration of glucose after diffusion in compartment A is 0.4mol and in part B is 0.8mol

Concentartion of NaCl on both sides become equal.

That is 0.45mol on each side.

This is obtained by taking the average concnetraton of both the compartments.

(0.5mol+0.4mol)/2 = 0.45mol

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