Suppose you had a balloon made of some highly flexible semipermeable membrane. The balloon is filled completely with a 0.2 M solution of some solute and is submerged in a 0.1 M solution of the same solute:

Initially, the volume of solution in the balloon is 0.25 L. Assuming the volume outside the semipermeable membrane is large, as the illustration shows, what would you expect for the solution volume inside the balloon once the system has come to equilibrium through osmosis? [Section 13.5]

Video transcript

Hello everyone today we have the falling problem. Assume that you have engineered to sell using a bio synthetic semi permeable membrane in a bioengineering laboratory instead of the usual 0.3 ionic strength. Inside normal cells. This cell has an internal ionic strength of 0.45 molar. The cell is submerged in a saline solution having an ionic strength of 0. molar. Initially the volume of the cell is five micro meters cubed, assuming there won't be any change in the volume of the solution outside the cell. What will be the volume of the cell after the system has come to equilibrium through osmosis. So as stated below, we have a semi permeable membrane that allows osmosis. And so since the inside of the cell has a higher concentration, so the inside of the cell has a higher concentration. This means that water will float inside higher concentration of salutes will cause water from the outside will flow inside as the definition of osmosis states until the concentration of both solutions is identical. And so the concentration Inside the cell, as indicated in the question Stem is .45. The volume is five μm cubed and the concentration of outside the cell Is .3 Molar. So we can simply use the dilution formula here to find the total volume of the cell after the dilution. And so that's gonna look a little bit. We're going to have a final volume which is V2. Is it what you are .45? Our initial malaria T. This is going to be multiplied by the volume five micro meters cubed and divided by the second polarity, Giving us an answer of our final volume, being 7.5 micro meters cubed. And with that we have answered the question. I hope this helped, and until next time.