Assume that a cell has: a) b) c) d) An initial volume of 5 nl Contains 0.6 M pro

Question

Assume that a cell has: a) b) c) d) An initial volume of 5 nl Contains 0.6 M protein, and therefore has an osmolarity of 0.6 Osm Protein, Na+ and Cl-cannot pass through the cell membrane The cell membrane is semi-permeable, so assume water passes freely for this problem Problem The cell is placed in a very large volume of 0.3 M NaCI. 1) What is the osmolarity of the NaCI solution? 2) Is the inside of the cell hypertonic, hypotonic or isotonic, relative to the outside of the cell? 3) Will water move into, not change, or out of the cell? 4) After this occurs, what will the osmolarity inside the cell be? For the inside of the cell: If C1 is the starting osmolarity, V1 is the starting volume, and C2 is the ending osmolarity, V2 is the ending volume. Use the equation C1xv1 C2xV2, to solve for V2, the final volume of the cell after being placed in a 0.4 M NaCI solution. 5)

Explanation / Answer

1) Osmolarity is the concentration of total number of ions or particles of solute present in a solution. 1 mole of Nacl dissociates into 1 mole of Na+ and 1 mole Cl- ions and the osmolarity becomes 2 osmoles. 0.3 M Nacl contains 0.3 moles of Na+ and 0.3 moles of Cl-. The osmolarity of 0.3 M Nacl solution is 2 x 0.3 =0.6 osm

2) Osmolarity inside the cell is 0.6 osm similar to osmolarity outside. Therefore the inside of cell is isotonic

3) Since the osmolarity is same inside and outside, no net movement of water molecules will take place.

4) Since there is no movement of water molecules, osmolarity inside will remain the same.

5) A 0.4 M Nacl solution will have a osmolarity of 2 x 0.4 =0.8 osm. The movement of water , i.e osmosis, will take place from inside to outside in order to equalize the osmolarity inside and outside the cell. C1 is the initial osmolarity inside cell = 0.6 osm , V1 is the starting volume = 5 nl , C2 is the final osmolarity the cell inside would try to reach =0.8 osm which is equal to the osmolarity outside , V2 = not known

C1 X V1 = C2 X V2

0.6 X 5 nl = 0.8 x V2

V2 = 3.75 nl

Final volume of cell after being place in a 0.4 M Nacl solution is 3.75 nl

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