Question

A cell has an initial volume 2 nl and osmolarity 0.2 Osm. The cell is placed...

A cell has an initial volume 2 nl and osmolarity 0.2 Osm. The cell is placed in a large volume of 0.05 M of CaCl2 solution. Is the 0.05 M of CaCl2 solution isotonic, hypotonic or hypertonic relative to osmolarity inside the cell? As the cell reaches equilibrium, what will be the final osmolarity and volume of the cell?

Homework Answers

Answer #1

The osmolarity inside the cell is 0.2 Osm. The osmolarity of the 0.05 M CaCl2 solution is 0.15 Osm. From this information, we can say that the CaCl2 solution is hypotonic relative to the osmolarity inside the cell.

-Since neither the Ca2+ nor the Cl- can cross the plasma membrane and enter the cell,

water will keep flowing into the cell until the osmolarity inside the cell is 0.15 Osm.

C1V1= C2V2

V2= (C1V1)/(C2)

V2= 2.67 nl

The final volume is going to be 2.67 nl.

The final osmolarity of the cell will have to be reached at 0.15 Osm.

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