Question

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?

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.