What happens to the red blood cell in CaCl₂ solution?
Here's the problem:One red blood cell is placed in a hypertonic solution of NaCl, another isplaced in a solution of CaCl<sub>2</sub> equimolar with the NaCl solution. Whatwould you expect to happen and why?My reasoning is that the red blood cell will shrink due to water loss by osmosis when placed in hypertonic NaCl. It will shrink even further due to the larger osmosis gradient due to there being such low concentration of Ca<sup>2+</sup> normally in the cell. However, I'm not sure if this is right so could someone please explain the correct answer?