What happens to the red blood cell in CaCl₂ solution?

Question

Here's the problem:One red blood cell is placed in a hypertonic solution of NaCl, another isplaced in a solution of CaCl₂ 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 Ca²⁺ normally in the cell. However, I'm not sure if this is right so could someone please explain the correct answer?

Answer 1

Right answer, wrong reason.

In water, CaCl2 forms three ions: Ca2+, and 2Cl-. Therefore if, for example, NaCl was 1 molar, then there would be a mole of Na+ and a mole of Cl-. But, if Cacl2 is 1 molar, there are 3 moles of particles. Therefore, the cell will shrink more in CaCl2 thn in NaCl