Ared blood cell placed in a hypertonic solution will shrink in a process called crenation. a red blood cell placed in a hypotonic solution will swell and potentially burst in a process called hemolysis. to prevent crenation or hemolysis, a cell must be placed in an isotonic solution such as 0.9% (m/v) nacl or 5.0% (m/v) glucose. this does not mean that a cell has a 5.0% (m/v) glucose concentration; it just means that 5.0% (m/v) glucose will exert the same osmotic pressure as the solution inside the cell, which contains several different solutes. a red blood cell is placed into each of the following solutions. indicate whether crenation, hemolysis, or neither will occur. solution
a: 3.21% (m/v) naclsolution
b: 1.65% (m/v) glucosesolution
c: distilled h2osolution
d: 6.97% (m/v) glucosesolution
e: 5.0% (m/v) glucose and 0.9%(m/v) nacldrag each solution to the appropriate bin.

The correct answers are A. "crenation", B. "hemolysis", C. "hemolysis", D. "crenation" and E. "neither will occur".

Explanation:

0.9% (m/v) NaCl or 5.0% (m/v) glucose are isotonic solutions at which the cells will not suffer any harmful consequence. A solution with a higher concentration than the isotonic conditions would result in the cells crenation, while a solution with a lower concentration would result in the cells hemolysis. Therefore the consequences of putting the red cells to the solutions stated in the question are as following:

A: 3.21% (m/v) NaCl Solution  = crenation (higher than 0.9% (m/v) NaCl)

B: 1.65% (m/v) glucose Solution  = hemolysis (lower than 5.0% (m/v) glucose)

C: distilled H2O Solution  = hemolysis (lower than 0.9% (m/v) NaCl or 5.0% (m/v) glucose)

D: 6.97% (m/v) glucose Solution  = crenation (higher than 5.0% (m/v) glucose)

E: 5.0% (m/v) glucose and 0.9%(m/v) NaCl = neither will occur (equal to 5.0% (m/v) glucose and 0.9% (m/v) NaCl)