Question

The answer is C. i just need to know how to solve.

What is approximately the osmolarity of a red blood cell? (a) 0.01 M (b) 0.03 M (c) 0.3 M (d) 1.0 M

Answer 1

A solution in which red blood cells neither swell nor shrink is the most accurate definition of a solution

which is isotonic to red blood cells. A 0.9% NaCl solution is said to be isotonic with red blood cells,

where the intracellular and extracellular fluids are in osmotic equilibrium across the cell membrane.

Osmolarity = molarity x n; where n= number of particles that dissociated from the solute molecule.

The osmolarity of solution containing a 1M solution of NaCl is: 1 x 2 = 2 osmol/L;

The osmolarity of solution containing a 1M solution of CaCl₂ is 1 x 3 = 3 osmol/L.

The osmolarity of solution containing a 1M solution of sucrose (non-electrolyte) is 1 x 1 = 1 osmol/L

So in our example, the osmolarity of the 0.9% NaCl solution is = (0.9/100) / 58.44 g/mol = 0.0001540041, but this is for 100 mL only, now we have to calculate it for 1 liter = 1000 mL

molarity = 0.0001540041 x 1000 = 0.15 M

osmolarity = 0.15M * 2 (Na+ & Cl- = 2 ions) = 0.15M * 2 = 0.3 Osm

Answer: (c) 0.3 M

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