Isotonic saline contains 0.9% NaCl, what is the osmolarity of isotonic solution?

Question

Isotonic saline contains 0.9% NaCl, what is the osmolarity of isotonic solution? the molecular weight of human serum albumin is 67,000 g/mol and the concentration of the protein in the serum is about 60 g/L. If you want to make an isotonic solution of serum albumin (60g/L or 6%) with NaCl for intravenous injection, what is the percentage of NaCl should be in the solution? why does teh protein contribute so little to the osmolarity of the solution? (hint- the osmolarity of NaCl plus albumin should equal to the osmolarity of isotonic solution. use molar for all calculations)

Explanation / Answer

Concentration of isotonic solution = 0.9%

This means 9 g NaCl is dissolved in 1 L water

Molar mass of NaCl = 58.44 g

Thus, moles of NaCl present = 9/58.44 = 0.154 moles

1 mole produces 2 moles of solute ions, thus number of osmoles present = 2*0.154 = 0.308 = 308 mOs

Thus, osmolarity = 308 mOs/L or 308 milli osmoles/L

Now, for the NaCl+protein solution :

concentration of protein = 6% = 60 g/L

Moles of protein present = 60/67000 = 8.95*10-4 moles = 0.895 millimoles

Since the protein doesn't ionise, this is also the number of osmoles of protein present in 1 L isotonic soultion

So, the number of osmoles of NaCl that must be present = total number of osmoles present in solution-number of osmoles of protein = 308-0.895 = 307.105 mOs

Thus, number of moles of NaCl required = number of osmoles of NaCl required/2 = 0.307105/2 = 0.1535

Thus, mass required = 0.1535*58.44 = 8.97 g

Thus, concentration of NaCl = 8.97%

Contribution of protein is so less because of the extremely large molecular weight of protein, due to which its number of osmoles become almost negligible.

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