Question

Prove to me that 0.85% NaCl (basic solution of an IV) is isosmotic. That is, that it is approximately 300 mosmo/L in concentration. You must show ALL your work, cancelling out all units. You must show units on each step. As a starter, the above concentration is equal to 0.85g Nacl/ 100 ml.

Answer 1

the osmolarity of the 0.85% NaCl solution is approximately equal to 300 mosmol/L.

Step-by-step explanation:

Molarity= number of moles/volume of the solution in liters

volume of the solution= 100 mL

1 L= 1000 mL

100 mL= 100/1000=0.1 L

molar mass of NaCl= 58.44 g/mol

number of moles= mass in gram/gram molecular mass

= 0.85/58.44

= 0.01455

molarity= number of moles/volume of the solution in liters

= 0.01455 / 0.1

= 0.1455 M

1 M= 1000 mM

so 0.1455 M = 0.1455 × 1000

= 145.5 mM

Osmolarity is the concentration of solutes in the solution.

NaCl dissociates to Na+ and Cl-

so osmolarity of 145.5 mM NaCl= 2 × 145.5

= 291 mosmol/L

Therefore, we conclude that the osmolarity of the 0.85% NaCl solution is approximately equal to 300 mosmol/L.