Question

How do you calculate the osmolarity of a solution?

Answer 1

Hint: The answer here is based on the basic concept of physical chemistry which states that osmolarity is the measure of a number of solute particles per litre of the solution. Write the calculation based on this definition.

Complete step by step solution:
In the lower classes, we have studied the basic concepts of physical chemistry which deals with the calculation of some of the physical quantities such as molarity, osmosis, osmolarity, osmolality etc.
Let us now discuss in detail about the osmolarity and its calculation.
- Osmolarity of a solution is defined as the measure of a number of osmoles of solute particles per litre of the solution. It is expressed in terms of osmol/L or Osm/L.
- An osmole is the one mole of particles which contribute to the osmotic pressure of a solution.
Let us consider a simple example of calculation of osmolarity of blood which has the given data as, concentrations of the solute $[N{{a}^{+}}]$ = 0.140 mol/L, [Glucose] = 180 mg/100mL, [Blood urea nitrogen] = 20 mg/100mL.
Now, since each sodium ion pair with a negative ion such as chloride ion, this gives 2 osmol of particles.
Thus, osmolarity of NaX$=\dfrac{0.140mol/L}{1L}\times \dfrac{2Osmol}{1mol}=0.280$ Osmol/L
Similarlt, osmolarity of the glucose solution will be$=\dfrac{0.180g}{100mL}\times \dfrac{1000mL}{1L}\dfrac{1mol}{180.2g}\times \dfrac{1Osmol}{1mol}=0.00999$ Osmol/L
Also, the osmolarity of blood urea nitrogen is$=\dfrac{0.20g}{100mL}\times \dfrac{1000mL}{1L}\times \dfrac{1mol}{28.01g}\times \dfrac{1Osmol}{1mol}=0.0071$ Osmol/L
Therefore, the blood osmolarity will be = $0.280+0.00999+0.0071=0.297$ Osmol/L
Thus, it can be written as osmolarity of blood solution is 297 mOsmol/L

Note:
Note that osmolarity and molarity vary slightly in the measurement and do not be confused because molarity measures the number of moles of solute per unit volume of the solution and osmolarity measures the number of osmoles of solute particles per unit volume of the solution.