Question

If an aqueous solution is 3.09% (w/v) in nickel(II) iodide,NiI2 , what is the osmolarity of the solution?

Osmolarity = _____osmol/L

and

The osmolarity of physiological saline solution is 0.300 osmol/L.

Assuming a membrane that is permeable only to water, what % (w/v) magnesium iodide,MgI2, solution is isotonic to physiological saline?

____% (w/v)

Answer 1

Sol 1.

As 3.09 % (w/v) NiI2 means 3.09 g of NiI2 present in 100 mL of solution .

As Molar Mass of NiI2 = 312.502 g/mol

So , Molarity of NiI2 solution

= ( Mass of NiI2 × 1000 ) / ( Molar Mass of NiI2 × Volume of solution )

= ( 3.09 × 1000 ) / ( 312.502 × 100 )

= 0.0988 M

Also , Dissociation Reaction of NiI2 :

NiI2 ----> Ni2+ + 2I-

So , Total no. of ions produced = 1 + 2 = 3

Therefore , Osmolarity of NiI2 solution

= Total No. of ions produced on dissociation × Molarity of NiI2 solution

= 3 × 0.0988 = 0.2964 osmol / L    

Sol 2 .

As MgI2 solution is isotonic with physiological saline solution.

So , Osmolarity of MgI2 = Osmolarity of physiological saline solution = 0.300 osmol / L

Also , Dissociation reaction of MgI2 :

MgI2 ----> Mg2+ + 2I-

So , Total no. of ions produced = 1 + 2 = 3

So , Molarity of MgI2 solution = Osmolarity of MgI2 solution / Total no. of ions produced  

= 0.300 / 3 = 0.1 M  

As Volume of solution = 1 L

So , Moles of MgI2 = Molarity of MgI2 solution × Volume of solution = 0.1 × 1 = 0.1 mol

Now , Molar Mass of MgI2 = 278.114 g/mol

So , Mass of MgI2 = Moles of MgI2 × Molar Mass of MgI2

= 0.1 × 278.114 = 27.8114 g  

Therefore , % (w/v) of MgI2

= ( Mass of MgI2 / Volume of solution ) × 100

= ( 27.8114 / 1000 ) × 100

= 2.78 % ( w/ v )