Question

What will be the abnormal molecular mass of ${{K}_{3}}\left[ Fe{{(CN)}_{6}} \right]$. If in an aqueous solution, it is 40% dissociated? (Given Normal molecular mass of ${{K}_{3}}\left[ Fe{{(CN)}_{6}} \right]$is 329).
149.54
723.8
82.25
329.0

Answer 1

Hint: Before solving this question, we should first know the formula to find the abnormal molecular mass and how different it is from the normal molecular mass. $Abnormal\,Molar\,Mass=\dfrac{Normal\,Molar\,Mass}{i}$, Now we have to find the Van’t Hoff factor and then put all the values in the formula to find the abnormal molecular mass.

Complete answer:
Van’t Hoff factor: The effect on the colligative properties of solutions by solutes is shown by the Van’t Hoff factor. It is represented by i. The ratio of the concentration of molecules formed when a substance is dissolved to the concentration of the substance by mass. The Van’t Hoff factors help us to know the number to which a substance dissociates or associates in a given solution. In the case of a nonelectrolyte substance that is dissolved in water, we take the value of i as 1. But in the case of an ionic compound that is dissolved in water, the value of I would be the number of ions present.
Abnormal Molar Masses: The colligative properties of solution gives theoretical values of molecular masses that come out to be different from the experimentally obtained values known as Abnormal Molar Masses.
When solutes are dissolved in a solvent they get dissociated and form ions, This was explained by Van’t Hoff. So colligative properties depend only on the number of the solute particle that is present in the solvent. So, when solute gets dissociated there is an increase in the number of ions and hence colligative properties get affected.
${{K}_{3}}\left[ Fe{{(CN)}_{6}} \right]\to \,3{{K}^{+}}+\,{{\left[ Fe{{(CN)}_{6}} \right]}^{3-}}$
1-0.4 $3\times 0.4$ 0.4
Van't Hoff Factor (i) = 1-0.4+($3\times 0.4$)+0.4
                                  =1+1.2
                                  =2.2
$i=\dfrac{Normal\,Molar\,Mass}{Abnormal\,Molar\,Mass}$
$Abnormal\,Molar\,Mass=\dfrac{Normal\,Molar\,Mass}{i}$
                                          = $\dfrac{329}{2.2}$
                                          = 149.54

So, Option (A) 149.54 is correct.

Note:
Some particles get associated in an aqueous state also, so the ions present there are less than the actual and the substance that gets dissociated in a solution, their observed molar mass will be less than the actual one. And the particles that are associated with solutions, their actual mass will be less than the observed one.