Question

In case of solute associate in solution, the Van't Hoff factor is:
(A) i > 1
(B) i = 1
 (C) i < 1
(D) None of the above

Answer 1

Hint: We should know the theory of the van't Hoff factor is entirely based on the number of constituent particles (ions or molecules) as the colligative properties depend upon the number of constituent particles.

Complete step by step answer:
In some cases the solute associates or dissociates in the solution where abnormal results are obtained since the colligative properties depend on the number of constituent particles of the solute. Here Van't Hoff came into picture and explained the effect of solutes on the colligative properties of solutions.
Let me explain solute association in solution,
Many organic solutes which undergo association in aqueous solution. When two or more molecules associate to form a bigger molecule. The number of constituent particles decreases consequently, the osmotic pressure, the elevation of boiling point or the depression of the freezing point will be less than the calculated one which is on the basis of a single constituent particle (single molecule) . The property varies inversely as the molar masses of the solute varies.
In order to account for this abnormality, van't Hoff introduced a factor i , which is known as van't hoff factor. The Van't Hoff factor is given by:
\[i = \dfrac{{Observed{\text{ }}osmotic{\text{ }}effect}}{{Normal{\text{ }}osmotic{\text{ effect}}}}\]…………………equation 1
Where, the osmotic effect implies to osmotic pressure, vapour pressure lowering, boiling point elevation and freezing point depression as the case may be,
We know that, the property varies inversely as the molar masses of the solute varies so the van't Hoff factor is given by:
\[i = \dfrac{{Normal{\text{ }}molar{\text{ }}mass}}{{Observed{\text{ }}molar{\text{ }}mass}}\]………………….equation 2
Now let's see the case of solute association of solution in dept:
Consider 1 mole of solute dissolved in a given volume of a solvent. Suppose, n number of molecules combine to form an associated bigger molecule.
\[nA\overset {} \leftrightarrows {(A)_n}\]
Let the degree of dissociation be $\alpha $
Then, the number of unassociated moles = $1 - \alpha $
The number of associated moles = $\dfrac{\alpha }{n}$
Hence, the number of effective moles = $(1 - \alpha ) + (\dfrac{\alpha }{n})$
Therefore i is given by:
Substituting the value in equation 1
\[i = \dfrac{{(1 - \alpha ) + (\dfrac{\alpha }{n})}}{1}\]
\[i = \dfrac{{n(1 - \alpha ) + \alpha }}{n}\]
\[i = \dfrac{{(n - 1)\alpha }}{n}\]
We can observe that there is n in the denominator and (n-1) in the numerator, so the value in the denominator will be less than than the numerator.
Let n=2
\[i = \dfrac{{(2 - 1)\alpha }}{2}\]
So, the i value is less than 1 but greater than 0.

Thus from the above discussion option A is the correct answer.

Note: The two important things to understand are:
-The n in the denominator and (n-1) in the numerator, so the value in the denominator will be less than than the numerator. The solute associates or dissociates in the solution where abnormal results are obtained since the colligative properties depend on the number of constituent particles of the solute.