Question

For a solution of a non- electrolyte in water, the Van’t Hoff factor is:
A. Always equal to 0
B.$\le 1$
C. always equal to 1
D. >1 but <2

Answer 1

Hint: there are two types of substances, electrolytes, that dissociate into their respective ions in water, and non- electrolytes which do not dissociate into respective ions in water. The Van't Hoff factor is used to determine the association and dissociation extent in solution.

Complete answer:
Van’t Hoff factor, represented by ‘i’ is the measure of the extent of dissociation and association of solutes in the solution. It can be called as the total number of particles after dissociation or association upon total number of particles before association or dissociation. Van’t Hoff factor can also be represented by, $i={\displaystyle \frac{no.ofparticlesafterdissociationorassociation}{no.ofparticlesbeforedissociationorassociation}}$ .
When any non- electrolyte solute is added in water, it does not dissolve to yield its respective ions, so no dissociation of ions occur in case of non – electrolytes in water. This means that particles remain the same as no dissociation or association takes place. So, the same particles after and before adding solute (non- electrolyte) will have a Van’t Hoff factor of 1.
So, i = 1 in case of non electrolytes in water because of no dissociation or association taking place.

Hence, option C is correct, which is always equal to 1.

Note:
When an electrolyte is added like KCl it has dissociation as, $KCl\to K^{+}+C{l}^{-}$ , so $i={\displaystyle \frac{2}{1}}$ = 1. So, for dissociation, the Van’t Hoff factor is always greater than 1. And when association occurs as in acetic acid, $2C{H}_{3}COOH\to {\left(C{H}_{3}COOH\right)}_2$ , i will be $i={\displaystyle \frac{1}{2}}$ , so for association the Van’t Hoff factor is always less than 1.