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# Predict whether van’t Hoff factor, (i) is less than one or greater than one in the following:(i) $C{{H}_{3}}COOH$ dissolved in water(ii) $C{{H}_{3}}COOH$ dissolved in benzene

Hint: For most non-electrolytes dissolved in water their Van 't Hoff factor is essentially 1. For most of the ionic compounds dissolved in water their Van 't Hoff factor is equal to the number of discrete ions in a formula unit of the substance.

Step-by-Step Solution:
Let us first get introduced to the van’t Hoff factor before moving on to the specifics of the given question.
The Van’t Hoff factor offers information on the effect of solutes on the colligative properties of solutions and it can be defined as the ratio of the concentration of particles formed when a substance is dissolved to the concentration of the substance by mass.
Van’t hoff factor also describes the extent to which a substance associates or dissociates in a solution. e.g. when a non-electrolytic substance is dissolved in water, the value of i is generally 1.
However, when a compound that is ionic, forms a solution in water, the value of i is equal to the total number of ions present in one formula unit of the ionic substance.
Now, let us try and apply this concept to the given question.
(i) $C{{H}_{3}}COOH$ dissolved in water
Acetic acid is a weak acid and will not completely dissociate into ions when gets dissolved in water. The dissociation reaction can be given as under.
$C{{H}_{3}}COO{{H}_{(aq)}}\xrightarrow{{{H}_{2}}O}C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}$
Its dissociation constant can be given by following equation.
$\text{Dissociation constant }\alpha \text{ = }\frac{[C{{H}_{3}}CO{{O}^{-}}][{{H}^{+}}]}{[{{H}_{2}}O]}$
There is a simple relationship between the dissociation constant and Van’t hoff factor. Dissociation constant expresses the degree of dissociation of the acid, so we can write the relation between dissociation constant and Van’t foff factor as
$i=\alpha n+(1-\alpha )$
Here, acetic acid yields 2 ions upon dissociation. So, we can say that
$i=1+\alpha$
So, we can say that van’t hoff factor for acetic acid in water will be more than 1.
ii) $C{{H}_{3}}COOH$ in benzene
In benzene, actually acetic acid dimerizes and so, its association reaction in benzene can be given as:
$2C{{H}_{3}}COOH\to 2(C{{H}_{3}}COOH)$
If we suppose that association factor is $\alpha$ , then we can find its Van’t hoff factor.
So, Van’t hoff factor can be given as
$i=n\alpha +(1-\alpha )$
$i=\frac{1}{2}\alpha +(1-\alpha )$
$i=1-\frac{\alpha }{2}$
Hence we can say that the Van't hoff factor for acetic acid in benzene will be less than one.
Thus we can conclude that the Van't hoff factor for acetic acid in water is more than one and for acetic acid in benzene will be less than one.

Note: Sometimes ion pairing occurs in a solution. At a given instant, a small percentage of the ions are paired and count as a single particle. Ion pairing occurs to some extent in all electrolyte solutions. This causes deviation from the van 't Hoff factor and tends to be greatest where the ions have multiple charges.