Question

Van’t Hoff factor, when benzoic acid is dissolved in benzene, will be:
(A)2
(B)1
(C)0.5
(D)1.5

Answer 1

Hint: Van’t Hoff factor of the molecules can be calculated by using the following formula,
\[\text{Van }\!\!'\!\!\text{ t Hoff factor i=}\dfrac{\text{n (Observed)}}{\text{n (Theoretical) }}\]
n (observed) = number solute particles present in the solution
n (Theoretical) = number of solute particles without considering association and dissociation.

Complete step by step answer:
>The structure of benzoic acid is as follows.

>The benzoic acid is soluble in water and benzene also.
>The molecular weight of benzoic acid is 122, but the observed molecular weight is 242.
>The observed molecular weight is double the expected molecular weight.
>This indicates that an association of benzoic acid in benzene solution into dimers.
>Therefore the Van’t Hoff factor of benzoic acid in benzene is
\[\begin{align}
  & \text{Van }\!\!'\!\!\text{ t Hoff factor i=}\dfrac{\text{n (Observed)}}{\text{n (Theoretical) }} \\
 & \text{ = }\dfrac{1}{2}=0.5 \\
\end{align}\]
>The Van’t Hoff factor for benzoic acid in benzene is 0.5.

So, the correct option is C.

Additional information:
>Benzoic acid is most regularly found in industries to manufacture a wide variety of products like perfumes, dyes, and as an insect repellent.
>Benzoic acid is available naturally in many plants and is involved in the biosynthesis of several secondary metabolites.

Note: Benzoic acid in the solution form dimers due to the presence of hydrogen bonding. Hydrogen bonding makes two molecules of benzoic acid into a single molecule by holding the two molecules together. The process of formation of a dimer is called dimerization. By using the Van't Hoff factor we can find the numbers of molecules present in the solution