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

Last updated date: 20th Jun 2024
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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.

>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.