Benzoic acid undergoes dimerisation in benzene solution, the van't Hoff factor ' i ' is related to the degree of association ' x ' of the acid as

Question

Benzoic acid undergoes dimerisation in benzene solution, the van't Hoff factor ' i ' is related to the degree of association ' x ' of the acid as (A) i=(1-x) (B) i=(1+x) (C) i=(1-x / 2) (D) i=(1+x / 2)

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/cd7a9dde14480dd93bc00065f7cfa21a-.png" /> Total =1- x +( x /2) =1-( x /2) i =(1- x / 2/1) as i =1-( x /2)

Q. Benzoic acid undergoes dimerisation in benzene solution, the van't Hoff factor ' $i$ ' is related to the degree of association ' $x$ ' of the acid as

$i = (1 - x)$

$i = (1 + x)$

$i = (1 - x /2)$

$i = (1 + x /2)$

Total $= 1 - x + \frac{x}{2} = 1 - \frac{x}{2}$
$i = \frac{1 - x /2}{1}$
as $i = 1 - \frac{x}{2}$

Q. Benzoic acid undergoes dimerisation in benzene solution, the van't Hoff factor ' i ' is related to the degree of association ' x ' of the acid as

i=(1−x)

i=(1+x)

i=(1−x/2)

i=(1+x/2)

Total =1−x+2x​=1−2x​ i=11−x/2​ as i=1−2x​