Question

if benzoic acid dimerise to 60% in benzene then vant hoff factor will be

Answer 1

here n=2 because acid dimerize.

Answer 2

Answer : The Van't Hoff factor will be, 0.7

Solution : Given,

Degree of association of benzoic acid = $\alpha$ = 60% = 0.60

The equilibrium reaction will be,

                            $2C_6H_5COOH\rightleftharpoons (C_6H_5COOH)_2$

Initial moles               1                             0

After association     $(1-\alpha)$                   $(\frac{\alpha}{2})$

The total number of moles after association = $1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2}$

Formula of Van't Hoff factor :

$i=\frac{\text{Calculated moles}}{\text{Theoretical moles}}$

$i=\frac{\frac{\alpha}{2}}{1}=\frac{\frac{0.60}{2}}{1}=0.7$

Therefore, the Van't Hoff factor will be, 0.7