Question

If benzoic acid dimerizes to 60% in benzene then van't hoff factor (i) will be

Answer 1

Answer:

The Van't Hoff factor will be, 0.7

Solution : Given,

Degree of association of benzoic acid = \alphaα = 60% = 0.60

The equilibrium reaction will be,

2C_6H_5COOH\rightleftharpoons (C_6H_5COOH)_22C

6

H

5

COOH⇌(C

6

H

5

COOH)

2

Initial moles 1 0

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

2

α

)

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

2

α

=1−

2

α

Formula of Van't Hoff factor :

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

Theoretical moles

Calculated moles

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

1

2

α

=

1

2

0.60

=0.7

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

Hope it helps u....

Mark my ans as brainliest....