# The van's Hoff factor i for a compound which undergoes dissociation in one solvent and association in other solvent is respectively: Option 1) Less than one and greater than one Option 2) Less than one and less than one. Option 3) Greater than one and less than one. Option 4) Greater than one and greater than one.

D Divya Saini

As learnt

Vant Hoff factor for dissociation -

$i=1-\alpha +\frac{\alpha }{n}$

$n=$ number of particles associated

$\alpha =$ degree of association

e.g. Carboxylic acid dimerized in non polar solvents like Hexane.

- wherein

n = 2 for dimerization

n = 3 for trimerization

AND

Vant Hoff factor for dissociation -

$i= 1+(n-1)\alpha$

Where

$n$ is the no. of dissociated particles

$\alpha =$ degree of association

- wherein

$NaC l \: \: \: n = 2$

$CaCl_{2} \: \: \: n = 3$

$K_{4}[F(CN_{6})]\: \: \: \: \: n=5$

Van't Hoff factor for association, $i=1-\alpha +\frac{\alpha }{n}$ ,where n=2 for dimerization

n=3 for trimerization

$i=1-\alpha \left ( 1-\frac{1}{n} \right )$

< 1

and Van't Hoff factor for dissociation, $i=1+(n-1)\alpha ,$ where n=2, 3 ....... (no. of dissociated particles)

>  1

Therefore, i <1  for association

and, i > 1 for dissociation

Option 1)

Less than one and greater than one

This option is incorrect

Option 2)

Less than one and less than one.

This option is incorrect

Option 3)

Greater than one and less than one.

This option is correct

Option 4)

Greater than one and greater than one.

This option is incorrect

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