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.

 

Answers (1)
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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