Question

If 60% of a substance (AB) dimerises in a solvent
and rest ionises, then its van't Hoff factor is

Answer 1

Its van't Hoff factor is 2.1.

Explanation:

1) Degree of association and van't Hoff factor (i) given by::

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

Where :

n = simple molecules of solute in solvent

We have :

Degree of association of Substance AB= 60% = 0.40

$2AB\rightarrow (AB)_2$

n = 2

$0.6=\frac{(1-i)\times 2}{2-1}$

i = 0.7

2) Degree of dissociation and van't Hoff factor (i) given by::

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

Where :

n = simple molecules of solute in solvent

We have :

Degree of dissociation of Substance AB = 40% = 0.40

AB → A + B

n = 2

$0.40=\frac{i-1}{2-1}$

i = 1.4

The final value of van't Hoff factor will be sum of particle coming from association and dissociation.

0.7 + 1.4 = 2.1

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