Question

van't Hoff factor of acetic acid which shows 10%
dimerization in water is
0.95
0.75
0.80
0.90​

Answer 1

Given:

Acetic Acid shows 10% dimerization in water.

To find:

Van't Hoff Factor ?

Calculation:

$\rm 2CH_{3}COOH \rightarrow (CH_{3}COOH)_{2}$

So, we can convert the Equation as :

$\rm CH_{3}COOH \rightarrow \dfrac{1}{2} (CH_{3}COOH)_{2}$

Initially we can say :

$\rm \: \: \: \: \: \: \: \: \: \: \: \: 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0$

After time t :

$\rm \: \: \: \: \: 1 - \alpha \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \dfrac{ \alpha }{2}$

Now, Van't Hoff Factor will be :

$\rm \: i = \dfrac{(1 - \alpha ) + \dfrac{ \alpha }{2} }{1}$

• putting $\alpha$ = 10%

$\rm \implies\: i = 1 -\dfrac{ \alpha }{2}$

$\rm \implies\: i = 1 -\dfrac{ 10\% }{2}$

$\rm \implies\: i = 1 -5\%$

$\rm \implies\: i = 1 - 0.05$

$\rm \implies\: i = 0.95$

So, Van't Hoff Factor is 0.95.