Question

The ratio of vant hoff factors of acetic acid in benzene and water (assuming complete association and dissociation respectively) is equal to?

Answer 1

reaction of acetic acid and benzene doesn't possible. but yes, acetic acid is dimerise in the presence of benzene.

hence, association happens ..

here two mole of acetic acid produces 1 mole of dimer acetic acid .

so, n = 2

so, Van't Hoff's factor, i = $1-\alpha+\frac{\alpha}{n}$

for completely association, $\alpha=1$

then, i = 1 - 1 + 1/2 = 0.5

reaction of acetic acid and water :

$CH_3COOH+H_2O\Leftrightarrow CH_3COO^-+H_3O^+$

so, n = 2

now Van't Hoff's factor, i' = $1-\alpha+n\alpha$

for completely dissociation, $\alpha=1$

then, i' = 1 - 1 + 2 × 1 = 2

now, ratio of i and i' = i/i' = 0.5/2 = 1/4

hence, answer is 1 : 4

Answer 2

Answer:

The ratio of van’t hoff factors of acetic acid in benzene and water is 1:1.

Explanation:

Van’t hoff factor for acetic acid in water.

The dissociation of acetic acid is as follows:

                 $\mathrm{CH}_{3} \mathrm{COOH} \leftrightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+}$

                 $\mathrm{CH}_{3} \mathrm{COOH}=1-\alpha$

                   $H^{+}=\alpha$

Hence, the “van’t hoff factor” will be in between 1 and 2

Van’t hoff factor for acetic acid in Benzene:  

The dissociation of acetic acid is as follows:

                    $2 \mathrm{CH}_{3} \mathrm{COOH} \leftrightarrow\left(\mathrm{CH}_{3} \mathrm{COOH}\right)_{2}$

                     $i=1-\left(1-\frac{1}{2}\right) \alpha$

                             $=1-\frac{\alpha}{2}$

For association of van’t hoff factor is 1.

Therefore, the ratio of “van’t hoff factors” of acetic acid in ‘benzene and water’ is 1:1.