Discussion Forum

The freezing point of benzene decreases by 0.45° C when 0.2 g of acetic acid is added to 20g of benzene. If acetic acid associates to form a  dimer in benzene, percentage association of acetic acid in benzene will be (K_f for benzene=5.12 K kg mol^-1 )

Answer:

Explanation:

Let the degree of association of acetic acid  (CH₃COOH) in benzene is $\alpha$  then

                   $2CH_{3}COOH\rightleftarrows (CH_{3}COOH)_{2}$

initial moles                         1                                   0

Moles at equilibrium           1- $\alpha$                      $\frac{\alpha}{2}$

$\therefore$    Total moles = $1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2}$

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

 Now, depression in freezing point (Δ T_f )  is given as

       Δ T_f = i K_f m                 ...........(i)

where, K_f = molal depression constant or cryoscopic constant

m=molality

   Molality = number of moles of solute / weight of solvent (in kg )

 $=\frac{0.2}{60}\times\frac{1000}{20}$

  Putting the values in Eq. (i)

$\therefore$    $0.45 =[1-\frac{\alpha}{2}](5.12)[\frac{0.2}{60}\times\frac{1000}{20}]$

$1-\frac{\alpha}{2}= \frac{0.45\times 60\times 20}{5.12\times0.2\times1000}$

$\Rightarrow$  $1-\frac{\alpha}{2}$ =0.527 $\Rightarrow$    $\frac{\alpha}{2}=1-0.527$

     $\therefore$      $\alpha$  =0.946

               The percentage of association  =94.6 %