From a measurement of the freezing-point depression of benzene, the molecular weight of acetic acid in a benzene solution was determined to be 100. The percentage association of acetic acid is:

Dear Student,

Please find below the solution of your asked query:

Normal molar mass of Acetic acid = 60 g/mol

Abnormal molar mass of Acetic acid = 100 g/mol

Vant Hoff factor (i) = Normal molar mass/Abnormal molar mass
                                = 60/100
                                = 0.6



So, now we can calculate the degree of association by using the following relation:

i = 1 - $\mathrm{\alpha }$ + $\mathrm{\alpha }$/2

0.6 = 1 - $\mathrm{\alpha }$/2

$\mathrm{\alpha }$/2 = 0.4

$\mathrm{\alpha }$ = 0.8
or
$\mathbf{\alpha }$ = 80 %

Hope this information clears your doubts about the topic.

Keep asking!!

Regards