The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g benzene is lowered by 0.45^@C. The degree of association of acetic acid in benzene is (Assume acetic acid dimerises in benzene and K_f for benzene = 5.12 K kg "mol"^(-1) ) ?

The freezing point of a solution containing 0.2 g of acetic acid in 20

1.	The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g benzene is lowered by 0.45^@C. The degree of association of acetic acid in benzene is (Assume acetic acid dimerises in benzene and K_f for benzene = 5.12 K kg "mol"^(-1) ) ?
Answer» `94.6%` `54.9%` `78.2%` `100%` Solution :Given : `w_2=0.2 G , w_1=20 g , DeltaT_f =0.45^@C` `DeltaT_f = (1000xxK_fxxw_2)/(w_1xxM) rArr 0.45=(1000xx5.12xx0.2)/(20xxM)` `THEREFORE M_"(observed)"`=113.78 (ACETIC acid ) As acetic acid dimerises in benzene , so, `{:(,2CH_3COOH hArr, (CH_3COOH)_2),("Before association",1,0),("After association", 1-alpha,alpha//2):}` (where `alpha` is degree of association) Molecular weight of acetic acid =60 `i="Normal molecular mass"/"Observed molecular mass"` `therefore M_"(normal)"/M_"(observed)"=1-alpha + alpha/2` `60/113.78 = 1-alpha + alpha/2 therefore alpha` = 0.946 or 94.6%