If $m$ is the normal molecular weight and $x$ is the degree of dissociation of $K_4[Fe(CN)_6]$,then the abnormal molecular weight in the solution of the salt will be $:-$

Three particles of a solute $A$ associate in benzene to form a trimer $A_3$. Calculate the freezing point of a $0.25 \ m$ solution if the degree of association of solute $A$ is $0.80$. The freezing point of benzene is $5.5 \ ^oC$ and $K_f$ is $5.12 \ K \ kg \ mol^{-1}$. (Answer in $K$)

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

The difference between the boiling point and the freezing point of a $0.2 \ m$ solution of acetic acid in benzene is $75.7 \ ^\circ C$. Calculate the value of the van't Hoff factor $i$. (For benzene,$K_b = 2.65 \ K \ m^{-1}$,$K_f = 5.12 \ K \ m^{-1}$,$T_b^o = 80 \ ^\circ C$,$T_f^o = 5.5 \ ^\circ C$)

When acetic acid is dissolved in benzene,its molecular mass:

$2 \ g$ of benzoic acid $(M_w = 122 \ g/mol)$ dissolved in $25 \ g$ of $C_6H_6$ shows a depression in freezing point equal to $1.62 \ K$. Given $K_f (C_6H_6) = 4.9 \ K \ kg \ mol^{-1}$. If the acid forms a dimer in the solution,then the percentage association of the acid is $......... \%$.