Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by

If the van't Hoff factor of a weak electrolyte $A_xB_y$ is $i$,then the degree of dissociation $(\alpha)$ is represented by which equation?

$2 \ g$ 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 $......... \%$.

In a solution with molality $m$,if the solute exists in the form of a dimer,which of the following cannot be the elevation in the boiling point of the solution?

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,$K_b = 2.65 \ K \ m^{-1}$,$K_f = 5.12 \ K \ m^{-1}$,$T_b^o = 80 \ ^\circ C$,$T_f^o = 5.5 \ ^\circ C$)

$HA_{(aq)} \rightleftharpoons H^{+}_{(aq)} + A^{-}_{(aq)}$
The freezing point depression of a $0.1 \ m$ aqueous solution of a monobasic weak acid $HA$ is $0.20^{\circ} C$. The dissociation constant for the acid is. Given: $K_{f}(H_2O) = 1.8 \ K \ kg \ mol^{-1}$,molality $\equiv$ molarity.